{ "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "AIME I" ], "Problem": "How many ppl qualify for USAMO and do I have a chance?\r\n\r\nI scored a 124 on the AMC 10 and got 8 right on the Alternate AIME.", "Solution_1": "AMC score means nothing almost, except that it allows you to qualify for AIME. Your 8 on the AIME Alter. might qualify you for the USAMO depending on what the floor score is. Every year around 250 people qualify for the USAMO.", "Solution_2": "The easiest way to qualify for USAMO is to move to Wyoming. ;)", "Solution_3": "lol on the Wyoming :) DPopov, I totally disagree. If you're in 11th or 12th grade, your AMC score means a lot, because your USAMO index is 10 times the AIME score, plus the SMC score. So even if you got an 11 on the AIME I this year, if you got only a 120 on the AMC, then you wouldn't qualify. On the other hand, if you're in 10th grade or below, then I agree :)", "Solution_4": "But spt10v is in 10th grade or under, so it doesn't matter for him." } { "Tag": [ "algebra solved", "algebra" ], "Problem": "a,b,c \\geq 1\r\nProve that:\r\n\r\n \\sqrt (a-1)+ \\sqrt (b-1)+ \\sqrt (c-1) \\leq \\sqrt (c(ab+1))", "Solution_1": "Any solution?", "Solution_2": "First of all I perform the substitutions a-1 = x^2 , b-1 = y^2 , c-1 = z^2\r\nAfter few calculation, the relationship becomes (x+y+z)^2 =< (1+z^2)*(1+(1+x^2)*(1+y^2)) \r\nIt's easy to show (brute calculation) (a+b)^2 =< (1+a^2)*(1+b^2) for all a,b real , so \r\n(x+y+z)^2 =< (1+z^2)*(1+(x+y)^2) =< (1+z^2)*(1+(1+x^2)*(1+y^2)) (q.e.d)\r\nDiogene" } { "Tag": [ "inequalities", "function" ], "Problem": "Suppose a+b=1 where a and b are positive numbers.\r\n(i) Show that $ab\\leq\\frac14$.\r\n(ii) Find the minimum value of $\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)$.", "Solution_1": "[quote=\"ifai\"]Suppose a+b=1 where a and b are positive numbers.\n(i) Show that $ab\\leq\\frac14$.\n(ii) Find the minimum value of $\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)$.[/quote]\r\n[hide=\"(i)\"]By AM-GM,\n$\\frac{a+b}{2} \\ge \\sqrt{ab}$ \n$\\frac{1}{2} \\ge \\sqrt{ab}$\n$\\frac{1}{4} \\ge ab$\n$ab \\le \\frac{1}{4}$[/hide]\n[hide=\"(ii)\"]Expanding the second expression, we have:\n$ab+ \\frac{1}{ab} + \\frac{a}{b} + \\frac{b}{a}$\nFor the expression to be of minimum value, $a=b=\\frac{1}{2}$\nSo we are left with $\\frac{1}{4}+ 4 + 1 + 1 = \\frac{25}{4}$\nSo the minimum value of $(a+\\frac{1}{a})(b+\\frac{1}{b}) = \\frac{25}{4}$[/hide]", "Solution_2": "(ii)\r\n\r\n[hide](a+1/a)(b+1/b)=(a^2+1)(b^2+1)/ab=((ab)^2+a^2+b^2+1)/ab=ab+(a^2+b^2+1)/ab\n\n=ab-2+(1+(a+b)^2)/ab=ab-2+2/ab\n\nFrom (i) and since a,b>0, 00. Then x+1/x>y+1/y.)\n\nso ab-2+2/ab>=(1/4)-2+2/(1/4)=25/4, where the minimum is obtained when a=b=1/2.[/hide]", "Solution_3": "[quote=\"ifai\"]Suppose a+b=1 where a and b are positive numbers.\n(i) Show that $ab\\leq\\frac14$.\n(ii) Find the minimum value of $\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)$.[/quote]\r\n\r\n(i) [hide]\nBy AM-GM, we have \\[ \\frac{a+b}{2} \\ge \\sqrt[2]{ab}, \\] so \\[ ab \\le \\left( \\frac{1}{2} \\right) ^2 = \\frac{1}{4}, \\] and we are done.\n[/hide]\n\n(ii)\n\n[hide]We need to minimize \\[ \\left( \\frac{a^2+1}{a} \\right) \\left( \\frac{b^2+1}{b} \\right) = \\frac{(a^2+1)(b^2+1)}{ab}. \\] The maximum of the denomenator is $\\frac{1}{4}$, so we need to minimize \\[ 4(a^2+1)(b^2+1). \\]\n\nI have to go...\n[/hide]", "Solution_4": "Continueing on Chess64's work... \r\n[hide]\nMinimization occurs when $a=b=\\frac{1}{2}$, so we have $4\\left(\\frac{1}{4}+1\\right)\\left(\\frac{1}{4}+1\\right)=4\\left(\\frac{5}{4}\\right)^2=\\frac{25}{4}$. :)\n[/hide]\r\n\r\nMasoud Zargar", "Solution_5": "Sad, most of you didn't prove the fact that, in (ii) we must have $a = b = 1/2$ to get the minimum value. Simple proofs do exist though. (Scorpius119 is the only one who tried to prove this.)", "Solution_6": "I have just derived something involving the minimization.\r\nWe know that\r\n$\\\\\\left(a+\\frac{1}{a}\\right)\\left(b+\\frac{1}{b}\\right)\\\\=\\left(\\frac{a^2+1}{a}\\right)\\left(\\frac{b^2+1}{b}\\right)\\\\=4(a^2+1)(b^2+1)$\r\nLet $a_1=a$, $a_2=b$, and $b_1=b_2=1$. If we use Cauchy-Schwarz, we get\r\n$\\left(\\sum_{i=1}^na_ib_i\\right)^2\\le\\left(\\sum_{i=1}^na_i^2\\right)\\left(\\sum_{i=1}^nb_i^2\\right)\\\\(a+b)^2=a^2+b^2+2ab=a^2+b^2+\\frac{1}{2}\\le 2(a^2+b^2)\\\\\\implies\\frac{a^2+b^2+\\frac{1}{2}}{a^2+b^2}=1+\\frac{1}{2(a^2+b^2)}\\le 2\\\\\\implies 2(a^2+b^2)=1\\implies a^2+b^2=\\frac{1}{2}\\\\\\implies (a+b)^2\\le 2(0.5)=1\\\\\\implies a^2+b^2+2ab=1\\le 1$.\r\nMinimization occurs when $a=b=\\frac{1}{2}$. This may not be a direct proof of the minimization, but it seems to me that it is the right way. Anyone have an idea. :(\r\n\r\nEDIT: Nope, this is messed up. :( \r\nMasoud Zargar", "Solution_7": "[hide=\"prompting\"]Try to think of the following equation: $a^2+b^2=1-2ab$. [/hide]", "Solution_8": "[hide=\"a nice solution\"]$\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)=\\frac{a^2b^2+a^2+b^2+1}{ab}=\\frac{a^2b^2+1-2ab+1}{ab}=\\frac{(1-ab)^2+1}{ab}$\n$\\geq4\\left[\\left(1-\\frac14\\right)^2+1\\right]=\\frac{25}4$[/hide]", "Solution_9": "${(a+\\frac 1a})(b+\\frac 1b)=\\frac{(a^2+1)(b^2+1)}{ab}=\\frac{(a^2+\\frac 14+\\frac 34)(b^2+\\frac 14+\\frac 34)}{ab}$\r\n\r\nBy AM-GM, $a^2+\\frac 14\\geq \\frac a2$, and similarly for $b$\r\n\r\nSo we get ${(a+\\frac 1a})(b+\\frac 1b)\\geq \\frac{(\\frac a2+\\frac 34)(\\frac b2+\\frac 34)}{ab}$\r\n\r\nIt shouldn't be too hard to finish from here...", "Solution_10": "[quote=\"ifai\"]Suppose a+b=1 where a and b are positive numbers.\n(i) Show that $ab\\leq\\frac14$.\n(ii) Find the minimum value of $\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)$.[/quote]\r\n\r\n(ii)\r\n\r\n[hide=\"not completely rigorous but oh well\"]Expanding, we find \\[ ab+\\frac{a}{b}+\\frac{b}{a}+\\frac{1}{ab}. \\] Since the maximum of $ab$ is $\\frac{1}{4}$, the minimum of $\\frac{1}{ab}$ is $4$. Now we have \\[ \\frac{17}{4} + \\frac{a}{b} + \\frac{b}{a} . \\] AM-GM on $\\frac{a}{b} + \\frac{b}{a}$ yields \\[ \\frac{a}{b} + \\frac{b}{a} \\ge 2\\sqrt{1} \\implies \\frac{a}{b} + \\frac{b}{a} \\ge 2, \\] so the minimum is \\[ \\frac{25}{4}. \\][/hide]", "Solution_11": "What you did is no better than any of the other incorrect posts on this topic, because you fail to justify\r\n\r\n$ab+\\frac 1{ab}\\geq \\frac{17}{4}$", "Solution_12": "[quote=\"ifai\"]Suppose a+b=1 where a and b are positive numbers.\n(i) Show that $ab\\leq\\frac14$.\n(ii) Find the minimum value of $\\left(a+\\frac1a\\right)\\left(b+\\frac1b\\right)$.[/quote]\r\n\r\n(i) $ab\\leq \\left(\\frac{a+b}{2}\\right)^2=\\frac{1}{4}.$\r\n\r\n(ii) $\\left(a+\\frac{1}{a}\\right)\\left(b+\\frac{1}{b}\\right)=ab+\\frac{1}{ab}+\\frac{b}{a}+\\frac{a}{b}$\r\n\r\n$\\geq ab+\\frac{1}{ab}+2\\sqrt{\\frac{b}{a}\\cdot \\frac{a}{b}}=ab+\\frac{1}{ab}+2\\geq \\frac{25}{4}.$\r\n\r\nThe equality holds when $a=b$ and $ab=\\frac{1}{4}$ and $a>0,b>0\\Longleftrightarrow a=b=\\frac{1}{2}.$\r\n\r\nAlternative solution for (ii)\r\n\r\n$\\left(a+\\frac{1}{a}\\right)\\left(b+\\frac1b\\right)=\\left\\{(\\sqrt{a})^2+\\left(\\frac{1}{\\sqrt{a}}\\right)^2\\right\\}\\left\\{(\\sqrt{b})^2+\\left(\\frac{1}{\\sqrt{b}}\\right)^2\\right\\}$\r\n\r\n$\\geq \\left(\\sqrt{ab}+\\frac{1}{\\sqrt{ab}}\\right)^2=\\frac{25}{4}.$", "Solution_13": "Most people just can get 25/4 but can't give the proof.", "Solution_14": "Why? Can't you understand the behaviour of the graph of $y=x+\\frac{1}{x}$ at $00$.So using those notations the given inequality is equivalent to: \r\n\\[\\sum_{cyc}(y+z)^{2}(z+x)(y-x)\\geq0\\leftrightarrow\\sum_{cyc}(y^{2}+2yz+z^{2})(yz+xy-x^{2}-xz)\\geq 0\\] $\\leftrightarrow$ \\[\\sum_{cyc}(y^{3}z+y^{2}xy-y^{2}x^{2}-y^{2}xz+2y^{2}z^{2}+2y^{2}zx-2yzx^{2}-2yz^{2}x+z^{3}y+z^{2}xy-z^{2}x^{2}-z^{3}x)\\geq 0\\] $\\leftrightarrow$ \\[\\sum_{cyc}xy^{3}\\geq\\sum_{cyc}x^{2}yz\\] $\\leftrightarrow$ $\\sum_{cyc}\\frac{xy^{3}}{xyz}\\geq \\sum_{cyc}x$ $\\leftrightarrow$ $\\sum_{cyc}\\frac{y^{2}}{z}\\geq \\sum_{cyc}x$ which is true by Cauchy, and we are done.", "Solution_4": "[quote=\"Svejk\"]As for the second inequality ,because a,b,c are the sides of a triangle we have: $a=x+y$,$b=y+z$ and $c=x+z$ ,where $x,y,z>0$.So using those notations the given inequality is equivalent to:\n\\[\\sum_{cyc}(y+z)^{2}(z+x)(y-x)\\geq0\\leftrightarrow\\sum_{cyc}(y^{2}+2yz+z^{2})(yz+xy-x^{2}-xz)\\geq 0 \\]\n$\\leftrightarrow$\n\\[\\sum_{cyc}(y^{3}z+y^{2}xy-y^{2}x^{2}-y^{2}xz+2y^{2}z^{2}+2y^{2}zx-2yzx^{2}-2yz^{2}x+z^{3}y+z^{2}xy-z^{2}x^{2}-z^{3}x)\\geq 0 \\]\n$\\leftrightarrow$\n\\[\\sum_{cyc}xy^{3}\\geq\\sum_{cyc}x^{2}yz \\]\n$\\leftrightarrow$ $\\sum_{cyc}\\frac{xy^{3}}{xyz}\\geq \\sum_{cyc}x$ $\\leftrightarrow$ $\\sum_{cyc}\\frac{y^{2}}{z}\\geq \\sum_{cyc}x$ which is true by Cauchy, and we are done.[/quote]\r\nDo you have other solutions for this inequality?", "Solution_5": "[quote=\"tunganh\"]Another inequality:\n$a,b,c$ are three sides of a triangle. Prove that:\n$T= \\sum a^{2}b(a-b) \\geq 0$[/quote]\r\nThis is [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/B3[/url]\r\nsearch, please! :D", "Solution_6": "[quote=\"N.T.TUAN\"][quote=\"tunganh\"]Another inequality:\n$a,b,c$ are three sides of a triangle. Prove that:\n$T= \\sum a^{2}b(a-b) \\geq 0$[/quote]\nThis is [url=http://www.kalva.demon.co.uk/imo/isoln/isoln836.html]IMO 1983/B3[/url]\nsearch, please! :D[/quote]\r\nBut the solution here is the same as Svejk's solution. Do you have other solutions?", "Solution_7": "I used search function and have http://www.mathlinks.ro/Forum/viewtopic.php?p=366624&sid=edff0df08e893dfe79c45897ce514418#p366624" } { "Tag": [ "algebra", "function", "domain", "conics", "parabola", "geometry", "rectangle" ], "Problem": "Let $f(x) = {\\sqrt{ax^2+bx}}$. For how many real values of a, is there atleast one positive value of b for which the domain of f and the range of f are the same set.\r\n\r\na) 1 b) 2 c) 3 d) 0 e) infinitely many", "Solution_1": "This is interesting...\r\n\r\n[hide]Clearly, $a = 0$ is a value that works.\n\nSince the range is always $[0, \\infty)$, so is the domain in this case.\n\nWe know that $ax^2 + bx$ is a parabola with x-intercepts at $0$ and $-\\frac{b}{a}$.\n\nBut we want $a$ to be negative because if the parabola is facing up, then we have a clash of domain and range because the range will include negatives...\n\nAnd so we are looking at the part of the parabola bound by the intercepts such that the maximum to $y = 0$ is $-\\frac{b}{a}$. \n\nWe know the value of $x$ which will allow this is $-\\frac{b}{2a}$, so we plug that into $f(x)$ and set $f(-\\frac{b}{2a}) = -\\frac{b}{a}$, setting domain equal to range, and we get $a = -4$ as the only other value for which the domain and range are the same. \n\nAnd $B$.\n[/hide]", "Solution_2": "[quote=\"white_horse_king88\"]This is interesting...\n\nClearly, $a = 0$ is a value that works.\n\nSince the range is always $[0, \\infty)$, so is the domain in this case.\n\n[/quote]\r\n\r\nIs it necessary that the range should be always $[0, \\infty)$.\r\n\r\nCan't we have a situation like, a part of the parabola whose square root function curve is joining the two opposite vertices of a rectangle y=m, y=n and x=m, x=n making the domain and range as [m, n]?\r\n\r\nWill it be possible to prove or disprove this idea wrong?", "Solution_3": "[quote=\"vidyamanohar\"][quote=\"white_horse_king88\"]This is interesting...\n\nClearly, $a = 0$ is a value that works.\n\nSince the range is always $[0, \\infty)$, so is the domain in this case.\n\n[/quote]\n\nIs it necessary that the range should be always $[0, \\infty)$.\n\nCan't we have a situation like, a part of the parabola whose square root function curve is joining the two opposite vertices of a rectangle y=m, y=n and x=m, x=n making the domain and range as [m, n]?\n\nWill it be possible to prove or disprove this idea wrong?\n[/quote]\r\n\r\nSorry for the confusion, I meant for $a=0$, the range is such. Clearly, the range is not always $[0, \\infty)$, as shown by the second solution.", "Solution_4": "[hide]\nHmm, Jarey...choice C was the correct answer on the AMC where this appeared. But in this post the answer should be B since the choices were switched around for some reason...[/hide]" } { "Tag": [], "Problem": "How many positive two-digit integers are there in which each of the two digits is prime?", "Solution_1": "There are $ 4$ prime digits: $ 2$, $ 3$, $ 5$, and $ 7$. Thus, there are $ 4$ choices for the first prime digit, and $ 4$ choices for the second prime digit. Multiplying yields $ 4 \\times 4 \\equal{} \\boxed{16}$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "My friend asked me about a hard problem in inequalities:\r\n[quote]a,b,c >0.\n$X=\\sum{\\frac{ba^2}{c}}$\n$Y=\\sum{\\frac{ca^2}{a}}$\nFind u min and 0 <= u <= 1 to make this inequality true: \n(1- u )X + u. Y <= $(\\sum{a^2})$[/quote]", "Solution_1": "noboby want to answer me! :( :(", "Solution_2": "Maybe $Y=\\sum{\\frac{ca^2}{b}}$ and$(1-u)X+uY\\geq \\sum{a^2}$?", "Solution_3": "a hard problem and nobody can solve", "Solution_4": "No! Because It's not nice and interesting ..." } { "Tag": [ "function", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Give an example of bounded continuous function f on (0, +inf), such that\r\nf(x)->0 as x->+inf. But f not in L^p(0, +inf) for any p>0.\r\n\r\n(1/x almost works, but 1/x is not bounded).", "Solution_1": "$\\frac{1}{x}$ not being bounded is not a big issue, since you can shift its singular point to the left, as in $\\frac{1}{x+1}$. However, this function is in $L^{p}(0,\\infty)$ for $p>1$. You want $f$ to go to $0$ VERY slowly.", "Solution_2": "Try $f(x)=\\frac1{\\log(x+2)}$. That 2 could be replaced by any real number greater than 1, of course.\r\n\r\nThe integral of $f^{p}$ diverges by limit comparison to $\\frac1x$ for all $p$." } { "Tag": [ "FTW", "factorial" ], "Problem": "Title says it all. Decide whether or not you want to consider yourself Asian.\r\n\r\nAsians: +\r\nNon-Asians: -\r\n\r\n(please don't take the +/- as a racist statement :wink: )\r\n\r\nrules same as before, no double posting, blah blah blah.\r\n\r\n+1", "Solution_1": "Non Asian ftw. 0. starting now b/c of survivor thing?", "Solution_2": "Asian.\r\n+1", "Solution_3": "dammit...0.", "Solution_4": "azns are so 1337\r\n\r\n+1", "Solution_5": "this can go on all day...0 :mad:", "Solution_6": "Aren't you not allowed to post twice in an hour? So yeah go back to the 0 before.", "Solution_7": "Aw...let's see someone not asian besides me post here...", "Solution_8": "Why do you guys argue about this?? \r\n\r\nDoes the race really matter??\r\nWell, I am Asian, but why would you want to have a tug of war with race?\r\nI mean we are same and created equal..\r\n\r\nI am going to go with Neutral.", "Solution_9": "-1", "Solution_10": "This isn't a competition to see what race is better, it is a friendly competition in which teams are easy to make. Also we can get sort of an estimate as to how many Asians vs. non-Asians there are on the board. If this is too racially insensitive of a game, a mod can delete it (I personally don't think it is, but that's not for me to decide).\r\n\r\n0", "Solution_11": "-1 haha. ha. now its longer.", "Solution_12": "0\r\nAsian here.", "Solution_13": "Not Asian.\r\n\r\n-1.", "Solution_14": "AZN PRIDE(and reflexes too!)\r\n\r\n0", "Solution_15": "[hide=\"uhh...what comes after 129 again?\"] [hide=\"hold on...im starting from one again. this might take a while\"]1,2,3,4... [hide=\"just kidding\"]130[/hide] [/hide][/hide]", "Solution_16": "this isn't a game anymore so much as you asians being irritating...", "Solution_17": "we like twinkies", "Solution_18": "DOWN WITH TWINKIES!\r\n\r\n129", "Solution_19": "$ 130$\r\n\r\nRubububububu.....wtf!", "Solution_20": "ah..\r\nwhatev...\r\ni'm not a twinkie..\r\ni'm a fob...\r\n\r\n(if you are korean, you may understand this, if you don't, pm me about it)\r\n\r\nanyway..\r\n131..\r\ngaspness\r\na palindrome!!!", "Solution_21": "kind of like pop, race car, and go hang a salami, I'm a lasagna hog :|", "Solution_22": "if we already posted, are we alowed to post again?\r\n\r\n[size=167][b]\u4e00\u767e\u4e09\u5341\u4e8c[/b][/size]\r\n\r\n[hide=\"BEAT THAT SUCKERS!!!!\"]JK :D [/hide]", "Solution_23": "131\r\n\r\nDOWN WITH TWINKIES!", "Solution_24": "Oh my god somebody lock/delete this thread now.", "Solution_25": "[b]\u4e00\u767e\u4e09\u5341\u4e8c[/b]", "Solution_26": "131\r\n\r\nAsians still rule the world.", "Solution_27": "132\r\n\r\nIs it just me or is 1=2 the only Non-Asian?", "Solution_28": "133\r\n\r\nAh ha $ 1\\plus{}3\\plus{}3\\equal{}7$.", "Solution_29": "the non asians have gone to do non asian things. come on 1=2, what are you still doing here?" } { "Tag": [ "inequalities", "algebra", "function", "domain", "calculus", "derivative" ], "Problem": "Let $ a,b,c>0, a\\plus{}b\\plus{}c\\equal{}1$. Prove that \r\n$ (a^2\\plus{}1)(b^2\\plus{}1)(c^2\\plus{}1) \\ge 8$", "Solution_1": "[quote=\"NguyenDungTN\"]Let $ a,b,c > 0, a \\plus{} b \\plus{} c \\equal{} 1$. Prove that \n$ (a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1) \\ge 8$[/quote]\r\n\r\nFirst, I think the condition you mean is $ a \\plus{} b \\plus{} c \\equal{} 3$. Second, even in this case, the problem is not true. For example, if $ a \\equal{} 2$ and $ b \\equal{} c \\equal{}1/2$, then\r\n\\[ (a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1) \\equal{} 5 \\cdot \\frac{5}{4} \\cdot \\frac{5}{4} \\equal{} \\frac{125}{16} < 8.\\]", "Solution_2": "[quote=\"nsato\"][quote=\"NguyenDungTN\"]Let $ a,b,c > 0, a \\plus{} b \\plus{} c \\equal{} 3$. Prove that \n$ (a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1) \\ge 8$[/quote]\n\nFirst, I think the condition you mean is $ a \\plus{} b \\plus{} c \\equal{} 3$. Second, even in this case, the problem is not true. For example, if $ a \\equal{} 2$ and $ b \\equal{} c \\equal{} 1/2$, then\n\\[ (a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1) \\equal{} 5 \\cdot \\frac {5}{4} \\cdot \\frac {5}{4} \\equal{} \\frac {125}{16} < 8.\n\\]\n[/quote]\r\n\r\nThank you, nsato! But this inequality is true if $ ab\\plus{}bc\\plus{}ca\\equal{}3$, problem from Agady.", "Solution_3": "$ (a^2\\plus{}1)(b^2\\plus{}1)(c^2\\plus{}1)\\ge \\frac{125}{16}$ is true for $ a\\plus{}b\\plus{}c\\equal{}3$, $ a,b,c\\ge 0$.\r\n\r\nThe domain is a closed and bounded subset of $ R^3$, and both the functions and their partial derivatives are continuous. Grad g is not zero... So we can use lagrange multipliers...\r\n\r\nIt ends up that we need to solve $ \\frac{a}{a^2\\plus{}1}\\equal{}\\frac{b}{b^2\\plus{}1}\\equal{}\\frac{c}{c^2\\plus{}1}$ and $ a\\plus{}b\\plus{}c\\equal{}3$.\r\n\r\nSubtract one from another, and clear the denominator which is nonzero, $ (a\\minus{}b)(ab\\minus{}1)\\equal{}0$. \r\n\r\ni) if $ a\\equal{}b\\equal{}c$, then $ a\\equal{}1$, giving $ 8$.\r\n\r\nii) if $ a\\equal{}b$, $ a\\ne c$, then $ a\\equal{}b$, $ ac\\equal{}bc\\equal{}1$. Then $ c(a\\plus{}b\\plus{}c\\minus{}3)\\equal{}0\\equal{}1\\plus{}1\\plus{}c^2\\minus{}3c\\equal{}(c\\minus{}2)(c\\minus{}1)$. If $ c\\equal{}1$, $ a\\equal{}b\\equal{}c$, so $ c\\equal{}2$. Then $ a\\equal{}b\\equal{}\\frac{1}{2}$, which gives $ \\frac{125}{16}$.\r\n\r\niii) if $ a\\ne b$, $ b\\ne c$, $ c\\ne a$. $ ab\\equal{}bc\\equal{}ca\\equal{}1$, so $ a\\equal{}b\\equal{}c\\equal{}1$, contradiction.\r\n\r\nThen we check the endpoints.\r\n\r\niv) If $ c\\equal{}0$, Then $ (a\\minus{}b)(ab\\minus{}1)\\equal{}0$, $ a\\plus{}b\\equal{}3$. This gives $ (a,b,c)\\equal{}(1.5,1.5,0),(\\frac{3\\minus{}\\sqrt{5}}{2},\\frac{3\\plus{}\\sqrt{5}}{2},0)$. which give $ \\frac{169}{16}$ and $ 9$, respectively.\r\n\r\nv) If $ c\\equal{}b\\equal{}0$, then $ a\\equal{}3$, giving $ 10$.\r\n\r\nSo the max and min are $ \\frac{169}{16}$, and $ \\frac{125}{16}$, respectively." } { "Tag": [ "limit", "function", "algebra", "domain", "calculus", "real analysis", "real analysis theorems" ], "Problem": "Hi all, I'm new.. I hope someone can help me, I don't understand this... my prof touched on it briefly, and I read the book but still don't quite understand it.\r\n\r\nOkay, why are the following both continous at a point?\r\n\r\nlim f(x) = x^2 - 9 = 0\r\nx --> 3\r\n\r\nand\r\n\r\nlim f(x) = | t - 2 | = 1\r\nx --> 3\r\n\r\nSo in both cases, if f(x) > 0 but <= 3, then it's continous? I know it's a simple way of trying to understand it, but I don't understand the defintion of continuity at a point. Can someone explain it to me, in the simplest/easiest way possible?\r\n\r\nThanks in advance...", "Solution_1": "Definition of continuity: Given $\\epsilon>0$, there exists a $\\delta>0$ such that whenever\r\n\\[ 0<|x-a|<\\delta\\Rightarrow |f(x)-f(a)|<\\epsilon \\]\r\nthen $f(x)$ is continuous at $x=a$.\r\nOther versions I have seen are like:\r\n$f(x)$ is continuous at $x=a$ if they meet the following conditions:\r\n1) $\\lim_{x\\to a}f(x)$ exists\r\n2) $\\lim_{x\\to a}f(x)=f(a)$\r\n\r\nFirst, to see the intuitive meaning, think of continuity like a line that has no 'broken gaps', no holes, that the graph is a 1-piece curve. For example, the graph of $y=x$ is a straight slanted line that has no holes, no matter how much you zoom in the curve, you cannot find a hole. That's what continuity is about. How is continuity linked to limits? Limits talks about approaching a certain value but continuity talks about approaching a certain value in a nice smooth curve with no holes. This is somewhat what I think continuity means.\r\n\r\nSecond, let's see the not-so beautiful side of continuity, a function that is not continuous. For example, let $f(x)=\\frac{x^2-1}{x-1}$. This function is not continuous at $x=1$ because it has a hole in it when you graph it. Why? Because you cannot divide by zero. If you try to graph it, it shows also a straight line similar to $y=x+1$ but with one exception: it has a hole at $x=1$. There are other kind's of discontinuities like the graph of $y=\\frac1x$\r\n\r\nThird, if you want to see how to work with $\\delta-\\epsilon$, I'll show you in a follow-up post. Hope you understand the intuitive meaning of continuity that way you can get more comfortable with it.", "Solution_2": "Hi, thanks for the reply. I think I get what you mean by continuity, graphically-speaking. But I'm not familiar with the delta/epsilon signs.. I guess I need to review on those ><\r\nBut I guess I'm trying to understand it numerically... so on the example you gave, since f(x) = 2 as it approaches 1, and f(1) = 2 and not 1, that's why it's not continuous? And on my examples from my first post, is the domain between 0 to 3..? So if it falls within that domain, then it's continuous?", "Solution_3": "The basic idea of the $\\epsilon - \\delta$ definition is this: a continuous function can't jump around too much in short spaces. In the definition of continuity, you can pick as small an $\\epsilon$ as you want, so that the function is only allowed to vary by a tiny bit, and it is still possible to find a little interval around $x$ where the function doesn't change by more than $\\epsilon$. \r\n\r\nAs 10000th user said, f a function jumps, then it obviously can't be continuous at that point. Can you think of any other ways that a function might not be continuous? \r\n\r\nIn his first example of a discontinuous function, the problem was that the function wasn't defined at $x = 1$. \r\n\r\nIn your example, the function is defined everywhere; it is a parabola. It is continuous because in a sense, the more you \"zoom in\" near $x=3$, the flatter the function will look. For example, in the interval $[3 - 1/1000, 3 + 1/1000]$, the function varies by about 3/250. Not a lot, eh? If you want to make this rigorous, you'd have to use the $\\epsilon-\\delta$ approach -- find $\\delta$ in terms of $\\epsilon$ so that inside the interval $[3 - \\delta, 3 + \\delta]$, the function varies by less than $\\epsilon$.", "Solution_4": "Well, I think you have got the meaning of continuity in the previous posts.\r\n \r\n a) In post 2(a typo): delete from the defintion this : $0<|x-a|.....$ and write $|x-a|, \\delta$\r\n \r\n b) You must imagine continuity like this\r\n << you have 10 balls which touch each other and through them on the space. If the keep touching with the initial order , then the function of their position is continuous. Otherwise, not. >>\r\n\r\n c) The delta - epsilon definition of the limit is really difficult. I had finished the Uni and only after it I could -easily- solve problems with this way. The best way is to see many solved problems in some good calculus books(spivak ect).\r\n\r\n[u] babis[/u]", "Solution_5": "Thanks guys... I do understand what it is now, but it's just when solving numerical problems I don't see it unless I graph it... the explanations were helpful, but I guess you all are thinking as mathematicians were I think as an average student trying to pass Calculus :D\r\n\r\n[quote=\"stergiu\"]\n c) The delta - epsilon definition of the limit is really difficult. I had finished the Uni and only after it I could -easily- solve problems with this way. The best way is to see many solved problems in some good calculus books(spivak ect).\n[/quote] \r\n\r\nYes, I haven't even learned this yet I think.. my prof didn't mention the delta-epsilon defn., all he said was that if \"f\" is defined on an open interval containing \"a,\" then \"f\" is continous at \"a\" \r\nif lim x --> a f(x) = f(a)\t\r\n\r\nSo I guess I'm trying to understand what he meant by this numerically, if I solve for f(x) and it's not equal to f(a) or if it doesn't fall within that domain, then it's continous... I don't even know if that's the correct way of thinking.. but that's how I understood it...\r\n\r\nAnyway, thanks again guys.. this forum is very helpful!", "Solution_6": "The definition with the limit is actually the same -- guess why? It's because the limit is formally defined using epsilons and deltas in a similar way. :P \r\n\r\nThose proofs are difficult to learn. Don't feel discouraged because math is difficult. If your work is hard, it means you're learning from it (usually)." } { "Tag": [ "geometry", "inequalities", "triangle inequality" ], "Problem": "176. Show that if $h_{a}$, $h_{b}$ and $h_{c}$ are the altitudes of a triangle, then\r\n\r\n$\\frac{1}{h_{a}}< \\frac{1}{h_{b}}+\\frac{1}{h_{c}}$\r\n\r\n194. A point is selected inside an equilateral triangle. From this point perpendiculars are dropped to each side. Show that the sum of the lengths of these perpendiculars is equal to the altitude length.", "Solution_1": "[hide=\"176\"]\nLet the triangle area be K\n$a * h_{a}= 2K$\n$b * h_{b}= 2K$\n$c * h_{c}= 2K$\n\n$\\frac{1}{h_{a}}+\\frac{1}{h_{b}}= \\frac{a+b}{2K}> \\frac{1}{h_{c}}= \\frac{c}{2K}$\n\nSince, $a+b > c$ by the triangle inequality\n[/hide]", "Solution_2": "[hide=\"194\"]Connecting the point to the vertices, we form 3 triangles (the point and two vertices for each). Let the altitude of the equilateral triangle be $h$, let the side length be $a$ and the altitudes of the three smaller triangles be $h_{1}$, $h_{2}$, and $h_{3}$. We can write the area of the equilateral triangle in two ways; in terms of $h$ and in terms of $h_{1}$, $h_{2}$, and $h_{3}$:\n\n$A=\\frac{a\\cdot h}{2}=\\frac{a(h_{1}+h_{2}+h_{3})}{2}$. Dividing out $\\frac{a}{2}$ from both sides, we have the equality\n\n$h_{1}+h_{2}+h_{3}=h$. This is what we desired and our proof is complete.\n\n$\\mathbb{QED}$[/hide]", "Solution_3": "[hide=\"176\"]From Triangle Inequality, $a3$, I see no reason why this case is easier.\r\n\r\nThe $n$-dimensional version is:\r\nGiven a convex region $R$ symmetrical to the origin in the $n$-dimensional space. Show that if $R$ has volume greater than $2^n$, then $R$ contains a lattice point different from the origin." } { "Tag": [ "function", "integration", "calculus", "real analysis", "real analysis solved" ], "Problem": "P8 \r\n\r\naR is a continuous function s.t. |f(x)| <= | \\int{a,x} f(t)dt| for all x in [a, b]. Prove that f(x)=0 for all x in [a, b].", "Solution_1": "Use only : |f| is bounded and its integral exists\r\n|f(x)| =< |Integral(a =< t =< x){f(t).dt}| =< Integral(a =< t =< x){|f(t)|.dt}| \r\nLet n be a natural such that n > b-a , and note u = (b-a)/n < 1. Note also M = sup(a =< t =< a+u){|f(t)|}\r\nManifestly, |f(x)| =< Integral(a =< t =< a+u){|f(t)|.dt}| =< u*M for all x in [a,a+u] , so :\r\n0 =< M =< u*M ===> M=0 ===> f(x)=0 on [a,a+u] , and f(x) =< Integral(a+u =< t =< x){|f(t)|.dt}| for all x in [a+u,b] \r\nBy using the same way it is easy to see f(x)=0 on [a+u,a+2*u], ... ,f(x)=0 on [a+(n-1)*u, a+n*u=b]\r\nFinally f(x)=0 everywhere on [a,b] .." } { "Tag": [ "ARML", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Can anyone give a list of big high school math competitons in Indiana besides AMC?", "Solution_1": "arml is one.", "Solution_2": "USAMO.\r\n\r\nEat this small message!", "Solution_3": "I think thats a little too hard. I dont know a single person that passed it.\r\n\r\n\r\nI mean like Indiana only competitons, those people from other states can do too.", "Solution_4": "in-state competitions aren't gonna get you anywhere, like ICTM.", "Solution_5": "But the point of this was INDIANA math competitons.\r\n\r\nLike I said, USAMO is a national one. Besides, its too hard and Ill probalby never pass it.\r\n\r\nDoes anyone have any advice or can post old amc probelmsA", "Solution_6": "[quote=\"vikas\"]\n\nDoes anyone have any advice or can post old amc probelmsA[/quote]\r\n\r\nYea whats 1+1.", "Solution_7": "i know there is a rose-something competition that exists in Indiana don't know the exact name", "Solution_8": "[quote=\"veryfriendly\"]i know there is a rose-something competition that exists in Indiana don't know the exact name[/quote]\r\n\r\n\r\nrose hulman? I will be going to that.", "Solution_9": "Behold the magical [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php#in]url[/url] that tells all!", "Solution_10": "hey do you guys know how/where you can take the\r\n\r\nUSNCO, USAPHO, or the USABO...\r\n\r\nit seems like my school doesnt offer it...", "Solution_11": "Indiana Mathleague and whoa!!!!! WHO ARE ALL THESE NEW PPL??\r\n\r\nMy sister took the usa chemistry olympiad\r\nnext year, im taking physics olymp.", "Solution_12": "erm so how can we take it? I want to take it too..", "Solution_13": "well, the semifinals are this saturday, so its probably too late to take it this year." } { "Tag": [ "algebra", "polynomial", "binomial theorem", "algebra unsolved" ], "Problem": "Let $P(x)$ be a polynomial of degree $n$, $n \\in \\mathbb{N}\\/$, such that $P(k)=2^{k}\\/$ for each $k =1,2, \\ldots, n+1\\/$. Determine $P(n+2)\\/$.", "Solution_1": "This does not really belong to the olympiad section (too easy: a complex polynomial of degree $n$ is entirely determined by $n+1$ of its values ...)", "Solution_2": "Moved to algebra.", "Solution_3": "Here's one approach. Define the polynomial $Q$ by the equation \\[Q(x) = 2 \\sum_{j=0}^{n}\\binom{x-1}{j}\\, .\\] By the binomial theorem, we have $Q(k) = 2^{k}= P(k)$ for all integers $k$ from $1$ to $n+1$. Furthermore, $Q$ is a polynomial of degree $n$. Thus, by julien_santini's observation, the polynomials $Q$ and $P$ must be identical.\r\n\r\nSo we have \\begin{eqnarray*}P(n+2) & = & Q(n+2) \\\\ & = & 2 \\sum_{j=0}^{n}\\binom{n+1}{j}\\\\ & = & 2 \\left[\\sum_{j=0}^{n+1}\\binom{n+1}{j}-1\\right] \\\\ & = & 2 \\left[ 2^{n+1}-1 \\right] \\\\ & = & 2^{n+2}-2 \\, . \\end{eqnarray*}", "Solution_4": "Nice solution. I give these solution half year ago in Russian forum." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that there exists exactly one subgroup order $n$ in the group $(\\mathbb{Q}/ \\mathbb{Z},+)$.", "Solution_1": "Let $H$ be a subgroup of order $n$.\r\nTake for every element the least positive representant and take the smallest element $x$ by this.\r\nThen all other elements are multiples of $x$ (otherwise it contradicts the choose of $x$), so $x$ generates $H$. Now the order of $x$ is exactly its denominator in reduced form, thus $x$ has denominator $n$ and then also $x=\\frac{1}{n}$ (since otherwise theres a smaller one).", "Solution_2": "Another way to put it: since any two elements are commensurable, every finitely generated subgroup is generated by one element. Since every element has finite order and every finite group is finitely generated, each group of order $n$ is generated by a single element of order $n$. All elements of order $n$ are of the form $\\frac kn$, and all such elements are contained in the group $\\left(\\frac1n\\mathbb{Z}\\right)/\\mathbb{Z}$, which has $n$ elements." } { "Tag": [ "geometry", "incenter", "circumcircle", "trigonometry", "trapezoid", "inradius", "ratio" ], "Problem": "Let $ABC$ be a scalene, acute-angled triangle with orthocenter $H$, incenter $I$. The lines $CH$ and $CI$ intersect the circumcircle of $\\triangle ABC$ at $D$ and $L$, respectively. Prove that $\\angle CIH$ is a right angle $\\longleftrightarrow$ $\\angle IDL$ is a right angle.", "Solution_1": "Let (O) be the circumcircle, R the circumradius, and r the inradius. For any acute triangle $\\triangle ABC$, $CH = 2R \\cos C.$ (If the angle $\\angle C$ was obtuse, we would have $CH = -2R \\cos C.$) WLOG, assume $\\angle A < \\angle B$. Then\r\n\r\n$\\angle ICH = \\angle ICB - \\angle HCB = \\frac{\\angle C}{2} - (90^\\circ - \\angle B) = \\frac{\\angle B - \\angle A}{2}$\r\n\r\nThe angle $\\angle CIH$ is right iff $\\frac{CI}{CH} = \\cos \\frac{B - A}{2}.$ Thus we have to show that $\\angle IDL = 90^\\circ$ is right if this condition is true. L is the midpoint of the circumcircle arc AB opposite to the vertex C. Let K be the other end of the circumcircle diameter KL. The line $KL \\perp AB$ is the perpendicular bisector of AB. The diameter KL forms an isosceles trapezoid CDLK with the altitude CD. Since the diagonal CL is the bisector of the angle $\\angle C$, the incenter $I \\in CL$. Since the angle $\\angle KDL = 90^\\circ$ on the diameter KL is right, the incenter $I \\in KD$ lies also on the other diagonal KD. Therefore, the incenter is identical with the diagonal intersection $I \\equiv CL \\cap KD$ of this isosceles trapezoid. Therefore, the diacentral line IO is the perpendicular bisector of both CD and KL, hence, it is parallel to the side AB (both perpendicilar to KL). Let $M \\equiv AB \\cap KL$ be the midpoint of AB. Then $r = OM = R \\cos C$ and $CH = 2R \\cos C = 2r.$ Using Euler's formula $\\frac{1}{r} = \\frac{1}{R + d} + \\frac{1}{R - d}$ or $R^2 - d^2 = 2rR,$ where\r\n\r\n$d = OI = OL \\tan \\widehat{OLI} = OL \\tan \\widehat{KLC} = R \\tan \\frac{B - A}{2}$\r\n\r\n$R^2 \\left(1 - \\tan^2 \\frac{B - A}{2}\\right) = 2rR$\r\n\r\n$\\frac r R = \\frac{\\cos^2 \\frac{B - A}{2} - \\sin^2 \\frac{B - A}{2}}{2\\cos^2 \\frac{B - A}{2}} = \\frac{\\cos (B - A)}{1 + \\cos (B - A)}$\r\n\r\nCombining the 2 expressions for the inradius-circumradius ratio,\r\n\r\n$\\frac{\\cos (B - A)}{1 + \\cos (B - A)} = \\cos C = -\\cos(B + A)$\r\n\r\n$\\cos (B - A) + \\cos (B + A) = -\\cos(B - A) \\cos(B + A)$\r\n\r\n$2 \\cos B \\cos A = -\\frac{\\cos (2B) + \\cos (2A)}{2} = 1 - \\cos^2 A - \\cos^2 B$\r\n\r\n$(\\cos B + \\cos A)^2 = 1,\\ \\ \\cos B + \\cos A = 1$\r\n\r\nLet F be the foot of the C-altitude. The parallel bases of the isosceles trapezoid CDLK are in the ratio\r\n\r\n$\\frac{CD}{KL} = \\frac{CH + 2 HF}{KL} = \\frac{2R \\cos C + 4R \\cos A \\cos B}{2R} =$\r\n\r\n$= \\cos C + 2 \\cos A \\cos B = \\cos (B - A)$\r\n\r\nThe diagonals CL = KD cut each other in the same ratio and their length is\r\n\r\n$\\frac{CL}{KL} = \\frac{KD}{KL} = \\cos \\widehat{DKL} = \\cos \\widehat{DCL} = \\cos \\frac{B - A}{2}$\r\n\r\nIn addition, the incenter I lies on a circle (L) with radius\r\n\r\n$LI = LB = LC = \\frac{c}{2 \\cos \\frac C 2} = \\frac{2R \\sin C}{2 \\cos \\frac C 2} = 2R \\sin \\frac C 2$\r\n\r\nConsequently,\r\n\r\n$\\cos (B - A) = \\frac{CI}{LI} = \\frac{CI}{2R \\sin \\frac C 2},\\ \\ CI = 2R \\sin \\frac C 2 \\cos (B - A)$\r\n\r\n$\\frac{CI}{CH} = \\frac{2R \\sin \\frac C 2 \\cos (B - A)}{2r} = [1 + \\cos (B - A)] \\sin \\frac C 2 =$\r\n\r\n$= 2 \\cos^2 \\frac{B - A}{2} \\cos \\frac{B + A}{2} = (\\cos B + \\cos A) \\cos \\frac{B - A}{2} =$\r\n\r\n$= \\cos \\frac{B - A}{2} = \\frac{KD}{KL}$\r\n\r\nAs a result, the triangles $\\triangle KDL \\sim \\triangle CIH$ with the equal angles $\\angle DKL = \\angle DCL \\equiv HCI$ are similar by SAS. Hence the corresponding angles $\\angle CIH = \\angle KDL \\equiv IDL = 90^\\circ$ are equal and right.", "Solution_2": "[color=darkred][b]Enunciation.[/b] Let $ABC$ be a scalene and acute triangle. Denote the intersections $D$, $L$ of the open rays $(CH$, $(CI$ respectively with the circumcircle of the given triangle. Prove that[/color]\r\n$\\boxed {\\ DI\\perp DL\\Longleftrightarrow IH\\perp IC\\Longleftrightarrow r=R\\cos C\\ }\\Longleftrightarrow\\cos A+\\cos B=1\\ .$\r\n\r\n[b]Proof.[/b] Suppose $a>b$. Remember some well-known relations in a triangle $ABC$ :\r\n$m(\\widehat {HCI})=m(\\widehat {CLO})=\\frac{A-B}{2}\\ ;$ $IC\\cdot IL=2Rr\\ ;$ $r=IC\\sin \\frac C2\\ ;$ $m(\\widehat {ILD})=90^{\\circ}-(A-B)\\ ;$ $CL=2R\\cos \\frac{A-B}{2}\\ ;$ $DL=2R\\sin \\frac{A-B}{2}\\ ;$ $IC^2=\\frac{ab(p-c)}{p}\\ ;$ $(p-a)(p-b)(p-c)=pr^2\\ ;$ $1-\\cos C=2\\sin^2\\frac C2\\ ;$ $\\sin \\frac C2=\\sqrt{\\frac{(p-a)(p-b)}{ab}}\\ ;$ $\\cos A+\\cos B+\\cos C=1+\\frac rR\\ .$\r\n\r\n$1.\\blacktriangleright$ $\\boxed {\\ DI\\perp DL\\ }$ $\\Longleftrightarrow$ $DL=IL\\cos \\widehat {ILD}$ $\\Longleftrightarrow$ $2R\\sin \\frac{A-B}{2}=IL\\sin (A-B)$ $\\Longleftrightarrow$ $R=IL\\cos \\frac{A-B}{2}$ $\\Longleftrightarrow$ $R\\cdot IC=2Rr\\cos \\frac{A-B}{2}$ $\\Longleftrightarrow$ $IC=2r\\cos \\frac{A-B}{2}$ $\\Longleftrightarrow$ $1=2\\sin \\frac C2\\cos\\frac{A-B}{2}$ $\\Longleftrightarrow$ $1=2\\cos \\frac{A+B}{2}\\cos \\frac{A-B}{2}$ $\\Longleftrightarrow$ $1=\\cos A+\\cos B$ $\\Longleftrightarrow$ $1+\\cos C=1+\\frac rR$ $\\Longleftrightarrow$ $\\boxed {\\ r=R\\cos C\\ }\\ .$\r\n\r\n$2.\\blacktriangleright$ $\\boxed {\\ IH\\perp IC\\ }$ $\\Longleftrightarrow$ $IC=HC\\cos \\frac{A-B}{2}$ $\\Longleftrightarrow$ $IC=2R\\cos C\\cos\\frac{A-B}{2}$ $\\Longleftrightarrow$ $IC=CL\\cos C$ $\\Longleftrightarrow$ $IC=(CI+IL)\\cos C$ $\\Longleftrightarrow$ $IC(1-\\cos C)=IL\\cos C$ $\\Longleftrightarrow$ $IC^2(1-\\cos C)=2Rr\\cos C$ $\\Longleftrightarrow$ $\\frac{ab(p-c)}{p}\\cdot 2\\sin ^2\\frac C2=2Rr\\cos C$ $\\Longleftrightarrow$ $\\frac{2ab(p-c)}{p}\\cdot \\frac{(p-a)(p-b)}{ab}=2Rr\\cos C\\Longleftrightarrow$ \r\n$\\prod (p-a)=Rrp\\cos C\\Longleftrightarrow pr^2=Rrp\\cos C$ $\\Longleftrightarrow$ $\\boxed {\\ r=R\\cos C\\ }\\ .$" } { "Tag": [ "real analysis", "limit", "abstract algebra" ], "Problem": "Oznaczmy przez $( a_n )_{n = 1}^{\\infty}$ rosnacy (strictly $>$) ciag liczb rzeczywistych.\r\nUdowodnij ze\r\n\\[ \\sum_{n = 1}^{\\infty} \\frac{1}{a_n} < + \\infty \\Rightarrow \\sum_{n = 1}^{\\infty} | \\frac{a_n}{a_{n + 1}} - \\frac{n + 1}{a_{n + 1}} | < + \\infty \\]\r\n\r\nP.S: Nie mam zielonego pojecia jak \"to\" zaatakowac.", "Solution_1": "Takie cos powinno dzialac:\r\n\r\n1) Korzystasz z tw. (cwiczenie 8, Rudin \"Podstawy analizy matematycznej\" - dzial ciagi i szeregi):\r\n\r\nJesli $\\{b_n\\}$ jest monotoniczny i ograniczony, a $\\sum a_n$ jest zbie\u017cny to $\\sum a_n b_n$ jest zbiezny\r\n\r\n2) Na mocy tego wystarczy wykazac, ze $\\sum |a_n - n-1|$ jest zbiezne (tym samym dostaniemy mocniejszy rezultat)\r\n\r\n3) $\\sum \\frac{1}{a_n} < + \\infty$ implikuje $a_n = \\Omega (n^{1+\\epsilon})$. I to na moj gust zalatwia sprawe. \r\n\r\nW razie czego gdyby powyzsze bylo totalnym bezsensem tlumacze sie pozna pora i ogolna sennoscia :P :D", "Solution_2": "W 3) miales na mysli $\\frac{1}{a_n}$ zamiast $a_n$ ? (wynikaloby to z uzycia $a_n$ w 2))\r\n2) nie zadziala : podstaw $a_n = n^2$...\r\nNatomiast z 1, i 3 chyba da sie jakies rozwiazanie sklecic ... :roll:\r\nCzekam jutro ... ;)", "Solution_3": "Jak tak teraz sie przyjrzalem to czy to zadanie jest w ogole prawdziwe? \r\n\r\nWezmy ten przyklad, ktory podales u gory $a_n=n^2$, a wtedy dostajesz $\\lim_{n \\to \\infty} \\frac{a_n -n-1}{a_{n+1}}=\\lim_{n \\to \\infty} \\frac{n^2-n-1}{n^2+2n+1}= \\lim_{n \\to \\infty} \\frac{1-\\frac{1}{n}-\\frac{1}{n^2}}{1+\\frac{2}{n}+\\frac{1}{n^2}}=1$", "Solution_4": "Najwyrazniej nie jest prawdziwe ;) (shame on me :oops:). Natomiast dalbym czesc mozgu odpowiedzialna za matematyczne myslenie uciac ze dokladnie to pytanie pojawilo sie w CMB (Canadian Mathematical Bulletin)...", "Solution_5": "To tak przy okazji sie zapytam skad ty wziales CMB ? Jest to jakos dostepne latwo w internecie, czy po prostu ruszyles do odpowiedniej biblioteki :)", "Solution_6": "Jest dostepne online ;) N.p take a look here : \r\nhttp://endeavor.macusa.net:591/macusa/mathpropress/FMPro?-db=problems.fp5&-format=mainresult.html&-lay=all&ProblemName=CMB&-find=\r\nNa owej stronie (problemcorner.org) znajdziesz problemy z innych czasopism, chocby z AMM ;)", "Solution_7": "Wielkie dzieki - problemy wygladaja ciekawie :D :D :D odnalazlem tez poprawna wersje powyzszego: powinno byc:\r\n\r\n$\\sum |\\frac{n}{a_n}-\\frac{n+1}{a_{n+1}}| < \\infty$", "Solution_8": "Oznaczmy $\\sum \\frac{1}{a_n} = M$\r\n\r\nSzacujemy:\r\n\r\n$\\sum |\\frac{n}{a_n}-\\frac{n+1}{a_{n+1}}| = \\sum |\\frac{n}{a_n}-\\frac{n+1}{a_n}+\\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}}| < \\sum |\\frac{n}{a_n}-\\frac{n+1}{a_n}|+|\\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}}| = \\sum |\\frac{1}{a_n}|+|\\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}}| = M + \\sum |\\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}}|$\r\n\r\nLiczba w module jest dodatnia, czyli\r\n\r\n$\\sum |\\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}}| = \\sum \\frac{n+1}{a_n}-\\frac{n+1}{a_{n+1}} = \\frac{1}{a_1} + \\sum \\frac{1}{a_n}$\r\n\r\nWi\u0119c ca\u0142a suma ograniczona jest przez $2M+\\frac{1}{a_1}$.", "Solution_9": "Jest tylko jeden malusienki bledzik na samym koncu:\r\n\\[ \\sum_{n = 1}^{\\infty} \\left( \\frac{n + 1}{a_n} - \\frac{n + 1}{a_{n + 1}} \\right) = \\left( \\frac{2}{a_1} - \\frac{2}{a_2} \\right) + \\left( \\frac{3}{a_2} - \\frac{3}{a_3} \\right) + \\ldots = \\frac{2}{a_1} + M \\]", "Solution_10": "Nie. Jedno z tych $\\frac{2}{a_1}$ wchodzi do $M$.", "Solution_11": "All right :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "hello i'm new here and don't know where this might go...but anywayshere is a good problem i just solved recently:\r\nif in a game there are only 2 possible scores $a$ and $b$(relativly primes), (u can choose to get either one each time) then:\r\ni)what is the largest score that can not be made by these numbers?(proof needed)\r\nii)what is the number of impossible scores?(proof needed)\r\n\r\nplease don't say for part (i), based on the Chicken McNugget Theorem it has to be $ab-a-b$--it would be unacceptable :) \r\n\r\ngood luck. :)", "Solution_1": "The coin exchange problem of Froebenius (what is the chicken MacNugget theorem???) is solved here :\r\n\r\nhttp://www.animath.fr/cours/arithm.pdf\r\n\r\nHave a look at pages 30-32. \r\n\r\nPierre.\r\n\r\nP.S. If you have solved this problem then post it in the proposed and own problem section, not in the unsolved problem one.", "Solution_2": "sorry about that, \r\nbut since i don't know any french, can u translate that? (i'm interested)\r\ni'm more interested in the solution for the second part\r\n\r\nthanx", "Solution_3": "i don't have much time right now but\r\nthe answer for the second part is:\r\n$\\frac{(a-1)(b-1)}{2}$" } { "Tag": [ "Support", "probability", "geometry", "calculus", "derivative", "linear algebra" ], "Problem": "This is an serious and sensitive question. Do you agree with induced abortion?", "Solution_1": "So, first, as moderator, I'd like to remind anyone who posts on this topic to be civil or I will be forced to use some combination of editing, deleting or locking, which I'd rather not do.\r\n\r\n\r\nSecond, I have no idea what you mean by \"induced abortion.\" Perhaps you could give a more detailed description of the circumstances of abortion you are asking about?", "Solution_2": "[quote=\"JBL\"]So, first, as moderator, I'd like to remind anyone who posts on this topic to be civil or I will be forced to use some combination of editing, deleting or locking, which I'd rather not do.\n\n\nSecond, I have no idea what you mean by \"induced abortion.\" Perhaps you could give a more detailed description of the circumstances of abortion you are asking about?[/quote]\r\n \r\n\r\n You don't understand? Maybe I've made mistake. I don't know English very much, so I looked it up in my Chinese-English dictionary, my dictionary told me that the word I wanted to say is the phrase \"induced abortion\", if that's wrong, that should be the problem of my dictionary, I should say sorry.\r\n For the phrase \"induced abortion\", the word \"abortion\" in it I think everyone may understand. For the word \"induced\", I've looked it up in the dictionary again, the meaning in Chinese is difficult to understand. So, I think if the reader have no idea at the word \"induced\", you can just ignore it and only look at the word \"abortion\".\r\n I'm sorry for my poor English. :blush:", "Solution_3": "I think that abortions are bad, but at the same time if somebody gets a child they don't want, they might take bad care of them so it's a tough issue. I don't believe though, that this is a very important issue in politics because I don't that it is a big enough problem in the current world.\r\n\r\nHow come youe sn is Alexander Lenin? just wondering.", "Solution_4": "[quote=\"hello\"]How come youe sn is Alexander Lenin? just wondering.[/quote]\r\n Could you say in complete sentence? I can't understand that because of my poor English :blush: .", "Solution_5": "I mean your name on the left side on the screen on this aops forum is Alexander Lenin, but you are from China. I was wondering how you came up with such a screen name?", "Solution_6": "[quote=\"hello\"]I mean your name on the left side on the screen on this aops forum is Alexander Lenin, but you are from China. I was wondering how you came up with such a screen name?[/quote]\r\n\r\nHmm......1, Just for fun; 2, for Alexander, just because I adore him; 3, for Lenin, I also respect him, and just give a last name", "Solution_7": "What do you think about abortions?", "Solution_8": "[quote=\"hello\"]What do you think about abortions?[/quote]\r\n\r\nI voted strongly disagree. Although in my country that's lawful. I can tell you, in my country, abortion is not only lawful, even that's advocated in certain extent, that's really terrible, I'm sad about the fact :(", "Solution_9": "Because China has a large population. Wasn't there a law before that the ideal family was with one male child in the family. Oh well, I think I will go sleep now, I am in Philly.", "Solution_10": "[quote=\"hello\"]Wasn't there a law before that the ideal family was with one male child in the family. [/quote]\r\n\r\nYou are wrong about with one male child, it should be changed into \"the more male children are, the better\"", "Solution_11": "for some reason, i strongly support it. i mean, if the parents don't want the child, how will they be able to raise it in a good way then? :? \r\nand what if that person later finds out his parents didn't want him at all. how should he feel then?", "Solution_12": "slightly support...\r\nI think if the parents all agree on that, then it is ok. After all, it is their right to do it.", "Solution_13": "STRONGLY SUPPORT. Now that doesn't mean that I am pro abortion and that everyone should get an abortion--NOT AT ALL! We need to make sure to take every precaution to prevent an abortion, BUT in the end, if one is necessary, I support the WOMAN'S RIGHT TO CHOOSE! \r\n\r\nI think that conservative right-to-life advocates are just being very silly--don't the 1500 soldiers who've died in Iraq have the right to life?", "Solution_14": "Although I do agree on a womens right to choose...this gets a little touchy.\r\n\r\nHaving sex has its consequences. Pregnancy is one of them. If they can't afford to have the baby... Are too young...etc. they should go with a much better alternative which is to put the child up for adoption.\r\n\r\nThe other thing is that the baby is considered \"unborn\" at that stage...So people think it really doesn't matter. But imagine what would life be like if you were aborted. :? \r\n\r\nI dunno...I have mixed feelings on this... I don't consider it murder, but I do find it morally incorrect.\r\n\r\nI don't see where all this stuff about a woman's right to choose is coming from though. If you really don't want kids go get a gonangiectomy. You have a right to choose there.\r\n\r\nTo have unprotected sex or not.... You have a right to choose there.\r\n\r\nNOW...\r\n\r\nIf the mother is in the condition in which the baby is in someway life threatening, causes psychological damage (rape, etc) I can understand why an abortion should be appropriate. This is the only case in which I completely support abortion.\r\n\r\nAbout those soldiers:\r\n\r\nWhen you sign up for the military, you should know that you are putting your life on the line. Regardless of whether it is wartime or not. It is a decision you make. It is a very honorable decision aswell. Becoming a soldier is a decision YOU make. It is NOT the decision your parents make.", "Solution_15": "what about third world countries where people get married when they are 12 or 13? (typically for the money the person they married with gives their family money)", "Solution_16": "The only thing I have found offensive on this forum was the person who said \"what's the difference between a rape-baby and a normal baby?\" (paraphrased, sorry). Honestly, I find this EXTREMELY offensive. This was obviously a male who had never had a female friend who was raped. What difference is it? What difference is being reminded of this pain, every day for the rest of your life? What difference is carrying the child of a rapist? What difference is resenting that baby, and hating it, because it is half... HIM?\r\nAnd if you don't think of it from the woman's point of view, from the child's then:\r\nWhat difference is it to bear the weight of having a rapist father? What difference is it not knowing your heritage? What difference is it, having a mother that, deep down, wants to cry when they look at you?\r\n\r\nThink about that.", "Solution_17": "[quote=\"bleumoose\"]The only thing I have found offensive on this forum was the person who said \"what's the difference between a rape-baby and a normal baby?\" (paraphrased, sorry). Honestly, I find this EXTREMELY offensive. [/quote]\nBleuMoose, I think that you are misunderstanding the intentions of Dark_Shadow, that specifically said,\n\n[quote=\"Dark_Shadow\"]You state that rape is rare, and abortion is okay in those cases. Otherwise, you're killing a child.\nWhat makes a \"rape baby\" any less valuable than a \"non-rape\" baby?\nEither it's about the morality of abortion, as aborting a fetus, or it's about killing a child, in which case the mother shouldn't matter.[/quote]\n\nI think that despite the emotional implications, from a moral point of view, he is right. If you have two kids (both have already born) and one is product of a rape, while the other is product of conscensual sex, would you kill the first one because of that reason? I presume your answer would be no, but then, what is the difference from before the birth and after the birth. Despite what the kid and the mother can feel afterwards, morally and phisically, what makes different one of the morulae from the other?\n\nWhat Dark_Shadow pretends (or what I understand that he pretends) is to argue that you should not make a difference in the law from an abortion product of a rape and an abortion for any other reason.\n\n[quote=\"bleumoose\"] This was obviously a male who had never had a female friend who was raped. What difference is it? What difference is being reminded of this pain, every day for the rest of your life? What difference is carrying the child of a rapist? What difference is resenting that baby, and hating it, because it is half... HIM?\nAnd if you don't think of it from the woman's point of view, from the child's then: What difference is it to bear the weight of having a rapist father? What difference is it not knowing your heritage? What difference is it, having a mother that, deep down, wants to cry when they look at you? Think about that.[/quote]\r\n\r\nI cannot answer for Dark_Shadow, but in my case, I know women as well as males that have been raped (yes, boys get raped once in a while too!), and I have personally seen the pain that a rape creates. I honestly believe that rape should be a capital crime, a crime punishable with death, because it produces a lot of damage, more than what a non-raped person can ever imagine. Nevertheless, there you are punishing the direct guilty of a crime, and rationally, the society has ever punish a criminal in a \"proportional\" measure to the damage that they produce to another person, institution or population.\r\n\r\nOn the other hand, when you argue about the abortion issue, your arguments are completely emotional, and you laws shouldn't be ruled by emotions but by rational arguments.\r\n\r\nAnother issue why there shouldn't be any difference between rape and no rape during abortion is because of the lying risk. I totally agree that rapist should be punish as far as law can punish them, THEY ARE CRIMINALS, nevertheless, let's think a moment about this: Suppose that the law rules that just deep health issues or rape can be motives to have an abortion and that Norman and Tracey (ficticius names, they came to my mind because of linear algebra: [b]Norm[/b]an and [b]Trace[/b]y) have been dating for a few months, they decide to have sex and Tracey got pregnate. She knows she is not allowed to have an abortion, even when she may want to, then she goes and tell Norman. Norman is a complete pusillanimous and he just try avoid the situation, probably puts in doubt the fidelity of Tracey and at the end, just tell her that it is not his child and that he is not going to talk to her again. Tracy in that moment feels a natural but unmesurable rage, she wants to get rid of \"that thing\" on her uterus, but legally she can't, unless... she had been raped... and the rapist would be Norman.... the perfect revange, don't you think. Now, I consider than Norman is a coward and a couple of other things that I don't mention because this is not the most appropiated place, but he is not a criminal. The lack of braveness and responsability is not a crime. But he may be treated as a criminal because if non-criminal actions. Probably if abortion had been legal on any situation, Tracey would just go and practice one, and afterwards just hate Norman for the rest of her life (what is totally acceptable) and probably tell the story to every new girlfriend that the guy has afterwards (what I also consider totally acceptable), but not try (and potentially succed) to send him to prision because of something he didn't do.", "Solution_18": "The fictitious story you just gave is EXACTLY the reason I believe all abortion should remain legal. That, and the fact that many women would be getting unsafe abortions anyways. My argument was meant to be nothing more than emotional. Many of the things I say on this forum are just something else for people to consider, and I want the people on this board that are discussing abortion to consider the emotional pain of rape and carrying a baby who is the product of a rape.", "Solution_19": "[quote=\"bleumoose\"]The fictitious story you just gave is EXACTLY the reason I believe all abortion should remain legal. That, and the fact that many women would be getting unsafe abortions anyways. My argument was meant to be nothing more than emotional. Many of the things I say on this forum are just something else for people to consider, and I want the people on this board that are discussing abortion to consider the emotional pain of rape and carrying a baby who is the product of a rape.[/quote]\r\n\r\nOK, then I also misunderstood you a little. As I told, I consider that the big people that write the laws should not be emotional when they write them, as well as the judges should not be emotional when they rule. Nevertheless, it is always acceptable to touch the sencitive fibers of the people to make a point.", "Solution_20": "Well... Maybe you shouldn't be emotional, but I wouldn't feel especially safe if the people writing the laws had all the emotional capacity of bricks. It's not just about making a point. It can be said the emotions and empathy is what prevents the ends from justifying the means.", "Solution_21": "Okay, sorry for misinterpreting you djimenez. It's just that it sounded like you were 'dumbing down' your arguments because of ash's age...ah well. I hope we have that cleared up then. Now on to more debate...\r\n\r\n[quote=\"ash101189\"]i am pro choice, but i do think it is killing an actual human. it doesnt matter if the baby is conscious or not of its surroundings! by killing it you are taking away the opportunity for it to be conscious.[/quote] \r\nThe thing is--this is going to sound weird--[i]I don't believe it's wrong[/i] to deprive something that, at the current moment, isn't and has never been conscious. I don't believe the future 'exists', I believe our choices create the future and that no one can know for sure what will happen in the future. So since the only state the fetus has as of now ever existed in is that of unconsciousness, it's not wrong to make sure it will never be conscious. And I know this will probably make no sense to you because we are approaching it from totally different viewpoints. You believe these fetuses have whole lives ahead of them. I don't believe there's anything ahead of us, there's only the past and the present. Arg, this is getting too philosophical. I hate this whole issue, it's so hard to phrase an argument for any side in a way that would justify a federal law.", "Solution_22": "[quote=\"LynnelleYe\"]The thing is--this is going to sound weird--[i]I don't believe it's wrong[/i] to deprive something that, at the current moment, isn't and has never been conscious.[/quote]Exactly my point. Now, the only problem that I see in that position, that is half practical and half philosophical, is that, even when I don't think that there is any wrong to do an abortion on an early stage (your only killing a few cells, like when you accidentally cut your finger while chopping onions... painful but not inmoral), the argument can be used to justify a libertine sexual behavior, and a [i]I don't have to worry about the consecuences[/i] argument to defend it.[quote=\"LynnelleYe\"]I believe our choices create the future and that no one can know for sure what will happen in the future. So since the only state the fetus has as of now ever existed in is that of unconsciousness, it's not wrong to make sure it will never be conscious.[/quote]You know, I had never thought it in that way, but it make a lot of sence.", "Solution_23": "[quote=\"MathFiend\"][/quote][quote=\"JBL\"]\nHowever, Mathfiend, you never really answered the question that I find most critical: you said that you believe foetuses are children. Fine. What about blastocysts? I want to know exactly where you draw the line between life and not life, before which my (hypothetical) wife can acceptably get rid of without crossing any moral lines.[\\quote]\nI can't answer your first question, what is a blastocyst? That would help me answer your question.[\\quote]\nA blastocyst is a group of 40 to 150 undifferentiated (well, mostly -- there are \"outer\" and \"inner,\" but they don't have distinct functions yet) cells. It is approximately the size of the dot you can make with a pencil on a piece of paper. It is the fourth stage that cells pass through after ovulation: ova, zygote (that being the sperm and egg joined, still just one cell), the morula (which goes up to about 32 or 40 cells) and then blastocyst.\n\n[quote][quote]As for your ring, you'd probably be better off carrying around a condom, as far as unwanted pregnancy and STDs are concerned. (That sentence is of course really about statistical realities, not about your own personal willpower, so please don't take it personally.)[/quote]\nOk, for the second part, I totally disagree, and it makes me a little angry. After all I've said, can't you see that my point is to discourage sex before marriage? How does carrying around a condom do that? It is condoning exactly what I am against! Why on earth would I carry a condom to make me resist the temptation, that is simply like tying your hands behind your back and holding a freshly baked still gooey chocolate chip cookie in front of your face, but just out of reach and saying \"want it?\" It simply puts the temptation right there. Let's (hypothetically) say I would carry a condom in my purse rather than wearing the ring on my finger. Someone says sex, I think [i]\"hey, there's a condom in my purse, I could go have sex and be safe about it.\"[/i] Now, if I have my ring, I am reminded that I will not have sex. See a difference? One completely disagrees with what I have said, and the other makes complete sense.[/quote][/quote][quote=\"MathFiend\"][/quote]\r\nSo, in an information vacuum, I would agree with you -- knowing nothing else about the world, what you just said sounds reasonable. And, in fact, if you look at studies made of teens who have taken virginity pledges and/or had abstinence only sex education versus those who have had a thorough sex education (info about contraception, etc.) the first group does tend to have sex for the first time slightly later than the second group. However, they are just as likely to get pregnant before marriage and more likely to acquire an STD. So, your ring (again, not literally yours), if examined as a tool to improve public health (and lower abortions), accomplishes exactly nothing. In fact, this is the single thing that most upsets me about the pro-life movement. They claim to want to prevent abortions, but they constantly set barriers in front of exactly those things (contraceptive education, morning-after pill, etc.) that would [i]reduce[/i] abortions. If they were really serious, they would be working with Planned Parenthood and NARAL, not against them.\r\n\r\nAlso, and this is really a question about you personally with a message about teens in general, do you really think that having a condom around would increase your libido so much that you would forget all about not having sex? I mean, that doesn't sound serious to me -- either you have the willpower and carrying the condom won't hurt it, or at some point you'll reach a situation where you want to have sex before you're married. If you find yourself in the second situation, how much better to have a condom handy? Besides, the process of getting the darn thing open and on might provide you just enough time to reconsider.", "Solution_24": "[quote=\"Psycho\"]for some reason, i strongly support it. i mean, if the parents don't want the child, how will they be able to raise it in a good way then? :? \nand what if that person later finds out his parents didn't want him at all. how should he feel then?[/quote]\r\nSo it's better to murder a person than to face hardship?", "Solution_25": "[quote=\"ZennyK\"][quote=\"Psycho\"]for some reason, i strongly support it. i mean, if the parents don't want the child, how will they be able to raise it in a good way then? :? \nand what if that person later finds out his parents didn't want him at all. how should he feel then?[/quote]\nSo it's better to murder a person than to face hardship?[/quote]\r\nZennyK, Define person. because as many of us had said before, we do not consider a few non-differenciated cells a human. I do think that it is better to smash a few cells like the ones you smash when you have an ich in your skin, or when you accidentally cut your finger when you are cutting the onion, than to sentence the human been that those cells can become to, to a misserable life or a hardship he/she didn't ask for!\r\n\r\nWell, if he realize that his/her parrents didn't want him when he is a child, well, that can be very traumatic, but I would say that in the adult age, it would not be nice, but it would not be that harsh either, I mean, it is just a matter of maturness!", "Solution_26": "[quote=\"djimenez\"][quote=\"ZennyK\"][quote=\"Psycho\"]for some reason, i strongly support it. i mean, if the parents don't want the child, how will they be able to raise it in a good way then? :? \nand what if that person later finds out his parents didn't want him at all. how should he feel then?[/quote]\nSo it's better to murder a person than to face hardship?[/quote]\nZennyK, Define person. because as many of us had said before, we do not consider a few non-differenciated cells a human. I do think that it is better to smash a few cells like the ones you smash when you have an ich in your skin, or when you accidentally cut your finger when you are cutting the onion, than to sentence the human been that those cells can become to, to a misserable life or a hardship he/she didn't ask for!\n\nWell, if he realize that his/her parrents didn't want him when he is a child, well, that can be very traumatic, but I would say that in the adult age, it would not be nice, but it would not be that harsh either, I mean, it is just a matter of maturness![/quote]\r\nWhen the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it. To terminate it would be to kill a human, regardless of whether or not it could \"survive.\" Many abortions take place after differentiation occurs.", "Solution_27": "[quote=\"ZennyK\"]\nWhen the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it.[/quote]\r\nWhy?", "Solution_28": "[quote=\"LynnelleYe\"][quote=\"ZennyK\"]\nWhen the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it.[/quote]\nWhy?[/quote]\r\nBecause then the embryo must develop into a human being.", "Solution_29": "[quote=\"ZennyK\"][quote=\"LynnelleYe\"][quote=\"ZennyK\"]When the cells are non-differentiated, it is probably not a person. As soon as the cells of the embryo differentiate, that's it.[/quote]Why?[/quote]Because then the embryo [b]must[/b] develop into a human being.[/quote]Why [b]must[/b] it? I mean, is it the differenciation of the cells in the embrion the thin line between a human been and a bunch human of cells? I am not so sure!" } { "Tag": [ "number theory", "greatest common divisor", "least common multiple", "inequalities", "relatively prime", "triangle inequality" ], "Problem": "Find all pairs of positive integers a,b so that gcd(a,b),lcm(a,b) and ab are sides of a triangle!\r\n\r\nTo avoid misunderstanding:\r\ngcd: greatest common divisor\r\nlcm: least common multiple", "Solution_1": "[quote=\"Misha123\"]Find all pairs of positive integers a,b so that gcd(a,b),lcm(a,b) and ab are sides of a triangle!\n\nTo avoid misunderstanding:\ngcd: greatest common divisor\nlcm: least common multiple[/quote]\r\nIf the numbers are relatively prime, then lcm(a,b) is ab and therefore it can work because the triangle is isosceles. Therefore there are an infinite amout?", "Solution_2": "[quote=\"Scrambled\"][quote=\"Misha123\"]Find all pairs of positive integers a,b so that gcd(a,b),lcm(a,b) and ab are sides of a triangle!\n\nTo avoid misunderstanding:\ngcd: greatest common divisor\nlcm: least common multiple[/quote]\nIf the numbers are relatively prime, then lcm(a,b) is ab and therefore it can work because the triangle is isosceles. Therefore there are an infinite amout?[/quote]\r\n\r\nThat makes sense, look at the rest though.\r\n\r\nwe have gcd(a,b)lcm(a,b) = ab\r\n\r\nSo therefore\r\n\r\ngcd(a,b) <= lcm(a,b) <= ab\r\n\r\nHence we must have gcd(a,b) + lcm(a,b) > ab\r\n\r\nWe then have: gcd(a,b) + lcm(a,b) = gcd(a,b) + ab/(gcd(a,b))\r\n\r\nIt is obvious that gcd(a,b) <= ab/2, hence if gcd(a,b) >= 2, ab/gcd(a,b) <= ab/2 and gcd(a,b) <=ab/2. Summing, we get LHS <= ab, which is a contradiction. It gets a tiny bit more complicated if we can have degenerate triangles (we have to include the triangle (2,2,2)) but Scrambled's answer encompasses all solutions.", "Solution_3": "[quote=\"blahblahblah\"]gcd(a,b) <= lcm(a,b) <= ab\n\nHence we must have gcd(a,b) + lcm(a,b) > ab[/quote]\r\n\r\nWhy is that?", "Solution_4": "[quote=\"Magnara\"][quote=\"blahblahblah\"]gcd(a,b) <= lcm(a,b) <= ab\n\nHence we must have gcd(a,b) + lcm(a,b) > ab[/quote]\n\nWhy is that?[/quote]\r\n\r\nApologies if it was not sufficiently clear. The first statement is obvious, the second is the triangle inequality, and is what we want to prove.", "Solution_5": "Okay, this is going to be obnoxious, but it's a brief English-usage lesson. The primary meaning of the word \"hence\" is \"because of this\" or \"as a result.\" (It also has a secondary meaning of \"in the future\" and an archaic usage of \"away from here,\" as in \"Get you hence!\") Your usage of the word \"hence\" implied that the triangle inequality was a direct consequence of the inequality you posted above it.\r\n\r\nScrambled -- often you will come across a math problem that asks you to find \"all of something\" when there are an infinite number of things. You were, of course, correct that there were an infinite number of solutions, but what you needed to do was give a description of all solutions (such as, \"All pairs of relatively prime integers\") and then also show that there are no other solutions. Knowing there are infinitely many of something is not necessarily very useful.", "Solution_6": "Thanks for your comments. And JBL: your comment about the infinity of solutions is exactly what I would say about this topic. I just post the solution I had: it's rather easy(I hope you will like it):\r\n\r\ngcd(a,b):=g.\r\nAs already said: ab=>lcm(a,b)=>max(a,b)=>min(a,b)=>g\r\nSo we must prove: ab/g+g=>ab. I make the substitution: a=a_1g,b=b_1g:\r\na_1b_1g+g=>a_1b_1g^2. You can rewrite it: g<=1+1/(a_1b_1). Now we have two cases: g=1 and g=2,a_1=b_1=1. It is easy to prove that both cases work.\r\n\r\nThanks,once more time, for your interest!", "Solution_7": "[quote=\"JBL\"]Okay, this is going to be obnoxious, but it's a brief English-usage lesson. The primary meaning of the word \"hence\" is \"because of this\" or \"as a result.\" (It also has a secondary meaning of \"in the future\" and an archaic usage of \"away from here,\" as in \"Get you hence!\") Your usage of the word \"hence\" implied that the triangle inequality was a direct consequence of the inequality you posted above it.\n\nScrambled -- often you will come across a math problem that asks you to find \"all of something\" when there are an infinite number of things. You were, of course, correct that there were an infinite number of solutions, but what you needed to do was give a description of all solutions (such as, \"All pairs of relatively prime integers\") and then also show that there are no other solutions. Knowing there are infinitely many of something is not necessarily very useful.[/quote]\r\n\r\nJBL, I'm well aware that my comment was unclear, but I don't believe that it was incorrect. I stated that gcd <= lcm <= ab.\r\n\r\nIt follows (or, hence) that[i] to solve the problem[/i], we need to show that these three qualities satisfy the triangle inequality. I was in a hurry, and didn't include that little bit, but I believe that it was otherwise correct. To better organize it, I should have said, hence a,b,c are sides of a triangle iff a+b > c by the triangle inequality, or whatever.", "Solution_8": "I'm not saying that what you said was incomprehensible or completely off the mark; I understood what you meant. However, lacking a phrase such as \"we need to show,\" the clear meaning of your sentence is that the triangle inequality follows from the ordering of those three numbers. I don't think I need to defend my comment beyond that -- I was attempting to alleviate a possible cause of the confusion." } { "Tag": [ "blogs", "\\/closed" ], "Problem": "What am I supposed to put there to show the number of comments in the blogs?\r\n\r\nI thought it was %, but that only deletes the next character.", "Solution_1": "[quote=\"emilgouliev\"]What am I supposed to put there to show the number of comments in the blogs?\n\nI thought it was %, but that only deletes the next character.[/quote]Small bug. It will be fixed very soon :)" } { "Tag": [ "LaTeX" ], "Problem": "Hello All,\r\n\r\nI was trying to answer [url=http://www.mathlinks.ro/viewtopic.php?t=208229]this post[/url]. My example is:\r\n\r\n\\[ f(x)=e^{-e^\\cot(x)}\\]\r\n\r\nAs you see, $ \\cot(x)$ appears very small, so people with not very good eyes have problems reading it. Please, suggest me, how to write this formula so that it will be larger.\r\n\r\nThank you!\r\nSergey Markelov\r\nmarkelov@mccme.ru", "Solution_1": "There are 4 sizes of characters in maths mode, in descending order of size: displaystyle, textstyle, scriptstyle, scriptscriptstyle. The cot x term is in scriptscriptstyle so you can make it one size larger by using scriptstyle: $ f(x) \\equal{} e^{ \\minus{} e^{\\scriptstyle\\cot x}}$", "Solution_2": "[quote=\"stevem\"]There are 4 sizes of characters in maths mode, in descending order of size: displaystyle, textstyle, scriptstyle, scriptscriptstyle. The cot x term is in scriptscriptstyle so you can make it one size larger by using scriptstyle: $ f(x) \\equal{} e^{ \\minus{} e^{\\scriptstyle\\cot x}}$[/quote]\r\n\r\nStevem,\r\n\r\nthank you very much!\r\n\r\nSergey Markelov" } { "Tag": [ "puzzles" ], "Problem": "If a daddy bull weighs 1,200 pounds and eats twelve bales of hay each day, and a baby bull, who weighs 300 pounds eats three bales of hay each day, how much hay then should a mommy bull eat if she weighs 800 pounds?", "Solution_1": "[hide]Since bulls are only male, the only way for there to be a mommy bull is if the bulls are homosexual, in which case the mommy bull would eat 8 bales, (assuming the equation relating weight and hay is linear) :rotfl:[/hide]", "Solution_2": "ur right...", "Solution_3": "Really??? It was a gay bull??? I was only joking :rotfl:", "Solution_4": "no there are no female bulls", "Solution_5": "Oh. Darn. I think my answer is better.", "Solution_6": "wait a minute does that mean bulls are gay or do they reproduce asexually or can they not reproduce except for theeir cells or only in this problem", "Solution_7": "No, they reproduce with cow. Cows are only femle and bulls are only male. :)", "Solution_8": "oh thanks i wasn't sure if i should put them on the list with those what are they...mules?" } { "Tag": [ "algebra solved", "algebra" ], "Problem": "x,y,z>1, 1/x+1/y+1/z=2\r\nshow that \r\n(x+y+z)^1/2 >= (x-1)^1/2+(y-1)^1/2+(y-1)^1/2", "Solution_1": "Using Cauchy - Schwarz we get \r\n\r\n(x+y+z)((x-1)/x + (y-1)/y + (z-1)/z) >= ((x-1)^(1/2) + (y-1)^(1/2) + (z-1)^(1/2))^2\r\n\r\nNow note that (x - 1)/x + (y - 1)/y + (z - 1)/z = 3 - (1/x + 1/y + 1/z) = 1 so the result follows immediately." } { "Tag": [ "inequalities", "trigonometry" ], "Problem": "[b]hi everybody,\ni am quite new to inequalities and got great help from the many marathons running here. I was more interested in marathon on triangle and trigonometry inequalities.so i was just wondering if we could sstart a marathon on the same.[/b]", "Solution_1": "Take a look at some of the problems being posted by jeez123 in [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]HSB[/url]. No inequalities, but some trigonometry." } { "Tag": [ "LaTeX" ], "Problem": "How to create new symbols that are required in $\\LaTeX$. I have to create new symbols which represent meet and join as used in the text book of \"Introduction to Lattice Theory\" by \"Gaber Szasz\" which resembles $\\smallfrown$, $\\smallsmile$ but has some more height than them. Or they are the symbols $\\cap$ and $\\cup$ but with extra space in between the open ends", "Solution_1": "I suggest you check out the [url=http://www.tug.org/tex-archive/info/symbols/comprehensive/]Comprehensive LaTeX Symbol List[/url]. It's possible your symbol already exists. If not, it has instructions how to create new symbols, but I should warn you that it is quite difficult to do so.", "Solution_2": "I have verified $\\LaTeX$ Symbol list. But, I am unable to find these symbols there. Also, there i am unable to find any procedure for creating new symbols.", "Solution_3": "Well, I hope this example will help:\r\n\r\n% put this in preambule\r\n\\def\\grpedeset{\\hbox{$\\Gamma$\\kern-5truept\\raise3.3truept\\hbox{\\tiny $\\Delta$}}}\r\n\r\n\r\n% somewhere in document you can use\r\n\\grpedeset\r\n\r\nTry to find more about \\kern, \\raise..... \r\nIt's very usefull." } { "Tag": [ "complex numbers" ], "Problem": "Calculate $\\sum_{k=0}^{n}sin (k)$", "Solution_1": "[hide=\"Hint\"] I have solved a general version of this problem a couple of times on Mathlinks, so I will just provide the following hint: Complex Numbers and Geometric Sums.[/hide]" } { "Tag": [], "Problem": "How many non-empty subsets does a three-element set have?", "Solution_1": "[quote=\"GameBot\"]How many non-empty subsets does a three-element set have?[/quote]\r\n\r\nLet the elements be a, b, and c. We have:\r\n\r\na\r\nb\r\nc\r\nab\r\nac\r\nbc\r\nabc\r\n\r\nThis makes a total of 7.", "Solution_2": "For each element we have 2 choices when making a set, in or out. Thus there are 2^3=8 sets. However, the null set is not an option so it is 8-1=7.", "Solution_3": "We use the formula which is (2^n)-1 where n is the number of numbers in the set.\n\nThe answer is 2^3-1=7" } { "Tag": [], "Problem": "how many terms are in the expansion of $ (2a\\minus{}5b\\plus{}3c)^{35}$", "Solution_1": "[hide=\"Hint\"]I think the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial Theorem[/url] would help.[/hide]\r\nBut I'm too busy/lazy to try it right now :lol:", "Solution_2": "All of the terms in the expansion are of the form $ a^{k_a}b^{k_b}c^{k_c}$, where $ 0\\leq k_a,k_b,k_c\\in\\mathbb{Z}$. We know that $ k_a \\plus{} k_b \\plus{} k_c \\equal{} 35$, and the number of solutions to this is $ \\dbinom{35 \\plus{} 3 \\minus{} 1}{3 \\minus{} 1} \\equal{} \\boxed{666}$. :diablo: :diablo: :diablo: says I'm correct.", "Solution_3": "Wait math154 how do you know to do that? Balls and urns?", "Solution_4": "Yes...search it if you don't know what it is. It is like the most asked topic on AoPS." } { "Tag": [ "puzzles" ], "Problem": "A farmer was cleaning out his big old dusty farm shed. \r\n\r\nHe dragged out an old water container which was full of old dirty water and it weighed 12 pounds. \r\n\r\nThe farmer put something in the container and then it weighed less. What did he put in the container?", "Solution_1": "A hole?\r\n\r\n(The message is too small. Please make the message longer before submitting.)", "Solution_2": "[hide=\"good one...\"]ACID!!!![/hide]", "Solution_3": "another good 1\r\n\r\n[hide]alkali metals it exploded and therefore weighed less[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "Cac ban Viet Nam oi,chi giup minh cach go LaTex tren mathlinks voi .Kem that day.vao tu hoi toi h ma chua biet \r\ncach go the nao ?", "Solution_1": "C\u00f3 hai m\u00f4i tr\u01b0\u1eddng c\u1ea7n ph\u00e2n bi\u1ec7t, m\u00f4i tr\u01b0\u1eddng ch\u1eef th\u01b0\u1eddng v\u00e0 m\u00f4i tr\u01b0\u1eddng to\u00e1n. N\u1ebfu l\u00e0 m\u00f4i tr\u01b0\u1eddng to\u00e1n th\u00ec c\u1ea7n \u0111\u1eb7t gi\u1eefa hai d\u1ea5u \u0111\u00f4 la. Chi ti\u1ebft xin xem t\u1ea1i LaTeX help t\u1ea1i di\u1ec5n \u0111\u00e0n n\u00e0y.", "Solution_2": "[quote=\"Nbach\"]Cac ban Viet Nam oi,chi giup minh cach go LaTex tren mathlinks voi .Kem that day.vao tu hoi toi h ma chua biet \ncach go the nao ?[/quote]\r\nXem o day nay http://www.mathlinks.ro/LaTeX/AoPS_L_About.php. Khong lam duoc thi vao day ma hoi http://www.mathlinks.ro/Forum/index.php?f=123." } { "Tag": [ "trigonometry" ], "Problem": "Evaluate $\\sum_{n=1}^{89} \\frac{1}{1+\\tan^{2006} n^\\circ}$.", "Solution_1": "[hide]\nLet $p=2006$, $S=\\sum_{n=1}^{89} \\frac{1}{1+\\tan^p n^\\circ}$. Reverse the summation to get\n\n\\[ S=\\sum_{n=1}^{89} \\frac{1}{1+\\tan^p (90-n)^\\circ}=\\sum_{n=1}^{89} \\frac{1}{1+\\cot^p n^\\circ}=\\sum_{n=1}^{89} \\frac{\\tan^p n^\\circ}{1+\\tan^p n^\\circ} \\]\n\nAdding the two term-by-term, $2S=\\sum_{n=1}^{89} 1=89\\Rightarrow S=\\boxed{\\frac{89}{2}}$ (and independent of $p$).\n[/hide]", "Solution_2": "What do you mean by \"reverse the summation\"?\r\n-Swapnil", "Solution_3": "\"Reversing the summation\" means writing the terms in reverse order. For example, if $A=1+2+3+\\cdots +n$, then $A$ also equals $n+(n-1)+(n-2)+\\cdots +1$.", "Solution_4": "that is an ingenious solution, can you point in the direction of where to get help with tricky sums to infinity?", "Solution_5": "nice solution scorpius :lol:", "Solution_6": "This is a great problem and I certainly wouldn't be able to get in short amount of time. But here is quick tip on approaching this type of problem.\r\n\r\n1. It's pretty hard to find $1 + \\tan^{2006} n$ for any angle without calculator. And even with it, it's not very fun. So, we need to consider general case of $\\tan^{p}$ or whatever you want to call the exponent.\r\n\r\n2. Note that summation goes from 1 to 89. Moreover, we have $\\tan (n) = \\cot (90-n)$ which fits the given interval. So, it might be a good idea to rewrite the sum BACKWARD.\r\n\r\n3. After you done #2, compare two sum (which are same) in terms of $\\tan$. Do you see anything that's nice? Find this by adding/subtracting if you need to. The desired sum should follow.\r\n\r\nNote that this is exactly what he did except I explained how you would approach this (without seeing his solution)." } { "Tag": [ "function", "calculus", "derivative", "combinatorics solved", "combinatorics" ], "Problem": "suppose for a>=b>=c>0 \r\nfind number of sols to a+b+c=n, for some integer n \r\ncan we generalize?", "Solution_1": "we claim that the number of ordered n+1-tuples (x1,...,xn+1) with x1>=x2>==...>=xn+1>0 and sum m equals: m-1 C n.\r\nproof 1. let yi =xi-1 then sum yi = m-(n+1) and yi>=0\r\nnow consider an m*n chessboard and consider a player to be at bottom left corner of this table. he can move either horizontally toward right or upward. the number of ways that he can reach the top right hand corner equals: m+n C m \r\nbut each path is a partition of the number m as the sum of n+1 nonnegative integers. \r\nhence the number of ordered partitions into nonnegative integers will be \r\nn+m C n.\r\nproof 2.\r\nwe use generating functions. the number of ordered partitions of m into n+1 non-negative integers is coefivient of x^m in the expansion of\r\n(1/1-x)^n+1 = 1-x)^-n-1 which equals m+n C n.", "Solution_2": "I do not think that order matters here..., because y_(n+1)>=y_n>=...>=y_1\r\nI wrote a small program to verify and the answers do not match your formula, Ali.", "Solution_3": "I think that the number of solutions is:\r\n\r\nNr=1/n! f^(n) (0) where\r\nf^(n) (0) is the n-th derivate of the function f(t)=1/(1-t)^3 in point t=0.\r\ncheers! :D :D", "Solution_4": "And of course the problem can be generalized! \r\n\r\nax+by+cz=n with x,y,z fixed, x,y,z are natural numbers!\r\nAnd you apply the same thing to this one, to find the number of it's solutions! \r\ncheers! :D :D", "Solution_5": "the derivitive thing is pretty kool!! :D \r\nif i computed things correctly, if a+b+c=n, for a>=b>=c>0,\r\n\r\nf^(-1)(t)=(-3)(1-t)^(-4)\r\n...\r\nf^(-n)(t)=(-3)(-4)...(n+2)(1-t)^(-n+3)\r\nso Nr=(n+2)!/((2!)(n!))=Nr=(n+2)C2. quiet similar to what Ali had, but i dont it is correct either.\r\n\r\nfor a+b+c=n, we have for n=3,4,...\r\nNr=1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,48,52\r\n\r\nanyways, Ali had a point, Nr should be the number of partitions with at most 3 parts\r\nso a(n) satisfy the recursion a(n)=1+a(n-2)+a(n-3)-a(n-5) (why??)", "Solution_6": "you may mean to say derivative :D:D", "Solution_7": "bugzpodder, my solution is corect iff we don't have the condition a>=b>=c(now I have seen it :( ) \r\nI computed all the solutions! for example n=4\r\nyour answer is 1, when mine is 3:\r\n2+1+1=4 (your only solution)\r\nand my solutions: 1+1+2,1+2+1,2+1+1. I think that all we need now to compute is which number of solutions we sould substract from my number! \r\n\r\ncheers! :D :D", "Solution_8": "[quote=\"bugzpodder\"]suppose for a>=b>=c>0 \nfind number of sols to a+b+c=n, for some integer n[/quote]\r\nFirst of all, we will compute number of solutions $x+y+z=n$, $x,y,z\\in\\mathbb{N}$.\r\nIt is a well known problem and answer is $A=C_{n-1}^2=\\frac{(n-1)(n-2)}{2}$. Now substract from $A$ threefold number of all solutions, for which $x,y,z$ are not pairwise distinct numbers. It is $[(n-1)/2]$ (take into consideration that if $n$ is divisible by 3 then we have solution with equal $x,y,z$.) Remaining triples can be uniquely ordered, and since each such triple has 6 different permutations, we obtain final answer:\r\n\\[\\frac{\\frac{(n-1)(n-2)}{2}-3\\left[\\frac{n-1}{2}\\right]+2\\xi_n}{6}+\\left[\\frac{n-1}{2}\\right],\\]\r\nwhere $\\xi_n=1$ if $n$ is divisible by 3, and $\\xi_n=0$ otherwise." } { "Tag": [ "calculus", "geometry", "3D geometry", "prism" ], "Problem": "Find the volume in cubic inches of the solid generated by revolving a region bounded by an equilateral triangle of side 4 inches about one side.", "Solution_1": "I would use calculus but I can't in this forum. When you revolve some region around an axis, you are essentially creating a cylindrical-type shape, which has volume $ V\\equal{}\\pi r^{2}h$ of course. Therefore, you want to do something similar here.", "Solution_2": "The problem is especially designed so that calculus is not necessary to solve this problem.\r\n\r\n[hide=\"Hint\"]Cones[/hide]", "Solution_3": "[hide=\"Solution.\"]\nYou are essentially creating two cones; the height of each is half the side length of the equilateral triangle and the radius of each is the same as the triangle's height. Therefore, the total volume is $ \\frac{4\\pi}{3}( 2\\sqrt{3} )^2$. That comes out to ${ \\boxed{16\\pi}}$.[/hide]", "Solution_4": "[hide]If we unravel the 3D shape, we obtain a triangular prism with the equilateral triangle as its base, and the height equal to the path that the centroid of the triangle travels through while revolving. The centroid of the triangle is $ \\frac{2\\sqrt{3}}{3}$ units away from the axis of revolution. The volume of the object is then just the area of the base times that length, or:\n\\[ \\frac{\\sqrt{3}}{4} \\cdot 4^2 \\cdot 2 \\pi \\cdot \\frac{2\\sqrt{3}}{3} \\\\\n\\equal{}\\boxed{16\\pi}\\][/hide]" } { "Tag": [ "calculus", "integration", "limit", "logarithms", "calculus computations" ], "Problem": "Find the value of $a$ such that $\\lim_{n\\to\\infty}\\frac{3}{2}\\int_{-\\sqrt[3]{a}}^{\\sqrt[3]{a}}\\left(1-\\frac{t^{3}}{n}\\right)^{n}t^{2}\\ dt=\\sqrt{2}\\ \\ (n=1,2,\\cdots).$", "Solution_1": "\\[a=\\log \\left(\\sqrt{2}\\pm\\sqrt{3}\\right)\\]", "Solution_2": "Your answer is incorrect. :(", "Solution_3": "I got $a = \\ln(\\sqrt2+\\sqrt3)$. \r\n\r\n[b]Benorin[/b], you have to discard the negative sign in your answer for an obvious reason.", "Solution_4": "Your answer is correct, FieryHydra. :)", "Solution_5": "well...this might seem a little unnecessary but could you please show how you arrived at your answers.", "Solution_6": "[quote=\"maokid7\"]well...this might seem a little unnecessary but could you please show how you arrived at your answers.[/quote]\r\n\r\nHere $\\lim_{n \\to \\infty }\\left( \\int f_{n}(x) \\, dx \\right)= \\int \\left( \\lim_{n \\to \\infty }f_{n}(x) \\right) \\, dx$.", "Solution_7": "o ok thanks i get it now.", "Solution_8": "[quote=\"Ramanujan\"][quote=\"maokid7\"]well...this might seem a little unnecessary but could you please show how you arrived at your answers.[/quote]\n\nHere $\\lim_{n \\to \\infty }\\left( \\int f_{n}(x) \\, dx \\right)= \\int \\left( \\lim_{n \\to \\infty }f_{n}(x) \\right) \\, dx$.[/quote]\r\n\r\n\r\nThat is one way... also notice that \r\n\r\n$\\frac{3}{2}\\int_{-\\sqrt[3]{a}}^{\\sqrt[3]{a}}\\left(1-\\frac{t^{3}}{n}\\right)^{n}t^{2}\\ dt=\\left[ \\frac{\\left(n-t^{3}\\right)^{n+1}}{2n^{n}(n+1)}\\right]_{t=-\\sqrt[3]{a}}^{\\sqrt[3]{a}}$" } { "Tag": [], "Problem": "What and how do you find the answer of:\r\n\r\nFind the coefficient of $ x^{25}$ in $ (x^2\\plus{}x\\plus{}3)^{10}$.\r\n\r\nThanks!", "Solution_1": "[hide]there is no $ x^{25}$ term in $ (x^2\\plus{}x\\plus{}3)^{10}$; the highest term is $ x^{20}$.\n\nbut if that was indeed the question, then it's $ \\boxed{0}$.[/hide]", "Solution_2": "[quote=\"HollyDrakes256\"]What and how do you find the answer of:\n\nFind the coefficient of $ x^{25}$ in $ (x^2 \\plus{} x \\plus{} 3)^{10}$.\n\nThanks![/quote]\r\n[hide]\nWell, just by looking at that, we see that the highest term of x is $ x^20$. Therefore, the coefficient of $ x^25$ is 0. >_>[/hide]", "Solution_3": "[quote=\"AdithyaGanesh\"][quote=\"HollyDrakes256\"]What and how do you find the answer of:\n\nFind the coefficient of $ x^{25}$ in $ (x^2 \\plus{} x \\plus{} 3)^{10}$.\n\nThanks![/quote]\n[hide]\nWell, just by looking at that, we see that the highest term of x is $ x^{20}$. Therefore, the coefficient of $ x^{25}$ is 0. >_>[/hide][/quote]\r\n\r\nCan't be lazy. You have to put the number in {}'s.", "Solution_4": "As for how you'd find the answer in a problem that isn't this stupid, let's try something like the coefficient of $ x^{12}$ in $ (x^2\\plus{}x\\plus{}3)^{10}$.\r\n\r\nIn this case, you have to take terms and bunch them together to create $ x^{12}$. With consideration to the powers of $ x$, we have something like $ 2a\\plus{}b\\equal{}12$, where $ a$ is the number of $ x^2$ terms and $ b$ is the number of $ x$ terms.\r\n\r\n$ a$ goes from $ 0$ to $ 6$, and for each $ a$ we get some $ b$ value that could leave constant terms to be multiplied in. At each value of $ a$, we have $ \\binom{10}{a}$ ways to pick the $ x^2$s, $ \\binom{10\\minus{}a}{b}$ ways to pick the $ x$s, and an extra $ 3^{10\\minus{}a\\minus{}b}$ constant term. That is, the coefficient is \\[ \\sum_{a\\equal{}0}^{6}\\binom{10}{a}\\binom{10\\minus{}a}{12\\minus{}2a}3^{a\\minus{}2}\\equal{}\\boxed{115965}\\]" } { "Tag": [ "probability" ], "Problem": "A number is randomly selected from the integers 1 through 9.\nWhat is the probability that the number is less than 7 and\ndivisible by 3? Express your answer as a common fraction.", "Solution_1": "The only ones that work are $ 3$ and $ 6$. There are $ 9$ total numbers, so $ \\boxed{\\frac{2}{9}}$" } { "Tag": [ "integration", "calculus", "function", "real analysis", "real analysis unsolved" ], "Problem": "iwhere $ F(x)\\equal{}\\int_a^x f$. \r\nAlso f:[a,b] -> R.\r\nI think my biggest problem here is that I can't say f if bounded (i.e. that it has a finite maximum).\r\nany ideas on how I can show this?\r\n\r\nthanks", "Solution_1": "Do you know how to show this:\r\nIf $ \\mu$ is some positive measure and $ f$ is integrable and $ \\lambda(E) \\equal{} \\int_E f \\,d\\mu$, then for any $ \\epsilon > 0$ there is a $ \\delta > 0$ such that $ \\mu(E) < \\delta$ implies $ |\\lambda(E)| < \\epsilon$?\r\n\r\nThere is a more general statement, namely that if $ \\nu$ and $ \\mu$ are measures such that $ |\\mu|(E) \\equal{} 0$ implies $ \\nu(E) \\equal{} 0$ then the above condition holds.", "Solution_2": "huh, not really...and this looks more complicated. :maybe: \r\nBasically, i'm supposed to get this from the definitions of uniform continuity.", "Solution_3": "Well, then, do you know that $ f$ can be approximated in the integral norm by functions that ARE bounded? For example, simple functions.", "Solution_4": "my take on this was to show that \r\n$ |F(y)\\minus{}F(x)| \\leq \\int_x^y |f|$\r\nwhich is easy to show using triangular inequality. Now, the right hand side is equal to zero when y=x, and it's an increasing function. But to get uniform continuity I need to show\r\n$ \\forall \\epsilon>0, \\exists \\delta>0, \\forall x,y, |x\\minus{}y|<\\delta \\Rightarrow |F(x)\\minus{}F(y)|<\\epsilon$\r\ni'm having trouble connecting these last two steps...", "Solution_5": "You indicated that you understand how to do with with $ f$ bounded... so do you see how to connect that with my previous post?", "Solution_6": "true.\r\nI'm sorry, I missed your argument. \r\nOk, so I know how to do with bounded ones. But what I don't know is how to extend that - i.e. I don't know how I could approximate\r\nthe function in the integral norm by functions that are bounded.", "Solution_7": "What do you mean by \"integrable\"? Do you mean \"Riemann integrable\" or \"Lebesgue integrable\"?\r\nHint: every Riemann integrable function is bounded.", "Solution_8": "This may help clear up a few things:\r\n\r\nhttp://forums.somethingawful.com/showthread.php?noseen=0&threadid=2843892&pagenumber=1" } { "Tag": [ "vector", "geometry", "parallelogram", "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Here is a quite hard problem:\r\n Let $ (E,||)$ be a real normed vector space with the property that for any $x$ in $E$ the set of those $y$ such that $||x+y||^2=||x||^2+||y||^2$ is stable my multiplication with real scalars. Then prove that the norm comes from a scalar product.", "Solution_1": "Oh, it's not that hard.. :)\r\n\r\nThe aim will be to show that for all $x,y\\in E$ the parallelogram identity holds, i.e. $\\|x+y\\|^2+\\|x-y\\|^2=2(\\|x\\|^2+\\|y\\|^2)$. Whenever this is true, we'll say that $P(x,y)$.\r\n\r\nFirst of all, let's fix some $x,y\\in E$ for which $\\|x+y\\|^2=\\|x\\|^2+\\|y\\|^2$ holds (in this case, we'll say that $Q(x,y)$). It's very easy to prove, by the given hypothesis, that for any $\\alpha,\\beta,\\gamma,\\delta\\in\\mathbb R$, if we put $u=\\alpha x+\\beta y,\\ v=\\gamma x+\\delta y$, then $P(u,v)$.\r\n\r\nNow assume we're given $x,y\\in E$. According to the paragraph above, in order to show that $P(x,y)$, it suffices to show that $x,y$ lie in a subspace of $E$ generated by two vectors $z,t$ s.t. $Q(z,t)$. To this end, we look for vectors of the form $z=x,t=\\alpha x+y$. It would all be over if we could prove the following: there is a real $\\alpha$ s.t. $Q(x,\\alpha x+y)$.\r\n\r\nConsider the function $f:\\mathbb R\\to\\mathbb R$, defined by $f(\\alpha)=\\|x+\\alpha x+y\\|^2-\\|\\alpha x+y\\|^2$. We want to show that $\\|x\\|^2$ lies in the range of $f$. However, it's easy to prove (exercise :)) that given $a\\in\\mathbb R$ and $x,y$ vectors in some normed vector space, we have $\\lim_{n\\to\\infty}(\\|(n+a)x+y\\|-\\|nx+y\\|)=a\\|x\\|\\ (*)$. it's clears now how to use $(*)$ to show that as $\\alpha\\to\\pm\\infty,\\ f(\\alpha)$ goes to $\\pm\\infty$ respectively, so, since $f$ is continuous, $\\|x\\|^2$ must lie in the range of $f$.\r\n\r\nI hope it's ok. \r\nI saw $(*)$ as an exercise in a book a long time ago. It's a very good book, actually, \"Teoreme si Probleme de Analiza Matematica\", by Sorin Radulescu and Marius Radulescu.", "Solution_2": "Ok if you say so it's not that hard. Yet, I had no chance in proving it... :(" } { "Tag": [ "topology", "Functional Analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "I have a question on some notation that appears in Hilbert Spaces.\r\n\r\nI came a cross a piece of a proof that says\r\n\r\n\r\n[quote]\nTo prove uniqueness it suffices to establish that if $ M_1$ and $ M_2$ satisfy (1) and (2), then so also $ M\\equal{}(M_1 \\plus{} M_2)^\\minus{}$ satisfies (1) and (2). ......\n[/quote]\r\n\r\nHere $ M_1$ and $ M_2$ are subspaces of the Hilbert space $ H.$ I do not understand what the symbol \"$ ^\\minus{}$\" means in $ (M_1 \\plus{} M_2)^\\minus{}.$\r\n\r\nYou can find the original paper in [url=http://www.springerlink.com/content/k367q0w08j702123/fulltext.pdf]here[/url] and this is the part on page 390, proof of Prop 1.7.", "Solution_1": "[quote=\"lenny\"]I do not understand what the symbol \"$ ^ \\minus{}$\" means in $ (M_1 \\plus{} M_2)^ \\minus{} .$[/quote] I did not look at the article but I would bet my house ;) that it means the topological closure of the subspace $ M_1\\plus{}M_2$ in your Hilbert space.", "Solution_2": "Ok, but what is the difference between the topological closure and just the \"closure\"? In other words, what role does the term topological play in here? What is the topology in the hilbert space?", "Solution_3": "Closure and topological closure are synonymous in this context. A Hilbert space is a metric space induced by its scalar product and as such a topological space." } { "Tag": [], "Problem": "Given system equation:\r\n$ \\left\\{ \\begin{array}{l}x \\plus{} xy \\plus{} y \\equal{} m \\plus{} 2 \\\\ x^2 y \\plus{} xy^2 \\equal{} m \\plus{} 1 \\\\\\end{array} \\right.$\r\nFind m such that the system equation above has one root.", "Solution_1": "hmm so\r\nfrom the equations, i know that the top one is 1 more than the bottom one\r\n(x^2y) + (xy^2) + 1 = x+ xy +y\r\nmove everything to the left\r\n(x^2y) + (xy^2) - x - y - xy + 1 = 0\r\nfactor\r\nxy(x+y-1) - (x+y-1) = 0\r\n(xy-1)(x+y-1) = 0\r\n\r\nif it only has 1 set of roots then\r\n\r\nxy-1=0, xy=1\r\nx+y-1=0, x+y=1\r\n\r\nsub these 2 back to the equations \r\n\r\ni got m=0", "Solution_2": "[quote=\"lgd0612\"]i got m=0[/quote]\r\nI don't think so because if m=0 then we have \r\n$ \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\left\\{ {\\begin{array}{*{20}c}{xy(x \\plus{} y) \\equal{} 1} \\\\{x \\plus{} y \\plus{} xy \\equal{} 2} \\\\\\end{array}} \\right. \\\\ \\Leftrightarrow \\left\\{ \\begin{array}{l}x \\plus{} y \\equal{} 1 \\\\ xy \\equal{} 1 \\\\ \\end{array} \\right.(*) \\\\ \\end{array}$\r\nBut system of equation (*) has no solution :?:\r\nMay be you forget $ (x \\plus{} y)^2 \\minus{} 4xy \\ge 0$", "Solution_3": "(x+y) + xy = m+2\r\nxy( x + y ) = m+1\r\n\r\na^2 - (m+2)a + (m+1) = 0\r\n\r\nD = m^2 = 0 \r\nm=0", "Solution_4": "[quote=\"Clear_sky\"]\n$ (x \\plus{} y) \\plus{} xy \\equal{} m \\plus{} 2$\n\n$ xy( x \\plus{} y ) \\equal{} m \\plus{} 1$\n\n$ a^2 \\minus{} (m \\plus{} 2)a \\plus{} (m \\plus{} 1) \\equal{} 0$\n\n$ D \\equal{} m^2 \\equal{} 0$ \n\n$ m \\equal{} 0$[/quote]\r\nAre you sure ?Have you read my reply above ?", "Solution_5": "oops. My fault" } { "Tag": [ "\\/closed" ], "Problem": "Congratulations to Arne for becoming a moderator in the Olympaid section.\r\nAnybody knows where is our dear [b]Myth[/b]?\r\nIt seems that he is not any more too interested in the forum.\r\nWe missed him a lot. :?", "Solution_1": "Actually I was a moderator in the past, I just had a break :D \r\n\r\nI guess Myth is very busy at the moment. Haven't heard from him for a long time.", "Solution_2": "Myth is still visiting this site. I saw his name on the \"who is online\" list.", "Solution_3": "yes indeed , i have just seen his name in the Who is online :D :roll: \r\n(at this very moment)", "Solution_4": "A new congratulation\r\nhey guys we have a new admit . \r\nLets say our loyalty to him :) \r\n[color=orange]nsato[/color]\r\n:clap2: :omighty:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46953[/url]", "Solution_5": "Hey I think Valentin idea has changed ,cause he had said it is impossible to became an admin :D \r\n\r\n\r\n[quote=\"math92\"]Can i be a administrator? [/quote]\n[quote=\"valentin vornicu\"]It is impossible.[/quote]\n[quote=\"colts18\"]What do you mean???(Why Not?)[/quote]\n[quote=\"valentin vornicu\"] :huh: Because it just is impossible.[/quote]\n[quote=\"bubala\"]But what if they make IMO and get a college degree and whatnot and AoPS hires them. Then they'd be admins[/quote]\n[quote=\"valentin vornicu\"]That's a totally different approach But for now math92 must be careful with his spam warnings \n\nTopic is officially closed[/quote]", "Solution_6": "you need to worry about not quadruple posting ;) :D", "Solution_7": "He was replying to math92. Thus, it is implied that he meant it's impossible for math92 to become an admin (which, btw is most likely correct :D ) .\r\n\r\nnsato is Naoki Sato, the new instructor at AoPS. So of course he becomes an administrator ;)" } { "Tag": [], "Problem": "What is the least positive integer with exactly five distinct\npositive factors?", "Solution_1": "so the number must be in form of $ x^4$.\r\n\r\nx is the smallest prime since the problem is asking for least positive integer.\r\n\r\n$ 2^4\\equal{}16$\r\n\r\nanswer : 16", "Solution_2": "Or you could realize that only square numbers have an odd number of distinct factors and easily see that the smallest perfect square with 5 factors is $ \\boxed{16}$." } { "Tag": [ "limit" ], "Problem": "First, you need to prove that \r\n\r\n$\\lim_{x \\rightarrow a}(f(x) + g(x)) = \\lim_{x \\rightarrow a}(f(x)) + \\lim_{x \\rightarrow a}(g(x)) $\r\nThis isn't too hard of a property to prove. Think about the definition of a limit.\r\n\r\nNow you also need these fairly simple limits:\r\n\r\n\r\n\r\n$\\lim_{x \\rightarrow a}(c) = c$ and\r\n$\\lim_{x \\rightarrow a}(2x) = 2a$ \r\n\r\nAgain, try to find an deltas for any epsilon.", "Solution_1": "[quote=\"tetrahedr0n\"]First, you need to prove that \n\n$\\lim_{x \\rightarrow a}(f(x) + g(x)) = \\lim_{x \\rightarrow a}(f(x)) + \\lim_{x \\rightarrow a}(g(x)) $\nThis isn't too hard of a property to prove. Think about the definition of a limit.\n\nNow you also need these fairly simple limits:\n\n\n\n$\\lim_{x \\rightarrow a}(c) = c$ and\n$\\lim_{x \\rightarrow a}(2x) = 2a$ \n\nAgain, try to find an deltas for any epsilon.[/quote]\r\n\r\nHint: Break it up into two limits (each with their own delta and epsilon) and use the triangle inequality.", "Solution_2": "If you think about the definition of a limit, what you are trying to prove is that when $x \\approx -3$, $5-2x \\approx 11$. Another property of limits that helps here is $\\lim_{x \\rightarrow a} f(x) = f(a)$ if $f(x)$ is defined at $a$.", "Solution_3": "[quote=\"Rep123max\"]And now I actually got one more question in proving that something is not a limit. It wants me to prove that $\\lim_{x\\rightarrow1}(3x+4)\\not=6.5$.\n\nAll I can do is show that 7 fits the definition, not that 6.5 does not.[/quote]\r\n\r\nWhy not pick delta = 0.1?\r\n\r\nThen |x-1| < 0.1 and 6.7 < |3(x-1) + 7| < 7.3.\r\n\r\nHence this fails to meet the definition of a limit. Furthermore, no matter what we say the limit is (as long as it's not 7), we can choose delta to be arbitrarily small so that |x-1| < delta implies |3(x-1) + 7| is contained within an interval not containing that number." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find the sum of all four-digit numbers whose all digits are odd.", "Solution_1": "This is indeed easy. :) \r\n\r\n[hide=\"Solution\"]\nThere are $ 5$ odd digits, so there are $ 5^4$ such numbers in the set (call it $ S$). Clearly, each number other than $ 5555$ corresponds to a single other number in $ S$ such that the average of the two numbers is $ 5555$ (a bijection from the $ S$ to $ S$, with a fixed point at $ 5555$). Adding all $ 5^4\\minus{}1$ such numbers, we see that there are $ \\frac{5^4\\minus{}1}{2}$ such correspondences in total, each contributing $ 5555\\cdot 2$ to the total sum. Thus, the total of the numbers (now including $ 5555$ itself) will be $ \\left(\\frac{5^4\\minus{}1}{2}\\right)\\cdot 5555\\cdot 2\\plus{}5555\\equal{}5^4\\cdot 5555\\equal{}3471875$.\n[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c>0$, $a+b+c=1$, Prove that $27(a-bc)(b-ca)(c-ab)\\leq 8abc$", "Solution_1": "[hide=\"maybe\"]Your problem equivalent to $27\\prod [a(a+b+c)-bc]\\leq 8\\prod a$. Now , you can use Muihard's theorem or Schur's theorem. I think so.[/hide]", "Solution_2": "Can you explain it a little more?", "Solution_3": "Another idea.\r\nPut $x=abc,y=ab+bc+ca$.\r\nThen $(a-bc)(b-ca)(c-ab)=4x-x^{2}-y^{2}-2xy$. So, we need prove $100x\\leq 27(x+y)^{2}$, or $f(a,b,c)=10\\sqrt{abc}-3\\sqrt{3}(abc+ab+bc+ca)\\leq 0$. We have $f(a,b,c)-f(a,t,t)=(\\sqrt{b}-\\sqrt{c})^{2}(\\frac{3\\sqrt{3}}{4}(\\sqrt{b}+\\sqrt{c})^{2}(a+1)-5\\sqrt{a})\\leq 0$, where $a=\\max{(a,b,c)}$ and $t=\\frac{b+c}{2}$. Done!.", "Solution_4": "[quote=\"Twins\"]$a,b,c>0$, $a+b+c=1$, Prove that $27(a-bc)(b-ca)(c-ab)\\leq 8abc$[/quote]\r\nMaybe the following proof better? :maybe: \r\nIf $a-bc<0$ and $b-ac\\geq0,$ $c-ab\\geq0$ or $a-bc,$ $b-ac$ and $c-ab$ are negative numbers its true.\r\n$a-bc<0$ and $b-ac<0$ it's can't be.\r\nLet $a-bc,$ $b-ac$ and $c-ab$ are non-negative numbers.\r\nThen $\\sqrt{(a-bc)(b-ac)}\\leq(1-c)\\sqrt{ab}.$\r\nHence remain to prove that $27(1-a)(1-b)(1-c)\\leq8,$ which triviality.", "Solution_5": "[quote=\"arqady\"][quote=\"Twins\"]$a,b,c>0$, $a+b+c=1$, Prove that $27(a-bc)(b-ca)(c-ab)\\leq 8abc$[/quote]\nMaybe the following proof better? :maybe: \nIf $a-bc<0$ and $b-ac\\geq0,$ $c-ab\\geq0$ or $a-bc,$ $b-ac$ and $c-ab$ are negative numbers its true.\n$a-bc<0$ and $b-ac<0$ it's can't be.\nLet $a-bc,$ $b-ac$ and $c-ab$ are non-negative numbers.\nThen $\\sqrt{(a-bc)(b-ac)}\\leq(1-c)\\sqrt{ab}.$\nHence remain to prove that $27(1-a)(1-b)(1-c)\\leq8,$ which triviality.[/quote]\r\nMy solution is similar to yours, but I want to find a nicer solution", "Solution_6": "We are to prove $\\left(1-\\frac{bc}{a}\\right)\\left(1-\\frac{ca}{b}\\right)\\left(1-\\frac{ab}{c}\\right)\\leq \\frac{8}{27}.$ Honestly I have finished, but my server has fade out.\r\n\r\nI am shocked. :(", "Solution_7": "We can use Mixing vaiaber Or Cauchy_svac ineq \r\nSorry it is verry long . I will post it on next sunday . Now i verry busy :maybe:" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Suppose $\\mu()$ is a Radon measure in $\\mathbb{R}^{n}$ on some $\\sigma$-algebra $\\Sigma$ of subsets of $\\mathbb{R}^{n}$. Let $f: \\mathbb{R}^{n}\\rightarrow \\mathbb{R}^{m}$ be continuous and proper (i.e. inverse images of open and compact sets are open and compact, resp.) with $f^{-1}(b)$ Borel for $b$ Borel. \r\n\r\nDo we know $f^{-1}(f(A)) \\in \\Sigma$ if $A \\in \\Sigma$?", "Solution_1": "Sorry. I don't know how to delete the other posts." } { "Tag": [ "algebra", "polynomial", "geometry", "incenter", "analytic geometry" ], "Problem": "1. Find a polynomial $ P(x)$ such that $ P(x)$ is divisible by $ x^2\\plus{}1$ and $ P(x)\\plus{}1$ is divisible by $ x^3\\plus{}x^2\\plus{}1$.\r\n\r\n(I got to the part where $ P(x)$ must have a degree of at least 3, and then I got stuck :oops: )\r\n\r\n2. Let $ P$ be an interior point of $ \\triangle{ABC}$, and let $ x,y,z$ denote the distances from $ P$ to $ BC$, $ AC$, and $ AB$ respectively. Where should $ P$ be located to maximize the product $ xyz$?\r\n\r\n(I have a hunch that $ P$ is at the incenter, but I'm not sure how to prove it using analytic geometry without diving into a pit of vicious variables.)", "Solution_1": "[hide=\"1\"] See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]this old post of mine[/url]. This is an application of the Chinese Remainder Theorem.\n\nWrite $ P(x) \\equal{} (x^3 \\plus{} x^2 \\plus{} 1) Q(x) \\minus{} 1$. Then $ P(x) \\equiv Q(x) \\minus{} 1 \\bmod (x^2 \\plus{} 1)$, so our general solution is $ Q(x) \\minus{} 1 \\equal{} R(x) (x^2 \\plus{} 1)$. Letting $ R(x) \\equal{} 1$, we have\n\n$ P(x) \\equal{} \\boxed{ (x^3 \\plus{} x^2 \\plus{} 1)(x^2 \\plus{} 2) \\minus{} 1 }$. [/hide]" } { "Tag": [ "limit", "trigonometry", "function" ], "Problem": "I don't know how challenging this is, but here's my \"homemade\" challenge question.\r\n\r\nEvaluate the limit as n approaches infinity of 10^n*100sin(9/(5*10^n)).", "Solution_1": "By subbing [tex]x=10^{-n}[/tex], the limit reduces to\r\n\r\n[tex]\\displaystyle \\lim_{x\\to0}\\frac{100\\sin\\left(9x/5\\right)}{x}[/tex].\r\n\r\nThis limit is indeterminant (since it approaches 0/0), so L'Hopital's rule says that this limit equals\r\n\r\n[tex]\\displaystyle \\lim_{x\\to0}\\frac{100\\cdot9/5\\cdot\\cos\\left(9x/5\\right)}{1}=180\\cos0=180[/tex].", "Solution_2": "Oops! I forgot part of the problem! The sine function in this problem is supposed to be in trigonometric form. So, converting the answer from degrees to radians, the limit equals :pi:.", "Solution_3": "[quote=\"ZennyK\"]Oops! I forgot part of the problem! The sine function in this problem is supposed to be in trigonometric form. So, converting the answer from degrees to radians, the limit equals :pi:.[/quote]\r\n\r\nWhat do you mean \"trigonometric form\"?", "Solution_4": "Instead of circular, i.e. its argument is in degrees rather than radians." } { "Tag": [ "pifinity is ze best" ], "Problem": "In how many ways can 8 people sit around a round table if 3 of the people -- Pierre, Rosa, and Thomas -- all want to sit together?", "Solution_1": "This is the same as arranging $ 6$ people around a table, and then multiplying by the $ 3!\\equal{}6$ permutations of Pierre, Rosa, and Thomas, so our answer is $ \\frac{6!}6\\cdot6\\equal{}\\boxed{720}$.", "Solution_2": "why isn't it 6!*3!=4320?", "Solution_3": "We need to count for rotations.", "Solution_4": "Keep in mind that we can rotate the group of 3 anywhere on the circle after placing them. Now we only need to account for the arrangement of the Pierre, Rosa, and Thomas in that group of 3, which is just $3!$. Now multiply by the number of arrangements of the other 5 people ($5!$), yielding $720$ as the answer.", "Solution_5": "I get where we get $3!*5!$, but why don't we divide by $8$ to account for rotations. Can someone explain? Thanks!", "Solution_6": "$3!$ is the number of arrangements for the three that want to be together and $5!$ is the number arrangements the other five that sit at the table.\n\n[hide]$3!\\cdot5!=720$[/hide]", "Solution_7": "I get that, but aren't there 8 rotations you got to account for, since you can rotate the table 8 times?", "Solution_8": "You want the rotations in the total.\n\nEdit: this is wrong oops", "Solution_9": "[quote=rubixsolver]You want the rotations in the total.[/quote]\nBut two rotations of the same seating are the same so wouldn't you divide?\n", "Solution_10": "[quote=jkliu][quote=rubixsolver]You want the rotations in the total.[/quote]\nBut two rotations of the same seating are the same so wouldn't you divide?[/quote]\n\nYes, why????", "Solution_11": "I have no clue what I was thinking when I said that, but that is wrong (lol).\n\nSo suppose you treat Pierre, Rosa, and Thomas as a superhuman. Once we place the superhuman, we place the remaining people relative to the superhuman. So we get $5!$ placements. Since you can permutate the humans inside the superhuman $3!$ ways, the answer is $5!\\times3!.$", "Solution_12": "I'm pretty sure the idea here is that you fix the table around the \"superhuman\" block", "Solution_13": "[quote=MathNerd555]I'm pretty sure the idea here is that you fix the table around the \"superhuman\" block[/quote]\n\nThat's EXACTLY what I did! There are 3! for the \"superhuman\" block, and 5! ways for the rest of the people, so our answer is 3!*5!.", "Solution_14": "[hide=sol]\n$\\frac{6!3!}{6}=3!5!=\\boxed{720}$\n[/hide]", "Solution_15": "Suppose we have A, B, C, D sitting around a table should we not consider both clockwise and anticlockwise arrangments as rotations .\n\nFor example arrangement ABCD shown below\n A \n D B\n C\naround a circle would also account for ADCB if we go anti clockwise from A but in circular counting this is considered a different arrangement. I am not able to get this.\n\nFor 4 people around a circle I get only \n\n4!/ (4 * 2) = 3 unique placments\n A\n D B\n C\n\n A\n C B\n D\n\n A\n D C\n B\n\nAll the other 21 permutations are rotations of these arrangments.\n\nBased on this answer to the question should be 360 and not 720\n\nCan someone please explain clockwise and counterclockwise are considered different arrangements when both are just rotations" } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Let $ABCD$ be a convex quadrilateral, with $\\angle BAC=\\angle DAC$ and $M$ a point inside such that $\\angle MBA=\\angle MCD$ and $\\angle MBC=\\angle MDC$. Show that the angle $\\angle ADC$ is equal to $\\angle BMC$ or $\\angle AMB$.", "Solution_1": "I see that nobody tryed it, so I will post my solution.\r\n\r\nDenote $\\angle{MBC}=\\angle{MDC}=\\alpha$, $\\angle{MCD}=\\angle{MBA}=\\beta$ and $\\angle{BAC}=\\angle{CAD}=t$.\r\n\r\nThus, $\\frac{MC}{\\sin{\\alpha}}=\\frac{BC}{\\sin{BMC}}$, and $\\frac{MC}{\\sin{\\alpha}}=\\frac{CD}{\\sin{(\\alpha+\\beta)}}$ by the sin law in $\\triangle{MBC}$, and $\\triangle{MCD}$.\r\n\r\nSo $\\frac{BC}{CD}=\\frac{\\sin{BMC}}{\\sin{(\\alpha+\\beta)}}$. $(1)$\r\n\r\nNow from the sin law in $\\triangle{ABC}$ and $\\triangle{ADC}$, we have that:\r\n$\\frac{BC}{\\sin{t}}=\\frac{AC}{\\sin{(\\alpha+\\beta)}}$ and $\\frac{CD}{\\sin{t}}=\\frac{AC}{\\sin{ADC}}$\r\n\r\nSo $\\frac{BC}{CD}=\\frac{\\sin{ADC}}{\\sin{(\\alpha+\\beta)}}$. $(2)$\r\n\r\nBy $(1)$ and $(2)$ we have that $\\sin{ADC}=\\sin{BMC}$, thus $\\angle{ADC}=\\angle{BMC}$, or $\\angle{ADC}=180-\\angle{BMC}$ etc.", "Solution_2": "[quote=\"pohoatza\"]I see that nobody tryed it, so I will post my solution.\n\nDenote $ \\angle{MBC} \\equal{} \\angle{MDC} \\equal{} \\alpha$, $ \\angle{MCD} \\equal{} \\angle{MBA} \\equal{} \\beta$ and $ \\angle{BAC} \\equal{} \\angle{CAD} \\equal{} t$.\n\nThus, $ \\frac {MC}{\\sin{\\alpha}} \\equal{} \\frac {BC}{\\sin{BMC}}$, and $ \\frac {MC}{\\sin{\\alpha}} \\equal{} \\frac {CD}{\\sin{(\\alpha \\plus{} \\beta)}}$ by the sin law in $ \\triangle{MBC}$, and $ \\triangle{MCD}$.\n\nSo $ \\frac {BC}{CD} \\equal{} \\frac {\\sin{BMC}}{\\sin{(\\alpha \\plus{} \\beta)}}$. $ (1)$\n\nNow from the sin law in $ \\triangle{ABC}$ and $ \\triangle{ADC}$, we have that:\n$ \\frac {BC}{\\sin{t}} \\equal{} \\frac {AC}{\\sin{(\\alpha \\plus{} \\beta)}}$ and $ \\frac {CD}{\\sin{t}} \\equal{} \\frac {AC}{\\sin{ADC}}$\n\nSo $ \\frac {BC}{CD} \\equal{} \\frac {\\sin{ADC}}{\\sin{(\\alpha \\plus{} \\beta)}}$. $ (2)$\n\nBy $ (1)$ and $ (2)$ we have that $ \\sin{ADC} \\equal{} \\sin{BMC}$, thus $ \\angle{ADC} \\equal{} \\angle{BMC}$, or $ \\angle{ADC} \\equal{} 180 \\minus{} \\angle{BMC}$ etc.[/quote]\r\nyou did'nt prove that A,M,C are collinear ???", "Solution_3": "If $A,M,C$ are collinear ,then we have easily $\\angle{BMA}=\\angle{CDA}$. Assume that $A,M,C$ aren't collinear. Let $N$ $K$ be reflections of $D$ and $B$ with $MC$. Then $NBMC$ and $KMCD$ are cyclic. Since $\\angle{MBA}=\\angle{MCD}=\\angle{MCN}$ We have: $N$ lies on the line $AB$. Let $E$ be intersection of $NB$,$DK$ and $CM$. Since $EM$ is bisector of $\\angle{AED}$, $C$ is excircle center of $\\triangle{EDA}$. (that touching to the side $AD$) hence $CD$ is external bisector of $\\angle{ADE}$ Then: $\\angle{BMC}=180^{\\circ}-\\angle{BNC}=180^{\\circ}-\\angle{CDK}=\\angle{ADC}$. So we are done! (U.J.B.O.)", "Solution_4": "[quote=pohoatza]Let $ABCD$ be a convex quadrilateral, with $\\angle BAC=\\angle DAC$ and $M$ a point inside such that $\\angle MBA=\\angle MCD$ and $\\angle MBC=\\angle MDC$. Show that the angle $\\angle ADC$ is equal to $\\angle BMC$ or $\\angle AMB$.[/quote]\nRemark:\nActually this configuration same as Miquel's Theorem\nLet points P,T,S are intersection of\nMD,MC,MB with BC,AB,CD respectively\nUsing above facts ,it can be shown that\nPBMT,TMDS,TSBC,TPCD,PBDS is cyclic\nAnd this happen if and only if \nT is Miquel Point of complete quadrilateral PBCDSM.", "Solution_5": "Denote $\\angle BAC = \\angle DAC = \\alpha$, $\\angle ABM = \\angle MCD = x$ and $\\angle MBC = \\angle MDC = y$. By the Sine Law we have\n\\[ \\frac{AC}{\\sin\\angle ADC} = \\frac{CD}{\\sin \\alpha} = \\frac{\\frac{CM}{\\sin y}\\sin(x+y)}{\\sin \\alpha} = \\frac{BC}{\\sin \\angle BMC}\\frac{\\sin \\angle ABC}{\\sin \\angle BAC} = \\frac{AC}{\\sin \\angle BMC} \\]\nthus $\\sin \\angle ADC = \\sin \\angle BMC$. \n\nSince both angles have measures strictly between $0^{\\circ}$ and $180^{\\circ}$, we either have $\\angle ADC = \\angle BMC$ (and we are done), or $\\angle ADC + \\angle BMC = 180^{\\circ}$. \n\nIn the latter case with $\\angle BMC = z$, $\\angle ADC = 180^{\\circ} - z$ we obtain $\\angle BCM = 180^{\\circ} - y - z$ and now summing all angles in the quadrilateral $ABCD$ leads to $z = \\alpha + x$. Now if $BM \\cap AC = M_1$, then $\\angle BM_1C = \\alpha + x = z = \\angle BMC$, so $M \\equiv M_1$, i.e. $M$ lies on $AC$. Hence we conclude $\\angle ADC = 180^{\\circ} - \\angle BMC = \\angle AMB$, as desired." } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra proposed", "Fixed point" ], "Problem": "Suppose that $a,b,c$ are real numbers and let $f(x)=ax^{2}+bx+c .$ \r\n [b] If $|f(10)| > 721$ show that there exists at least a point $x_0$ in $[0,1]$ such that $|f(x_0)| > 1$ . [/b]", "Solution_1": "I like the problem! can u post your solution? :)", "Solution_2": "[quote=\"flavian\"]I like the problem! ... your solution ? :)[/quote]\r\n[color=green] [b] Suppose that $a,b,c$ are real numbers and let $f(x)=ax^{2}+bx+c .$ If $|f(10)| > 721$ show that there exists\n at least a point $x_0$ in $[0,1]$ such that $|f(x_0)| > 1 .$[/b] [/color]\r\n[b]Solution.[/b] Assume $| f(r) | \\le 1$ for all real (rational it's enough) points $r$ from $[0,1] .$ Because\r\n\\[ f(x)=(2x-1)(x-1)f(0)-4x(x-1)f\\left(\\frac{1}{2}\\right)+x(2x-1)f(1) , \\]\r\nfor $x\\in (-\\infty,0)\\cup (1,\\infty)$ we find $\\displaystyle |f(x)|\\le \\displaystyle (2x-1)(x-1)+4x(x-1)+x(2x-1)=\\displaystyle 8x^2-8x+1=: Q(x) .$\r\n Therefore $|f(10)| \\le Q(10)= 721$ [b] which is false ! [/b] [size=75] Remark: observe that $|Q(t)|=|1-8t(1-t)|\\le 1\\; ,\\; \\forall t\\in [0,1] .$ [/size]", "Solution_3": "[hide=\"Flip2004 and Flavian, see here !\"]\n$f\\in R[X],\\ gr f=p\\in N,\\ n>1\\Longrightarrow$\n\n$f=\\frac{1}{p!}\\prod\\limits_{k=0}^p (pX-k)\\cdot \\sum\\limits_{k=0}^p {p\\choose k} (-1)^{p-k} f\\left(\\frac kp\\right)\\frac{1}{pX-k}$.\n\n$\\left( \\forall \\right) x\\in [0,1],\\ |{f(x)}|\\le 1\\Longrightarrow \\left(\\forall\\right)x\\in[0,1],\\ |f(x)|\\le g(x)$ where\n\n$g=\\frac{1}{p!}\\prod\\limits_{k=0}^p (pX-k)\\cdot \\sum\\limits_{k=0}^p {p\\choose k}\\frac{1}{pX-k}$.\n\n$721=6!+1$. But $10=\\ ?$ a.s.o. $\\Longrightarrow$ [b]Generalization ...[/b] [u]I wish you every success ![/u] [/hide]", "Solution_4": "[quote=\"Virgil Nicula\"][hide=Flip2004 and Flavian, see here !] [/quote]\r\nThank you very much ! The generalization is the following : \r\n[color=green][b] Let $f(x)=\\sum\\limits_{k=0}^{n}a_kx^k$ , $a$ fixed in $(1,\\infty),$ and \n\\[ \\begin{array}{|c|} \\hline \\\\ K_n(a)= \\displaystyle \\frac{\\left(2a-1+2\\sqrt{a(a-1)}\\right)^n +\\left(2a-1-2\\sqrt{a(a-1)}\\right)^n }{2} \\\\ \\\\ \\hline \\end{array} \\]\nIf we assume $\\begin{array}{|c|} \\hline \\\\ \\displaystyle |f(a)|>\\displaystyle K_n(a) \\\\ \\\\ \\hline \\end{array}\\; ,\\;$ then there exists a point $x_0$ in $[0,1]$ such that $|f(x_0)| >1 \\; .$ [/b][/color]\r\n[size=75]\nRemarks: \n1) $K_n(a) = T_n(2a-1)$ where $T_n(z)$ is Chebychev polynomial of first kind, defined for $|z| \\le 1$ by $T_n(z)=\\cos\\left(n\\cdot\\arccos{z}\\right) .$ \n2) The constant $K_n(a)$ cannot be improved: to see this it's sufficient to select \\[ f(x): =T_n(2x-1)=\\sum\\limits_{k=0}^{n}(-1)^{n-k}\\frac{n}{n+k} {n+k\\choose 2k}2^{2k}x^k\\; , \\; n\\ge 1, T_0(x): =1 . \\]\nIt's known that $| T_n(2x-1) |\\le 1 \\; ,\\; x\\in{\\mathbb R} ,$ iff $x\\in [0,1] \\; .$\n3) $K_2(10) = \\frac{(19+6\\sqrt{10})^2+(19-6\\sqrt{10})^2 }{2}=721\\; .$\n[/size]\r\n\r\nThanks for interest,Alex[=flip2004]", "Solution_5": "What is this Alex collection and why are the problems in it so difficult :surrender:", "Solution_6": "[quote=\"t0rajir0u\"]What is this Alex collection and why are the problems in it so difficult ?[/quote]\r\nI try to answer you by means of a private e-mail message . / flip2004" } { "Tag": [ "geometry" ], "Problem": "[img]http://i706.photobucket.com/albums/ww68/jellymuffins2009/triangke.jpg[/img]\r\n\r\n\r\nAngle AE$", "Solution_2": "The question I have comes from 5.18 in Intro to Geo. It says, \"Since 0$ so that $ a\\plus{}b\\plus{}c\\equal{}1$ then prove that:\r\n\r\n\\[ \\frac{1\\plus{}a}{1\\minus{}a} \\cdot \\frac{1\\plus{}b}{1\\minus{}b} \\cdot \\frac{1\\plus{}c}{1\\minus{}c} \\ge 8\\]", "Solution_1": "[quote=\"moldovan\"]If $ a,b,c > 0$ so that $ a \\plus{} b \\plus{} c \\equal{} 1$ then prove that:\n\\[ \\frac {1 \\plus{} a}{1 \\minus{} a} \\cdot \\frac {1 \\plus{} b}{1 \\minus{} b} \\cdot \\frac {1 \\plus{} c}{1 \\minus{} c} \\ge 8\\]\n[/quote]\r\n\r\nTake $ x \\equal{} a \\plus{} b$ and use $ (x \\plus{} y)(y \\plus{} z)(z \\plus{} x) \\ge 8xyz$", "Solution_2": "Very nice problem!\nMy solution:\nThe problem is equivalent to:\n$ a,b,c>0,a+b+c=3\\Rightarrow\\prod\\frac{3+a}{3-a}\\ge 8\\Leftrightarrow $\n$ \\Leftrightarrow\\prod(1+\\frac{2a}{b+c})\\ge 8 $\nWith substitutions $ x=\\frac{2a}{b+c},y=\\frac{2b}{c+a},z=\\frac{2c}{a+b} $ we obtain\n1)\n$ x,y,z>0,\\sum\\frac{1}{x+2}=1\\Leftrightarrow xy+yz+zx+xyz=4 $\nYou must show that:\n2)\n${ x,y,z>0,xy+yz+zx+xyz=4\\Rightarrow (1+x)(1+y}(1+z)\\ge 8 $\nAfter calculations, (2) is equivalent to:\n3)\n$ x,y,z>0,xy+yz+zx+xyz=4\\Rightarrow x+y+z\\ge 3 $\nwhich is known! (or not!)\nWe conclude this proof! Remains as possibly to post proof of (3)!\n____________\nSandu Marin" } { "Tag": [ "ratio", "puzzles" ], "Problem": "Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this cryptarithmetic equation. It is known that [b]A[/b] > [b]D[/b] > [b]G[/b], [b]B [/b]> [b]C[/b], [b]E [/b]> [b]F[/b], [b]H [/b]> [b]I[/b], and [b]C[/b] is neither 1 nor 5.\r\n\r\n[b]A[/b]/([b]B[/b]+[b]C)[/b] + [b]D[/b]/([b]E[/b]+[b]F[/b]) + [b]G[/b]/([b]H[/b]+[b]I[/b]) = 1", "Solution_1": "so E could still be F and H could still be I", "Solution_2": "[quote=\"U\u00a7ername\"]so E could still be F and H could still be I[/quote]\r\n\r\nEdit: I understand your query now, and I have edited the problem text to address the said concern.", "Solution_3": "[quote] Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this cryptarithmetic equation. It is known that A > D > G, B > C, E > F, H > I, and C is neither 1 nor 5.\n\nA/(B+C) + D/(E+F) + G/(H+I) = 1 [/quote]\r\n\r\nWell, I had almost solved the above problem, but I had committed an error somewhere which leads to a 'misplaced digit' (I apparently 'miss' the answer by a single digit).\r\n\r\nHere's my near-complete solution.\r\n\r\n[b] 8 + 2 + 1\n \n 7+4 6 + 5 9 + 3 \n \nwhere\n\nA = 8; D =2 ; G =1; B = 7; C = 4; E = 6; F = 5; H = 9, and I = 3 (this is incorrect)\n\n[/b]" } { "Tag": [ "quadratics", "function" ], "Problem": "Let f(x)=$ ax^2\\plus{}bx\\plus{}c$ be any quadratic with real coefficients such that $ |f(x)|\\leq 1$ for $ 0\\leq x \\leq 1$. Find the smallest M such that $ |a|\\plus{}|b|\\plus{}|c|\\leq M$.", "Solution_1": "[quote=\"kyyuanmathcount\"]Let f(x)=$ ax^2 \\plus{} bx \\plus{} c$ be any quadratic with real coefficients such that $ |f(x)|\\leq 1$ for $ 0\\leq x \\leq 1$. Find the smallest M such that $ |a| \\plus{} |b| \\plus{} |c|\\leq M$.[/quote]\r\nI'm not sure how rigorous this is...\r\n[hide]\nIf $ |a|\\plus{}|b|\\plus{}|c|>1$, then we have that $ f(x)\\equal{}|a|\\plus{}|b|\\plus{}|c|$ (for x=1), but this violates $ |f(x)| \\leq 1$, so M=1?[/hide]", "Solution_2": "[hide=\"Response\"]Consider $ f(x)\\equal{}(x\\minus{}1)^2$.[/hide]", "Solution_3": "[hide]Consider the function 8x^2-8x+1[/hide]", "Solution_4": "[quote=\"kyyuanmathcount\"]Let f(x)=$ ax^2 \\plus{} bx \\plus{} c$ be any quadratic with real coefficients such that $ |f(x)|\\leq 1$ for $ 0\\leq x \\leq 1$. Find the smallest M such that $ |a| \\plus{} |b| \\plus{} |c|\\leq M$.[/quote]\r\nMy answer is $ M\\equal{}17.$\r\n[hide]If this is correct, I will post the detailed answer.[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "What is the maximum number of parts into which a plane can be divided by n circles?", "Solution_1": "Can the same be generalised to convex figures as such??\r\n(Recursion seems to be the key)." } { "Tag": [], "Problem": "$ x$ and $ y$ are two numbres from $ \\mathbb{N}^2$\r\n\t\r\nprove that the smallest numbre $ t$ from $ \\mathbb{N}$ ($ t>0$) such as $ y$ divides $ xt$ is a divisor for$ y$.\r\n\r\n;)", "Solution_1": "[quote=\"sami.d\"]$ x$ and $ y$ are two numbres from $ \\mathbb{N}^2$\n\t\nprove that the smallest numbre $ t$ from $ \\mathbb{N}$ ($ t > 0$) such as $ y$ divides $ xt$ is a divisor for$ y$.\n\n;)[/quote]Let me get this straight... so $ x,y\\in\\mathbb{N}^2$? How is that a number if it is the Cartesian product $ \\mathbb{N}\\times\\mathbb{N}$? I guess $ \\mathbb{N}\\equal{}\\{0,1,2,\\dots\\}$?\r\nAssuming $ x,y\\in\\mathbb{N}$, prove that the smallest number $ t\\in\\mathbb{N}$ ($ t>0$) such that $ y$ divides $ xt$, is a divisor of $ y$, ie $ t|y$. :ninja:", "Solution_2": "Maybe it is positive squares?", "Solution_3": "I'm guessing he means $ (x, y) \\in \\mathbb{N}^2$." } { "Tag": [ "Support", "AMC", "USA(J)MO", "USAMO" ], "Problem": "[quote=\"MysticTerminator\"]Zach Abel\nZarathustra Brady\nRyan Ko\nYi Sun\nArnav Tripathy\nAlex Zhai[/quote]\r\nAre there here any real american, except maybe Ryan Ko?", "Solution_1": "The first two names are caucasian, I believe. Beyond that I'm not sure what you mean by real american.", "Solution_2": "[quote=\"pavel kozlov\"][quote=\"MysticTerminator\"]Zach Abel\nZarathustra Brady\nRyan Ko\nYi Sun\nArnav Tripathy\nAlex Zhai[/quote]\nAre there here any real american, except maybe Ryan Ko?[/quote]\n\nRyan Ko is of Asian descent. The name they have listed on the AMC website is Taehyeon which would have been more of a giveaway :P \n\n[quote=\"s1ck_psych0\"][quote=\"amir2\"]\n[quote=\"Orl\"]\nIt is a pretty smart trick to increase one's post count by asking questions all the time which can be figured out easily by searching them on the web[/quote]\n\n\n\nThere are smarter tricks too. Mr. Green[/quote]\n\n\n\nI guess you're right! Wink[/quote]\r\n\r\nlol, this could start an infinite chain of posts :rotfl:", "Solution_3": "[quote=\"pavel kozlov\"][quote=\"MysticTerminator\"]Zach Abel\nZarathustra Brady\nRyan Ko\nYi Sun\nArnav Tripathy\nAlex Zhai[/quote]\nAre there here any real american, except maybe Ryan Ko?[/quote]\r\n\r\nZach is American\r\nI know because he graduated from the school I go to\r\nwhose math team will definitely not be as succesful next year without him\r\n :(", "Solution_4": "Maybe, this is somehow off topic but Zarathustra was an Iranian Prophet with the relegion of Zarathustrian (In farsi(Iranian) we pronounce it like Zartosht)\r\nIt's not a very popular relegion in Iran (But I think it's the second most relegion after Islam in Iran), in the world too (No offence to Zarathustrians)\r\nI'm very interested about Zarathustra Brady, any one knows wether he is Zarathustrian or his parents just picked this name for him.\r\nDoes any one know about it? (Maybe Zarathustra himself is in this Forum)", "Solution_5": "I'm American!", "Solution_6": "[quote=\"Nima Ahmadi Pour\"]the relegion of Zarathustrian (In farsi(Iranian) we pronounce it like Zartosht)[/quote]\r\n\r\nIn English, \"Zoroastrian.\"\r\n\r\nThe question, \"Are there any real Americans?\" is a meaningless question. A better question might be, \"How many of those team members were born in America?\"", "Solution_7": "[quote=\"pavel kozlov\"][quote=\"MysticTerminator\"]Zach Abel\nZarathustra Brady\nRyan Ko\nYi Sun\nArnav Tripathy\nAlex Zhai[/quote]\nAre there here any real american, except maybe Ryan Ko?[/quote]\r\n\r\nPavel, the situation is similar to that of USSR olympiad teams: the\r\ngeneral population and the IMO population of the country look very different. In the USSR, a huge proportion of olympiad winners were Jewish (not all were permitted to travel to the IMO) and the same for scientists, mathematicians, engineers, chess-masters, and intelligentsia in general. In the USA, immigrants from Asia (and ex-USSR), children of such immigrants, children of white American scientists and intelligentsia, and Jews form a large proportion of the IMO population. Of the 6 names listed above for the USA IMO team, at least one sounds Jewish, three are obviously East Asian (whether US-born or not is unclear), and one Indian.", "Solution_8": "[quote=\"Nima Ahmadi Pour\"]I'm very interested about Zarathustra Brady, any one knows wether he is Zarathustrian or his parents just picked this name for him.[/quote]\r\nThe latter. He goes by Zeb (his initials), and his AOPS username is also zeb.\r\nI think the name is a reference to Nietzsche's work, possibly indirectly. Zarathustra is the German form.", "Solution_9": "Being named after a Nietzsche work is pretty awesome, actually.", "Solution_10": "Wow, I'm famous :P \r\n\r\nThe name is just there to throw you off... I'm German/Russian/Irish/Jewish.", "Solution_11": "The name was selected because of a paternal fascination with \"Thus Spoke Zarathustra\", and because of a paternal dislike for common names, such as Kevin, John, Daniel, etc.", "Solution_12": "[quote=\"JBL\"] The question, \"Are there any real Americans?\" is a meaningless question. [/quote]\r\n\r\nThe questioner is from a country where \"Politically Correct\" is no longer an official phrase and where, unlike the Real America, one can ask and answer such questions openly. There is, of course, a variety of obvious and non-trivial interpretations of the question. By the way, it is a near-universal presumption by mathematically educated immigrants to the USA, that American IMO teams are dominated by immigrants (which is partially correct, and becomes more correct if one adds US-born children of the immigrant intelligentsia).", "Solution_13": "I'm not sure what the purpose of your first sentence is.\r\n\r\n[quote=\"fleeting_guest\"]There is, of course, a variety of obvious and non-trivial interpretations of the question.[/quote]\r\n\r\nWhich makes the question effectively meaningless, lacking further elaboration.", "Solution_14": "[quote=\"JBL\"]I'm not sure what the purpose of your first sentence is.\n\n[quote=\"fleeting_guest\"]There is, of course, a variety of obvious and non-trivial interpretations of the question.[/quote]\n\nWhich makes the question effectively meaningless, lacking further elaboration.[/quote]\r\n\r\nThe question and its answers are clear and obviously meaningful, notwithstanding politically expedient deconstructions. It is simply a demographic fact that in the US (and Canada, Australia, and elsewhere) the IMO population differs dramatically, and racially, from the general population. Some racial groups have, so far, been nonexistent within that population, whereas others have been been hugely overrepresented compared to their share of national population. \r\nIt is of course possible to try and further deconstruct these unpleasant facts by \"controlling\" for various factors, but the discrepancies remain enormous.\r\nI do not think any service is done by Political Correction of phrases such as \"real American\", particularly as used by non-US natives.\r\n\r\nAdded:\r\nlet me mention, e.g. for those reading this outside USA, what type of political considerations I refer to. \r\n\r\nThere remains in the US an ideology of \"immigrant melting pot\", as opposed to multiculturalism. It has broken down somewhat in recent times, but differentiating between which citizens are more American than the others is still considered in many social situations as making an invidious and politically incorrect distinction. In theory, Yanks are supposed to be one big family, just don't ask about the color of the garbage collectors or what language they speak. Because of its more overtly competitive, zero-sum ethos as compared to other countries and also because of its troubled racial history (oppression of blacks), in the US when obvious patterns appear such as Asians/Jews/immigrants dominating science, this is not necessarily treated as good news (\"look at those talented Chinese, isn't it nice they are here!\") but rather as something to be covered over by either avoiding the subject or by invoking the We Are All Americans ideology.", "Solution_15": "If a question has a variety of obvious interpretations (as you wrote), then it is a poor question. A good question has only one obvious interpretations (and hopefully no non-obvious interpretations).\r\n\r\nYou think the question is clear and well-posed? Is a third-generation student of Japanese descent (say) a \"real American?\" What about a second-generation Indian whose parents spoke English from childhood (in India, before emigrating)? What about a twelfth-generation African-American?\r\n\r\nPerhaps more importantly, the intent of the question was almost certainly closer to my interpretation (with immigration the relevant category) than yours (race). I cannot be certain of this, because the question is not a good one. But your last two posts are, most likely, much further off the point than the post of mine you are criticizing.", "Solution_16": "[quote=\"JBL\"]If a question has a variety of obvious interpretations (as you wrote), then it is a poor question. A good question has only one obvious interpretations (and hopefully no non-obvious interpretations). [/quote] Most questions, and most reasonable questions, contain some ambiguity. That is not a problem, much less something that renders the question \"meaningless\".\n\n[quote] You think the question is clear and well-posed? [/quote]\nI think it was a reasonably clear question and that it was reasonable to ask it. A PhD thesis on the definition of \"real American\" is not needed in order to grasp that someone was asking about an apparent difference between the composition of IMO team(s) from the USA and the composition of the USA national population. Whether the question was specifically pointing to differences in birthplace, citizenship, immigrant status, racial group, social stratum, or other such paramaters was not assumed in the answers. \n\n[quote] Is a third-generation student of Japanese descent (say) a \"real American?\" What about a second-generation Indian whose parents spoke English from childhood (in India, before emigrating)? What about a twelfth-generation African-American? [/quote]We don't need to answer those questions in order to answer Pavel's question. If, as you claim, I have misunderstood his question, I am sure that Pavel can clarify what was meant, or whether he had such a specific interpretation in mind.", "Solution_17": "This is absurd. We agree the question contained ambiguity, yet you are criticizing me for attempting to disambiguate. It is not necessary to answer my questions to answer Pavel's question, but it is most certainly necessary to answer them to support a claim that the question of \"real Americanness\" is clear and well-posed. I did have not called Pavel's question \"unreasonable,\" so I don't see why you think it necessary to note that Pavel's question was a reasonable one.\r\nMoreover, Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.", "Solution_18": "[quote=\"JBL\"] the question contained ambiguity, yet you are criticizing me for attempting to disambiguate. [/quote]\n\nI disputed the idea that ambiguous questions (either in general or this particular question) are meaningless or poor. There was, obviously, no criticism of question-disambiguation as such.\n\n[quote] It is not necessary to answer my questions to answer Pavel's question, but it is most certainly necessary to answer them to support a claim that the question of \"real Americanness\" is clear and well-posed. [/quote]\n\nYou concede that Pavel's question (are there any real Americans on the American IMO team) was answerable, so presumably it was not meaningless, after all. Some other question that you introduced and that only you have discussed here (\"the question of real Americanness\") may or may not be meaningful, clear or well-posed, but that is, literally, a different question.\n\n[quote] Moreover, Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.[/quote]\r\n\r\nPavel did not falsify any of the interpretations of the question by at least five different respondents, so his \"neglect\" does not privilege the JBL interpretation over the others, but it does work against your statement that some non-JBL interpretations are worse than the JBL interpretation.", "Solution_19": "I think the purpose of this topic has been fulfilled. Please stop arguing before it needs to be locked.", "Solution_20": "Uh, JBL is the moderator of this forum - I don't think you can tell him when the topics should be locked.", "Solution_21": "When you say Real American, do you refer to those whose family has lived in American ever since 16th century?", "Solution_22": "So that means no americans(whose parents are americans),who have no other country connections n stuff?", "Solution_23": "I thought the IMO rules made pretty clear that anyone who is a member of a national team to the IMO has to have some connection to the country sending out the national team. It's my understanding that the United States team to the IMO consists of citizens or permanent residents of the United States (as contrasted with persons who live in the United States with temporary visitor or diplomatic visas). The United States team might be glad to pick up some of the Canadians who score well on the USAMO :D but it's my understanding that those math competitors can only compete for Canada. \r\n\r\ntokenadult \r\n\r\n(a \"real American\" born in the United States and descended from pre-revolutionary settlers in New England and married to a \"real American\" first-generation immigrant who is now a naturalized citizen)", "Solution_24": "With any luck, this will be my final post on this topic.\r\n\r\nThank you, eryaman, shobber, shady and tokenadult for your various responses, all of which I like and all of which I will ignore from here onwards, as the rest of this post is addressed to fleeting_guest.\r\n\r\nThe initial question of Pavel's was, paraphrased, \"Are there any real Americans on the American IMO team?\" I said, \"The question, \"Are there any real Americans?\" is a meaningless question. A better question might be, \"How many of those team members were born in America?\"\"\r\nThe next day, you gave a perfectly reasonable response to Pavel's question, noting the discernable ethnicities/nations of origin of several of the USIMO team members. More than 2 weeks later, you responded to my initial post, beginning with a sentence whose purpose I still don't understand (and which you did not explain when I requested)*. You also allowed that the question \"Are there real Americans?\" has \"a variety of obvious and non-trivial interpretations,\" one of which (presumably) is the question I had suggested in its place, \"Are there any non-immigrant members of the US team?\" You argued that \"the IMO population differs dramatically, and racially, from the general population,\" a proposition which is true (if perhaps unfortunate) and which has nothing to do with what I said. Nor, I think, does it have very much to do with the question Pavel asked. If it did, your initial (June 20) response to Pavel's question would seem to be sufficient to cover that territory. \r\n\r\nNow, the position I have been arguing is, more or less, that Pavel's original question was so ambiguous so as to be effectively meaningless. I therefore assume, since you have continued to joust with me, that you disagree with this contention. It is unlikely that we will come to an agreement over this point. I would note that in your first (June 20) response to Pavel's question, you noted that one of the team members was Indian. That same team member, two days earlier, had posted a one-line response to Pavel's question, saying \"I'm American!\" You could make it your business to tell him that he is not a \"real American.\" I suppose also that if the question were truly unambiguous, you would be able to tell me, in each of the three hypothetical cases I gave you, whether those individuals were \"real Americans.\" While it may not be necessary to answer my three hypothetical questions to answer Pavel's question, it is certainly necessary that [i]the ability to answer my question[/i] exists in order for Pavel's question to be meaningful. The only way Pavel's question can mean anything is if, in the case of specific individuals with specific racial and national backgrounds, you can decide who is and who is not a \"real American.\" Now, if you think you can create a satisfactory delineation between those of us who really are Americans and the rest (\"fake Americans?\" \"Imported Americans?\" \"racial minorities?\"), go ahead.\r\n\r\nNow, to nit-pick at your most recent post, I had (and have) no intention of conceding that Pavel's question is meaningful -- I was merely following your own construction.\r\n\"The question of real Americanness,\" that is whether or not such a thing is meaningful, is obviously fundamental to answering Pavel's question. If the notion is meaningless, his question is as well. It is indeed a different question, but it is a question whose answer is fundamental to the meaning of Pavel's question.\r\n\r\nThe question I offered in place of Pavel's question, that of immigrant status, has a key advantage over Pavel's: it is not ambiguous at all. Moreover, I chose that question because I believed that it captured the issue that Pavel was actually interested in. I may be wrong about that -- we don't know because Pavel hasn't re-entered the conversation. On this point, we had the following exchange:[quote=\"fleeting_guest\"]If, as you claim, I have misunderstood his question, I am sure that Pavel can clarify what was meant, or whether he had such a specific interpretation in mind.[/quote][quote=\"JBL\"]Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.[/quote][quote=\"fleeting_guest\"]Pavel did not falsify any of the interpretations of the question by at least five different respondents, so his \"neglect\" does not privilege the JBL interpretation over the others, but it does work against your statement that some non-JBL interpretations are worse than the JBL interpretation.[/quote] I don't think I really need to say much about this -- you took my remark to imply that Pavel's silence supports my view, immediately after making an identical implication (that Pavel's silence would support your view, by not correcting it) and then criticized me for that implication. I fully agree that Pavel's silence indicates nothing, and it certainly isn't the reason that I believe myself to be correct. \r\n\r\nYou may consider this my honest attempt to establish what we've been arguing about and what my position is. If I have erred, please feel free to point out where.\r\n\r\n* You also later implicitly accused me of \"politically expedient deconstructions.\" Miriam-Webster's Collegiate Dictionary tells me that \"expedient\" means 1: \"suitable for achieving a particular end in a given circumstance\" and 2: \"characterized by concern with what is opportune; especially : governed by self-interest\" with the additional note that \"EXPEDIENT usually implies what is immediately advantageous without regard for ethics or consistent principles .\" This also puzzles me, because I don't know what political agenda I am being accused of pursuing (at the expense of my ethics and consistent principles, no less) by attempting to disambiguate a clearly ambiguous question.", "Solution_25": "This has turned into a topic from the twilight zone or something...\r\n\r\nEryaman: mysmartmouth never told JBL to lock the topic. He attempted to put the conversation back on track and said that it [i]might[/i] be locked (therefore implying that it would be up to JBL's discretion) if the conversation continued in that manner.\r\n\r\nAnyway, the thread has gone far off topic...", "Solution_26": "re: tokenadult comment on nationality requirements for the IMO, I don't think the rules were as specific as permanent residency, but most countries impose such requirements in their selection procedures. I think there was some requirement (or assumption within the language of the regulations) of affiliation with the school system of a given country. This may have been rationalized and specified more in recent years in light of home-schooling and IMO contestants who exploit multiple citizenships to qualify in a less selective country than the one where they live or study.\r\nre: JBL, \r\n\r\n\r\n[quote=\"JBL\"] [60-line made-for-PM metadiscussion] [/quote]It would take a 200-line dissertation to analyze it in full, which is unlikely to happen. Some things need correction on the record, though:\n\n\n[quote] More than 2 weeks later, you responded to my initial post, [/quote]My posting appeared less than 5 days after its predecessor; longer gaps of 5 and 13 days had already occurred in the thread and are apparently OK with you, so it is unclear what nefarious behavior you point to here. \n\n[quote] beginning with a sentence whose purpose I still don't understand (and which you did not explain when I requested)*. [/quote]\n\nPlease retract the falsehood. You did not post any request for explanation. As it happens, I did post a long explanatory paragraph about the political correctness of \"real Americans\" discussions in the United States, which is exactly the subject matter of the sentence you call unexplained. It is outrageous to allege evasions of questions that were never asked but happen to have been answered anyway. \n\n\n[quote] You argued that \"the IMO population differs dramatically, and racially, from the general population,\" a proposition which is true (if perhaps unfortunate) and which has nothing to do with what I said. Nor, I think, does it have very much to do with the question Pavel asked. [/quote]\n\nThe fact that 2-3 additional posters independently provided ethnic/racial/religious data on the American IMOers in response to Pavel's question contradicts both your points.\n\n\n[quote] in your first (June 20) response ... you noted that one of the team members was Indian. That same team member, two days earlier, had posted a one-line response to Pavel's question, saying \"I'm American!\" You could make it your business to tell him that he is not a \"real American.\" [/quote]\n\nAn outrageous fabrication. Please retract the falsehood and quote my actual words rather than rhetorically convenient straw-men. I did not suggest that any team member \"was Indian\" or speculate on his level of Americanism. \n\n[quote] [I suppose also that if the question were truly unambiguous, you would be able to tell me, in each of the three hypothetical cases I gave you, whether those individuals were \"real Americans.\" While it may not be necessary to answer my three hypothetical questions to answer Pavel's question, it is certainly necessary that [i]the ability to answer my question[/i] exists in order for Pavel's question to be meaningful. The only way Pavel's question can mean anything is if, in the case of specific individuals with specific racial and national backgrounds, you can decide who is and who is not a \"real American.\" [/quote]\n\nPavel's question about the IMO, it is not itself an IMO question and need not reach that level of clarity. Certainly it is possible to discuss the \"average American\" with his 2.2 children even though no particular American has that number of kids, or whether this year's World Cup 2nd place team consisted of \"typical Frenchmen\" or whether there are any \"real Frenchmen\" on the team. What the question means in all its subtleties, ramifications, vagaries (or even its offensiveness) are interesting subjects but they are not grounds to simply deconstruct it as meaningless. \n\n[quote] you took my remark to imply that Pavel's silence supports my view, immediately after making an identical implication (that Pavel's silence would support your view, by not correcting it) [/quote]\r\n\r\nThat was not my implication. My implication was that he could speak for himself without your interposition, not that one view or another was preferred.", "Solution_27": "This thread has certainly gone off topic and has spun into a debate that currently offers no closure for the original poster..\r\n\r\n[quote]this will be my final post on this topic. [/quote]\r\n\r\nWhy didn't you make it [i]the[/i] final post? :lol:", "Solution_28": "There is no implicit expectation that all threads should conclude with a result satisfactory to the original poster (in particular, we don't know exactly what would constitute a satisfactory response, see the thread), nor should there be any problem with threads that wander off of their original topic. The discussion is relevant to the question; that it fails to actually answer the question is mostly immaterial.\r\n\r\nWhat is annoying is the useless and mostly contentless posts along the line of 'stop arguing', 'this thread should be locked', 'this thread is off topic'. PM a moderator (lol), or better yet say/do nothing about it.", "Solution_29": "I refuse to stop posting reviews of the entire discussion that may or may not contain responses to other people's written statements.\r\n\r\nWhat is annoying is people who reply to posts which are \"useless and mostly contentless\" with a request to employ the PM system (which they may or may not fail to use themselves) and recommend a course of inaction for any further doubts..", "Solution_30": "You're probably the kind of guy who thinks that backseat moderating and being passive aggressive is cool, but really, it isn't. Go away.", "Solution_31": "I'm (still) American!\r\n\r\n\r\n[i]Moderator's note: What passed for discussion here has moved to PM. This thread is locked.[/i]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Funny Strings\r\n\r\nLet's consider a string of non-negative integers, containing N elements. Suppose these elements are $S_1, S_2, .. ,S_N$, in the order in which they are placed inside the string. Such a string is called 'funny' if the string $S_1+1, S_2, S_3, .., S_{N-1}, S_N -1$ can be obtained by rotating the first string (to the left or to the right) several times. For instance, the strings 2 2 2 3 and 1 2 1 2 2 are funny, but the string 1 2 1 2 is not. Your task is to find a funny string having N elements, for which the sum of its elements $S_1+S_2+..+S_N$ is equal to K.\r\n\r\nProve: GCD(N,K)=1 is a necessary condition for a string to be funny.", "Solution_1": "Is there anyone who has some ideas? I doubt its truth and is there anyone who can give a counterexample?" } { "Tag": [ "logarithms", "inequalities" ], "Problem": "Given that the equation $\\lg (ax) \\lg(ax^2)=4$ has real solution in x, and all\r\nthe solutions are greater than 1. Find the range of a.\r\n\r\nps: $\\lg x=\\log_{10} x$", "Solution_1": "Is the level of the problem is so low that no one wants to post the solution?", "Solution_2": "It probably has more to do with the time at which you posted it.\r\n$log(ax)+log(ax^2)=4$\r\n$log(ax+ax^2)=4$\r\n$ax+ax^2=10^4$\r\n[if a\u22600]\r\n$x^2+x-\\frac{10^4}{a}=0$\r\n$x=\\frac{-1\\pm\\sqrt{1+\\frac{10^4}{a}}}{2}$\r\nTwo requirements must be met: $ax>0$ and ${1+\\frac{10^4}{a}}\\geq0$\r\nIf a>0 then the second condition is met. Then needed is\r\n$\\frac{-1+\\sqrt{1+\\frac{10^4}{a}}}{2}>0$\r\n$\\sqrt{1+\\frac{10^4}{a}}>1$\r\n${1+\\frac{10^4}{a}}>1$\r\n${\\frac{10^4}{a}}>0$ which is true.\r\n\r\nIf a<0 then ax^2 will be negative and log(ax^2) will not exist.\r\nSo the answer is:\r\n(0,infinity)", "Solution_3": "[quote=\"Wumbate\"]It probably has more to do with the time at which you posted it.\n$log(ax)+log(ax^2)=4$\n$log(ax+ax^2)=4$\n[/quote]\r\n\r\nI don't know how you got that. I think there is no such rule like that in logarithms. I believe you have mistaken it with $\\lg a+\\lg b=\\lg ab\\neq\\lg (a+b)$. Btw, the problem is $\\lg ax\\lg ax^2$. Please tell me if I'm mistaking.\r\n\r\nMasoud Zargar", "Solution_4": "I think 'Wumbate' made some mistakes,its supposed to be product of logs,not [b]sum[/b] of logs...anyway...\r\n\r\n[hide]$(\\lg a + \\lg x)(\\lg a + 2.\\lg x)=4$ for:$a,x>0$ and doing $\\lg a=s$ and $\\lg x=l$ we get:\n\n$2l^2+3.ls+s^2-4=0$ solving for the positive real numbers:\n\n$l=\\frac{\\sqrt{s^2+32}-3s}{4}$,its was given that $l>1$ consequently:$\\frac{\\sqrt{s^2+32}-3s}{4}>1 \\implies 8s^2+24s-16<0$ solving again the inequation for the positive real numbers:\n\n$11$ [/quote]\r\n$x>1$ is NOT the same as $l=\\lg x >1$.\r\nMoreover, $s=\\lg a$ may be less than or equal to 0.", "Solution_6": "[quote=\"ifai\"]Given that the equation $\\lg (ax) \\lg(ax^2)=4$ has real solution in x, and all\nthe solutions are greater than 1. Find the range of a.\n\nps: $\\lg x=\\log_{10} x$[/quote]\r\n\r\nThe right solution\r\n[hide]$(\\lg a + \\lg x)(\\lg a + 2.\\lg x)-4=0$ for:$a>0, x>1$ \n\nwe need to have both roots positive for this equation.\n\nso, the sum and product of the roots must be positive(discriminant is always positive for this equation)\n\nhence we should have\n$- 3\\lg a > 0 \\Rightarrow \\lg a < 0$\n\nand\n\n$(\\lg a)^2 - 4 > 0 \\Rightarrow \\log a \\in ( - \\infty ,\\, - 2) \\cup (2,\\,\\infty )$\n\nimplying $\\lg a < \\, - 2 \\Rightarrow 0 < a < \\frac{1}{{100}}$[/hide]", "Solution_7": "Correcting Gabriel's solution:\r\n$l = \\frac{\\pm\\sqrt{s^2+32}-3s}{4} > 0$\r\n$\\pm\\sqrt{s^2 + 32} > 3s$.\r\n\r\nCase 1: $\\sqrt{s^2 + 32} > 3s$\r\nIf $s \\le 0$ this is obviously true. Otherwise we square both sides and get that $0 < s < 2$. Thus, $s < 2$ which means that $0 < a < 100$.\r\n\r\nCase 2: $-\\sqrt{s^2 + 32} > 3s$\r\nIf $s \\ge 0$ this has no solutions. Otherwise we must have that $s^2 + 32 < 9s^2$. This is equivalent to $s < -2$, but this doesn't really add any solutions that weren't in case 1.\r\n\r\nAnswer: $0 < a < 100$.\r\n\r\nIt can be seen that the solution above mine is incorrect. For example, there are solutions for a = 1.", "Solution_8": "[quote=\"Kalle\"]Correcting Gabriel's solution:\r\n$l = \\frac{\\pm\\sqrt{s^2+32}-3s}{4} > 0$\r\n$\\pm\\sqrt{s^2 + 32} > 3s$.\r\n\r\nCase 1: $\\sqrt{s^2 + 32} > 3s$\r\nIf $s \\le 0$ this is obviously true. Otherwise we square both sides and get that $0 < s < 2$. Thus, $s < 2$ which means that $0 < a < 100$.\r\n\r\nCase 2: $-\\sqrt{s^2 + 32} > 3s$\r\nIf $s \\ge 0$ this has no solutions. Otherwise we must have that $s^2 + 32 < 9s^2$. \r\n\r\neverything is right till here.\r\nThis is equivalent to $s<-2$, \r\nAnswer: $0 < a < 1/100$.\r\n\r\nThink whether case1 is necessary.", "Solution_9": "Yeah, I turned the inequality the wrong way around in Case 2, but the solution is still right. You don't need two real solutions, only one, so it is Case 1 which decides the answer. As I said, there is a solution for $x > 1$ for the equation if $a = 1$, so your answer is wrong.\r\n\r\nedit: oh, \"all the solutions are greater than 1\". Yeah, ok. I didn't read the question properly.", "Solution_10": "Kalle your answer is incorrect, and when a=1 the solutions are $x_1=10^{\\sqrt 2}>1$, but $x_2=10^{-\\sqrt 2}<1$.", "Solution_11": "Yeah, as I said, I thought the question only required one root to be greater than 1." } { "Tag": [ "calculus", "integration", "inequalities", "calculus computations" ], "Problem": "Prove $ \\int_{1 \\minus{} t}^{1 \\plus{} t} x^{\\frac {3}{x}}\\ dx \\geq 2t \\ (0\\leq t < 1)$.", "Solution_1": "[hide=\"My solution\"]Consider $ F(t) \\equal{} \\int_{1 \\minus{} t}^{1 \\plus{} t}x^{\\frac {3}{x}}\\ dx \\minus{} 2t$. We know that $ F(t) \\equal{} 0$ at $ t \\equal{} 0$. Thus, we need only to show that $ F(t)$ is increasing on $ [0,1)$. Differentiating we get $ F'(t) \\equal{} (1 \\plus{} t)^{\\frac {3}{1 \\plus{} t}} \\plus{} (1 \\minus{} t)^{\\frac {3}{1 \\minus{} t}} \\minus{} 2.$\nHowever, by [url=http://www.mathlinks.ro/viewtopic.php?t=302774]an inequality of Vasc[/url], $ F'(t) \\ge 0$ on $ [0,1)$. Hence $ F(t)$ is increasing, and the inequality has been proven.\n\nNote: The inequality states that for positive reals $ a,b$ such that $ a \\plus{} b \\equal{} 2$, $ a^{\\frac {3}{a}} \\plus{} b^{\\frac {3}{b}}\\ge 2$. The proof of that inequality is hideous, though, and I'm not sure you had it in mind.[/hide]", "Solution_2": "Anyone else have a solution to this?", "Solution_3": "For $ 0\\leq t\\leq 1$\r\n\t\r\n$ \\int_{1\\minus{}t}^{1\\plus{}t}x^{3/x}dx\\geq \\int_{1\\minus{}t}^{1\\plus{}t}x^{3/2}dx\\equal{}\\frac{2}{5}\\left( \\left( 1\\plus{}t\\right) ^{5/2}\\minus{}\\left( 1\\minus{}t\\right) ^{5/2}\\right) \\equal{}\\frac{2}{5}\\left( \\frac{\\allowbreak 10t\\plus{}20t^{3}\\plus{}2t^{5}}{\\left( 1\\plus{}t\\right) ^{5/2}\\plus{}\\left( 1\\minus{}t\\right) ^{5/2}}\\right) \\allowbreak \\geq \\frac{2}{5}\\left( \\frac{\\allowbreak 10t}{2}\\right) \\equal{}\\allowbreak 2t$" } { "Tag": [ "group theory", "superior algebra", "superior algebra solved" ], "Problem": "I wonder if anyone in this forum knows/ has come across the following. It may be trivial.\r\n\r\nLet $G$ be a finite group, and let $k$ be the order of an arbitrary element $g \\in G$. Is it necessarily true that $G$ has at least $k$ equivalence classes? (in other words, does it follow that the number of irreducible complex characters of $G$ is at least $k$?)", "Solution_1": "i think this is trivial because you may send k to any k'th root of unity, so you get k different characters...i think one may even prove that the number of complex characters is divisible by k...but i may have misunderstood something.", "Solution_2": ":D :D :D \r\nTrivial, indeed --- there are always at least $k$ explicitly given linear characters, and these are irreducible. And what an incredible proof had I found to this triviality!! :cool: It's always a shame when you've thought you've developed an ingenious proof to something as simple as $1+1=2$. :lol: :lol: :lol: Cool-cool. :cool: \r\n\r\nBy the way, no Peter, $k$ does not necessarily divide the number of equivalence classes. Take, for example, the dihedral group of order $2p$, where $p$ is an odd prime (for the example, only $S_3$ would suffice, but anyway :) ). This group has both elements of order $2$ and $p$, but the number of the equivalence classes is less than $2p$, and thus not divisible by both $2$ and $p$.\r\n\r\nAlso note that, if $G$ is nilpotent, then $k$ may be replaced with the exponent of the group." } { "Tag": [ "geometry", "3D geometry", "sphere", "induction", "combinatorics solved", "combinatorics" ], "Problem": "(1) What is the maximum number of regions that n plane divide the 3-dim. space into ?\r\n\r\n(2) What is the maximum number of regions that n spheres divide the 3-dim. space into ?", "Solution_1": "for [1] I got to this answer 1/6(n^3+5n+6)! \r\n\r\ncheers!!", "Solution_2": "and for the 2nd one I got : n/3(n^2-3n+8)\r\n\r\ncheers! :D :D", "Solution_3": "The problems are similar to the ones posted before with lines and circles in the plane. I didn't check them out, but I assume the idea is the same? (induction; we see how many new regions are formed when we add a new sphere/plane)", "Solution_4": "Why don't you provide some arguments ? It would quite nice to get a more complete solution. But I guess it's too time-consuming...", "Solution_5": "Full solution for number one: :D\r\n\r\n[1]First we have to compute the maximum dn of regions that a line is divided by n points.\r\n\r\nWe have that dn=n+1. Indeed, because a point divides the line d1 in 2 regions. After k points have divided the line in dk regions, then the \"k+1\"th is also in region which divides it in to 2 so d_(k+1)=dk+1 and induction.\r\n\r\n[2]Now we must compute the maximum pn of regions which the plane is divided by n lines!\r\n\r\nWe have first the formula: p_(n+1)=pn+dn. Indeed, because if n lines have divided the plane pn in regions Ri (1<=i<=pn) then we consider the a_(n+1) line d, which intersects all the other points in n distinct points. From [1] we get that the n points divide d in a number of regions rj which can be equal to dn: (1<=j<=n). Each of these regions rj separetes an \"old\" region Ri in two : Ri=Ri'+Ri\". So the number pn of the old regions Ri adds to the number dn to obtain p_(n+1). WE have the following equalities p_(i+1)-pi=di, and we add all these relations from 2 to n-1 and get that pn=(n^2+n+2)/2.\r\n\r\n\r\nNow we get to our problem :D :D and we see that the number of regions that can be added by the (n+1) plane, this means that s_(n+1)-sn equals the number dn of the regions in which the n+1 plane is divided by the intersection lines of the \"old\" planes! So we have that s_(n+1)-si=di. we sum all these form i=2 to n-1 and we get that sn=1/6(n^3+5n+6). This maximum is obtained if there are no paralel planes or there are 3 secantes! \r\nhope I made my self clear! :)\r\ncheers! :D :D :D", "Solution_6": "for the 2nd go and check:\r\n\r\nhttp://www.mathlinks.ro/phpBB/viewtopic.php?t=1205\r\n\r\ncheers! :D :D", "Solution_7": "Could someone elaborate lagrangia's solution? .. \nthank you." } { "Tag": [ "inequalities", "linear algebra", "matrix", "blogs" ], "Problem": "Let $ A_1,A_2,...,A_n \\in H^{ \\plus{} \\plus{} }_n(\\mathbb{C})$ Such that $ \\sum_{k \\equal{} 1}^{n}{A_k} \\equal{} nI_n$\r\n1)Prove that : $ \\det(\\sum_{k \\equal{} 1}^{n}{(A_k \\plus{} A^{ \\minus{} 1}_k)^2}) \\geq \\frac {(n^2 \\plus{} 1)^{2n}}{n^n}$\r\n2) Prove that : $ \\det(A_1^k \\plus{} A_2^k \\plus{} ... \\plus{} A_n^k) \\geq \\det(A_1^{k \\minus{} 1} \\plus{} A_2^{k \\minus{} 1} \\plus{} ... \\plus{} A_n^{k \\minus{} 1})$ , $ \\forall k \\in \\mathbb{N^{*}}$\r\n $ I_n$ is identity matrix.\r\n $ H^{ \\plus{} \\plus{} }_n(\\mathbb{C})$ are set $ n$x$ n$ Hermitian matrices and positive definite.\r\n[hide=\"Answer of me \"]Very interesting ! :P [/hide]", "Solution_1": "Here I prove for (1) , for (2) don't dificult \r\nLemma1) Let $ A_1^{ \\minus{} 1} \\plus{} A_2^{ \\minus{} 1} \\plus{} ... \\plus{} A_m^{ \\minus{} 1} \\geq m^2(A_1 \\plus{} A_2 \\plus{} ... \\plus{} A_m)^{ \\minus{} 1}$ for all $ A_i ,\\forall i \\in \\{1,2,..,m\\}$ are Hermitian matrices and positive define. you can read proof in http://www.mathlinks.ro/viewtopic.php?t=275033\r\nLemma2) $ m^2(A_1^2 \\plus{} A_2^2 \\plus{} ... \\plus{} A_m^2) \\geq (A_1 \\plus{} A_2 \\plus{} .. \\plus{} A_m)^2$ , for all $ A_i ,\\forall i \\in \\{1,2,..,m\\}$ are Hermitian matrices . You can read small proof in blog of me .\r\nand a some lemma here .....\r\n Use lemma1) we have :\r\n $ n^2\\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})^{2}}\\geq (\\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})})^2$\r\n by $ \\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})} \\equal{} nI_n \\plus{} \\sum_{k \\equal{} 1}^{n}{A^{ \\minus{} 1}_{k}} \\geq nI_n \\plus{} n^2(A_1 \\plus{} A_2 \\plus{} .. \\plus{} A_n)^{ \\minus{} 1} \\equal{} 2nI_n$\r\n then $ n^2\\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})^{2}}\\geq 4n^2I_n$\r\n from here $ \\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})^{2}}\\geq 4I_n$ or $ \\det(\\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})^{2}})\\geq 4^n$\r\n Sorry :blush: for 1) is true if $ A_1 \\plus{} A_2 \\plus{} ... \\plus{} A_n \\equal{} I_n$ then\r\n$ \\det(\\sum_{k \\equal{} 1}^{n}{(A_{k} \\plus{} A^{ \\minus{} 1}_{k})^{2}})\\geq\\frac {(n^{2} \\plus{} 1)^{2n}}{n^{2n}}$" } { "Tag": [ "trigonometry" ], "Problem": "Solve for $x \\in R$:\r\n\\[ \\sin^3{x}(1+\\cot{x})+\\cos^3{x}(1+\\tan{x})=\\cos{2x} \\]", "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52500]the solution[/url].", "Solution_2": "first of all $x\\neq \\frac{k\\pi}{2},k\\in Z$ ..... [b](*)[/b]\n\nequation is equivalent with:\n\n$\\sin x+\\cos x=\\cos 2x$ \n\n$\\sin x+\\cos x=-(\\sin x+\\cos x)(\\sin x-\\cos x)$ \n\n$(\\sin x+\\cos x)(\\sin x-\\cos x+1)=0$\n\n$(\\sin x+\\cos x)\\left(\\cos\\left(x+\\frac{\\pi}{4}\\right)-\\frac{\\sqrt{2}}{2}\\right)=0$\n\nsecond factor of LHS is $\\neq 0$ (condition [b](*)[/b]), so solution is \n\n$\\boxed{x=m\\pi-\\frac{\\pi}{4}\\ ,\\ m\\in Z}$", "Solution_3": "[hide=good one]mark $\\sin{x}=a$ and $\\cos{x}=b$\\\\\nthe equation is equivalent to $a^3\\left(1+\\frac{b}{a}\\right)+b^3\\left(1+\\frac{a}{b}\\right)=b^2-a^2$\\\\\n we consider two cases $a+b=0$ and $a+b\\neq 0$\\\\\nfor $a+b \\neq 0$ we get:\\\\\n$a^2+b^2=a-b \\implies \\sin2x=0$\\\\\nclearly this don't works\\\\\nso we have $a+b=0$\\\\\nwhich gives $\\tan{x}=-1$\\\\\nnow this gives $x=\\varphi\\pi-\\frac{\\pi}{4} \\qquad \\forall \\qquad \\varphi \\in \\mathbb{Z}$[/hide]", "Solution_4": "[hide]$\\sin x+\\cos x=\\cos 2x\\Rightarrow {{(\\sin x+\\cos x)}^{2}}={{\\cos }^{2}}2x\\Leftrightarrow 1+\\sin 2x=1-{{\\sin }^{2}}2x$thus $\\sin 2x(1+\\sin 2x)=0$[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$a_1,...,a_n$ are non-negative reals. Let $x_i=\\frac{a_i+a_{i+1}}{2}$ and $y_i=\\frac{a_i+a_{i+1}+a_{i+2}}{3}$. Prove the following inequality:\r\n\r\n\\[\\sum _{i=1}^n x_i ^{x_{i+1}} \\geq \\sum _{i=1}^n y_i ^{y_{i+1}}\\]\r\n\r\nwhere $a_{n+1}=a_1$, $a_{n+2}=a_2$\r\n\r\nP.S. It may be easy though I'm not sure about my solution so I'm posting.", "Solution_1": "[quote=\"Megus\"]\n\nwhere $a_{n+1}=a_1$, $a_{n+2}=a_2$\n\n[/quote]\r\n\r\nDoesn't that make all the a's the same?", "Solution_2": "No, I just put it to reassure that indices are taken modulo n which should be obvious in such inequalities\r\n\r\nSomebody, any solution?" } { "Tag": [ "geometry", "MATHCOUNTS", "AMC", "AIME", "ARML", "USA(J)MO", "USAMO", "\\/closed" ], "Problem": "The Art of Problem Solving [b][i]Introduction to Geometry[/i][/b] textbook is now available.\r\n\r\n[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9]Click here for excerpts and to order.[/url]\r\n\r\n[b][i]Introduction to Geometry[/i][/b] is a complete introductory geometry text, covering all topics of a standard geometry curriculum, plus most of the problem solving tactics required for success in programs such as MATHCOUNTS and the AMC. The textbook is 496 pages, and the solutions manual is 192 pages. Full solutions to all problems are in either the solutions manual or the text itself.", "Solution_1": "Wow! it looks awesome, I can't wait to get it!!! :jump: :jump:", "Solution_2": "I guess it's too late to buy now :( :( \r\n\r\nMy states are on March 11...", "Solution_3": "[quote=\"robinhe\"]I guess it's too late to buy now :( :( \n\nMy states are on March 11...[/quote]\r\n\r\nIt's never too late to buy! Even if you have hardly any time left for Mathcounts, there is still AMC, AIME, ARML, USAMO and more to prepare for :)", "Solution_4": "It's certainly not too late for national MATHCOUNTS.", "Solution_5": "The books are an extention on the AoPS book 1 material, so as long as you know that, you should be fine for states :D .", "Solution_6": "[quote=\"joml88\"][quote=\"robinhe\"]I guess it's too late to buy now :( :( \n\nMy states are on March 11...[/quote]\n\nIt's never too late to buy! Even if you have hardly any time left for Mathcounts, there is still AMC, AIME, ARML, USAMO and more to prepare for :)[/quote]\r\n\r\nYeah, I was planning to buy it after states... :D :D", "Solution_7": "My states are on March 17-18 so it might be too late to buy.", "Solution_8": "Im buying it tonight!\r\n(would I be the first order) :?:", "Solution_9": "Nope, I win :winner_first::winner_first:!", "Solution_10": "*steals kyyuanmathcount's book* I WIN!", "Solution_11": "Swipes it back :ninja: .\r\n\r\nThese books are also great review, not only for MATHCOUNTS, but I feel they can help AIME's on some problems.", "Solution_12": "Yeah, like the book is just going ot randomly appear on your front doorstep tonight so you can study for like 5 minutes before going to bed...\r\n\r\nI will buy it sometime in my lifetime (most likely in a few weeks) :D", "Solution_13": "Or you could make it your lifelong goal to someday own it. \r\n\r\nThis book is going to be awesome, like the counting one.", "Solution_14": "[quote=\"biffanddoc\"]Im buying it tonight!\n(would I be the first order) :?:[/quote]\r\n\r\nNot the first order, but close! First order was the same as last time - AAST.", "Solution_15": "Don't they own like 600 copies (or some other tremendous number) of your originals??", "Solution_16": "[quote=\"bubala\"]Don't they own like 600 copies (or some other tremendous number) of your originals??[/quote]\r\n\r\nWe're big fans of AAST. :)", "Solution_17": "who's AAST??", "Solution_18": "I believe this is their website:\r\n[url=http://www.bergen.org/AAST/]www.bergen.org/AAST/[/url]I googled it and just didnt think it was american association for surgery of trauma.\r\nAll hail aops for the geometry book! :omighty:", "Solution_19": "Just bought it ", "Solution_20": "When will the Intro to Geometry course next be offered?", "Solution_21": "Anyone has red the book? What you think about it? It can help also a future physic's student? What existensial questions! :P", "Solution_22": "I placed the order 20 min after seeing the announcement, and paid extra for expedited shipping. Hope it arrives tomorrow :D", "Solution_23": "who can buy it and give it to me?? :lol: :lol:", "Solution_24": "[quote=\"G-UNIT\"]When will the Intro to Geometry course next be offered?[/quote]\r\n\r\nWe have not yet set the summer schedule, but we will probably not offer that class before the Fall at earliest.", "Solution_25": "[quote=\"easyas3.14159...\"]I believe this is their website:\n[url=http://www.bergen.org/AAST/]www.bergen.org/AAST/[/url]I googled it and just didnt think it was american association for surgery of trauma.\nAll hail aops for the geometry book! :omighty:[/quote][b] :roll: :roll: [/b]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $ p $ be a prime number , $ x\\in Z $ . {$ a_n $} : \r\n $ a_n = 1 +\\frac{x}{n+1} + \\frac{x^2}{2n+1} + ... + \\frac{x^p}{pn+1} $ \r\n Prove that : $ a_n $ doesn't equal to 0 for all $ n=1,2,..,p-1 $", "Solution_1": "Suppose the contrary.\r\nLet $P=(n+1)(2n+1)...(pn+1)$. We see that $p\\mid P$, more exactly: one of the numbers $kn+1$ is divisible by $p$ and all other numbers are not. Consider $a_nP$. It is a sum of $p+1$ summands and we know that $p$ of them are divisible by $p$, so $p\\mid \\frac{P}{kn+1}x^k$. It follows that $p\\mid x$. It is clear $x<0$, so $x\\leq p-1$. But then \\[0=a_n \\leq 1+\\frac{|x|}{n+1}+\\frac{|x|^2}{2n+1}+...+\\frac{|x|^{p-1}}{n(p-1)+1}-\\frac{|x|^p}{pn+1}=S.\\] We have $\\frac{|x^s|}{ns+1}<\\frac{|x|^{s+1}}{n(s+1)+1}$ for $|x|\\geq p$, therefore $0$\r\n$\\nabla z(60,100)=\\left<-1.2, -4\\right>$\r\n\r\nb) should be easy too, but I am drawing a complete blank. I thinks it's something related to the $tan(\\theta)$ and the $\\vec{r}'$", "Solution_1": "The slope in the direction of the most rapid increase is the largest possible directional derivative.\r\n\r\nThat is, $\\max_{||u||=1}D_uz=\\max_{||u||=1}\\nabla z\\cdot u=||\\nabla z||.$\r\n\r\nIn this particular instance, $||\\nabla z||=\\sqrt{(-1.2)^2+(-4)^2}=\\frac{2\\sqrt{109}}5\\approx 4.176.$\r\n\r\nThat's the slope. The relationship of a slope to an elevation angle is that $m=\\tan \\theta$ or $\\theta=\\arctan m\\approx \\arctan 4.176\\approx 76.5^{\\circ}.$\r\n\r\nVery steep hill - probably best not to go there without ropes.", "Solution_2": "lol,\r\n\r\nI was right about the $tan(\\theta)$ part...heh." } { "Tag": [ "Euler", "function", "LaTeX", "number theory", "number theory proposed" ], "Problem": "What are some questions I could use to test fundamentals in topics of Euler (0) - function, formulae for d(n) and funky symol (n) (means sum of divisors of n), multiplicative arithmetic functions, and the mobius inversion formula?", "Solution_1": "That's the Euler $\\phi$ function. To write that in LaTeX, try (dollar) \\phi(n) (dollar). The sum-of-divisors function is $\\sigma(n)$ (that's (dollar) \\sigma(n) (dollar) - the thing you called \"funky symbol\" is sigma. Some other useful multiplicative functions include $\\tau(n)$, the number of divisors (that's (dollar) \\tau(n) (dollar)), $\\delta(n)$, which is 1 at 1 and zero everywere else, there's 1 as a function, and there's the M\u00f6bius function $\\mu(n)$ (\\mu(n)), which is zero if $n$ is divisible by any square and $(-1)^r$ if $n$ has $r$ prime factors taken one at a time.\r\n\r\nI'm looking at a copy of Rosen's [i]Elementary Number Theory[/i]; the multiplicative functions are in chapter 7. He's got lots of exercises. Here are a few:\r\n\r\n1. For which positive integers $n$ is $\\phi(n)$ a power of 2?\r\n\r\n2. Show that the equation $\\sigma(n)=k$ has at most a finite number of solutions when $k$ is a positive integer.\r\n\r\n3. For how many consecutive integers can the M\u00f6bius function $\\mu(n)$ take the value $0$?", "Solution_2": "Thank you for questions. I think I'll look up the book too if it has alot of questions for number theory.", "Solution_3": "Is this the book you have?" } { "Tag": [ "inequalities", "function", "logarithms", "inequalities unsolved" ], "Problem": "Prove the inequality\r\n\\[ \\sqrt {a^{1 \\minus{} a}b^{1 \\minus{} b}c^{1 \\minus{} c}} \\le \\frac {1}{3}\r\n\\]\r\nholds for all positive real numbers $ a$, $ b$ and $ c$ with $ a \\plus{} b \\plus{} c \\equal{} 1$.", "Solution_1": "[quote=\"April\"]Prove the inequality\n\\[ \\sqrt {a^{1 \\minus{} a}b^{1 \\minus{} b}c^{1 \\minus{} c}} \\le \\frac {1}{3}\n\\]\nholds for all positive real numbers $ a$, $ b$ and $ c$ with $ a \\plus{} b \\plus{} c \\equal{} 1$.[/quote]\r\nIt is equivalent : $ (1\\minus{}a)ln(a)\\plus{}(1\\minus{}b)ln(b)\\plus{}(1\\minus{}c)ln(c) \\le \\minus{}2ln(3)$\r\n$ f(x) \\equal{} (1 \\minus{} x)ln(x)$ - concave function for $ x > 0$ and by Jensen it is true", "Solution_2": "[hide=\"Alternative Solution\"]\n\nWe have $ a^a b^b > \\equal{} a^b b^a$ for all positive real numbers a and b (trivial),\nBy multiplying this inequality for (a,b) (b,c) (c,a), $ a^{1 \\minus{} a} b^{1 \\minus{} b} c^{1 \\minus{} c} \\equal{} a^{b \\plus{} c} b^{c \\plus{} a} c^{a \\plus{} b} \\leq a^{2a} b^{2b} c^{2c}$\nMultiplying $ a^{2\\minus{}2a} b^{2\\minus{}2b} c^{2\\minus{}2c}$ on both sides yields\n$ (a^{1 \\minus{} a} b^{1 \\minus{} b} c^{1 \\minus{} c} )^3 \\leq a^2 b^2 c^2 \\leq \\frac {1}{64}$ (AM-GM)\nTaking the sixth roots of both side yields the desired result.\n\n[/hide]", "Solution_3": "By using the weighted AM-GM-inequality we get \\[ a^\\frac{1\\minus{}a}2b^\\frac{1\\minus{}b}2c^\\frac{1\\minus{}c}2 \\leq \\frac{(1\\minus{}a)a\\plus{}(1\\minus{}b)b\\plus{}(1\\minus{}c)c}2\\equal{}\\frac{1\\minus{}(a^2\\plus{}b^2\\plus{}c^2)}2\\]\r\nApplying Cauchy-Schwartz-inequality yields $ 1\\equal{}(a\\plus{}b\\plus{}c)^2 \\leq 3(a^2\\plus{}b^2\\plus{}c^2)$\r\nSo we get \\[ \\sqrt {a^{1 \\minus{} a}b^{1 \\minus{} b}c^{1 \\minus{} c}} \\leq \\frac{1\\minus{}\\frac1{3}}2\\equal{}\\frac1{3}\\]", "Solution_4": "[quote=\"April\"]Prove the inequality\n\\[ \\sqrt {a^{1 \\minus{} a}b^{1 \\minus{} b}c^{1 \\minus{} c}} \\le \\frac {1}{3}\n\\]\nholds for all positive real numbers $ a$, $ b$ and $ c$ with $ a \\plus{} b \\plus{} c \\equal{} 1$.[/quote]\r\nIt's equivalent to:\r\n$ \\sum_{cyc} (1\\minus{}x) \\cdot \\ln(x) \\le \\ln \\left ( \\frac{1}{9} \\right )$\r\n$ \\ln(x)$ is concave so applying Weighted Jensen gives us:\r\n$ \\frac{\\sum_{cyc} (1\\minus{}x) \\cdot \\ln(x)}{\\sum_{cyc} (1\\minus{}x)} \\le \\ln \\left ( \\frac{\\sum_{cyc} x \\cdot (1\\minus{}x)}{\\sum_{cyc} (1\\minus{}x)} \\right )$ $ \\iff$\r\n$ \\sum_{cyc} (1\\minus{}x) \\cdot \\ln(x) \\le 2 \\cdot \\ln \\left ( \\frac{1 \\minus{} \\sum_{cyc} x^2}{2} \\right ) \\le 2 \\cdot \\ln \\left ( \\frac{1 \\minus{} \\frac{1}{3}}{2} \\right ) \\equal{} \\ln \\left ( \\frac{1}{9} \\right )$. QED", "Solution_5": "[quote=\"qwerty414\"]\n$ a^2 b^2 c^2 \\leq \\frac {1}{64}$ (AM-GM)\n[/quote]\r\nI think you have a mistake.\r\nIt will be modified $ a^2b^2c^2 \\le \\frac {1}{729}$, I think.\r\n\r\nSorry about my bad english.", "Solution_6": "$ a\\log bc \\plus{} b\\log ca \\plus{} c\\log ab\\leq \\log (ab \\plus{} bc \\plus{} ca)\\ \\because a \\plus{} b \\plus{} c \\equal{} 1$\r\n\r\n$ \\leq \\log \\frac {(a \\plus{} b \\plus{} c)^2}{3} \\equal{} \\log \\frac {1}{9}$\r\n\r\n$ \\Longleftrightarrow (b \\plus{} c)\\log a \\plus{} (c \\plus{} a)\\log b \\plus{} (a \\plus{} b)\\log c\\leq \\log \\frac {1}{9}$\r\n\r\n$ \\Longleftrightarrow \\log a^{b \\plus{} c}\\log b^{c \\plus{} a}\\log c^{a \\plus{} b}\\leq \\log \\frac {1}{9}$\r\n\r\n$ \\Longleftrightarrow a^{b \\plus{} c}b^{c \\plus{} a}c^{a \\plus{} b}\\leq \\frac {1}{9}$\r\n\r\n$ \\Longleftrightarrow \\sqrt {a^{b \\plus{} c}b^{c \\plus{} a}c^{a \\plus{} b}}\\leq \\frac {1}{3}$ Q.E.D.", "Solution_7": "Use the weighted version of AM-GM." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "The points on a circle are colored in three different colors.\r\n\r\nProve that there exist infinitely many isosceles triangles with vertices on the circle and of the same color.", "Solution_1": "See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5434[/url]" } { "Tag": [ "function", "Functional Analysis", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $ X \\equal{} C([0,1])$ using the sup norm. Prove that the closure of every non-empty open subset of $ X$ is not compact.\r\n\r\nI've been fighting with similar type theorems and problems since almost all references to Arzela-Ascoli assume that you are quite good with functional analysis already.\r\n\r\nWhat I've picked up is that (since the set is closed) we either need to show not bounded or not equicontinuous by Arzela-Ascoli. My guess would be the latter.\r\n\r\nNon-empty open: If I interpret this correctly I think it means if $ A\\subset X$ is nonempty and open then for any $ f\\in A$, there exists $ \\varepsilon > 0$ such that whenever $ \\sup_x |f \\minus{} g| < \\varepsilon$ we have $ g\\in A$.\r\n\r\nOh darn. As I was typing this, it clarified where my misunderstanding was. I'll still post this, but I'll let other people try to answer it. (At the very least it will provide a good reference if someone is searching for this topic).", "Solution_1": "$ C[0,1]$ with the sup-norm is an infinite dimensional Banach space, hence the unit ball is not compact. Note that any closed subset of a compact set is also compact to finish the proof.", "Solution_2": "That was far easier than I was thinking. \r\n\r\nI was trying to explicitly construct something along the lines of, if $ f$ is in some open set then there is some $ \\epsilon$ neighborhood of continuous functions. So $ f_n(x) \\equal{} f(x) \\plus{} \\frac{\\epsilon}{2} \\sin(nx)$ (or some variant, the idea that I thought of while typing is to have some oscillating function in the \"wiggle room\" such that no subsequence converges uniformly) are all in that open set. Thus they are all in the closure of that open set, but no subsequence converges uniformly, so the set is not compact.\r\n\r\nDoes this also work (in principle, I'm not certain that particular class of functions works)?" } { "Tag": [], "Problem": "An antique dealer buys a chair for $ \\$20$ and sells it for $ \\$85$. The dealer later buys back the same chair for $ \\$110$ and sells it for $ \\$135$. How many dollars total profit did the dealer make on this chair?", "Solution_1": "First, the dealer gets $ \\minus{}20$ dollars. Then, he gets $ 85$ dollars. Then he gets $ \\minus{}110$ dollars, and then he gets $ 135$. Adding, the dealer made $ \\minus{}20\\plus{}85\\minus{}110\\plus{}135\\equal{} \\boxed{90}$ dollars.", "Solution_2": "The answer is 65-110+135=90." } { "Tag": [ "geometry", "rectangle", "number theory proposed", "number theory" ], "Problem": "Given a rectangle in plane. Prove that for every integer n one can part the origin rectangle into some sub-rectangles such that any line parallel to the side of the original rectangle that does not contain any side of any sub-retangles cuts exactly n sub-rectangles", "Solution_1": "The partition I described [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26382]here[/url], in post #2, should do the trick." } { "Tag": [ "calculus", "articles", "probability", "blogs", "conics", "ellipse", "Support" ], "Problem": "Solafidefarms and I are having a friendly argument, and he invited others to join in. To see the backstory please visit his blog.\r\n\r\nI wrote:\r\n[quote]\nSorry for not taking up your challenge, my ISP has been acting funny the last few days, and learning Perl is too much fun.\n\nI have found one good article on biblical contridictions, it sort of glorifies atheism, sorry for being offended a bit, I was too.\n\n\nContradictions\nThe Bible is riddled with repetitions and contradictions, things that the Bible bangers would be quick to point out in anything that they want to criticize. For instance, Genesis 1 and 2 disagree about the order in which things are created, and how satisfied God is about the results of his labors. The flood story is really two interwoven stories that contradict each other on how many of each kind of animal are to be brought into the Ark--is it one pair each or seven pairs each of the \"clean\" ones? The Gospel of John disagrees with the other three Gospels on the activities of Jesus Christ (how long had he stayed in Jerusalem--a couple of days or a whole year?) and all four Gospels contradict each other on the details of Jesus Christ's last moments and resurrection. The Gospels of Matthew and Luke contradict each other on the genealogy of Jesus Christ' father; though both agree that Joseph was not his real father. Repetitions and contradictions are understandable for a hodgepodge collection of documents, but not for some carefully constructed treatise, reflecting a well-thought-out plan.\n\nOf the various methods I've seen to \"explain\" these:\n1. \"That is to be taken metaphorically\" In other words, what is written is not what is meant. I find this entertaining, especially for those who decide what ISN'T to be taken as other than the absolute WORD OF GOD--which just happens to agree with the particular thing they happen to want...\n\n2. \"There was more there than....\" This is used when one verse says \"there was a\" and another says \"there was b\", so they decide there was \"a\" AND \"b\"--which is said nowhere. This makes them happy, since it doesn't say there WASN'T \"a+b\". But it doesn't say there was \"a+b+litle green martians\". This is often the same crowd that insists theirs is the ONLY possible interpretation (i.e. only \"a\") and the only way. I find it entertaining they they don't mind adding to verses.\n\n3. \"It has to be understood in context\" I find this amusing because it comes from the same crowd that likes to push likewise extracted verses that support their particular view. Often it is just one of the verses in the contradictory set is suppose to be taken as THE TRUTH when if you add more to it it suddenly becomes \"out of context\". How many of you have goten JUST John 3:16 (taken out of all context) thrown up at you?\n\n4. \"there was just a copying/writing error\" This is sometimes called a \"transcription error\", as in where one number was meant and an incorrect one was copied down. Or that what was \"quoted\" wasn't really what was said, but just what the author thought was said when he thought it was said. And that's right--I'm not disagreeing with events, I'm disagreeing with what is WRITTEN. Which is apparently agreed that it is incorrect. This is an amusing misdirection to the problem that the bible itself is wrong.\n\n5. \"That is a miracle\". Naturally. That is why it is stated as fact.\n\n6. \"God works in mysterious ways\" A useful dodge when the speaker doesn't understand the conflict between what the bible SAYS and what they WISH it said.\n\n\n\nThe following is from the KJ version. If you have something different, let me know.\nQuote\n\nGod good to all, or just a few?\nPSA 145:9 The LORD is good to all: and his tender mercies are over all his works.\n\nJER 13:14 And I will dash them one against another, even the fathers and the sons together, saith the LORD: I will not pity, nor spare, nor have mercy, but destroy them.\nWar or Peace?\nEXO 15:3 The LORD is a man of war: the LORD is his name.\n\nROM 15:33 Now the God of peace be with you all. Amen.\nWho is the father of Joseph?\nMAT 1:16 And Jacob begat Joseph the husband of Mary, of whom was born Jesus, who is called Christ.\n\nLUK 3:23 And Jesus himself began to be about thirty years of age, being (as was supposed) the son of Joseph, which was the son of Heli.\nWho was at the Empty Tomb? Is it:\nMAT 28:1 In the end of the sabbath, as it began to dawn toward the first day of the week, came Mary Magdalene and the other Mary to see the sepulchre.\n\nMAR 16:1 And when the sabbath was past, Mary Magdalene, and Mary the mother of James, and Salome, had bought sweet spices, that they might come and anoint him.\n\nJOH 20:1 The first day of the week cometh Mary Magdalene early, when it was yet dark, unto the sepulchre, and seeth the stone taken away from the sepulchre.\nIs Jesus equal to or lesser than?\nJOH 10:30 I and my Father are one.\n\nJOH 14:28 Ye have heard how I said unto you, I go away, and come again unto you. If ye loved me, ye would rejoice, because I said, I go unto the Father: for my Father is greater than I.\nWhich first--beasts or man?\nGEN 1:25 And God made the beast of the earth after his kind, and cattle after their kind, and every thing that creepeth upon the earth after his kind: and God saw that it was good.\nGEN 1:26 And God said, Let us make man in our image, after our likeness: and let them have dominion over the fish of the sea, and over the fowl of the air, and over the cattle, and over all the earth, and over every creeping thing that creepeth upon the earth.\n\nGEN 2:18 And the LORD God said, It is not good that the man should be alone; I will make him an help meet for him.\nGEN 2:19 And out of the ground the LORD God formed every beast of the field, and every fowl of the air; and brought them unto Adam to see what he would call them: and whatsoever Adam called every living creature, that was the name thereof.\nThe number of beasts in the ark\nGEN 7:2 Of every clean beast thou shalt take to thee by sevens, the male and his female: and of beasts that are not clean by two, the male and his female.\n\nGEN 7:8 Of clean beasts, and of beasts that are not clean, and of fowls, and of every thing that creepeth upon the earth, GEN 7:9 There went in two and two unto Noah into the ark, the male and the female, as God had commanded Noah.\nHow many stalls and horsemen?\nKI1 4:26 And Solomon had forty thousand stalls of horses for his chariots, and twelve thousand horsemen.\n\nCH2 9:25 And Solomon had four thousand stalls for horses and chariots, and twelve thousand horsemen; whom he bestowed in the chariot cities, and with the king at Jerusalem.\nIs it folly to be wise or not?\nPRO 4:7 Wisdom is the principal thing; therefore get wisdom: and with all thy getting get understanding.\n\nECC 1:18 For in much wisdom is much grief: and he that increaseth knowledge increaseth sorrow.\n\n1 Cor.1:19: \"For it is written, I will destroy the wisdom of the wise, and wil bring to nothing the understanding of the prudent.\"\nHuman vs. ghostly impregnation\nACT 2:30 Therefore being a prophet, and knowing that God had sworn with an oath to him, that of the fruit of his loins, according to the flesh, he would raise up Christ to sit on his throne;\n\nMAT 1:18 Now the birth of Jesus Christ was on this wise: When as his mother Mary was espoused to Joseph, before they came together, she was found with child of the Holy Ghost.\nThe sins of the father\nISA 14:21 Prepare slaughter for his children for the iniquity of their fathers; that they do not rise, nor possess the land, nor fill the face of the world with cities.\n\nDEU 24:16 The fathers shall not be put to death for the children, neither shall the children be put to death for the fathers: every man shall be put to death for his own sin.\nThe bat is not a bird\nLEV 11:13 And these are they which ye shall have in abomination among the fowls; they shall not be eaten, they are an abomination: the eagle, and the ossifrage, and the ospray,\nLEV 11:14 And the vulture, and the kite after his kind;\nLEV 11:15 Every raven after his kind;\nLEV 11:16 And the owl, and the night hawk, and the cuckow, and the hawk after his kind,\nLEV 11:17 And the little owl, and the cormorant, and the great owl,\nLEV 11:18 And the swan, and the pelican, and the gier eagle,\nLEV 11:19 And the stork, the heron after her kind, and the lapwing, and the bat.\n\nDEU 14:11 Of all clean birds ye shall eat.\nDEU 14:12 But these are they of which ye shall not eat: the eagle, and the ossifrage, and the ospray,\nDEU 14:13 And the glede, and the kite, and the vulture after his kind,\nDEU 14:14 And every raven after his kind,\nDEU 14:15 And the owl, and the night hawk, and the cuckow, and the hawk after his kind,\nDEU 14:16 The little owl, and the great owl, and the swan,\nDEU 14:17 And the pelican, and the gier eagle, and the cormorant,\nDEU 14:18 And the stork, and the heron after her kind, and the lapwing, and the bat.\nRabbits do not chew their cud\nLEV 11:6 And the hare, because he cheweth the cud, but divideth not the hoof; he is unclean unto you.\n\n'Gerah', the term which appears in the MT means (chewed) cud, and also perhaps grain, or berry (also a 20th of a sheckel, but I think that we can agree that that is irrelevant here). It does *not* mean dung, and there is a perfectly adequate Hebrew word for that, which could have been used. Furthermore, the phrase translated 'chew the cud' in the KJV is more exactly 'bring up the cud'. Rabbits do not bring up anything; they let it go all the way through, then eat it again. The description given in Leviticus is inaccurate, and that's that. Rabbits do eat their own dung; they do not bring anything up and chew on it.\nInsects do NOT have four feet\nLEV 11:21 Yet these may ye eat of every flying creeping thing that goeth upon all four, which have legs above their feet, to leap withal upon the earth;\nLEV 11:22 Even these of them ye may eat; the locust after his kind, and the bald locust after his kind, and the beetle after his kind, and the grasshopper after his kind.\nLEV 11:23 But all other flying creeping things, which have four feet, shall be an abomination unto you.\nSnails do not melt\nPSA 58:8 As a snail which melteth, let every one of them pass away: like the untimely birth of a woman, that they may not see the sun.\nFowl from waters or ground?\nGEN 1:20 And God said, Let the waters bring forth abundantly the moving creature that hath life, and fowl that may fly above the earth in the open firmament of heaven.\nGEN 1:21 And God created great whales, and every living creature that moveth, which the waters brought forth abundantly, after their kind, and every winged fowl after his kind: and God saw that it was good.\n\nGEN 2:19 And out of the ground the LORD God formed every beast of the field, and every fowl of the air; and brought them unto Adam to see what he would call them: and whatsoever Adam called every living creature, that was the name thereof.\nOdd genetic engineering\nGEN 30:39 And the flocks conceived before the rods, and brought forth cattle ringstraked, speckled, and spotted.\nThe shape of the earth\nISA 40:22 It is he that sitteth upon the circle of the earth, and the inhabitants thereof are as grasshoppers; that stretcheth out the heavens as a curtain, and spreadeth them out as a tent to dwell in:\n\nMAT 4:8 Again, the devil taketh him up into an exceeding high mountain, and sheweth him all the kingdoms of the world, and the glory of them;\n\nAstromical bodies are spherical, and you cannot see the entire exterior surface from anyplace. The kingdoms of Egypt, China, Greece, Crete, sections of Asia Minor, India, Maya (in Mexico), Carthage (North Africa), Rome (Italy), Korea, and other settlements from these kingdoms of the world were widely distributed.\nSnakes, while built low, do not eat dirt\nGEN 3:14 And the LORD God said unto the serpent, Because thou hast done this, thou art cursed above all cattle, and above every beast of the field; upon thy belly shalt thou go, and dust shalt thou eat all the days of thy life:\nEarth supported?\nJOB 26:7 He stretcheth out the north over the empty place, and hangeth the earth upon nothing.\n\nJOB 38:4 Where wast thou when I laid the foundations of the earth? declare, if thou hast understanding.\n\nHeaven supported too\nJOB 26:11 The pillars of heaven tremble and are astonished at his reproof.\nThe hydrological cycle\nECC 1:7 All the rivers run into the sea; yet the sea is not full; unto the place from whence the rivers come, thither they return again.\n\nJOB 38:22 Hast thou entered into the treasures of the snow? or hast thou seen the treasures of the hail,\n\nStorehouses are not part of the cycle\n\nOrder of creation\nHere is the order in the first (Genesis 1), the Priestly tradition:\n\nDay 1: Sky, Earth, light\nDay 2: Water, both in ocean basins and above the sky(!)\nDay 3: Plants\nDay 4: Sun, Moon, stars (as calendrical and navigational aids)\nDay 5: Sea monsters (whales), fish, birds, land animals, creepy-crawlies (reptiles, insects, etc.)\nDay 6: Humans (apparently both sexes at the same time)\nDay 7: Nothing (the Gods took the first day off anyone ever did)\n\nNote that there are \"days\", \"evenings\", and \"mornings\" before the Sun was created. Here, the Deity is referred to as \"Elohim\", which is a plural, thus the literal translation, \"the Gods\". In this tale, the Gods seem satisfied with what they have done, saying after each step that \"it was good\".\n\nThe second one (Genesis 2), the Yahwist tradition, goes:\n\nEarth and heavens (misty)\nAdam, the first man (on a desolate Earth)\nPlants\nAnimals\nEve, the first woman (from Adam's rib)\nHow orderly were things created?\n#1: Step-by-step. The only discrepancy is that there is no Sun or Moon or stars on the first three \"days\".\n#2: God fixes things up as he goes. The first man is lonely, and is not satisfied with animals. God finally creates a woman for him. (funny thing that an omniscient god would forget things)\nHow satisfied with creation was he?\n#1: God says \"it was good\" after each of his labors, and rests on the seventh day, evidently very satisfied.\n#2: God has to fix up his creation as he goes, and he would certainly not be very satisfied with the disobedience of that primordial couple. (funny thing that an omniscient god would forget things)\nMoses' personality\nNum.12:3: \"Now the man Moses was very meek, above all the men which were upon the fact of the earth.\"\n\nNum.31:14, 17, 18: \"And Moses was wroth...And Moses said unto them, \"Have ye saved all the women alive? ... Now therefore kill every male among the little ones, and kill every woman, ... But all the women children ... keep alive for yourselves.\"\nRighteous live?\nPs.92:12: \"The righteous shall flourish like the palm tree.\"\n\nIsa.57:1: \"The righteous perisheth, and no man layeth it to heart.\"\nActs 1:18: \"Now this man (Judas) purchased a field with the reward of iniquity; and falling headlong, he burst asunder in the midst, and all his bowels gushed out.\"\n\nMatt. 27:5-7: \"And he (Judas) cast down the pieces of silver in the temple, and departed, and went and hanged himself. And the chief priests...bought with them the potter's field.\"\nJesus' first sermon plain or mount?\nMatt.5:1,2: \"And seeing the multitudes, he went up into a mountain: and when he was set, his disciples came unto him: And he opened his mouth, and taught them, saying....\"\nLuke6:17,20: \"And he came down with them, and stood in the plain, and the company of his disciples, and a great multitude of people...came to hear him.. And he lifted up his eyes on his disciples and said...\"\nJesus' last words\nMatt.27:46,50: \"And about the ninth hour Jesus cried with a loud voice, saying, \"Eli, eli, lama sabachthani?\" that is to say, \"My God, my God, why hast thou forsaken me?\" ...Jesus, when he cried again with a loud voice, yielded up the ghost.\"\n\nLuke23:46: \"And when Jesus had cried with a loud voice, he said, \"Father, unto thy hands I commend my spirit:\" and having said thus, he gave up the ghost.\"\n\nJohn19:30: \"When Jesus therefore had received the vinegar, he said, \"It is finished:\" and he bowed his head, and gave up the ghost.\"\nYears of famine\nII SAMUEL 24:13: So God came to David, and told him, and said unto him, shall SEVEN YEARS OF FAMINE come unto thee in thy land? or will thou flee three months before thine enemies, while they pursue. thee?\n\nI CHRONICLES 21:11: SO God came to David, and said unto him, Thus saith the LORD, Choose thee. Either THREE YEARS OF FAMINE or three months to be destryed before thy foes, while that the sword of thine enemies overtaketh thee;\nMoved David to anger?\nII SAMUEL 24: And again the anger of the LORD was kindled against Israel, and he moved David against them to say, Go, number Isreal and Judah.\n\nI CHRONICLES 21: And SATAN stood up against Isreal, and provoked David to number Israel.\nThe GENEALOGY OF JESUS?\nIn two places in the New Testament the genealogy of Jesus son of Mary (PBUH) is mentioned. Matthew 1:6-16 and Luke 3:23-31. Each gives the ancestors of Joseph the CLAIMED husband of Mary and Step father of Jesus(PBUH). The first one starts from Abraham(verse 2) all the way down to Jesus. The second one from Jesus all the way back to Adam. The only common name to these two lists between David and Jesus is JOSEPH, How can this be true? and also How can Jesus have a genealogy when all Muslims and most Christians believe that Jesus had/has no father.\nGod be seen?\nExod. 24:9,10; Amos 9:1; Gen. 26:2; and John 14:9\nGod CAN be seen:\n\"And I will take away my hand, and thou shalt see my backparts.\" (Ex. 33:23)\n\"And the Lord spake to Moses face to face, as a man speaketh to his friend.\" (Ex. 33:11)\n\"For I have seen God face to face, and my life is preserved.\" (Gen. 32:30)\n\nGod CANNOT be seen:\n\"No man hath seen God at any time.\" (John 1:18)\n\"And he said, Thou canst not see my face; for there shall no man see me and live.\" (Ex. 33:20)\n\"Whom no man hath seen nor can see.\" (1 Tim. 6:16)\nCRUEL, UNMERCIFUL, DESTRUCTIVE, and FEROCIOUS or KIND, MERCIFUL, and GOOD:\n\"I will not pity, nor spare, nor have mercy, but destroy.\" (Jer. 13:14) \"Now go and smite Amalek, and utterly destroy all that they have, and spare them not, but slay both man and woman, infant and suckling.\"\n\n\"The Lord is very pitiful and of tender mercy.\" (James 5:11)\n\"For his mercy endureth forever.\" (1 Chron. 16:34)\n\"The Lord is good to all, and his tender mercies are over all his works.\" (Ps. 145:9)\n\"God is love.\" (1 John 4:16)\nTempts?\n\"And it came to pass after these things, that God did tempt Abraham.\" (Gen 22:1)\n\n\"Let no man say when he is tempted, I am tempted of God; for God cannot be tempted with evil, neither tempteth he any man.\" (James 1:13)\nJudas died how?\n\"And he cast down the pieces of silver into the temple and departed, and went out and hanged himself.\" (Matt. 27:5)\n\n\"And falling headlong, he burst asunder in the midst, and all of his bowels gushed out.\" (Acts 1:18)\nAscend to heaven\n\"And Elijah went up by a whirlwind into heaven.\" (2 Kings 2:11)\n\n\"No man hath ascended up to heaven but he that came down from heaven, ... the Son of Man.\" (John 3:13)\nWhat was Jesus' prediction regarding Peter's denial?\nBefore the - crow - Matthew 26:34\n\nBefore the - crow twice - Mark 14:30\nHow many times did the - crow?\nMAR 14:72 And the second time the - crew. And Peter called to mind the word that Jesus said unto him, Before the - crow twice, thou shalt deny me thrice. And when he thought thereon, he wept.\n\nMAT 26:74 Then began he to curse and to swear, saying, I know not the man. And immediately the - crew.\nMAT 26:75 And Peter remembered the word of Jesus, which said unto him, Before the - crow, thou shalt deny me thrice. And he went out, and wept bitterly.\n\nLUK 22:60 And Peter said, Man, I know not what thou sayest. And immediately, while he yet spake, the - crew.\nLUK 22:61 And the Lord turned, and looked upon Peter. And Peter remembered the word of the Lord, how he had said unto him, Before the - crow, thou shalt deny me thrice.\n\nJOH 13:38 Jesus answered him, Wilt thou lay down thy life for my sake? Verily, verily, I say unto thee, The - shall not crow, still thou hast denied me thrice.\n\nJOH 18:27 Peter then denied again: and immediately the - crew.\nWho killed Saul\nSA1 31:4 Then said Saul unto his armourbearer, Draw thy sword, and thrust me through therewith; lest these uncircumcised come and thrust me through, and abuse me. But his armourbearer would not; for he was sore afraid. Therefore Saul took a sword, and fell upon it.\nSA1 31:5 And when his armourbearer saw that Saul was dead, he fell likewise upon his sword, and died with him.\nSA1 31:6 So Saul died, and his three sons, and his armourbearer, and all his men, that same day together.\nSA2 1:15 And David called one of the young men, and said, Go near, and fall upon him. And he smote him that he died.\nHow many beatitudes in the Sermon on the Mount\nMAT 5:3 Blessed are the poor in spirit: for theirs is the kingdom of heaven.\nMAT 5:4 Blessed are they that mourn: for they shall be comforted.\nMAT 5:5 Blessed are the meek: for they shall inherit the earth.\nMAT 5:6 Blessed are they which do hunger and thirst after righteousness: for they shall be filled.\nMAT 5:7 Blessed are the merciful: for they shall obtain mercy.\nMAT 5:8 Blessed are the pure in heart: for they shall see God.\nMAT 5:9 Blessed are the peacemakers: for they shall be called the children of God.\nMAT 5:10 Blessed are they which are persecuted for righteousness' sake: for theirs is the kingdom of heaven.\nMAT 5:11 Blessed are ye, when men shall revile you, and persecute you, and shall say all manner of evil against you falsely, for my sake.\n\nLUK 6:20 And he lifted up his eyes on his disciples, and said, Blessed be ye poor: for yours is the kingdom of God.\nLUK 6:21 Blessed are ye that hunger now: for ye shall be filled. Blessed are ye that weep now: for ye shall laugh.\nLUK 6:22 Blessed are ye, when men shall hate you, and when they shall separate you from their company, and shall reproach you, and cast out your name as evil, for the Son of man's sake.\nLUK 6:23 Rejoice ye in that day, and leap for joy: for, behold, your reward is great in heaven: for in the like manner did their fathers unto the prophets.\nDoes every man sin?\nKI1 8:46 If they sin against thee, (for there is no man that sinneth not,) and thou be angry with them, and deliver them to the enemy, so that they carry them away captives unto the land of the enemy, far or near;\n\nCH2 6:36 If they sin against thee, (for there is no man which sinneth not,) and thou be angry with them, and deliver them over before their enemies, and they carry them away captives unto a land far off or near;\n\nPRO 20:9 Who can say, I have made my heart clean, I am pure from my sin?\n\nECC 7:20 For there is not a just man upon earth, that doeth good, and sinneth not.\n\nJO1 1:8 If we say that we have no sin, we deceive ourselves, and the truth is not in us.\nJO1 1:9 If we confess our sins, he is faithful and just to forgive us our sins, and to cleanse us from all unrighteousness.\nJO1 1:10 If we say that we have not sinned, we make him a liar, and his word is not in us.\n\nJO1 3:9 Whosoever is born of God doth not commit sin; for his seed remaineth in him: and he cannot sin, because he is born of God.\nWho bought potter's field\nACT 1:18 Now this man purchased a field with the reward of iniquity; and falling headlong, he burst asunder in the midst, and all his bowels gushed out.\nACT 1:19 And it was known unto all the dwellers at Jerusalem; insomuch as that field is called in their proper tongue, Aceldama, that is to say, The field of blood.\n\nMAT 27:6 And the chief priests took the silver pieces, and said, It is not lawful for to put them into the treasury, because it is the price of blood.\nMAT 27:7 And they took counsel, and bought with them the potter's field, to bury strangers in.\nMAT 27:8 Wherefore that field was called, The field of blood, unto this day.\nWho prophesied the potter's field?\nMatthew 27:9-10 (mentions Jeremy but no such verse in Jeremiah) is in Zechariah 11:12-13\nWho bears guilt?\nGAL 6:2 Bear ye one another's burdens, and so fulfil the law of Christ.\n\nGAL 6:5 For every man shall bear his own burden.\nDo you answer a fool?\nPRO 26:4 Answer not a fool according to his folly, lest thou also be like unto him.\n\nPRO 26:5 Answer a fool according to his folly, lest he be wise in his own conceit.\nHow many children did Michal, the daughter of Saul, have?\nSA2 6:23 Therefore Michal the daughter of Saul had no child unto the day of her death.\n\nSA2 21:8 But the king took the two sons of Rizpah the daughter of Aiah, whom she bare unto Saul, Armoni and Mephibosheth; and the five sons of Michal the daughter of Saul, whom she brought up for Adriel the son of Barzillai the Meholathite:\nHow old was Jehoiachin when he began to reign?\nKI2 24:8 Jehoiachin was eighteen years old when he began to reign, and he reigned in Jerusalem three months. And his mother's name was Nehushta, the daughter of Elnathan of Jerusalem.\n\nCH2 36:9 Jehoiachin was eight years old when he began to reign, and he reigned three months and ten days in Jerusalem: and he did that which was evil in the sight of the LORD.\nMarriage?\nProverbs 18:22\n1 Corinthians 7 (whole book. See 1,2,27,39,40)\nDid those with Saul/Paul at his conversion hear a voice?\nACT 9:7 And the men which journeyed with him stood speechless, hearing a voice, but seeing no man.\n\nACT 22:9 And they that were with me saw indeed the light, and were afraid; but they heard not the voice of him that spake to me.\nWhere was Jesus three days after his baptism?\nMAR 1:12 And immediately the spirit driveth him into the wilderness.\n\nJOH 1:35 Again the next day after John stood, and two of his disciples;\n\n(various trapsing)\nHow many apostles were in office between the resurection and ascention?\n1 Corinthians 15:5 (12)\nMatthew 27:3-5 (minus one from 12)\nActs 1:9-26 (Mathias not elected until after resurrection)\n\nMAT 28:16 Then the eleven disciples went away into Galilee, into a mountain where Jesus had appointed them.\nJudging\n1 Cor 3:15 \" The spiritual man makes judgments about all things, but he himself is not subject to any man's judgment:\" (NIV)\n\n1 Cor 4:5 \" Therefore judge nothing before the appointed time; wait till the Lord comes. He will bring to light what is hidden in darkness and will expose the motives of men's hearts. At that time each will receive his praise from God.\"\nGood deeds\nMatt 5:16 \"In the same way, let your light shine before men, that they may see your good deeds and praise your Father in heaven.\" (NIV)\n\nMatt 6:3-4 \"But when you give to the needy, do not let your left hand know what your right hand is doing, so that your giving may be in secert. Then your Father, who sees what is done in secret, will reward you.\" (NIV)\nFor or against?\nMAT 12:30 He that is not with me is against me; and he that gathereth not with me scattereth abroad.\n(default is against)\n\nMAR 9:40 For he that is not against us is on our part.\n(default is for)\n\nLUK 9:50 And Jesus said unto him, Forbid him not: for he that is not against us is for us.\n(default is for)\nWhom did they see at the tomb?\nMAT 28:2 And, behold, there was a great earthquake: for the angel of the Lord descended from heaven, and came and rolled back the stone from the door, and sat upon it.\nMAT 28:3 His countenance was like lightning, and his raiment white as snow:\nMAT 28:4 And for fear of him the keepers did shake, and became as dead men.\nMAT 28:5 And the angel answered and said unto the women, Fear not ye: for I know that ye seek Jesus, which was crucified.\n\nMAR 16:5 And entering into the sepulchre, they saw a young man sitting on the right side, clothed in a long white garment; and they were affrighted.\n\nLUK 24:4 And it came to pass, as they were much perplexed thereabout, behold, two men stood by them in shining garments:\n\nJOH 20:12 And seeth two angels in white sitting, the one at the head, and the other at the feet, where the body of Jesus had lain.\nGod change?\nmalachi 3:6\njames 1:17\n1 samuel 15:29\njonah 3:10\ngenesis 6:6\nDestruction of cities (what said was jeremiah was zechariah)\nMAT 27:9 Then was fulfilled that which was spoken by Jeremy the prophet, saying, And they took the thirty pieces of silver, the price of him that was valued, whom they of the children of Israel did value;\n\nzechariah 11:11-13\n(nothing in Jeremiah remotely like)\nWho's sepulchers\nacts 7:16\ngenesis 23:17,18\nStrong drink?\nproverbs 31:6,7\njohn 2:11-11\nWhen second coming?\nMAT 24:34 Verily I say unto you, This generation shall not pass, till all these things be fulfilled.\n\nMAR 13:30 Verily I say unto you, that this generation shall not pass, till all these things be done.\n\nLUK 21:32 Verily I say unto you, This generation shall not pass away, till all be fulfilled.\n\n1 thessalonians 4:15-18\nSolomon's overseers\n550 in I Kings 9:23\n250 in II Chron 8:10\nThe mother of Abijah:\nMaachah the daughter of Absalom 2 Chron 9:20\n\nMichaiah the daughter of Uriel 2 Chron 13:2\nWhen did Baasha die?\n26th year of the reign of Asa I Kings 16:6-8\n\n36th year of the reign of Asa I 2 Chron 16:1\nHow old was Ahaziah when he began to reign?\n22 in 2 Kings 8:26\n\n42 in 2 Chron 22:2\nWho was Josiah's successor?\nJehoahaz - 2 Chron 36:1\n\nShallum - Jeremiah 22:11\nThe differences in the census figures of Ezra and Nehemiah.\nWhat was the color of the robe placed on Jesus during his trial?\nscarlet - Matthew 27:28\n\npurple John 19:2\nWhat did they give him to drink?\nvinegar - Matthew 27:34\n\nwine with myrrh - Mark 15:23\nHow long was Jesus in the tomb?\nDepends where you look; Matthew 12:40 gives Jesus prophesying that he will spend \"three days and three nights in the heart of the earth\", and Mark 10:34 has \"after three days (meta treis emeras) he will rise again\". As far as I can see from a quick look, the prophecies have \"after three days\", but the post-Resurrection narratives have \"on the third day\".\n\n\n\n\nPlease give counter arguments that we can discuss. I am in no way attempting to make a cheap shot or smunge christianity. I am just trying to examine the faith under the vision of logic and other peoples views.\n[/quote]\n\nHe replied\n[quote]\nCould we please make this a topic in G&FF? I'm completely willing to post the topic, and this to be a reply.\n\nYou have made a lot of accusations. I feel that I cannot answer all of them. If I cannot post a public topic, may I enlist the help of some fellow believers on this site?(MathFiend, JesusFreak, darktreb, etc.)\n\nBilly\n[/quote]\r\n\r\n\r\nI have done what Billy has requested. Let the thought provoking discourse begin.", "Solution_1": "I have moved this thread to Round Table as I felt the topic was a little heavy for G&FF.", "Solution_2": "I would like to make a very bold statement:\r\n\r\n[u][b]The second coming of Jesus has already taken place![/b][/u]\r\n\r\n[hide]\nFirst we must understand the prophecy about his first coming.\n\n[b]ELIJAH AND HIS SECOND COMING[/b]\n\n2,000 years ago, all Isrealites were waiting for the return of the prophet Elijah, to be followed by the Messiah. (Matt. 11:14; 17:10; John 1:21) It was commonly believed, however, that Elijah would descend bodily from heaven, just as he was lifted up.\n\nWhen Jesus claimed to be the Messiah, the question arose: Where was Elijah?\n\n
\"And Jesus answered and said unto them, Elias truly shall first come, and restore all things. But I say unto you, that Elias is come already... Then the disciples understood that he spake unto them of John the Baptist.\" (Matt. 17:10-13)
\n\nNotice that John the Baptist was not the same person as the earlier prophet Elijah, and he did not descend from the sky on a chariot of fire, as was commonly expected (II Kings 2:9-11) However, John the Baptist was identified as a person who had come [b]in the power and spirit[/b] of Elijah (Luke 1:13-17)\n\nIf the original Prophet, Elijah, had actually bodily gone up, then we should have expected the self-same person to also descend from heaven to fulfill the prophecy of his second advent. The fact that someone else was appointed to represent Elijah in power and spirit proves that the ascent of Elijah to heaven should be taken in a spiritual sense, rather than a physical sense.\n\nIt is not ironic that history has repeated itself and yet a majority of the Christians are not willing to learn from history!\n\n[b]THE RETURN OF JESUS CHRIST[/b]\n\nThere are many parallels in the stories of the return of Elijah and the return of the Messiah. It is commonly believed that Jesus Christ was bodily lifted up to heaven (just like Elijah was) and he will return from heaven in great glory (like Elijah was supposed to come in a chariot of fire). \n\n[b]What if the return of the Messiah is also very different from expectations?[/b] Will people reject him also like John the Baptist was rejected by most of his people?\n\nSuppose the return of the Messiah is not a coming in a cloud with power and glory. (In a cloud he will be too far for us to be able to see him; it obviously has a different interpretation.) Suppose he comes suddenly and very quietly, unobserved by most people, as the Bible says:\n\n
\"Watch therefore: for you know not what hour your Lord doth come... therefore be ye also ready: for in such an hour as ye think not, the son of man cometh.\" (Matt. 24:42-44)
\n\n
\"But the day of the Lord will come as a thief in the night.\" (2 Peter 3:10)
\n\n
\"...know perfectly that the Day of the Lord so cometh as a thief in the night.\" (1 Thess. 5:2)
\n\nThe Bible says that when the Messiah comes, he will be rejected by the world:\n\n
\"But first must he suffer many things, and be rejected this generation. And as it was in the days No-e, so shall it be also in the days of the Son of man.\" (Luke, 17:25-26)
\n\n[b]PUNISHMENT FOR HIS REJECTION[/b]\n\nRejection of Noah by his people resulted in a punishment from God in the form of the flood. Rejection of the Messiah in our age has also to result in punishment from God. The punishment was predicted by Jesus himself. This was to be wars, floods, earthquakes and pestilence.\n\nThe 20th century was full of wars, earthquakes, floods and pestilience like plague and AIDS. There was a terrible earthquake in San Fransisco in 1906. There was a similar devastating earthquake in India around the same time. Then came World War I in 1914, followed by many more wars.\n\nGod is just and merciful! Why would God inflict all these punishments for no reason at all. [b]The Messiah had to come first and be rejected.[/b] Then and only then would the world be justifiably punished. [b]We must look for the Messiah before the beginning of the twentieth century.[/b]\n\n[b]HE HAS ALREADY COME[/b]\n\nThus it came to pass. The Messiah came quietly near the end of the nineteenth century; unobserved by the world. He did not make a grand entrance with power and great glory. He came very quietly.\n\nThe original Elijah did not return. John the Baptist appeared in the power and spirit of Elijah. In exactly the same manner, the original Jesus of Nazareth did not return. Mirza Ghulam Ahmad of Qadian, India, came [b]in the power and spirit of Jesus[/b]. But we did not learn from history. The Jews had rejected John the Baptist because he was not the same original Elijah; and so the world rejected the new Messiah because he was not the same Jesus. Jesus himself said:\n\n\"Ye shall not see me henceforth, till ye shall say, [b]blessed is he that cometh in the name of the Lord.[/b]\" (Matt. 23:39)\n\nThe Messiah of our age proclaimed that he had been sent in the name of Jesus, but he was rejected by the majority just as predicted. (Luke 17:25)\n[/hide]\r\n\r\nIf anyone is interested, I can post some more information.", "Solution_3": "this is very interesting. please post more information. :)", "Solution_4": "I tend to go with the transcription error folks, as it is pretty well known that the Bible has many versions, all of them say different things, and none of them are in the original language. Your quotes on a \"wrathful\" verses a \"merciful\" God are, if you look carefully, from the Old (wrathful) and the New (merciful) Testaments. It is widely believed in the Christian world that the coming of Jesus heralded a new, more forgiving version of the Jewish (etc) faith, as can be seen in these comparisons. Outside of that, I really don't have too much to say, other than my beliefs are that the Bible isn't meant to be taken literally in most senses. Who cares in what order God created the universe, or how long it took? The important message is that God created the universe. Who cares how long Jesus stayed in one place or another? The important messages are the ones he imparted on how to live your life: \"Love your neighbor\" and \"Love the Lord thy God with all thy heart, with all thy soul, and with all thy mind\" and so on. THAT'S what's important; not the small details of His time here on earth.", "Solution_5": "bleumoose, are you talking to me, or penguinintegral? :?\r\n\r\nThank you for showing your interest, hl78reg. I will post more later. :)", "Solution_6": "[quote=\"bleumoose\"]I tend to go with the transcription error folks, as it is pretty well known that the Bible has many versions, all of them say different things, and none of them are in the original language. Your quotes on a \"wrathful\" verses a \"merciful\" God are, if you look carefully, from the Old (wrathful) and the New (merciful) Testaments. It is widely believed in the Christian world that the coming of Jesus heralded a new, more forgiving version of the Jewish (etc) faith, as can be seen in these comparisons. Outside of that, I really don't have too much to say, other than my beliefs are that the Bible isn't meant to be taken literally in most senses. Who cares in what order God created the universe, or how long it took? The important message is that God created the universe. Who cares how long Jesus stayed in one place or another? The important messages are the ones he imparted on how to live your life: \"Love your neighbor\" and \"Love the Lord thy God with all thy heart, with all thy soul, and with all thy mind\" and so on. THAT'S what's important; not the small details of His time here on earth.[/quote]\r\n\r\nSo, in otherwords, you now wish me to prove things using ideas now?\r\n\r\nAlso, jesus wasn't the only prophet/messiah trying to make a name for themselves. There were as many messiahs back then as there are crazy people today claiming to prove fermat's last theorm with pre-calculus methods.(note the parallel drawn)\r\n\r\nPlease, do not resort to name calling. Every serious discussion on this topic I have seen has quickly disinigrated into a flame war.", "Solution_7": "[quote=\"chess64\"]I would like to make a very bold statement:\n\n[u][b]The second coming of Jesus has already taken place![/b][/u]\n\n[hide]\nFirst we must understand the prophecy about his first coming.\n\n[b]ELIJAH AND HIS SECOND COMING[/b]\n\n2,000 years ago, all Isrealites were waiting for the return of the prophet Elijah, to be followed by the Messiah. (Matt. 11:14; 17:10; John 1:21) It was commonly believed, however, that Elijah would descend bodily from heaven, just as he was lifted up.\n\nWhen Jesus claimed to be the Messiah, the question arose: Where was Elijah?\n\n
\"And Jesus answered and said unto them, Elias truly shall first come, and restore all things. But I say unto you, that Elias is come already... Then the disciples understood that he spake unto them of John the Baptist.\" (Matt. 17:10-13)
\n\nNotice that John the Baptist was not the same person as the earlier prophet Elijah, and he did not descend from the sky on a chariot of fire, as was commonly expected (II Kings 2:9-11) However, John the Baptist was identified as a person who had come [b]in the power and spirit[/b] of Elijah (Luke 1:13-17)\n\nIf the original Prophet, Elijah, had actually bodily gone up, then we should have expected the self-same person to also descend from heaven to fulfill the prophecy of his second advent. The fact that someone else was appointed to represent Elijah in power and spirit proves that the ascent of Elijah to heaven should be taken in a spiritual sense, rather than a physical sense.\n\nIt is not ironic that history has repeated itself and yet a majority of the Christians are not willing to learn from history!\n\n[b]THE RETURN OF JESUS CHRIST[/b]\n\nThere are many parallels in the stories of the return of Elijah and the return of the Messiah. It is commonly believed that Jesus Christ was bodily lifted up to heaven (just like Elijah was) and he will return from heaven in great glory (like Elijah was supposed to come in a chariot of fire). \n\n[b]What if the return of the Messiah is also very different from expectations?[/b] Will people reject him also like John the Baptist was rejected by most of his people?\n\nSuppose the return of the Messiah is not a coming in a cloud with power and glory. (In a cloud he will be too far for us to be able to see him; it obviously has a different interpretation.) Suppose he comes suddenly and very quietly, unobserved by most people, as the Bible says:\n\n
\"Watch therefore: for you know not what hour your Lord doth come... therefore be ye also ready: for in such an hour as ye think not, the son of man cometh.\" (Matt. 24:42-44)
\n\n
\"But the day of the Lord will come as a thief in the night.\" (2 Peter 3:10)
\n\n
\"...know perfectly that the Day of the Lord so cometh as a thief in the night.\" (1 Thess. 5:2)
\n\nThe Bible says that when the Messiah comes, he will be rejected by the world:\n\n
\"But first must he suffer many things, and be rejected this generation. And as it was in the days No-e, so shall it be also in the days of the Son of man.\" (Luke, 17:25-26)
\n\n[b]PUNISHMENT FOR HIS REJECTION[/b]\n\nRejection of Noah by his people resulted in a punishment from God in the form of the flood. Rejection of the Messiah in our age has also to result in punishment from God. The punishment was predicted by Jesus himself. This was to be wars, floods, earthquakes and pestilence.\n\nThe 20th century was full of wars, earthquakes, floods and pestilience like plague and AIDS. There was a terrible earthquake in San Fransisco in 1906. There was a similar devastating earthquake in India around the same time. Then came World War I in 1914, followed by many more wars.\n\nGod is just and merciful! Why would God inflict all these punishments for no reason at all. [b]The Messiah had to come first and be rejected.[/b] Then and only then would the world be justifiably punished. [b]We must look for the Messiah before the beginning of the twentieth century.[/b]\n\n[b]HE HAS ALREADY COME[/b]\n\nThus it came to pass. The Messiah came quietly near the end of the nineteenth century; unobserved by the world. He did not make a grand entrance with power and great glory. He came very quietly.\n\nThe original Elijah did not return. John the Baptist appeared in the power and spirit of Elijah. In exactly the same manner, the original Jesus of Nazareth did not return. Mirza Ghulam Ahmad of Qadian, India, came [b]in the power and spirit of Jesus[/b]. But we did not learn from history. The Jews had rejected John the Baptist because he was not the same original Elijah; and so the world rejected the new Messiah because he was not the same Jesus. Jesus himself said:\n\n\"Ye shall not see me henceforth, till ye shall say, [b]blessed is he that cometh in the name of the Lord.[/b]\" (Matt. 23:39)\n\nThe Messiah of our age proclaimed that he had been sent in the name of Jesus, but he was rejected by the majority just as predicted. (Luke 17:25)\n[/hide]\n\nIf anyone is interested, I can post some more information.[/quote]\r\n\r\nI can make the same argument about the 21th century. 9/11, london bombings, madrid \r\nbombings, plane crashes, infomercials, etc.\r\n\r\nBad things do not equal messiah.", "Solution_8": "[quote=\"PenguinIntegral\"]I can make the same argument about the 21th century. 9/11, london bombings, madrid bombings, plane crashes, infomercials, etc.[/quote]\r\n\r\nVery true! But think about it. Does God stop punishing the world after one century? Has the whole world accepted him yet? No! You are actually just supporting my claim. ;)", "Solution_9": "[quote=\"chess64\"][quote=\"PenguinIntegral\"]I can make the same argument about the 21th century. 9/11, london bombings, madrid bombings, plane crashes, infomercials, etc.[/quote]\n\nVery true! But think about it. Does God stop punishing the world after one century? Has the whole world accepted him yet? No! You are actually just supporting my claim. ;)[/quote]\r\n\r\nUm, give me a point at which the comming happened, and I will give you some really bad stuff that happened before then.", "Solution_10": "[quote=\"bleumoose\"]I tend to go with the transcription error folks, as it is pretty well known that the Bible has many versions, all of them say different things, and none of them are in the original language. Your quotes on a \"wrathful\" verses a \"merciful\" God are, if you look carefully, from the Old (wrathful) and the New (merciful) Testaments. It is widely believed in the Christian world that the coming of Jesus heralded a new, more forgiving version of the Jewish (etc) faith, as can be seen in these comparisons. Outside of that, I really don't have too much to say, other than my beliefs are that the Bible isn't meant to be taken literally in most senses. Who cares in what order God created the universe, or how long it took? The important message is that God created the universe. Who cares how long Jesus stayed in one place or another? The important messages are the ones he imparted on how to live your life: \"Love your neighbor\" and \"Love the Lord thy God with all thy heart, with all thy soul, and with all thy mind\" and so on. THAT'S what's important; not the small details of His time here on earth.[/quote]\r\n\r\nSo the bible is not infallible?", "Solution_11": "[quote=\"PenguinIntegral\"]Um, give me a point at which the comming happened...[/quote]\r\n\r\nhttp://www.alislam.org/library/messiah.html", "Solution_12": "This is not meant to be an argument against Christianity, but at what point do we stop blaming God or giving God credit for natural disasters, all of which have modern, verifiable, scientific explanations? I understand that it's one thing to give credit to the creator for [i]everything[/i], but why do we intertwine our actions with the existence of natural phenonema? Can we not view the creator with a modern, scientific, skeptical eye?\r\n\r\nIf religion does not jive with what we actually observe, do we try harder to observe, or do we alter the tenets of our religion?", "Solution_13": "actually, thousands of years ago, i think it was okay in some polytheistic religions to be angered by a god and just pray to another. can anyone confirm that?", "Solution_14": "[quote=\"chess64\"][quote=\"PenguinIntegral\"]Um, give me a point at which the comming happened...[/quote]\n\nhttp://www.alislam.org/library/messiah.html[/quote]\r\n\r\nWell, for a start, natural disasters, wars, and sickness?", "Solution_15": "It's not a subjectivist like me who is interested in truth. I'm not. I'm interested in what affects me and everything I subjectively value. It is religious people who seem interested in truth. That is why I ask the questoins I ask, to explore both their psychological need for truth and also question their assumptions.", "Solution_16": "If suppose that god will sent some people to the hell fire is the \"TRUTH\", doesn't you value your own safety???\r\n\r\nSo in my opinion without knowing the \"TRUTH\" then we can not know if our action indeed will benefit or against our interest.", "Solution_17": "[quote=\"oldman\"]\nSo in my opinion without knowing the \"TRUTH\" then we can not know if our action indeed will benefit or against our interest.[/quote]\r\n\r\nExactly. That's why I don't worry about religious axioms. How could I ever establish their truth?\r\n\r\nIf anyone had ever been smart enough to establish a verifiable ontological argument (something that could be checked by computer were it valid), it would be quite famous. Then I would start paying attention.\r\n\r\nUntil then, I feel quite comfortable setting my own standards of action and I am comfortable that they are good standards -- to my own subjective set of values.", "Solution_18": "I think we should all be good and live life like we're supposed to :)", "Solution_19": "[quote=\"MCrawford\"][quote=\"oldman\"]\nSo in my opinion without knowing the \"TRUTH\" then we can not know if our action indeed will benefit or against our interest.[/quote]\n\nExactly. That's why I don't worry about religious axioms. How could I ever establish their truth?\n\nIf anyone had ever been smart enough to establish a verifiable ontological argument (something that could be checked by computer were it valid), it would be quite famous. Then I would start paying attention.\n\nUntil then, I feel quite comfortable setting my own standards of action and I am comfortable that they are good standards -- to my own subjective set of values.[/quote]\r\n\r\nFor my self ontological argument will always have a problem. First, we can not make a well definittion for the word \"god\", because people who believe in him only know Him by his attributes (all knowing, omnipresent, etc). Once we make a god definition then we bound god subject to our definition, which is not good if indeed that god is boundless.\r\n\r\n(here is an attempt on impossibility of succesfull ontological argument, http://cogprints.org/2492/01/ontarg-en.pdf)\r\n\r\nThe second, any ontological argument will always based on certain axiom that might not all people agree that they are really an axiom (btw have you see Godel ontological argument, some logician says that if the axiom is really accepted than the conclusion follows).\r\n\r\nMaybe knowing of god can only obtained through gnosis. The necessay condition for this if you really really seek the god, and the sufficient condition is if god willing to revealed himself.\r\nAnd since only a few who really seek him very very hard, then there isn't a surprise that only a few who believe in him.\r\n\r\nJust for thought exercise: suppose a three dimensional man is consider a higher being by a one dimensional bacteria. How do you make the bacteria aware of the three dimensional receptivity???", "Solution_20": "[quote=\"oldman\"]\nAnd since only a few who really seek him very very hard, then there isn't a surprise that only a few who believe in him.\n[/quote]\r\n\r\nI disagree that few seek God very very hard. In fact, I think almost everyone does at some point who has any inkling of belief.", "Solution_21": "Has anyone heard of the oscillating universe theory?", "Solution_22": "no.\r\n\r\nnot to spam or anything. :)\r\n\r\ni think this religion topic is getting out of control. :D", "Solution_23": "[quote=\"hwenterprise\"]no.\n\nnot to spam or anything. :)\n\ni think this religion topic is getting out of control. :D[/quote]\r\n\r\nWell, take a look at oscillating universes, and it can show just how this universe is not the only or the first of the force's creation :) Or just read my blog...", "Solution_24": "I believe in God and Jesus as our savior(christian) This thread is getting out of hand, don't you think :D", "Solution_25": "[quote=\"jhollenbeck\"]I believe in God and Jesus as our savior(christian) This thread is getting out of hand, don't you think :D[/quote]\r\n\r\nThere was a little spamming, which is rude, but I think the serious participants kept the discussion at a respectful level. Was there any post in particular you want me to take a look at as a moderator?", "Solution_26": "I did not have time to read through the entire discussion (just first 4 and a half pages) but it seems many of the arguments in this thread have gone off topic to the existence of a creator, which we already have a thread for, and that is not what we are discussing here. For a discussion on whether there is a creator or not go to that thread. Religion is separate from their being a creator, because religion implies that the already existing creator wants worshippers. The thread started out as a debate over contradictions in holy books and I believe it should stay that way. \r\n\r\nAlso, none of the contradictions that I read in the Old Testament (or for me, the Torah) work, and if they would be brought up instead of in a list, in an organized fashion..\r\n\r\nAlso, by Abraham and Isaac, couldn't it also be that God is setting an example that if he tells you to do something you do it? It is still Abraham's will that he did it, God didnt force him, God only knew that he would choose to do it, so it is still in the merit of Abraham.\r\n \r\nBy Jephtah I believe there are also sources which say he exiled her for 3 months, which is biblically considered like killing them. The \"never marry\" and \"die a virgin\" emphasis is also important. In other reports, I dont believe it even says that he sacrificed her..\r\n\r\n\r\nAlso, being Jewish, I believe in the Torah (Old Testament for Christians) which encompasses the 5 books and many others, about 40, of course, none related to Jesus. Also, I believe in the original Hebrew text (which is original because they found scripts thousands of years old that are the exact same. The english translation of the Old Testament is I believe not 100% accurate, any of them, since they are (or atleast some of them I think, I think all of them) based on the Saptaguent, the Greek translation of the Torah which was translated by 70 rabbis who translated parts differently so that the Greeks wouldnt take those parts as offensive against them (the greeks) and attack them (the rabbis). I'm not 100% sure on that though, that it is based on the Saptaguent. However, I am sure that the english translation isnt 100% because Hebrew cant fully be translated into english.. there are problems w/ it.\r\n\r\n(P.S. I love the Round Table arguments.. 'dunno why' (heh) but it seems that its so much better to argue w/ people who actually know what they're talking about than w/ \"lots of other people\" who dont)", "Solution_27": "I think the discussion moved to tangents because those tangents are relevant to the discussion.\r\n\r\nIn fact, I think the discussion of holy books was tangential to the discussion of religion in the first place, though I'll let the original poster correct me if I am mistaken.", "Solution_28": "Well, not really, because the first post is about contradictions in holy books, so discussing holy books is somewhat tangential, but discussing contradictions in holy books is on the topic.l", "Solution_29": "So, I was just wondering. After reading a bunch of tangents I have one question I must ask because I wish to participate in this discussion.\r\n\r\nAre we talking about generally, the validity/infallibility of the Holy Bible? Or just of the concept of religion?" } { "Tag": [ "\\/closed" ], "Problem": "1.) Who teaches the online classes?\r\n\r\n2.) I heard the classes are in \"online classrooms/chatrooms\". If you have a slow internet connection, would it take a long time to \"load\" the classroom/chatroom.\r\n\r\n3.) If the chatroom/classroom suddenly gets spammed by some person (like they keep on posting the same thing over and over again, so it gets really annoying), would the person get kicked out? (I find this really annoying in AIM chatrooms)\r\n\r\nThanks! :)", "Solution_1": "1) Depends on what class it is. See the Enroll page, I think it says who's teaching each class.\r\n\r\n2) It shouldn't take [i]that[/i] long...\r\n\r\n3) The classrooms are moderated. You should go to MathJams to see how it is. Whatever you type goes to the teacher, who then makes it available to the rest of the class if he wishes.\r\n\r\nHope that helped. :)", "Solution_2": "[quote=\"Drunken_Math\"]1.) Who teaches the online classes?\n[/quote]\r\n\r\nThe instructors are listed on the classes list.", "Solution_3": "[quote=\"Drunken_Math\"]1.) Who teaches the online classes?[/quote]The classes are all taught by either Richard Rusczyk, Mathew Crawford, Naoki Sato, or myself. [url=http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php]Read our bios.[/url] \n\n[quote=\"Drunken_Math\"]2.) I heard the classes are in \"online classrooms/chatrooms\". If you have a slow internet connection, would it take a long time to \"load\" the classroom/chatroom.[/quote]The classroom itself is a Java applet -- it may be initially slow to load at the start of class, but once the class starts it shouldn't matter whether you're on dialup or broadband. You should test it out by attending one of our free [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jams[/url].\n\n[quote=\"Drunken_Math\"]3.) If the chatroom/classroom suddenly gets spammed by some person (like they keep on posting the same thing over and over again, so it gets really annoying), would the person get kicked out? (I find this really annoying in AIM chatrooms)[/quote]Our classroom is moderated, meaning that you only see what the instructor chooses to let you see. So no \"spamming\" of the classroom is possible.\n\n[quote=\"Drunken_Math\"]Thanks! :)[/quote]You're welcome. Thanks for being a part of AoPS!", "Solution_4": "Yes, when I was on dialup, the chatroom would take several minutes to load.\r\n\r\nI highly recommend MCrawford and RRusczyk because I've taken classes under them. I can't say for the others. I hope to take a class under Naoki later this year\r\n\r\nNo spam allowed.", "Solution_5": "I must echo solafidefarms's sentiments. Mr. Crawford, Mr. Rusczyk, and Mr. Patrick are all fantastic instructors. I've had the pleasure of working with all of them during my IS, and each session is always engaging, lots of fun, and even uplifting sometimes.\r\n\r\nUnless you are extremely fortunate, I can't imagine anything short of a math summer program parallelling the opportunities provided here.", "Solution_6": "[quote=\"mel\"]I must echo solafidefarms's sentiments. Mr. Crawford, Mr. Rusczyk, and Mr. Patrick are all fantastic instructors. I've had the pleasure of working with all of them during my IS, and each session is always engaging, lots of fun, and even uplifting sometimes.\n\nUnless you are extremely fortunate, I can't imagine anything short of a math summer program parallelling the opportunities provided here.[/quote]\r\n\r\nOh, for this summer Intro B IS... I got to work with wdragyn and joejia!!! Like mel said \"each session is always engaging, lots of fun, and even uplifting sometimes\"!!!! \r\n\r\nI got a question though... ok... if I would like to send them thank you cards :D ... where should I sent it to? I thought that the instructors are from different places and I hope that my thank-yous are received. I think I should sent it to the AoPS central place but I'm not sure.", "Solution_7": "If you send it to AoPS central, we'll make sure it finds the right home.\r\n\r\nAoPS Incorporated\r\nPO Box 2185\r\nAlpine, CA 91903-2185" } { "Tag": [], "Problem": "Suppose we have the letters M, I and U we can use\r\nto form \"words\". The challenge is, based on the\r\n\"word\" MI, transforming it in word MU applying in each\r\nstep, one of the following rules:\r\n\r\n1) If the word ends I add a U (eg MI -> MIU)\r\n2) Duplicate all the text after an M (eg MIU -> MIUIU)\r\n3) Replace any sequence III by U (eg MUIIIU -> MUUU)\r\n4) Remove any UU (eg MUUU -> MU)\r\n\r\n :)", "Solution_1": "Is it possible?\r\nSuppose I=1, U=3, M=0\r\n01>03\r\n1: Add 3 (no change mod 3)\r\n2: Duplicate (if original string is not 0 mod 3, next will neither)\r\n3: Replace 111 by 3 (no change mod 3)\r\n4: remove 33 (no change mod 3)\r\n\r\nSo, with an original string \"not 0 mod 3\" (like 01), you can't arrive to a string \"0 mod 3\" (like 03)\r\n\r\nDid I commit a mistake? :(", "Solution_2": "This is a puzzle from Godel, Escher, Bach, which happens to be a good book.\r\nSee: http://en.wikipedia.org/wiki/MU_puzzle\r\nPacman's solution is pretty much the same thing as the wikipedia solution." } { "Tag": [], "Problem": "The mean of five numbers is 5. If one of the numbers is removed from the group, the mean of the remaining four numbers is 4.25. What number was removed from the group?", "Solution_1": "[quote=\"GameBot\"]The mean of five numbers is 5. If one of the numbers is removed from the group, the mean of the remaining four numbers is 4.25. What number was removed from the group?[/quote]\r\n\r\nSince originally there were 5 numbers, and the mean was 5, that means the sum was $ 5 \\times 5 \\equal{} 25$. When one of the numbers was removed, there were four numbers, with a mean of 4.25, so a sum of $ 4 \\times 4.25 \\equal{} 17$. The difference between the sums is 8, so that's the number that was removed." } { "Tag": [ "search", "ceiling function", "logarithms", "induction", "algorithm", "combinatorics proposed", "combinatorics" ], "Problem": "We have two players, a question asker and a question answerer. The question answerer has selected a subset of size $ k$ of an $ n$-set, and the question asker's questions take the form of arbitrary subset $ S$ of the $ n$-set. In all cases, we want to know the minimum number of questions for the question-asker to be sure of learning the selected $ k$-set.\r\n\r\n1) $ k \\equal{} 2$; answerer answers \"yes\" if the selected pair of elements are both contained in $ S$\r\n\r\n2) $ k \\equal{} 2$; answerer answers \"yes\" if at least one of the selected elements is contained in $ S$\r\n\r\n3) $ k \\equal{} 2$; answerer answers \"yes\" if exactly one of the selected elements is contained in $ S$\r\n\r\n4) $ k \\equal{} 3$; answerer answers \"yes\" if at least two of the selected triple of elements is contained in $ S$\r\n\r\n5) arbitrary (fixed) $ k$; answerer answers \"yes\" if the selected subset is contained in $ S$\r\n\r\n6) arbitrary (fixed) $ k$; answerer answers \"yes\" if the selected subset has nonempty intersection with $ S$\r\n\r\n7) arbitrary (fixed) $ k$; answerer answers with the number of selected elements contained in $ S$", "Solution_1": "Approximate solution to question 1:\r\nFirst, we have a search space of size $ \\binom{n}{2}$ and we're asking yes-no questions, so we certainly can't do better than $ \\left\\lceil \\log_2 \\binom{n}{2} \\right\\rceil$ questions. We can achieve $ 2 \\left\\lceil \\log_2 n\\right\\rceil$, as follows: divide the $ n$-set as $ A \\cup B$ with $ |A|$ and $ |B|$ both equal to $ \\frac{n}{2}$. \r\nFirst ask, \"Is the pair in $ A$?\".\r\nIf the answer is \"yes,\" by induction you need only $ 1 \\plus{} 2 \\left\\lceil \\log_2 \\frac{n}{2}\\right\\rceil \\leq 2 \\left\\lceil \\log_2 n\\right\\rceil$ questions total. \r\nIf the answer is \"no,\" then ask \"Is the pair in $ B$?\" \r\nIf the answer is \"yes,\" by induction you need only $ 1 \\plus{} 2 \\left\\lceil \\log_2 \\frac{n}{2}\\right\\rceil \\leq 2 \\left\\lceil \\log_2 n\\right\\rceil$ questions total. \r\nIf the answer is \"no,\" then we have one number in each pair. Now do binary search in each half (for example, asking \"is the pair in this half of $ A$ or in $ B$?\" as the first question) to get $ 2 \\plus{} 2 \\left\\lceil \\log_2 \\frac{n}{2}\\right\\rceil \\leq 2 \\left\\lceil \\log_2 n\\right\\rceil$ questions total.\r\n\r\nThe difference $ \\left\\lceil \\log_2 \\binom{n}{2} \\right\\rceil \\minus{} 2 \\left\\lceil \\log_2 n\\right\\rceil$ is never more than 2 (and in fact this algorithm gives exactly 1 more than our theoretical bound more than half the time), and it seems likely that this is the true value (though maybe the distance from the theoretical bound can be reduced to 1 in all cases?).\r\n\r\n\r\nThe answer to (2) is identical to the answer from (1).\r\n\r\n\r\n(3) is cute, and an answer within a constant of the theoretical bound is still possible (was discussing this with D. Speyer and P. McNamara, and I don't think we worked out the precise details).\r\n\r\n\r\nEven for $ k \\equal{} 2$, the answer to question 7 seems challenging.\r\n\r\n\r\nOne further variant:\r\n\r\n8) $ k \\equal{} 3$; answerer answers \"yes\" if exactly two of the selected triple of elements is contained in $ S$." } { "Tag": [ "number theory", "relatively prime", "real analysis", "real analysis theorems" ], "Problem": "Does the set $ \\{a\\sqrt {p} \\plus{} b\\sqrt {q}|a,b\\in \\mathbb{Q}\\}$ dense in $ \\mathbb{R}$ for relatively prime integers $ p$ and $ q$?\r\n\r\nthx.", "Solution_1": "I assume the question is meant to say $ a,b\\in\\mathbb{Z}.$ Allowing either $ a$ or $ b$ to take on all rational values doesn't make for an interesting question.\r\n\r\nBut to answer that question: a necessary and sufficient condition for $ \\{au\\plus{}bv\\mid a,b\\in\\mathbb{Z}\\}$ to be dense in $ \\mathbb{R}$ is for $ \\frac{u}{v}$ to be irrational.", "Solution_2": "upps :blush: , it is related to the combinatorial problems below:\r\n\r\nIf $ x$ is irrational and $ n \\in \\mathbb{N}$, from the sequences $ x,2x,\\cdots,nx$, two of them with fractional part less than $ \\frac{1}{n}$.\r\n\r\nthx for your reply." } { "Tag": [], "Problem": "I'm not sure what the book is talking about two holidays. \r\n\r\nThe first holiday is mentioned as:\r\n\r\n[quote=\"My book\"]\nThe fifteenth birthday for many Hispanic girls is a coming of age celebration with a party at home. This party is called [b]una fiesta de quinceanera[/b](n in cean has ~ symbol above it) and can range from a small, informal gathering to a celebration resembling a wedding. In most cases a local [b]conjunto[/b](group of musicians) plays. It is customary for the [b]padrino[/b] or [b]madrina[/b] to present the honoree with a special gift.[/quote]\n\nThe second holiday is mentioned as:\n\n[quote=\"My book\"]\nIn small towns and cities throughout the Spnaish speaking world, there are special celebrations for all kinds of occasions. You've already learned that in Spanish-speaking countries many people celebrate not only their birthday, but also their [b]Dia de santo[/b](I didn't learn this.. What is this??). Many cities and countries have saint's days which they often celebrate as holidays. Spain's patron saint, for example, is Santiago. July 25 is the Feast of Santiago and is an important holiday in Spain. Juan Carlos, the king of Spain, has his saint's day on June 24, [b]el Dia de San Juan[/b], making it a national holiday[/quote]\r\n\r\nMaybe some parts are descriptions of holidays but I still don't get it. :? \r\n\r\nOh and the bold parts are just bolded because that's how it's shown on the book.\r\n\r\nMuchas gracias! :lol: :lol: :lol: :lol:", "Solution_1": "[quote=\"Silverfalcon\"]I'm not sure what the book is talking about two holidays. \n\nThe first holiday is mentioned as:\n\n[quote=\"My book\"]\nThe fifteenth birthday for many Hispanic girls is a coming of age celebration with a party at home. This party is called [b]una fiesta de quinceanera[/b](n in cean has ~ symbol above it) and can range from a small, informal gathering to a celebration resembling a wedding. In most cases a local [b]conjunto[/b](group of musicians) plays. It is customary for the [b]padrino[/b] or [b]madrina[/b] to present the honoree with a special gift.[/quote]\n\nMaybe some parts are descriptions of holidays but I still don't get it. :? \n\nOh and the bold parts are just bolded because that's how it's shown on the book.\n\nMuchas gracias! :lol: :lol: :lol: :lol:[/quote]\r\n\r\ni don't kno anything about the second one, but we discussed the first one (quinceanera) in my spanish class. you know how for girls in America, their 16th birthday (or \"sweet 16\") is a big deal? well, for Mexicans, the 15th birthday is their big deal. however, it is even more so than for Americans.An all out party is thrown. The girl usually wears a pink dress, normally pale pink. LOTS of people are invited, one girl at my school had over 300 people attend her quinceanera. there is an organised dance that the family of the girl practices before hand, and then performs. basically, the quinceanera is a big deal. \r\n\r\nthis is what my spanish teacher told us in class, and what i have heard from the girl at my school who had her quinceneara a couple years ago. if i got anything wrong, anyone who knows about these things please correct me!", "Solution_2": "Why in this world when someone says something about spanish speaking people, some american comes with the comparation with Mexico. Mexican are just a small part of the spanish speaking population in the world. \r\n\r\nThe [b]quncea\u00f1os[/b] how it is usually called, instead of [b]fiesta de quincea\u00f1era[/b] is basically one of the biggest parties that a woman would ever have on her honor in her life, and yes, it is quite similar to a wedding party, except for the small fact that there is no groom (the wedding party, any way, is not about the groom!). Usually is a BIG BIG party. But lately in some countries there has been a change on the uses. At least in Costa Rica, some families decide to spend the money of the party paying a travel for the girl alone, or may be the girl and some adult going with her, to some place in the world. I know people who have gone to specific cities and spend a few weeks there, or having a tour, say, for all western Europe.\r\n\r\nWith respect to the [b]D\u00eda de tu Santo[/b], each day on the Catholic calendar is dedicated to some saints (today, April 15th, is Saint Anastasia and Saint Damian, and some more, you can check some [url=http://www.terra.es/personal/angerod/calendar2.htm]here[/url]), then the chances that your name has some saint in the calendar is pretty high, at least if your name is a common name, as in my case, David, is in six different days, because there have been six different saints with my name and each of them has its own day (of course, it is not the only saint on each of those days).\r\n\r\nBy the way, non of those is what I would call A HOLIDAY, it is not a day you don't work or something like that. Both are somehow similar to any birthday, just that a little different :D", "Solution_3": "I think that what Silverfalcon migth be talking about the Santo Patrono\u00b4s day of a city or a Country. My city, for example, has it Santo Patrono\u00b4s Days on January 20th and a movile date on May, and no one works on that days. M\u00e9xico\u00b4s Santa Patrona is the Virgen de Guadalupe (December 12) and almost no one works on that day. (by the way it is my birthday and I have never gone to my school on my birthday)", "Solution_4": "I get the first one now. Thanks.\r\n\r\nFor the second one, I still don't quite get what those \"saint days\" do. Are they just no working days? I mean all of them? Or just some of them?\r\n\r\nIs it like Martin Luther King Day(well, there are days with president but this is one I can think of with person's name)?\r\n\r\nOr is it just like a day with someone's name with nothing really special?\r\n\r\nI'm bit confused. :blush:", "Solution_5": "[quote=\"Silverfalcon\"]I get the first one now. Thanks.\n\nFor the second one, I still don't quite get what those \"saint days\" do. Are they just no working days? I mean all of them? Or just some of them?\n\nIs it like Martin Luther King Day(well, there are days with president but this is one I can think of with person's name)?\n\nOr is it just like a day with someone's name with nothing really special?\n\nI'm bit confused. :blush:[/quote]\r\n\r\nOK, I thought that you were talking about the [b]Dia de tu santo[/b] instead of [b]Dia del Patron(a)[/b] of the city, but it has sence (hehe, thanks Dave).\r\n\r\nIt is not quite like Martin Luther King Day. This day is in honor to an specific person and it is celebrated at the same time in all the US (Even when here in Georgia is a lot bigger the celebration than basically everywhere else). El [b]dia del patron[/b] is quite different. Remember: LatinAmerica was conquered by Spain, that was mainly catholic in that time. Therefore, many many places in Latin America has name of catholic saints (I am from Costa Rica, and the capital of the country is in the city called San Jose, that means, Saint Joseph). Hence, in the Saint Joseph day, it was a holiday, just in that city (it was, but no longer is, at least not for the goverment employees, the goverment no longer respect the eclesiastic holidays, except for, may be, Hollyweek). For around ten years I lived in a city called San Juan, then, in Saint John's day, I had hollyday, and also, in Saint Peter Day (because I use to travel a few miles and study in San Pedro!).\r\n\r\nDoes it make more sence? if not, I can try to reexplain.", "Solution_6": "[quote=\"djimenez\"]Why in this world when someone says something about spanish speaking people, some american comes with the comparation with Mexico. Mexican are just a small part of the spanish speaking population in the world. [/quote]\r\n\r\nmy bad...I guess I keep forgetting that there is more than just Mexico out there. My spanish teacher married a Mexican, so everything she tells us is in the context of Mexico, rather than Spain etc. (and yes I kno there's more, I just can't think atm). So, I guess that whenever someone mentions Spanish, I think of Senora Castillo, and in return think of Mexico. my bad. :blush:", "Solution_7": "[quote=\"MathFiend\"][quote=\"djimenez\"]Why in this world when someone says something about spanish speaking people, some american comes with the comparation with Mexico. Mexican are just a small part of the spanish speaking population in the world. [/quote]\n\nmy bad...I guess I keep forgetting that there is more than just Mexico out there. My spanish teacher married a Mexican, so everything she tells us is in the context of Mexico, rather than Spain etc. (and yes I kno there's more, I just can't think atm). So, I guess that whenever someone mentions Spanish, I think of Senora Castillo, and in return think of Mexico. my bad. :blush:[/quote]\r\n\r\nNo problem about that, somehow it is not exactly your fault. There are 23 countries where Spanish spoken for at least half a million people, and the native spanish speaking population, depending on the source, varies from 350 to 500 millions. The most populated spanish countries is Mexico, no doubht about that, the population there is around 75 millions, but anyway, that is around 20% of us, who are hispanic. \r\n\r\nNevertheless, the hispanic culture is very wide, nevertheless, most of the hispanic here in the United states feel that we are all thought to be Mexican. Do not misunderstand me, I don't have anything against mexicans, but I am not mexican, nevertheless, the hispanic television on the US is mainly mexican, the \"hispanic food places\" usually sell mexican style food, the \"hispanic products\" that you find in most of the supermarkets (at least Kroger and Publix) use to be from the mexican culture, and many of us don't use it on our home countries.\r\n\r\nI remember the son of a friend that was complaining because the first time he went to the school (fourth grade) when they arrived to the US (his father was a visitor here) the teacher noted his hispanic accent and (T: teacher, K: kid)\r\n[list]T: Then, where in Mexico are you from?\nK: No, I am not from Mexico, I am from Spain!\nT: Yeah, yeah! it is the same thing![/list]\nThe kid standed up and went to the World Map in one side of the classroom and said\n[list]K: No... you see, this one is Mexico, this one is Spain... THERE IS AN OCEAN IN THE MIDDLE! (My adendum: AND VERY WIDE BY THE WAY!!!).\nT: Yes, but you are hispanic, you are all the same.... [/list]\r\nThat kind of mentality is the one that sometimes offend to several hispanic people (and I include myself!).", "Solution_8": "Hi I lived in Spain all my childhood so I have a pretty good sense of what [b]el dia del santo[/b] is like.\r\n\r\nWhen it was my saints' day I usually got a phonecall from everybody that I knew wishing me a \"happy saints day\" . It's also a good way to catch up on friendships if you haven't spoken with someone for a while.\r\n\r\nWithin families people usally get together and have a big family lunch or dinner, sometimes on the weekend (Sunday) because everybody is quite busy.\r\n\r\nLittle kids sometimes get gifts from family members but friends do not bring gifts and a party with freinds is not customary either.\r\n\r\nYes there are some important saints days like St. Joseph, St George orf St. Jhon. The big festival and celebrations attached to those days are due to pagan reasons often not related to catholicity at all. Alot of this festivals were present before and someone (probably one of the hundreds of popes in the middle ages) named that particular festival after a saint.\r\n\r\nHope this helps you understand \"el dia del santo\" in Spain. [/b]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "let $u_{n}$ be sequence determinate by : \r\n $u_{1}=1 and u_{n+1}= \\sqrt{u_{n}^{2}+\\frac{1}{2^{n}}}$forall$n \\geq 1$\r\nset a limit $lim_{n \\to \\infty}u_{n}$", "Solution_1": "We have $u_{n+1}^{2}-u_{n}^{2}=\\frac{1}{2^{n}}$, now use telescopic sum." } { "Tag": [ "geometry", "rectangle" ], "Problem": "[quote=\"Mathcounts State Sprint 2005\"]What is the number of ways to completely cover a 2x5 rectangle with 2x1 rectangles?[/quote]\r\n\r\nCould someone please explain how to solve this kind of problem and be sure to count every case? I missed a point because of this problem.", "Solution_1": "I hate those too. You have to visiualize it, but I sometimes get it wrong too. :(" } { "Tag": [ "algebra", "polynomial", "Ring Theory", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "show that if $ f\\equal{}a_{0}\\plus{}a_{1}x\\plus{}...\\plus{}a_{n}x^{n}$is nipotent then $ a_{0},a_{1},...,a_{n}$ are nilpotents. Ring R is commutative with 1", "Solution_1": "$ R[x]$ is a subring of $ R[[X]]$ hence the claim follows from the statement of your last thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=305331", "Solution_2": "could you give me any other way to prove it", "Solution_3": "I could but why? Everything else would only complicate matters here and I doubt that it would give any significant insight.", "Solution_4": "[quote=\"wiosna\"]could you give me any other way to prove it[/quote]Induction on $ n$ - but for the induction step you will certainly bump into some kind of identity given in -oo-'s proof in one of your other postings." } { "Tag": [], "Problem": "In triangle $ABC$, angle $B$ is a right angle. Points $P$ and $Q$ are on side $BC$, with $P$ between $B$ and $Q$, so that segments $AP$ and $AQ$ trisect angle $BAC$. The measure of angle $APC$ is $114^\\circ$. What is the number of degrees in the measure of angle $ACB$?", "Solution_1": "[hide=\"Solution\"]The number of degrees in angle APB is 180-114=66. The number of degrees in angle BAP is 90-66=24. The number of degrees in angle BAC is 3*24=72. The number of degrees in angle ACB is 90-72=18.[/hide]", "Solution_2": "[hide=\"My Answer.\"]We are given $\\angle BAP=\\angle PAQ=\\angle QAC$. If we find one of these angles, we can find $\\angle BAC.$ We know that $\\angle APC=114$, so $\\angle APB=66$, because of a straight angle. Now, we know $\\angle BAP=24$, because the angles of a triangle add to 180. Now we know $\\angle BAQ=3\\cdot 24=72$. Therefore, $\\angle ACB=180-(72+90)=180-162=18$.[/hide]" } { "Tag": [ "geometry", "Pythagorean Theorem" ], "Problem": "Let ABC be a triangle with hypotenuse 4. On each side of triangle ABC an equilateral triangle is constructed outward (\"as shown in the figure\" but not really). Find the sum of the the areas of the equilateral triangles.", "Solution_1": "my algebraically devoid answer: \r\n[hide] I just thought of the pythagorean theorem and extended it. The two triangles on the unknown sides should equal the area of the triangle on the known side. So, I got $(1/2*4*4*2) = 16$. I'm not sure if that is correct though... oops I meant *2...[/hide]", "Solution_2": "[hide]\nIf we denote sides as $a,b,c=4$ then areas of constructed triangles are: \n$a^2{\\sqrt 3\\over 4},b^2{\\sqrt 3\\over 4},c^2{\\sqrt 3\\over 4}$\n\nSo we have sum equal to ${\\sqrt 3\\over 4}(a^2+b^2+c^2)={\\sqrt 3\\over 4}(c^2+c^2)=8\\sqrt 3$\n[/hide]", "Solution_3": "How did you get the $\\sqrt(3)$ ? I would love to know what I did wrong. Please explain :)", "Solution_4": "If we have equilateral triangle $\\Delta ABC$ with side length $x$, then area of the triangle is $\\frac{x^2\\sqrt 3}{4}$.\r\n\r\nProof:\r\nDenote with $D$ intersection of $AB$ and perpendicular from $C$.\r\nWe know that area is equal to $\\frac{AB\\cdot CD}{2}$ and since we have $x=AB$ it suffices to show that $CD={x\\sqrt 3\\over 2}$.\r\n\r\nUsing pythagorean theorem we see that $CD^2=x^2-(\\frac{x}{2})^2=x^2-\\frac{x^2}{4}=\\frac{3x^2}{4}$. \r\nThus, $CD=\\frac{x\\sqrt 3}{2}$", "Solution_5": "Oh, I feel so stupid, that's right!" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "The surface area of a particular cube is 600 square inches. When the edges of the cube are doubled in length, what is the volume of the new cube, in cubic inches?", "Solution_1": "[hide] We have $ 6e^2\\equal{}600 \\implies e\\equal{}10$. Twice this is $ 20$, so $ 20^3\\equal{}\\boxed{8000}$.[/hide]" } { "Tag": [], "Problem": "Most people think that [tex]0^2 = 0 \\times 0 = 0.[/tex] Zero squared is also\r\n[tex]\\displaystyle{0^{5-3} = \\frac{0^5}{0^3}}[/tex] and if zero to a power is zero, then the expression equals 0/0, and division by zero is undefined! So what does 0 squared equal?", "Solution_1": "You can't use laws of exponents if the base is 0.", "Solution_2": "Oh.", "Solution_3": "To elaborate a little -- while it is true that 02 = 05 - 3, it is [i]not[/i] true that this also equals 05/03. Any time you arrive by arithmetic procedure at an expression with a denominator of 0, it means you've done something wrong." } { "Tag": [ "geometry", "combinatorics proposed", "combinatorics" ], "Problem": "On a plane, call the band with width $d$ be the set of all points which are away the distance of less than or equal to $\\frac{d}{2}$ from a line.Given four points $A,\\ B,\\ C,\\ D$ on a plane.If you take three points among them, there exists the band with width $1$ containing them. Prove that there exist the band with width $\\sqrt{2}$ containing all four points .", "Solution_1": "Ok,I'll try to give it a shot.\r\nIt's easy to see that if a triangle with three perpendicular with length >1,then there's no band with width 1 containing them.\r\nWe suppose that triangle ABC has the max area among the four triangles.\r\nWe consider that Hb=1 ( Why? the reason is obvious ... )\r\nE is belong to the line AC and BE is perpendicular to AC\r\nif AE<=1 or CE<=1 then AB<=$\\sqrt{2}$ or CB<=$\\sqrt{2}$\r\nit's easy to see that there exist a band with width 1 containg them\r\nso we consider that AE>1 or CE>1 now we just consider BD>$\\sqrt{2}$ ( Why? Quite simply ... ) then either triangle BAD or triangle BCD has three perpendicular with length >1(Here is a little bit complicated),which means that one of them cannot be covered by a band with width 1. so we done.", "Solution_2": "A solution is given on page 266 of the March 2008 issue of the American Mathematical Monthly.", "Solution_3": "where the magazine can be found?" } { "Tag": [ "linear algebra", "matrix", "vector", "quadratics", "algebra", "linear algebra unsolved" ], "Problem": "Can someone help me prove this:\r\n\r\nGiven a complex mxm matrix A and complex m-vector b, prove that the complex n-vector x is the solution of the least squares problem if and only if A*(Ax-b)=0.\r\n\r\nThanks in advance.", "Solution_1": "e.g., write $ f(\\bar{x})\\equal{}\\parallel{}A\\bar{x}\\minus{}\\bar{b}\\parallel{}^2\\equal{}\\parallel{}b\\parallel{}^2\\plus{}\\bar{x}^TA^TA\\bar{x}\\minus{}2\\bar{x}^TA^T\\bar{b}$ as a quadratic equation, then determine the gradient as $ \\nabla f(\\bar{x})\\equal{}2[A^TA\\bar{x}\\minus{}A^T\\bar{b}]$ and the Hesse matrix as $ H_f(\\bar{x})\\equal{}2A^TA$. Since the Hesse matrix is positive semi-definite, we have a minimum at $ \\nabla f(\\bar{x})\\equal{}\\bar{0}\\Longleftrightarrow A^T(A\\bar{x}\\minus{}\\bar{b})\\equal{}\\bar{0}$", "Solution_2": "Note: JimmyT did say his matrices were complex valued. And that asterisk in what he wrote as \"A*(Ax-b)\" isn't a multiplication symbol, it's the Hermitian conjugate.\r\n\r\nIt looks like hjbrasch answered the question is a real context. Some adjustments in language are needed." } { "Tag": [ "geometry", "trigonometry", "geometry proposed" ], "Problem": "Let $ABCDE$ be a convex pentagon, such that $AB=BC$, $CD=DE$, $\\angle B+\\angle D=180^{\\circ}$, and it's area is $\\sqrt2$. \r\n\r\na) If $\\angle B=135^{\\circ}$, find the length of $[BD]$.\r\n\r\nb) Find the minimum of the length of $[BD]$.", "Solution_1": "Put $ \\angle D \\equal{}\\alpha \\Rightarrow \\angle B\\equal{} \\pi \\minus{} \\alpha$, $ AB\\equal{}BC\\equal{}x$,$ CD\\equal{}DE\\equal{}y$ and $ \\angle ACE \\equal{}\\theta$.\r\n\r\nWe have: $ [ABCDE ]\\equal{} [ABC]\\plus{}[ACE]\\plus{}[CDE]$. \r\n\r\n$ [ABC]\\equal{}\\frac{1}{2}\\cdot x^2\\cdot\\sin(\\pi \\minus{}\\alpha)$ and $ [CDE]\\equal{}\\frac{1}{2}\\cdot y^2 \\cdot \\sin\\alpha$.\r\n$ AC\\equal{} 2x \\cos\\frac{\\alpha}{2}$ and $ CE\\equal{}2y\\sin\\frac{\\alpha}{2}$. \r\nTherefore $ [ACE]\\equal{}\\frac{1}{2}\\sin\\theta\\cdot(2x\\cos\\frac{\\alpha}{2}) \\cdot (2y\\sin\\frac{\\alpha}{2}) \\equal{} xy\\sin \\theta \\cdot\\sin\\alpha$.\r\nHence $ [ABCDE]\\equal{}\\frac{\\sin\\alpha}{2}\\cdot(x^2\\plus{}y^2\\plus{}2xy\\sin\\theta)$.\r\nBut $ \\angle BCD \\equal{} \\frac{\\pi}{2}\\plus{}\\theta$, so : $ [ABCDE]\\equal{} \\frac{\\sin\\alpha}{2}\\cdot(x^2\\plus{}y^2\\minus{}2xy\\cos\\angle BCD)\\equal{}\\frac{BD^2}{2}\\cdot \\sin \\alpha$. \r\n\r\nSo we have the equation: $ BD^2\\equal{}\\frac{2\\sqrt{2}}{\\sin \\alpha}$\r\n\r\na) if $ \\angle B \\equal{} \\frac{3\\pi}{4} \\Rightarrow \\alpha\\equal{}\\frac{\\pi}{4}\\Rightarrow BD\\equal{}2$.\r\nb) $ BD$ is minimal if $ \\frac{2\\sqrt{2}}{\\sin \\alpha}$ is minimal, i.e. if $ \\alpha\\equal{}\\frac{\\pi}{2}$. In this case $ BD\\equal{}\\sqrt{2\\sqrt{2}}$." } { "Tag": [], "Problem": "a train traveling from aytown to beetown meets with an accident after 1 hour. the train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at beetown 2 hours late. if the train had covered 80 miles more before the accident, it would have been just one hour late. what is the usual rate of the train?", "Solution_1": "[quote=\"iamagenius\"]a train traveling from aytown to beetown meets with an accident after 1 hour. the train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at beetown 2 hours late. if the train had covered 80 miles more before the accident, it would have been just one hour late. what is the usual rate of the train?[/quote]\r\n\r\nThere are two ways, I think\r\n\r\n[hide=\"slower way\"]Let's say it traveled total $d$ miles and that its normal speed is $a$ mph.\n\nWe can make the time equations:\n\n$\\frac{d}{a}+2=1+0.5+\\frac{d-a}{\\frac{4}{5}a}$\n\n$\\frac{d}{a}+1=1+\\frac{80}{a}+0.5+\\frac{d-80-a}{\\frac{4}{5}a}$\n\nSolve and get $a=20$ and $d=140$. So the usual rate is $\\boxed{20}$ mph.[/hide]\n\n[hide=\"faster way\"]Let's say the usual speed is $a$ mph. If it travels at four-fifths speed, it takes five-fourth of usual time, which is one-fourth more than usual.\n\nSo travelling $80$ miles in 1/4 more time than what it used to made a difference of $1$ hour, which is equivalent to travelling $a$ miles. So\n\n$\\frac{80}{4}=a$\n\nAnd $a=\\boxed{20}$(mph).[/hide]", "Solution_2": "Let the normal speed be x miles/hr\r\n normal time duration be y hrs\r\ntotal distance= xy miles\r\nin the unfortunate day everything changed but the total distance remained constant.\r\nIn the 1st 1hr it travelled x miles\r\nIn the next 30 mins it travelled 0 miles\r\nin the 1st 1.5 hrs it travelled x miles\r\ntotal time taken that day=y+2\r\nafter 1st 1.5 hrs time left=(y+2)-1.5=y+0.5, which it travelled with speed 4/5x miles/hr\r\ndistance covered=4/5x(y+0.5)\r\ntotal distance= x+4/5x(y+0.5)\r\nx+4/5x(y+0.5)=xy-------------------(1)\r\nsimilarly we get 2nd equation\r\n(x+80)+4/5x(y-0.5)=xy ---------(2)\r\nsolve to get the answer." } { "Tag": [ "LaTeX" ], "Problem": "Three circles of radius r are inscribed in a circle\r\nof radius 2 + (radical 3). Find r. (excuse my lack of LaTeX)\r\n\r\nWhat approaches would you recommend?", "Solution_1": "If those three circles are mutually externally tangent, then all you have to do is connect their centers to form an equilateral triangle whose centroid happens to be the center of the larger circle.", "Solution_2": "It doesn't say they're mutually tangent so....\r\n\r\n1) Let's assume that they're tangent how do we solve? I've got a diagram drawn but I don't know what to do next.\r\n\r\n2) Assume that they're not tangent then what?", "Solution_3": "2) It's impossible to solve without more information", "Solution_4": "Since it is inscribed, it's mutually tangent.\r\n\r\nTry connecting the centers of the circles and drawing the radii of the bigger circle.", "Solution_5": "Well, I've tried messing around with different diagrams, etc but I'm not getting anywhere. So feel free to post a solution :P", "Solution_6": "Did you know that $ 2/3$ times the altitude of the equilateral triangle sunehra told you to draw is equal to the big circle's radius? :wink:", "Solution_7": "No I didn't know that :) How can we prove that?", "Solution_8": "Draw all three medians and use $ 30\\minus{}60\\minus{}90$ triangles. The rest will follow through. :)", "Solution_9": "What answer did you get?\r\n\r\nI got r = $ 2\\sqrt{3} \\plus{} 3$ $ /4$ \r\n\r\n\r\nBut the answer is r = $ /sqrt{3}$ (why didn't that come out right?)\r\n\r\n\r\nSo how did you solve it? I would post my solution, but I suck at LeTax. \r\nBasically, I split up the equilateral triangle into 5 congruent 30-60-90 triangles by drawing the medians and dropping down a perpendicular line.", "Solution_10": "My above hint was incorrect (I was thinking of something else), but by using $ 30\\minus{}60\\minus{}90$ angles, the answer is extremely easy. I got $ \\sqrt{3}$ also. I'll post a solution a few minutes later.", "Solution_11": "[quote=\"Evolved\"]But the answer is r = $ / sqrt3$ (why didn't that come out right?)[/quote]\r\n\r\nthe code is [code]$\\sqrt{3}$[/code]", "Solution_12": "[geogebra]59302c7085fd4e0be10fc8543343ac2439b17acb[/geogebra] Now, we have the equation $ \\frac {2r\\sqrt {3} \\plus{} 3r}{3} \\equal{} 2 \\plus{} \\sqrt {3}\\Rightarrow{r(2\\sqrt {3} \\plus{} 3)} \\equal{} 6 \\plus{} 3\\sqrt {3}$. We get $ r \\equal{} \\frac {6 \\plus{} 3\\sqrt {3}}{2\\sqrt {3} \\plus{} 3}$. This ultimately simplifies to $ \\sqrt {3}$.", "Solution_13": "Nice diagram :) My diagram was pretty much the same expect I mistakenly labeled r as r$ \\sqrt{3}$", "Solution_14": "[quote=\"Evolved\"]What answer did you get?\n\nI got r = $ 2\\sqrt {3} \\plus{} 3$ $ /4$ \n\n\nBut the answer is r = $ /sqrt{3}$ (why didn't that come out right?)\n\n\nSo how did you solve it? I would post my solution, but I suck at LeTax. \nBasically, I split up the equilateral triangle into 5 congruent 30-60-90 triangles by drawing the medians and dropping down a perpendicular line.[/quote]\r\n\r\nIt's supposed to be : $ \\sqrt{3}$.\r\nNot /, it's \\." } { "Tag": [ "vector" ], "Problem": "this is kinda random, but i was wondering... how do you prove Archimedes' Lemmas? i've tried, but they seem to be nigh impossible! :huh: \r\nyou can find them with diagrams here:\r\n\r\nhttp://www.gogeometry.com/ArchBooLem00.htm\r\n\r\nhow do you prove this stuff? man, i feel like an idiot,\r\ncuz so old guy who lived more than 2,000 years ago is beating the crap outta my brain... T_T", "Solution_1": "Think of vectors and similar triangles(I don't have enough time to prove this right now)." } { "Tag": [ "geometry", "circumcircle", "trapezoid", "trigonometry", "trig identities", "Law of Sines", "geometry solved" ], "Problem": "My English is not well.So I translated the problem on net.Maybe there are some mistakes in it.\r\n\r\n\r\n\r\nThere's a point O in equilateral triangle ABC.Angle AOB =110 degrees. Angle BOC =135 degrees.\r\nQuestions:\r\n(1)Can it form a triangle to regard OA , OB , OC as? If can,ask not appearing this triangle reading of every mao; If cannot, please give reasons. \r\n(2)If horn size of AOB keep intact, act as horn when how many degrees BOC mean , is the triangle taking OA , OB , OC as side a right triangle?", "Solution_1": "Just rewording the post:\r\n\r\n$O$ is a point in an equilateral triangle $ABC$ such that $\\angle AOB = 110^\\circ$, $\\angle BOC = 135^\\circ$.\r\n\r\n1) Can $OA$, $OB$, $OC$ be the sides of a triangle? If so, what are the angles of the triangle formed? If not, explain why.\r\n\r\n2) If we keep the size of $\\angle AOB$ as $110^\\circ$ and allow the size of $\\angle BOC$ to vary, what must the size of $\\angle BOC$ be when the triangle with side lengths $OA$, $OB$, $OC$ is a right-angled triangle?", "Solution_2": "Pompeiu's Theorem states that when $ABC$ is an equilateral triangle and $O$ is a point in the plane, $OA,OB,OC$ are the sides of a triangle, which is degenerate only when $O$ lies on the circumcircle of $ABC$, so the answer is yes.\r\n\r\nIf $O$ lies inside the triangle $ABC$, then the angles of the triangle formed by $OA,OB,OC$ are $\\angle BOC-60^{\\circ},\\angle COA-60^{\\circ},\\angle AOB-60^{\\circ}$. This is found by considering points $P,S\\in BC,\\ N,R\\in CA,\\ Q.M\\in AB$ s.t. $O\\in MN,PQ,RS$ and $MN\\|BC,PQ\\|CA,RS\\|AB$. We now use the isosceles trapezoids $OMBP,OPCR,ORAM$ to construct our triangle: $OA=RM,OB=MP,OC=PR$, so the triangle we want is $RMP$. It's easy to compute its angles in terms of $\\angle BOC,\\angle COA,\\angle AOB$ now.\r\n\r\nTry to show that the locus of $O$ s.t. $OA^2=OB^2+OC^2$ is an arc subtending an angle of $150^{\\circ}$ and passing through $B,C$.", "Solution_3": "Just for the sake of completeness (it's clear that similar problems will regularly occur on ML) let me state a more general fact:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let O be a point inside an equilateral triangle ABC. Then, there exists a triangle XYZ with the sidelengths YZ = OA, ZX = OB and XY = OC; moreover, the angles of such a triangle XYZ are\n\n< ZXY = < BOC - 60\u00b0;\n< XYZ = < COA - 60\u00b0;\n< YZX = < AOB - 60\u00b0.\n\nParticularly, this triangle XYZ is right-angled if and only if one of the angles < BOC, < COA and < AOB equals 150\u00b0.[/color]\r\n\r\n[i]Proof.[/i] Let D, E, F be the orthogonal projections of the point O on the sides BC, CA, AB of triangle ABC. Then, since < OEA = 90\u00b0 and < OFA = 90\u00b0, the points E and F lie on the circle with diameter OA. In other words, the circumcircle of triangle AEF has diameter OA. But by the extended law of sines, a side of a triangle equals the diameter of the circumcircle, multiplied with the sine of the opposite angle; thus, in triangle AEF, we have $EF=OA\\cdot\\sin\\measuredangle EAF$. Since < EAF = < CAB, and < CAB = 60\u00b0 (since triangle ABC is equilateral), this becomes $EF=OA\\cdot\\sin 60^{\\circ}$. Similarly, $FD=OB\\cdot\\sin 60^{\\circ}$ and $DE=OC\\cdot\\sin 60^{\\circ}$. Hence, EF : FD : DE = OA : OB : OC. Thus, there exists a triangle XYZ with the sidelengths YZ = OA, ZX = OB and XY = OC, and any such triangle XYZ must be similar to the triangle DEF. Hence, < XYZ = < DEF.\r\n\r\nNow, since the points E and F lie on the circle with diameter OA, the quadrilateral AEOF is cyclic, so that < FEO = < FAO. In other words, < FEO = < BAO. Similarly, < DEO = < BCO. Thus,\r\n\r\n < DEF = < FEO + < DEO = < BAO + < BCO\r\n = (180\u00b0 - < ABO - < AOB) + (180\u00b0 - < CBO - < BOC)\r\n\r\nby the sum of angles in triangles AOB and BOC. After a rearrangement of terms, this becomes\r\n\r\n < DEF = (360\u00b0 - < AOB - < BOC) - (< ABO + < CBO) = < COA - < ABC = < COA - 60\u00b0\r\n\r\n(since < ABC = 60\u00b0, as triangle ABC is equilateral). Hence, < XYZ = < DEF = < COA - 60\u00b0. Similarly, < YZX = < AOB - 60\u00b0 and < ZXY = < BOC - 60\u00b0.\r\n\r\nFinally, the triangle XYZ is right-angled if and only if one of its angles < ZXY = < BOC - 60\u00b0, < XYZ = < COA - 60\u00b0, < YZX = < AOB - 60\u00b0 equals 90\u00b0. This holds if and only if one of the angles < BOC, < COA, < AOB equals 60\u00b0 + 90\u00b0 = 150\u00b0.\r\n\r\nThus, Theorem 1 is completely proven.\r\n\r\nOf course, Theorem 1 easily solves the problem.\r\n\r\n Darij" } { "Tag": [ "function", "integration", "college contests" ], "Problem": "Let $f(x)$ be a continuous and decrease function on interval [0, b] and a$\\in$[0, b]. Prove that:\r\n$b.\\int\\limits_{0}^{a}f(x)dx\\geq a.\\int\\limits_{0}^{b}f(x)dx$", "Solution_1": "We have that $ \\int_{0}^{a}f(x)dx=\\frac{a}{b}\\int_{0}^{b}f\\left(\\frac{a}{b}x\\right)dx$ so it remains to prove that $a\\int_{0}^{b}f\\left(\\frac{a}{b}x\\right)dx \\geq a\\int_{0}^{b}f(x)dx$ which is true because $f(\\frac{a}{b}x)\\geq f(x)$." } { "Tag": [ "calculus", "integration", "algebra", "partial fractions", "calculus computations" ], "Problem": "Now this is tricky to write...bear with me.\r\n\r\nIntegrate (2x^2 - 9x - 6)/(x(x^2 - x - 6)) with respect to x between the lower limit x = 4 and upper limit x = 6. Given that this definite integral = Ln(m/n) determine the values for the inegers m and n.", "Solution_1": "[quote=\"sludgethrower\"]\n\\[\\int_{4}^{6}\\frac{2x^{2}-9x-6}{x(x^{2}-x-6)}\\ dx=\\ln \\frac{m}{n}\\]\nDetermine the values for the integers $m$ and $n$.\n[/quote]\r\n[hide=\"hint\"]\n$x^{2}-x-6=(x-3)(x+2)$ so partial fractions can easily be used\n[/hide]", "Solution_2": "I know the answer...so i don't need hints...", "Solution_3": "If you don't need hints why have you posted that problem?" } { "Tag": [ "inequalities", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let we are given the sequence of nonnegative real numbers $ a_n$ and assume the following inequality holds for all $ n\\in N$:\r\n$ a_n\\geq\\frac{n^2a_{n\\plus{}1}\\minus{}1}{n^2\\minus{}n}$.\r\nShow that $ a_n$ is converges.", "Solution_1": "[quote=\"Ibrogimov\"]Let we are given the sequence of nonnegative real numbers $ a_n$ and assume the following inequality holds for all $ n\\in N$:\n$ a_n\\geq\\frac {n^2a_{n \\plus{} 1} \\minus{} 1}{n^2 \\minus{} n}$.\nShow that $ a_n$ is converges.[/quote]\r\n$ (n \\minus{} 1)a_n\\geq na_{n \\plus{} 1} \\minus{} \\frac {1}{n}$\r\nLet $ x_n \\equal{} (n \\minus{} 1)a_n$\r\nso \r\n$ x_n\\geq x_{n \\plus{} 1} \\minus{} \\frac {1}{n}$\r\n$ x_{n \\plus{} 1} \\minus{} x_n\\leq \\frac {1}{n}$\r\nSo \r\n$ x_{n \\plus{} 1} \\minus{} x_1\\leq \\sum_{i \\equal{} 1}^n\\frac {1}{i}$\r\n$ \\Rightarrow x_{n \\plus{} 1}\\leq x_1 \\plus{} \\sum_{i \\equal{} 1}^n\\frac {1}{i}$\r\nBut $ \\lim_{n\\to\\infty}\\frac {1}{n}.({\\sum_{i \\equal{} 1}^n\\frac {1}{i}} \\plus{} x_1) \\equal{} 0$\r\nBut $ x_n\\geq 0$ \r\nSo $ \\lim_{n\\to\\infty}x_n \\equal{} 0$", "Solution_2": "TTsphn, a lot of thanks for your help!" } { "Tag": [ "linear algebra", "matrix", "trigonometry", "induction", "algebra", "polynomial" ], "Problem": "Q1 show that DETERMINANT\r\n\r\ntan (A+P) , tan (B+P) , tan( C+P )\r\ntan (A+Q) , tan ( B+Q) , tan( C+Q ) =O\r\ntan ( A+R ) , tan ( B+R) , tan( C+R )\r\n\r\nwhere A+B+C+P+Q+R=0", "Solution_1": "Pretty version:\r\n[quote=\"ayush_2008\"]Choose $ A,B,C,P,Q,R$ such that $ A\\plus{}B\\plus{}C\\plus{}P\\plus{}Q\\plus{}R\\equal{}0$. Show that\n$ \\det\\begin{bmatrix}\\tan(A\\plus{}P)&\\tan(B\\plus{}P)&\\tan(C\\plus{}P)\\\\ \\tan(A\\plus{}Q)&\\tan(B\\plus{}Q)&\\tan(C\\plus{}Q)\\\\ \\tan(A\\plus{}R)&\\tan(B\\plus{}R)&\\tan(C\\plus{}R)\\end{bmatrix}\\equal{}0$[/quote]\r\n\r\nI would also add that none of the pairwise sums are $ \\frac{\\pi}{2}$ mod $ \\pi$. You get an $ \\infty\\minus{}\\infty$ indeterminate form in that case, and you don't get any linear dependence relation unless you allow infinitesimal coefficients.", "Solution_2": "[hide=\"Hint\"]\nHint: If three angles $ x$, $ y$, $ z$ are such that $ x \\plus{} y \\plus{} z \\equal{} 0$, then $ \\tan x \\plus{} \\tan y \\plus{} \\tan z \\equal{} \\tan x\\tan y\\tan z$.\nAnd\n$ \\det\\left(\\begin{array}{ccc}a_1 & a_2 & a_3 \\\\\nb_1 & b_2 & b_3 \\\\\nc_1 & c_2 & c_3\\end{array}\\right) \\equal{} a_1b_2c_3 \\plus{} a_2b_3c_1 \\plus{} a_3b_1c_2 \\minus{} a_1b_3c_2 \\minus{} a_2b_1c_3 \\minus{} a_3b_2c_1$.[/hide]\r\n\r\nWhat is way more interesting is to find a simpler proof for the canonical generalization:\r\n\r\n[color=blue][b]Problem.[/b] Let $ n\\in\\mathbb{N}$ be odd, and let $ A_1$, $ A_2$, ..., $ A_n$, $ X_1$, $ X_2$, ..., $ X_n$ be $ 2n$ angles such that $ \\sum_{k \\equal{} 1}^nA_k \\plus{} \\sum_{k \\equal{} 1}^nX_k \\equal{} 0$. Define a matrix $ M\\in\\mathbb{R}^{n\\times n}$ by\n\n$ M \\equal{} \\left(\\tan\\left(A_i \\plus{} X_j\\right)\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$.\n\nShow that $ \\det M \\equal{} 0$.[/color]\r\n\r\n darij", "Solution_3": "Better: drop the odd $ n$ condition, and make it $ \\sum A_k\\plus{}\\sum X_k\\equiv (n\\minus{}1)\\frac{\\pi}{2}\\mod \\pi.$", "Solution_4": "you try with $ n \\equal{} 5$\r\n If $ x_1 \\plus{} x_2 \\plus{} .. \\plus{} x_5 \\equal{} 0$ then $ tan((x_1 \\plus{} x_2) \\plus{} (x_3 \\plus{} x_4) \\plus{} x_5) \\equal{} \\frac {tan(x_1 \\plus{} x_2) \\plus{} tan(x_3 \\plus{} x_4) \\plus{} tan(x_5) \\minus{} tan(x_1 \\plus{} x_2)tan(x_3 \\plus{} x_4)tan(x_5)}{P(x_1,..,x_5)}$ \r\nthen $ 0 \\equal{} tan(x_1 \\plus{} x_2) \\plus{} tan(x_3 \\plus{} x_4) \\plus{} tan(x_5) \\minus{} tan(x_1 \\plus{} x_2)tan(x_3 \\plus{} x_4)tan(x_5)$ .... \r\n $ \\sum_{k \\equal{} 1}^{5}{tan(x_k)} \\plus{} \\prod_{k \\equal{} 1}^{5}tan(x_k) \\minus{} \\sum{tan(x_i)tan(x_j)tan(x_k)} \\equal{} 0$\r\n That $ \\sum_{k \\equal{} 1}^{5}{tan(x_k)}\\neq \\prod_{k \\equal{} 1}^{5}tan(x_k)$ it isn't easy to answer that you are correct . .. I hope that you can give more information with $ n \\equal{} 5$ and example about it . Thanks !", "Solution_5": "That's why I was asking for a simpler proof.\n\nMine goes, roughly, as follows (sorry for using other notations than you):\n\nFor any angles $ \\phi_1$, $ \\phi_2$, ..., $ \\phi_n$ with tangents $ x_1$, $ x_2$, ..., $ x_n$, a straightforward induction shows that\n\n$ \\tan\\left(\\phi_1 + \\phi_2 + ... + \\phi_n\\right) = \\frac {\\sum_{i = 0}^{\\infty} \\left( - 1\\right)^i e_{2i + 1}\\left(x_1,x_2,...,x_n\\right) } {\\sum_{i = 0}^{\\infty} \\left( - 1\\right)^i e_{2i}\\left(x_1,x_2,...,x_n\\right) }$,\n\nwhere $ e_i$ denotes the $ i$-th elementary symmetric polynomial (note that the sums are not really infinite, because $ e_m\\left(x_1,x_2,...,x_n\\right) = 0$ for all $ m > n$).\n\nThus, if $ \\phi_1 + \\phi_2 + ... + \\phi_n\\equiv \\left(n - 1\\right)\\frac {\\pi}{2}\\mod\\pi$, then\n\n$ \\sum_{i = 0}^{\\infty} \\left( - 1\\right)^i e_{2i + 1}\\left(x_1,x_2,...,x_n\\right) = 0$ if $ n$ is odd;\n$ \\sum_{i = 0}^{\\infty} \\left( - 1\\right)^i e_{2i}\\left(x_1,x_2,...,x_n\\right) = 0$ if $ n$ is even.\n\nThese two results can be unified: If $ \\phi_1 + \\phi_2 + ... + \\phi_n \\equiv \\left(n - 1\\right)\\frac {\\pi}{2}\\mod\\pi$, then\n\n$ \\sum_{j = 0}^{\\infty} \\left( - 1\\right)^j e_{n - 2j}\\left(x_1,x_2,...,x_n\\right) = 0$,\n\nwhere $ e_k$ is defined to be the $ 0$ polynomial for all $ k < 0$.\n\nNow, this yields\n\n$ e_n\\left(x_1,x_2,...,x_n\\right) = - \\sum_{j = 1}^{\\infty} \\left( - 1\\right)^j e_{n - 2j}\\left(x_1,x_2,...,x_n\\right)$\n\nwhenever $ \\phi_1 + \\phi_2 + ... + \\phi_n\\equiv \\left(n - 1\\right)\\frac {\\pi}{2}\\mod\\pi$.\n\nAs a consequence, if $ \\sum_{k = 1}^nA_k + \\sum_{k = 1}^nX_k \\equiv \\left(n - 1\\right)\\frac {\\pi}{2}\\mod\\pi$ (I am solving Jmerry's generalization here), then\n\n$ \\det M = \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}\\tan\\left(A_1 + X_{\\sigma\\left(1\\right)}\\right)\\tan\\left(A_2 + X_{\\sigma\\left(2\\right)}\\right)...\\tan\\left(A_n + X_{\\sigma\\left(n\\right)}\\right)$\n$ = \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}e_n\\left(\\tan\\left(A_1 + X_{\\sigma\\left(1\\right)}\\right),\\tan\\left(A_2 + X_{\\sigma\\left(2\\right)}\\right),...,\\tan\\left(A_n + X_{\\sigma\\left(n\\right)}\\right)\\right)$\n$ = \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}\\left( - \\sum_{j = 1}^{\\infty} \\left( - 1\\right)^j e_{n - 2j}\\left(\\tan\\left(A_1 + X_{\\sigma\\left(1\\right)}\\right),\\tan\\left(A_2 + X_{\\sigma\\left(2\\right)}\\right),...,\\tan\\left(A_n + X_{\\sigma\\left(n\\right)}\\right)\\right)\\right)$\n(because $ \\left(A_1 + X_{\\sigma\\left(1\\right)}\\right) + \\left(A_2 + X_{\\sigma\\left(2\\right)}\\right) + ... + \\left(A_n + X_{\\sigma\\left(n\\right)}\\right)\\equiv \\left(n - 1\\right)\\frac {\\pi}{2}\\mod\\pi$)\n$ = - \\sum_{j = 1}^{\\infty}\\left( - 1\\right)^j \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}e_{n - 2j}\\left(\\tan\\left(A_1 + X_{\\sigma\\left(1\\right)}\\right),\\tan\\left(A_2 + X_{\\sigma\\left(2\\right)}\\right),...,\\tan\\left(A_n + X_{\\sigma\\left(n\\right)}\\right)\\right)$\n$ = 0$,\n\nbecause for any $ j > 1$, the term $ \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}e_{n - 2j}\\left(\\tan\\left(A_1 + X_{\\sigma\\left(1\\right)}\\right),\\tan\\left(A_2 + X_{\\sigma\\left(2\\right)}\\right),...,\\tan\\left(A_n + X_{\\sigma\\left(n\\right)}\\right)\\right)$ equals $ 0$, which in turn follows from the more general (and easy) fact that whenever $ U = \\left(u_{i,j}\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$ is an $ n\\times n$ matrix, and $ P$ is a polynomial of $ n$ variables with total degree $ < n - 1$, then\n\n$ \\sum_{\\sigma\\in S_n}\\left( - 1\\right)^{\\sigma}P\\left(u_{1,\\sigma\\left(1\\right)},u_{2,\\sigma\\left(2\\right)},...,u_{n,\\sigma\\left(n\\right)}\\right) = 0$.\n\n Darij", "Solution_6": "A proof isn't simple . I will read many times to understand it and I hope that I can make more ... :read: \r\n To add interesting If you are interested a some $ \\sin,\\cos,ch,sh$ (a short proof) .....\r\n[color=blue][b]Problem.[/b] Let $ n\\in\\mathbb{N}$ , and let $ A_1$, $ A_2$, ..., $ A_n$, $ X_1$, $ X_2$, ..., $ X_n$ be $ 2n$ complex number : \n$ M\\in\\mathbb{C}^{n\\times n}$, $ N\\in\\mathbb{C}^{n\\times n}$, $ P\\in\\mathbb{C}^{n\\times n}$ , $ Q\\in\\mathbb{C}^{n\\times n}$ by\n $ M \\equal{} \\left(\\sin^m\\left(A_i \\plus{} X_j\\right)\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$,$ N \\equal{} \\left(\\cos^m\\left(A_i \\plus{} X_j\\right)\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$,$ P \\equal{} \\left(ch^m\\left(A_i \\plus{} X_j\\right)\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$,$ Q \\equal{} \\left(Sh^m\\left(A_i \\plus{} X_j\\right)\\right)_{1\\leq i\\leq n,\\ 1\\leq j\\leq n}$.\n\nShow that $ \\det(M) \\equal{} \\det(N) \\equal{} \\det(P) \\equal{} \\det(Q) \\equal{} 0$ if $ n > m \\plus{} 1$.[/color]\r\n ........ :play_ball: ...... :play_ball:", "Solution_7": "thanks to all masters" } { "Tag": [ "trigonometry" ], "Problem": "1. find $ \\tan x$ if $ \\sin x \\plus{}\\cos x\\equal{}\\frac{1}{5}$ and $ \\frac{\\pi}{2}\\le x\\le\\pi$.\r\n2. A parabolic dish is 10 deep and 100 in diameter. How far away from its vertex is its focus?", "Solution_1": "[hide=\"1?\"]Wouldn't you just divide the whole equation by $ \\cos{x}$?\n\nThen, $ \\tan x\\equal{}\\frac{1}{5}\\sec x\\minus{}1$\n\nOr did you want a numerical answer?[/hide]", "Solution_2": "[hide=\"Number 1\"]$ \\sin x \\plus{} \\cos x \\equal{} \\sqrt{2} \\sin\\left(x \\plus{} \\frac{\\pi}{4}\\right)$, and so we have\n\n$ x \\plus{} \\frac{\\pi}{4} \\equal{} \\pi \\minus{} \\arcsin\\left(\\frac{\\sqrt{2}}{10}\\right)$ => $ x \\equal{} \\frac{3 \\pi}{4} \\minus{} \\arcsin\\left(\\frac{\\sqrt{2}}{10}\\right)$[/hide]", "Solution_3": "[hide=\"#1\"]$ \\sin x\\plus{}\\cos x\\equal{}{1\\over 5}$\n\nSquaring, we get\n\n$ 1\\plus{}2\\sin x\\cos x\\equal{}{1\\over 25}\\implies \\sin x\\cos x\\equal{}\\minus{}{12\\over 25}$\n\nHence $ \\sin x, \\cos x$ are the solutions to the equation\n\n$ t^2\\minus{}{1\\over 5}t\\minus{}{12\\over 25}\\equal{}0\\iff t\\in\\left\\{{4\\over 5},\\minus{}{3\\over 5}\\right\\}$\n\nBut as $ x$ lies in the second quadrant, its sine is positive and the cosine negative, hence $ \\sin x\\equal{}{4\\over 5}\\land\\cos x\\equal{}\\minus{}{3\\over 5}\\implies\\tan x\\equal{}\\minus{}{4\\over 3}$[/hide]", "Solution_4": "[hide=\"maybe Im wrong for number 1\"]Square and simplify to get $ sin2x = - \\frac {24}{25}$. So, you know it is a 7-24-25 triangle with $ 2x$ being in quadrants III or IV. Then, $ cos2x = \\frac {7}{25}$.\n$ \\cos^2x - sin^2x = \\frac {7}{25}$\n$ \\cos^2x + sin^2x = 1$.Add and solve for $ cosx$, and it is $ \\frac {4}{5}$. It's a 3-4-5 triangle, and $ \\tanh = \\frac {3}{4}$.[/hide]\r\n\r\nEdit:Didn't read the $ \\frac{\\pi}{2}$ part." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "I just wanted to know how all of you practice for CMO. For example, I do the past contests, and I try to do some of the easier USAMOs, but are there any other olympiads that have comparable problem types? Also, what topics are usually more heavily covered for CMO? What basic theorems should I know? :maybe: :maybe:", "Solution_1": "[quote=\"Nawahd Werdna\"]I just wanted to know how all of you practice for CMO. For example, I do the past contests, and I try to do some of the easier USAMOs, but are there any other olympiads that have comparable problem types? Also, what topics are usually more heavily covered for CMO? What basic theorems should I know? :maybe: :maybe:[/quote]\r\n\r\n\r\n\r\nI want to know too!", "Solution_2": "Many people see CMO in fact, as a \"speed contest\". Compared to other olympiads obviously, it has a lot less time... for a lot more questions too! So practising speed is a huge factor. \r\n\r\nGetting used to thinking ideas up quickly is something I do for CMO." } { "Tag": [], "Problem": "Let $a$ be a real number such that $2a^6+a^2=\\frac{3}{2}+2a^4$. Prove that $a^8>1$.", "Solution_1": "[color=blue][b]Let $a$ be a real number such that $2a^6 + a^2 = \\frac{3}{2} + 2a^4$ Prove that $a^8 > 1$[/b][/color]\r\n\r\n\r\n[u][b]Solution:[/b][/u]\r\n$2a^6 + a^2 = \\frac{3}{2} + 2a^4 \\Rightarrow 4a^6 - 4a^4 + 2a^2 - 3 = 0 \\Rightarrow$\r\n\r\n$\\Rightarrow 4a^4 \\left( {a^2 - 1} \\right) + 2\\left( {a^2 - 1} \\right) = 1 \\Rightarrow \\left( {a^2 - 1} \\right)\\left( {4a^4 + 2} \\right) = 1 \\Rightarrow$\r\n\r\n$\\Rightarrow a^2 - 1 > 0 \\Rightarrow a^2 > 1 \\Rightarrow a^8 > 1$" } { "Tag": [ "inequalities solved", "inequalities" ], "Problem": "Let a,b,c \\in [0,1] Find the minimum value of K such that the following ineq holds:\r\n(1-a)(1-b)(1-c) \\leq K(1-(a+b+c)/3)", "Solution_1": "Kmin =1 . Use AM-GM .." } { "Tag": [ "calculus", "integration", "inequalities", "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0,1]\\rightarrow\\mathbb{R}$ is a decrasing function. For all $ x\\in(0,1)$ prove that $ x\\int_{0}^{1}f(t)dt\\leq\\int_{0}^{x}f(t)dt.$ Thank you", "Solution_1": "Hallo friend :) \r\n\r\nLet consider the function $ g(x)\\equal{}\\frac{\\int_0^xf(t)dt}{x}$ on $ (0;1]$. We have $ g'(x)\\leq 0$ on $ (0;1)$. So $ g$ is decreasing on $ (0;1)$. This implies for any $ x\\in(0;1)$ we have\r\n$ g(x)\\geq \\lim_{t\\to1^{\\minus{}}}g(t)\\equal{}g(1)\\equal{}\\int_0^1f(t)dt$" } { "Tag": [ "geometry", "geometric transformation", "angle bisector" ], "Problem": "D\u1ea1 em mu\u1ed1n chuy\u1ec3n m\u1ea5y c\u00e1i \u0111\u1ec1 n\u00e0y sang ti\u1ebfng anh. Ai gi\u00fap em v\u1edbi.\r\n1. Cho tam gi\u00e1c ABC kh\u00f4ng c\u00e2n. \u0110\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp I ti\u1ebfp x\u00fac v\u1edbi AB,AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i M v\u00e0 N. MN c\u1eaft BC t\u1ea1i D. Ti\u1ebfp tuy\u1ebfn song song v\u1edbi BC v\u00e0 ti\u1ebfp x\u00fac v\u1edbi (I) c\u1eaft AB,AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i E,F. K l\u00e0 trung \u0111i\u1ec3m EF.CMR: KD ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n (I) .\r\n2. Cho tam gi\u00e1c ABC c\u00f3 ph\u00e2n gi\u00e1c AD (D thu\u1ed9c BC). (O_1) v\u00e0 (O_2) l\u00e0 2 \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c ABD v\u00e0 ACD. AD c\u1eaft 2 ti\u1ebfp tuy\u1ebfn chung c\u1ee7a (O_1) v\u00e0 (O_2) t\u1ea1i P,Q.CMR: PQ^2=AB.AC. \r\n3. Cho (O). 6 \u0111\u01b0\u1eddng tr\u00f2n nh\u1ecf ngo\u00e0i nhau v\u00e0 \u1edf mi\u1ec1n trong c\u1ee7a (O) l\u1ea7n l\u01b0\u1ee3t ti\u1ebfp x\u00fac v\u1edbi (O). \u0110\u01b0\u1eddng tr\u00f2n Oi ti\u1ebfp x\u00fac v\u1edbi (O) t\u1ea1i Ai. V\u00e0 \u0111\u01b0\u1eddng tr\u00f2n (Ok) ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n (O(k+1)) (qui \u01b0\u1edbc(O7)=(O1)). CMR: A1A4,A2A5,A3A6 \u0111\u1ed3ng qui", "Solution_1": "1. Let be given a non-isoscles triangle ABC. The incircle I touch AB, AC at M, N. MN intersect BC at D. The tangent // and touch (I) intersect AB, AC at E, F, respectively. K is the midpoint of EF. Prove that KD is a tangent of (I).\r\n2. Given triangle ABC, AD is angle-bisector. (O_1) and (O_2) are scribed-circle ABD and ACD, respectively. AD intersect common tangent of (O_1) and (O_2) at P, Q. Prove that: PQ^2= AB.AC\r\n3.Given (O). 6 small-circles non-intersect are touch with (O),. (Oi) touch (O) at Ai, and (Ok) touch (Ok+1). Prove that A1A4,A2A5,A3A6 intersect at common point.!@!\r\nDo you agree??? :lol:", "Solution_2": "I haven't learn these things (both Vietnamese and English) yet so I probably don't understand what you guys are talking about but I know that your translation has some punctuation problem.\r\n\r\n1. Let ABC is an isosceles triangle. The incircle I touches AB and AC at M, N. MN intersects BC at D. The parallel tangent and touch (I) intersect AB, AC at E, F. K is the midpoint of EF. Prove that KD is a tangent of (I).\r\n\r\nThis sentence sounds awkward and I don't know how to change it. \"The tangent // and touch (I) intersect AB, AC at E, F, respectively.\" :noo: \r\n\r\n2. Given triangle ABC and AD is the angle bisector of angle A. (O_1) and (O_2) are scribed-circle ABD and ACD. AD intersects common tangent of (O_1) and (O_2) at P, Q. Prove that: PQ^2= AB.AC\r\n\r\n3.Given (O). 6 small non-intersect circles are touch with (O),. (Oi) touch (O) at Ai, and (Ok) touch (Ok+1). Prove that A1A4,A2A5,A3A6 intersect at common point\r\n\r\nThis one souds weird to me too but I don't know how to fix it. \"6 small-circles non-intersect are touch with (O),. (Oi) touch (O) at Ai, and (Ok) touch (Ok+1).\" :wink:\r\n\r\nBesides, the word \"touch\" you are using here is misused. I think it has to be another word." } { "Tag": [ "function" ], "Problem": "this seems real challenging (atleast to me)\r\ngiven a number n(n<=1000) find a minimum number which is a multiple of n with the sum of digits equal to n..\r\nlike for\r\n10 its 190\r\n11 209\r\n12 48\r\n20 3980\r\netc\r\nthe output will definitely not fit into the long long datatype for n>100 it seems .take care abt that\r\nplease help me with this\r\nbest regards\r\nvasu", "Solution_1": "I hope this shows up well... this code is in C++ and should help no matter which language you are using. There may be one or two mistakes, and it's not perfect, but it's a good guideline. It's not completed; I'll tell you where it needs to be worked on via a comment.\r\n\r\n[code]#include \r\nusing namespace std;\r\n\r\nint tentothe(int y);\r\n\r\nint main()\r\n{\r\n int n, x = 0, y, runningtotal, helper, current, solution;\r\n\r\n cout << \"Enter your number: \";\r\n cin >> n;\r\n\r\n for (solution = 0; solution == 0; x++)\r\n {\r\n current = n * x; //set the current number you're trying\r\n\r\n for (y = 1; tentothe(y) <= current; y++);\r\n\r\n for (helper = 0 ; y >= 0; --y)\r\n {\r\n runningtotal = (current / tentothe(y)) - helper;\r\n helper = runningtotal; //Not sure about this...\r\n //NOT FINISHED!!! PLEASE FINISH IF YOU SEE WHAT I'M GETTING AT. : )\r\n }\r\n }\r\n}\r\n\r\nint tentothe(int y)\r\n{\r\n// make 10 go to the yth power. There should be an easier way to do this than a\r\n//recursive function, but I'm blanking on it\r\n}[/code]", "Solution_2": "To .cpp: I'm not sure exactly what you are trying to do, but it looks like you are going through every single multiple of n, finding the sum of digits, and seeing if it works. That isn't going to work at all here; not only because the multiple of n won't fit into a 'long long' variable, but even if it did or you use some arbitrary precision arithmetic, it will take far too long to find a solution.\r\n\r\nTo galois: this looks very similar to a problem from the recent icpc world finals. Checking every multiple will be too slow; but I'm guessing that checking every single number whose digits add to n will run fast enough. This shouldn't be too hard to do (just store powers of 10 mod n, and repeatedly check for length ceil(n/9), ceil(n/9)+1, ..). I'd expect it to run fast enough as well, since the number of possibilities is increasing very fast, so you should end up finding the correct multiple reasonably quickly.", "Solution_3": "hey stephen\r\na doubt here.... i was trying sumthing similar so to say...but what's the point in computing powers of 10 mod n whel all u r checking for are numbers with digital sum n and on finding one check if it is divisible by n..if it does we are done else continue..\r\nwhy wud we require the modulo values for the powers..??\r\n \r\nkind regards\r\nvasu", "Solution_4": "If you are using arbitrary-precision arithmetic you won't have to; I was just assuming you would use, say, an integer array of digits, in which case you'd need to check if 1*last digit + 10 * secondtolast + 100 * thirdtolast + ... is divisible by n, and thus can just use powers mod n to reduce the calculations to fit inside an int.\r\n\r\n[edit] Actually, you don't even need that; using an array, you just keep multiplying by 10, adding the next digit, then modding by n.\r\n\r\n[edit again]Although, depending on how you are producing partitions of n, you're probably just changing a couple of digits at a time, so it would be faster to just add/subtract a power of 10 or so from a running total.", "Solution_5": "Sorry about that messy code, I had to turn off the computer more quickly than I expected and had to stop writing without a completed program or many comments. I think that I had an ok idea, but TripleM's [i]does[/i] seem faster once the program can fill in the array with integers... I might work on that\r\n\r\nTripleM: When you speak of a 'long long integer', do you just mean the long int type? That should be enough for standard purposes... And unless you do a lot of programming using an array, it'll be the same limit either way...\r\n\r\ngalois: Did you mean me when you said Stephen? Anyways, I used the powers of 10 in an attempt to isolate digits... 1234 / 1000 = 1, 1234 / 100 = 12, etc. I was trying to use that to my advantage. I might keep working on this problem later, it seems pretty interesting...", "Solution_6": "well long long means 64-bit...\r\nand i see some pattern in the patterns dude\r\nfor example n=28;\r\nminimum number will be 4 digits\r\n1999\r\nnext is 2899,2989,2998,\r\nthen 3799,3889,3898,3979,3988,3997\r\nthen 4699,4789,4798,4879,4888,4897,4969,4978,4987,4996\r\nthen ...\r\ndo u notice the patterns the difference are all of the form 9.10^k... except in certain cases where it's to figure out as well\r\ni guess this help in generating the new number in the sequence as strings...\r\n\r\nso u c .cpp the long long wont help here .u might have answers running into more than 20 odd digits easily..\r\nand 64-bit allows maximum 18 digits...\r\n\r\n\r\nbest regards\r\nvasu", "Solution_7": "looks like something wrong with the above post...guys pay little heed to the above for the moment", "Solution_8": "Hmm...which programming language are you using?", "Solution_9": "Well, I've tried coding this using my method, and while it works on most inputs <=1000, there are still several where it takes too long.\r\nI've also tried a dynamic programming approach; the time limit at SPOJ is 7 seconds, so this way is working for me up to about input size 600 or so (I'm using java, so it will be faster for you), but 1000 is still too slow. \r\n\r\nI have one more idea of how to speed this up, but I'll have to think about it for a while." } { "Tag": [ "geometry", "rectangle", "trigonometry", "trig identities", "Law of Sines", "AMC" ], "Problem": "Source: MOCP January 2001, #57\r\n\r\nLet ABCD be a rectangle and let E be a point on the diagonal BD with angle DAE = 15:^o:. Let F be a point in AB with EF perpendicular to AB. It is known that EF = 1/2*AB and AD=a. Find the measure of angle EAC and the length of segment EC (in terms of a).", "Solution_1": "[quote=\"mathfanatic\"]Source: MOCP January 2001, #57\n\nLet ABCD be a rectangle and let E be a point on the diagonal BD with angle DAE = 15:^o:. Let F be a point in AB with EF perpendicular to AB. It is known that EF = 1/2*AB and AD=a. Find the measure of angle EAC and the length of segment EC (in terms of a).[/quote]\r\n\r\nDo you mean F is [i]on[/i] AB?", "Solution_2": "mathfanatic wrote:Source: MOCP January 2001, #57\n\nLet ABCD be a rectangle and let E be a point on the diagonal BD with angle DAE = 15:^o:. Let F be a point in AB with EF perpendicular to AB. It is known that EF = 1/2*AB and AD=a. Find the measure of angle EAC and the length of segment EC (in terms of a).\n\n\n\n[hide]Alright. From right triangle EAF, we have cos 15 = EF/AE, or 2 cos 15 = AB/AE. Now, using the law of sines in triangle ABE, we fine that equivalent to sin AEB/sin ABD. Ok, now let EAC be x. Then x+15 =CAD=ADB, and AEB=15+FEB=15+ADB=30+x. So, our equation is equivalent to 2 cos 15= sin (30+x)/sin(15+x). Expanding, we have the RHS equal to (sin 30 cos x +cos 30 sin x)/(sin 15 cos x + cos 15 sin x). Using the double angle formulas in the numerator and multiplying the denominator to the other side, we get 2 sin 15 cos 15 cos x + 2 cos :^2: 15 sin x = 2 sin 15 cos 15 cos x +\n\n 2 cos :^2: 15 sin x - sin x. Cancelling like terms, we get sin x =0, or E lies on AC. Thus, angle EAC = 0, and using right triangle ACD, we get EC= AC/2 = a/(2 cos 15). [/hide]\n\n\n\nThanks for your help ComplexZeta. I think I may actually have solved a geometry problem!", "Solution_3": "So that's where that trig problem you gave me came from." } { "Tag": [ "ARML" ], "Problem": "Who is going to ARML 2009?\r\nI know it early...\r\nAnyone from Oklahoma???", "Solution_1": "ehh, wrong forum.\r\n\r\nyou should post this in ARML section." } { "Tag": [ "ratio", "geometry", "circumcircle", "geometry unsolved" ], "Problem": "In triangle $ ABC$, $ \\angle A \\equal{} 90^{\\circ}$. $ M$ is the midpoint of $ BC$. Choose $ D$ on $ AC$ such that $ AD \\equal{} AM$. The\r\ncircumcircles of triangles $ AMC$ and $ BDC$ intersect at $ C$ and at a point $ P$. \r\nWhat is the ratio of angles: $ \\frac{\\angle ACB}{\\angle PCB}$? (Find with complete proof)", "Solution_1": "It is surprising that no one sent a solution to this problem...(Probably it was quite easy for this section... :huh: )\r\nOK...here is my solution.\r\nWe have $ CDPB$ cyclic. So, $ \\angle ADP \\equal{} \\angle MBP$. Also, $ CMPA$ cyclic and $ \\angle CAP \\equal{} \\angle PMB$. And $ AD \\equal{} CM \\equal{} MB$. \r\nSo $ \\triangle APD \\cong \\triangle MPB$. So, $ DP \\equal{} BP \\iff \\text{arc } DP \\equal{} \\text{arc } BP$ and the result follows.", "Solution_2": "Oh...I did not answer correctly :P \r\nSo...$ \\angle ACB \\equal{} 2\\angle PCB$ and the ratio is certainly $ 2$ :wink:" } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometry proposed" ], "Problem": "Given are $\\triangle ABC$ with any $M$ inside triangle, circumradius $R$ prove that\r\n$\\frac{MBMC}{a}+\\frac{MCMA}{b}+\\frac{MAMB}{c}\\ge \\frac{\\sqrt{3}(R^{2}-OM^{2})}{R}$\r\n$\\frac{a}{MBMC}+\\frac{b}{MCMA}+\\frac{c}{MAMB}\\le \\frac{3\\sqrt{3}R}{R^{2}-OM^{2}}$ :) :)", "Solution_1": "what point is O?", "Solution_2": "sorry $O$ is circumcenter ok?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given two distinct numbers $ b_1$ and $ b_2$, their product can be formed in two ways: $ b_1 \\times b_2$ and $ b_2 \\times b_1.$ Given three distinct numbers, $ b_1, b_2, b_3,$ their product can be formed in twelve ways: \r\n\r\n$ b_1\\times(b_2 \\times b_3);$ $ (b_1 \\times b_2) \\times b_3;$ $ b_1 \\times (b_3 \\times b_2);$ $ (b_1 \\times b_3) \\times b_2;$ $ b_2 \\times (b_1 \\times b_3);$ $ (b_2 \\times b_1) \\times b_3;$ $ b_2 \\times(b_3 \\times b_1);$ $ (b_2 \\times b_3)\\times b_1;$ $ b_3 \\times(b_1 \\times b_2);$ $ (b_3 \\times b_1)\\times b_2;$ $ b_3 \\times(b_2 \\times b_1);$ $ (b_3 \\times b_2) \\times b_1.$ \r\n\r\nIn how many ways can the product of $ n$ distinct letters be formed?", "Solution_1": "If I haven't miss understood the question\r\n\r\nIf $ a_n$ is the number of ways to multiply $ n$ numbers then\r\n\r\n$ a_n \\equal{} 2^{n\\minus{}2} n! \\quad n>1$ and $ a_1\\equal{}1$\r\n\r\n[b]Proof[/b]\r\n\r\n$ a_n \\equal{} 2 \\times n \\times a_{n\\minus{}1}$ - reasoning, there are two ways to order multiplying by k by the rest and n ways to choose k" } { "Tag": [ "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "given d(x,y) is a metric on X\r\nshow that d'(x,y) = d(x,y) / [1+d(x,y)] \r\nalso define a metric.\r\n\r\nI am struggling in the details of triangular inequility....can someone help me out????", "Solution_1": "What you want to show is the following: If $a \\le b+c,$ then $\\frac{a}{1+a}\\le \\frac{b}{1+b}+\\frac{c}{1+c}.$ Here $a,\\ b,\\ c \\ge 0.$\r\n\r\nThe simplest way to show the above (without fancy inequalities) is by multiplying all the terms, getting rid of the denominators and cancelling terms on either side of the resulting inequality. This actually turns out to be quite easy.", "Solution_2": "Hi,\r\nI think this method should work:\r\n\r\nPlease observe that the function $f(x)=\\frac{x}{1+x}$ is an increasing function on $(0,\\infty)$.\r\nThus we get , via $d(x,y)\\leq d(x,z)+d(z,y)$, that \r\n\r\n$\\frac{d(x,y)}{1+d(x,y)}\\leq \\frac{d(x,z)+d(y,z)}{1+d(x,z)+d(y,z)}=\\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\\frac{d(y,z)}{1+d(x,z)+d(z,y)}$\r\n\r\n$\\leq \\frac{d(x,z)}{1+d(x,z)}+\\frac{d(y,z)}{1+d(y,z)}$,\r\nand you are done.\r\nDidi" } { "Tag": [], "Problem": "find the velocity v in the figure assuming that all the pulleys and ropes are of negligible mass in the file attached to the post..", "Solution_1": "$ v \\equal{} \\frac{4}{3}\\;m/s$\r\n\r\nAttach images directly, you need not compress them. If they are too large, save in .gif format.", "Solution_2": "How u got this one...???", "Solution_3": "Consider the displacement of the point on different parts of the string." } { "Tag": [ "geometry", "circumcircle", "power of a point", "geometry proposed" ], "Problem": "Given are three circles $\\omega_{i}$ ($i=1,2,3$) with equal radius, who have a common point $P$. Of these three circles $PA, PB$ and $PC$ are diameters of $\\omega_{1}, \\omega_{2}$ and $\\omega_{3}$ respectively.\r\n\r\nDraw tangents $l_{1}, l_{2}$ and $l_{3}$ to the three circles, through the points $A,B$ and $C$ respectively. Call $l_{1}\\cap l_{2}= \\left\\{G\\right\\}, l_{1}\\cap l_{3}= \\left\\{H\\right\\}$ and $l_{2}\\cap l_{3}= \\left\\{I\\right\\}$.\r\n\r\nCall $X$ and $Y$ the midpoints of $GH$ and $GI$ respectively. Now, call $PX \\cap \\omega_{1}= \\left\\{J\\right\\}$ and $PY \\cap \\omega_{2}= \\left\\{K\\right\\}$.\r\n\r\n(a) Prove that $YKJX$ is a cyclic quadrilateral.\r\n\r\nIf we call $L$ the midpoint of $XY$, then the circle through $L,K$ and $Y$ intersects $PC$ again in $M$.\r\n\r\n(b) Prove that $PA$ is tangent to the circumcircle of $\\Delta ALM$\r\n\r\nAgain, this is a self-made problem. I don't think it is very nice, because it is too transparant. If anybody has any comments or suggestions to improve my problems or any other remarks, they would be really appreciated. Please tell me (by posting or PM) what you think, even if you think this is a lousy problem (Especially if you think it is lousy!) \r\nThe same goes for http://www.mathlinks.ro/Forum/viewtopic.php?t=124552 :)", "Solution_1": "[quote=\"Jan\"]\n(a) Prove that $YKJX$ is a cyclic quadrilateral.\n[/quote]\n\nLet $\\omega$ be the circle with centre $P$ and radius $r : = PA = PB = PC$.\n\nThen circle $\\omega_{1}$ is the inverse of line $GH$ with respect to $\\omega$, as $\\omega_{2}$ is the inverse of line $GI$. In particular, $J$ is the inverse of $X$, and $K$ is the inverse of $Y$. Hence, $PX \\cdot PJ = PY \\cdot PK = r^{2}$. Then by power of a point, quadrilateral $XJKY$ is cyclic.\n\n[quote=\"Jan\"]\nIf we call $L$ the midpoint of $XY$, then the circle through $L,K$ and $Y$ intersects $PC$ again in $M$.\n[/quote]\r\n\r\nAre the variables correct here? The circumcircle of triangle $LKY$ appears to intersect $PC$ twice.", "Solution_2": "Oh I'm so sorry nsato, that is indeed a typo. $L$ is not the midpoint of $XY$, but the midpoint of the circle segment formed by XY, ie the intersection point of $PC$ and $XY$ is called $L$.\r\n\r\nSorry for the mistake, now it should be correct. \r\nI fear you will find part (b) too easy, too transparent.\r\n\r\nJust a note: Another statement, equivalent to (b) could be: Prove that the circumcentre of $\\Delta ALM$ lies on the line $GH$." } { "Tag": [ "floor function", "number theory", "prime factorization", "number theory unsolved" ], "Problem": "Show that for all prime numbers $p$, \\[Q(p) = \\prod^{p-1}_{k=1}k^{2k-p-1}\\] is an integer.", "Solution_1": "[hide]For any prime $q$, let $x_{q}=\\lfloor{\\frac{p-1}{q}}\\rfloor$\nThen the power of $q$ in the prime factorization of this number is $\\sum_{i=1}^{x_{q}}{2qi-p-1}=x_{q}(q(x_{q}+1)-p-1)$\nBut $qx_{q}\\ge p-q+1$, because were it p-q, that would mean $q|p$, a contradiction. Then the exponent of $q$ is nonnegative, as desired...[/hide]" } { "Tag": [ "Why the bump" ], "Problem": "If you walk for $ 45$ minutes at a rate of $ 4$ mph and then run for $ 30$ minutes at a rate of $ 10$ mph, how many miles have you gone at the end of one hour and $ 15$ minutes?\r\n\r\n\\[ \\textbf{(A)}\\ 3.5 \\text{ miles} \\qquad\r\n\\textbf{(B)}\\ 8 \\text{ miles} \\qquad\r\n\\textbf{(C)}\\ 9 \\text{ miles} \\qquad\r\n\\textbf{(D)}\\ 25 \\frac{1}{3} \\text{ miles} \\qquad\r\n\\textbf{(E)}\\ 480 \\text{ miles}\r\n\\]", "Solution_1": "[hide]You walked $ \\frac {3}{4} \\times 4$ or $ 3$ miles. You ran $ \\frac12 \\times 10$ or $ 5$ miles. Thus, the answer is $ \\boxed{\\textbf{(B)}\\ 8 \\text{ miles}}$.[/hide]", "Solution_2": "I did $\\frac{45}{60} \\times 4 + \\frac{30}{60} \\times 10 = 3+5 = 8$." } { "Tag": [ "MATHCOUNTS", "search", "function", "Purple Comet" ], "Problem": "I am looking for a good book with great MC problems and easy to understand and well-written solutions and i am stuck between the handbooks and the all time greatest mathcounts hits. Please help soon thank you", "Solution_1": "By \"stuck between\", do you mean you can't decide between to two of them? If so, I'd get the all-time greatest MathCounts problems.", "Solution_2": "thank you izzy\r\nhas anyone read both the books and has an opinion about this?", "Solution_3": "Personally, I believe that you shouldn't get either of these books. Really, doing problems is the best way to improve. I'm assuming your primary goal is to do well in MC. You can't really get good at it by doing thousands and thousands of MC problems. There just aren't enough hard problems compared to the loads of easy problems. Of course, doing MC problems will increase your speed, but you shouldn't worry about that until the competitions actually start. Basically, you should just do many hard problems that challenge you, and there really are many problems out there (AMCs, Purple Comet, etc.) that can do that and prepare you for the hardest MC problems. So that's my suggestion: first challenge yourself, and then work on actual MC problems.\r\n\r\nEdit: Why neither of the books are really that good: the handbooks CAN be found on Google if you are sophisticated with the search function and the \"Greatest Problems\" book really doesn't have such great problems.", "Solution_4": "i would say handbook\r\n\r\nbut try to get all the past tests, and if you can solve all the mc problems fairly quickly, move on to harder stuff", "Solution_5": "if i could have only one or the other, i would take the all time greatest problems. i have the handbook and most of the problems are school level, while the greatest problems are state to national problems-with solutions. so i guess it depends on whether you are striving to make your school team, or to qualify for nationals.\r\n\r\ngoodluck with either(or both) :D" } { "Tag": [ "function", "algebra", "polynomial", "ratio", "integration", "calculus", "abstract algebra" ], "Problem": "Suppose $ f$ is entire and that $ |Ref|\\leq M|z|^k$ for sufficiently large $ |z|$. Show that $ f$ is a polynomial of degree at most $ k$.\r\n\r\nIt seems like you can replace a part of the hypothesis of lots of standard problems with $ Ref$ instead of $ f$ and still get the same strength conclusion.", "Solution_1": "for me, i would think about $ e^{f}$, but I am not sure", "Solution_2": "finish using the Hadamard factorization theorem.", "Solution_3": "[quote=\"Albanian Eagle\"] the Hadamard factorization theorem.[/quote]\r\n\r\nwhat is that??", "Solution_4": "That's a big overkill. ;) Just notice that the gradient of a harmonic function is not much greater than the function itself and that the conjugate harmonic functions $ \\text{Re\\,}f$ and $ \\text{Im\\,}f$ have essentially the same gradients. So, any reasonable growth restriction on the real part automatically transfers to the imaginary part and, thereby, to $ |f|$. \r\n\r\nThe alternative is just to use the [url=http://en.wikipedia.org/wiki/Schwarz_formula]Schwarz formula[/url].", "Solution_5": "I've been using the Schwarz formula on problems like these, but it seems cheating. I was looking to see if anyone had a way of doing it without just recovering $ f$ as the first step.\r\n\r\nThanks. I'll try to figure out the details of that gradient method.\r\n\r\nMy concern with relating growth of a harmonic function with its harmonic conjugate was that you can define an analytic branch $ f(z)\\equal{}\\log(1\\minus{}z)$ on $ |z\\minus{}1|<1$. Then as I approach 1 along the real axis $ |Im(f)|\\equal{}|arg(1\\minus{}z)| \\equiv 0$ but its conjugate $ |Re(f)|\\equal{}|log|1\\minus{}z\\parallel{}\\to \\infty$ blows up. \r\n\r\nI guess this has to do with continuity on the boundary or something?", "Solution_6": "Yes, that's about the boundary. That logarithmic singularity shows up as a jump discontinuity in the imaginary part.\r\n\r\nWith the Schwarz formula, you're not using it at values anywhere near the boundary; you want to apply it at some fixed ratio between the point of interest's radius and the outer circle for this application.\r\n\r\nSince I just built this for another question, I'll post my version, which doesn't [i]exactly[/i] recover $ f$:\r\n\r\nLet $ g$ be the real part of $ f$. It's harmonic, so we have the Poisson formula $ g(z)\\equal{}\\frac1{2\\pi R}\\int_{|w|\\equal{}R}g(w)\\frac{R^2\\minus{}|z|^2}{|z\\minus{}w|^2}\\,d|w|$. Now, $ g(z)\\equal{}\\frac12f(z)\\plus{}\\frac12\\overline{f(z)}$, so $ f'(z)\\equal{}2\\frac{\\partial g}{\\partial z}$. We differentiate that integral formula:\r\n$ 2g(z)\\equal{}\\frac1{\\pi R}\\int_{|w|\\equal{}R}g(w)\\frac{R^2\\minus{}z\\overline{z}}{(z\\minus{}w)(\\overline{z}\\minus{}\\overline{w})}\\,d|w|$\r\n$ 2\\frac{\\partial g}{\\partial z}\\equal{}\\frac1{\\pi R}\\int_{|w|\\equal{}R}g(w)\\left(\\frac{\\minus{}\\overline{z}}{(z\\minus{}w)(\\overline{z}\\minus{}\\overline{w})}\\minus{}\\frac{R^2\\minus{}z\\overline{z}}{(z\\minus{}w)^2(\\overline{z}\\minus{}\\overline{w})}\\right)\\,d|w|$\r\n$ f'(z)\\equal{}\\frac1{\\pi R}\\int_{|w|\\equal{}R}g(w)\\frac{\\minus{}z\\overline{z}\\plus{}w\\overline{z}\\minus{}R^2\\plus{}z\\overline{z}}{(z\\minus{}w)^2(\\overline{z}\\minus{}\\overline{w})}\\,d|w|$\r\n$ f'(z)\\equal{}\\frac1{\\pi R}\\int_{|w|\\equal{}R}g(w)\\frac{w\\overline{z}\\minus{}R^2}{|z\\minus{}w|^2(z\\minus{}w)}\\,d|w|$\r\n\r\nNow, choose $ |z|\\equal{}r\\equal{}\\frac12R$. We have $ |w\\overline{z}|\\equal{}2r^2$ and $ |z\\minus{}w|\\ge r$, so $ \\left|\\frac{w\\overline{z}\\minus{}R^2}{|z\\minus{}w|^2(z\\minus{}w)}\\right|\\le \\frac{6r^2}{r^3}\\equal{}\\frac{6}{r}$ and\r\n$ |f'(z)|\\le\\frac1{2\\pi r}\\int_0^{2\\pi}2^kMr^k\\cdot\\frac{6}{r}\\cdot 2r\\,d\\theta\\equal{}6M\\cdot 2^kr^{k\\minus{}1}$\r\nTherefore $ f'$ must be polynomial of degree $ k\\minus{}1$ or less; integrating, $ f$ is polynomial of degree $ k$ or less.\r\n\r\nAs a side note, if we merely start with a one-sided bound $ \\text{Re}(f(z))\\le M|z|^k$, we can derive a two-sided bound $ |\\text{Re}(f(z))|\\le M'|z|^k\\plus{}C$ of the same order. Away from the boundary, the Poisson kernel doesn't vary that much." } { "Tag": [ "calculus", "integration", "function", "limit", "logarithms", "topology", "real analysis" ], "Problem": "$h : [0,\\infty) \\to [0,\\infty)$ is a nonnegative measurable function with $\\lim_{x\\to\\infty}\\frac{h(x)}{x}= \\infty$. \r\nProve that for each $\\epsilon > 0$ there exists a nonnegative continuous function $f: (0,1) \\to [0,\\infty)$ satisfying :\r\n\\[\\int_{0}^{1}f = \\epsilon \\ ,\\quad \\int_{0}^{1}h \\circ f = \\infty \\]\r\n\r\n\r\nInspired by the problem of finding a function $f \\ge 1$ satisfying :\r\n\\[\\int_{0}^{1}f < \\infty \\ ,\\quad \\int_{0}^{1}f \\cdot \\log{f}= \\infty \\]", "Solution_1": "Some hint.\r\n\r\nWe can divide the problem into three.\r\n\r\nStep 1. solve easier problem:\r\nStrengthen the hypothesis part by assuming that h(x) = x g(x) for all x with g(x) a continuous non-negative strictly-increasing function. Weaken the conclusion part by replacing the requirement $\\int_{0}^{1}f = \\epsilon$ with the easier requirement $\\int_{0}^{1}f < \\infty$.\r\n\r\nStep 2. Weaken the hypothesis back.\r\n\r\nStep 3. Strengthen the conclusion part back.", "Solution_2": "[b]NOTATION[/b]\r\n1. $\\int f$ = $\\int_{0}^{1}f(t)\\,dt$. (this is for convenience)\r\n2. $f: (a,b)\\uparrow(c,d)$ means that $f$ is a continuous strictly-increasing function with $f(a+) = c$ and $f(b-)=d$, i.e. $f$ is a orientation-preserving homeomorphism from $(a,b)$ onto $(c,d)$.\r\n\r\n\r\n[b]LEMMA[/b]\r\nLet $N>0$ and $\\ell: (N,\\infty)\\uparrow(0,\\infty)$. Then there is a function $f: (0,1)\\uparrow(N,\\infty)$ with $\\int f < \\infty$ and $\\int f \\cdot (\\ell\\circ f) = \\infty$.\r\n\r\n\r\n[b]PROOF OF LEMMA[/b]\r\nIt is easy to find a continuous positive function $w$ defined on $(0,1)$ and a function $g: (0,1)\\uparrow(N,\\infty)$ satisfying the following two equations:\r\n\\[\\int w \\cdot g < \\infty \\]\r\n\r\n\\[\\int w \\cdot g \\cdot (\\ell\\circ g) = \\infty \\]\r\n(Hint: try $w = 1/g$).\r\nWe may assume $\\int w = 1$ and define $W: (0,1)\\uparrow(0,1)$ to be an antiderivative of $w$.\r\nLet $f = g\\circ W^{-1}$. Then change of variable shows:\r\n\\[\\int f = \\int w \\cdot g \\]\r\n\r\n\\[\\int f \\cdot (\\ell\\circ f) = \\int w \\cdot g \\cdot (\\ell\\circ g) \\]\r\n[b]THEOREM[/b]\r\nLet $h: [0,\\infty)\\to[0,\\infty)$ be a measurable function with $\\lim_{t\\to\\infty}h(t)/t = \\infty$.\r\nFor any $\\epsilon >0$ there is $f: (0,1)\\uparrow(0,\\infty)$ with $\\int f = \\epsilon$ and $\\int h \\circ f = \\infty$.\r\n\r\n[b]PROOF OF THEOREM[/b]\r\nIt is easy to find a positive real $N>0$ and a function $\\ell: (N,\\infty)\\uparrow(0,\\infty)$ such that $\\ell(t) \\le h(t)/t$ for all $t \\in (N,\\infty)$.\r\nBy the LEMMA, there is $g: (0,1)\\uparrow(N,\\infty)$ with $\\int g < \\infty$ and $\\int g \\cdot (\\ell\\circ g) = \\infty$.\r\nWe may assume $\\int g$ is large enough so that it is greater than the given $\\epsilon$. \r\nThen we can find $b\\in(0,1)$ and $f: (0,1)\\uparrow(0,\\infty)$ such that $f \\le g$, $\\int f = \\epsilon$, and that $f=g$ on $(b,1)$.\r\nWe have $\\int_{0}^{1}h\\circ f \\ge \\int_{b}^{1}h\\circ g \\ge \\int_{b}^{1}g \\cdot (\\ell\\circ g) = \\infty$.\r\n\r\n\r\nI'm still interested in other's solution. Also what about a discrete version of the theorem (sequence version)?", "Solution_3": "Discrete version: $h\\colon [0,\\infty) \\to [0,\\infty)$ is a nonnegative measurable function with $\\lim_{x\\to 0}\\frac{h(x)}{x}= \\infty$. For each $\\epsilon > 0$ there exists a series with positive terms $a_{n}$ such that $\\sum_{n}a_{n}\\le \\epsilon$ but $\\sum_{n}h(a_{n})$ is divergent.\r\n\r\nProof: for $k\\ge 1$ find an integer $M_{k}$ such that $h(x_{k})/x_{k}\\ge 2^{k}$ where $x_{k}=\\epsilon 2^{-k}/M_{k}$. Consider the sum of $M_{k}$ equal terms $\\epsilon 2^{-k}/M_{k}+\\dots+\\epsilon 2^{-k}/M_{k}$. This sum is equal to $\\epsilon 2^{-k}$. Combining such blocks of $M_{k}$ elements $(k\\ge 1)$ into an infinite series, we get a series with $\\sum_{n}a_{n}=\\epsilon$. On the other hand, the $k$th block of the series $\\sum_{n}h(a_{n})$ sums up to at least $\\epsilon/M_{k}+\\dots+\\epsilon/M_{k}=\\epsilon$. Hence $\\sum_{n}h(a_{n})$ diverges. \r\n\r\nP.S. I speculate that the lack of responses to the original problem has something to do with somewhat annoying requirements ($f$ must be continuous, $\\int f=\\epsilon$) that add little, if anything, to the substance of the problem." } { "Tag": [ "inequalities", "algebra", "polynomial", "trigonometry", "induction", "triangle inequality", "inequalities unsolved" ], "Problem": "Let $n$ be a positive integer, and $a_j$, for $j=1,2,\\ldots,n$ are complex numbers. Suppose $I$ is an arbitrary nonempty subset of $\\{1,2,\\ldots,n\\}$, the inequality $\\left|-1+ \\prod_{j\\in I} (1+a_j) \\right| \\leq \\frac 12$ always holds. \r\n\r\nProve that $\\sum_{j=1}^n |a_j| \\leq 3$.", "Solution_1": "The trick is in figuring out how to handle $|ab + a + b| \\leq 1/2$.. I still haven't figured it out yet.\r\n\r\n\r\nHowever, one can write $\\sum |a_i| \\leq \\Bigg| \\sum a_i \\Bigg| \\leq \\frac{1}{2} + Z$, and this looks like it could reasonably do it, if good enough bounds can be found for the symmetric polynomials.\r\n\r\n(where Z is the modulus of the 2nd symmetric polynomial of the $a_i$, plus the 3rd, 4th, .. to the nth.)", "Solution_2": "[quote=\"Singular\"]The trick is in figuring out how to handle $|ab + a + b| \\leq 1/2$.. I still haven't figured it out yet.\n\n\nHowever, one can write $\\sum |a_i| \\leq \\Bigg| \\sum a_i \\Bigg| \\leq \\frac{1}{2} + Z$, and this looks like it could reasonably do it, if good enough bounds can be found for the symmetric polynomials.\n\n(where Z is the modulus of the 2nd symmetric polynomial of the $a_i$, plus the 3rd, 4th, .. to the nth.)[/quote]\r\n\r\ndid you reverse the sign of triangle inequality?", "Solution_3": "yes i did, oops.", "Solution_4": "oh,is this so hard?", "Solution_5": "Did anyone figure out how to do it ?", "Solution_6": "Yes, maybe someone has the official solution ?", "Solution_7": "yes,I have .and I will post it soon :)", "Solution_8": "sorry.I can't find the office solution...\r\nand I 'm very sorry for breaking one's promise :( :(", "Solution_9": "It doesnot mind but there are so many Chinese in the forum, noone has the solution?", "Solution_10": "Fortunately,I find the offical solution now.And I think this problem is hard.Some problem is easy if you know the solution,but I think this one not competely is.\r\nnow let me post the offical solution.\r\nlet $a_j+1=r_j e^{i \\theta_j},|\\theta_j| \\leq \\pi$ for $j=1,2,\\cdots,n$\r\nwith the condition we have \r\n$|\\prod_{j \\in I} r_j e^{i \\sum_{j \\in I} \\theta_j} -1 |\\leq \\frac 1 2$ $(*)$\r\nwe will prove a lemma first.\r\n$lemma$,if $r,\\theta$ are two real number,$r>0,|\\theta| \\leq \\pi$ and $|r e^{i \\theta}-1|<\\frac 1 2$ \r\nthen we have : \r\n(1)$\\frac 1 2 \\leq r \\leq \\frac 3 2$\r\n(2)$|\\theta| \\leq \\frac{\\pi}6$ \r\n(3)$|re^{i \\theta}-1| \\leq |r-1|+|\\theta|$\r\nproof of lemma (1),(2) can be proved in a graph easily(also you can prove it by algebra)\r\n(3) is because \r\n$|re^{i \\theta}-1=|r \\cos \\theta +r\\sin \\theta-1|=|(r-1)(\\cos \\theta+i \\sin \\theta)+(\\cos \\theta-1+i \\sin \\theta)| \\leq |r-1|+\\sqrt{(\\cos \\theta-1)^2+\\sin^2 \\theta}=|r-1|+2|\\sin \\frac{\\theta}{2}| \\leq |r-1|+|\\theta|$\r\nso the lemma is proved.\r\nreturn to the orginal problem:\r\nwith the lemma we have:\r\n$\\frac 1 2 \\leq |\\prod_{j \\in I} r_j | \\leq \\frac 3 2$ ,$|\\sum_{j \\in I}| \\leq \\frac{\\pi}6$ \r\nand $|a_j|=|r_j e^{i \\theta_j}-1| \\leq |r_j-1|+|\\theta_j|$,\r\nso $\\sum^n_{j =1}|a_j| \\leq \\sum^n_{j =1}|r_j-1|+\\sum^n_{j=1}|\\theta_j| =\\sum_{r_j \\geq 1}|r_j-1|+\\sum_{r_j < 1}|r_j-1|+\\sum_{\\theta_j \\geq 0}|\\theta_j|+\\sum_{\\theta_j < 0}|\\theta_j|$\r\nbut we have \r\n$\\sum_{r_j \\geq 1}|r_j-1| =\\sum_{r_j \\geq 1}(r_j-1) \\leq \\prod_{r_j \\geq 1}(1+r_j-1)-1(**) \\leq \\frac 3 2 -1=\\frac 1 2$\r\n$\\sum_{r_j <1 }|r_j-1| =\\sum_{r_j \\geq 1}(1-r_j) \\leq \\prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \\leq(***) \\frac 3 2 -1 \\leq 2-1=1$\r\n$\\sum^n_{j=1}|\\theta_j-1| =\\sum_{\\theta_j \\geq 0}|\\theta_j| -\\sum_{\\theta_j < 0}|\\theta_j| \\leq \\frac{\\pi} 6 + \\frac{\\pi} 6 = \\frac{\\pi} 3$\r\nso $\\sum^n_{j =1}|a_j| \\leq \\frac 1 2 +1+\\frac{\\pi} 3<3$\r\n$(**)$ and $(***)$ can be proved by induction easily.\r\nso the problem are proved.\r\nmaybe this is my longest post in mathlink. :lol:\r\nand I hope it is clear for you. :)", "Solution_11": "[quote=\"zhaobin\"]$\\sum_{r_j <1 }|r_j-1| =\\sum_{r_j \\geq 1}(1-r_j) \\leq \\prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \\leq(***) \\frac 3 2 -1 \\leq 2-1=1$[/quote]\r\n\r\nZhaobin, I don't how to prove this inequality. (i'm refering to this one : $\\sum \\ldots \\leq \\prod_{r_j<1} \\ldots$) It doesn't seem obvious by induction. Can you help me?", "Solution_12": "ok,with pleasure. :) \r\nbut an another nice proof is coming now :) \r\n just notice $\\frac 1 {1-x} \\ge 1+x$ for $00$, there is just an additional constant term and some adjustment to $x$ in the expansion. Try it!" } { "Tag": [ "limit", "function", "calculus", "trigonometry" ], "Problem": "$ \\lim_{x\\to\\infty}2^{x}\\sqrt{2-\\sqrt{2+\\sqrt{2+\\sqrt{2+...+\\sqrt{2+\\sqrt{3}}}}}}$\r\n\r\nWhen I solve it, I get $ 2^{x}*0$. Then that becomes $ \\infty*0$ which is indeterminable. I don't know the graph and don't want to apply l'hospital's rule. Is there an alternative way to do this problem?", "Solution_1": "Are there $ x$ surd symbols?", "Solution_2": "No it's just $ 2^{x}$ times the square root part", "Solution_3": "Which doesn't answer t0r's question. How many square roots are there meant to be in \"the square root part\" -- presumably, some function of $ x$, yes? (Also, this question would have been much better placed in the Calculus Computations section.)", "Solution_4": "If there are always an infinite number of squareroots, the big squareroot term converges to 0. Since it is a constant, you evaluate it first, and then take the limit in terms of x, so it's just 0. Just like $ \\lim_{x \\to \\infty} (x \\cdot 0) \\equal{} 0$, even though x increases without bound.\r\n\r\n\r\nIf there are x surd symbols, and the squareroot term [u]is approaching[/u] (as apposed to \"equal to\") 0 while 2^x is increasing, you have a much more interesting problem. Let the nth term be f(n). Then:\r\n\r\n$ f(n\\plus{}1) \\equal{} 2^{n\\plus{}1} \\sqrt{2 \\minus{} \\sqrt{4\\minus{} \\left( \\frac{f(n)}{2^n} \\right) ^2}}$\r\n\r\nLetting $ g(x) \\equal{} f(x)^2$, we can square and rearrange to get:\r\n\r\n$ g(x\\plus{}1) \\minus{} g(x) \\equal{} \\frac{1}{4} \\left( \\frac{g(x\\plus{}1)}{2^{x\\plus{}1}} \\right) ^2$\r\n\r\nSo, it's clear by telescoping that the limit is positive, if it exists.", "Solution_5": "[quote=\"The Zuton Force\"]So, it's clear by telescoping that the limit is positive, if it exists.[/quote]\r\n\r\nYou could also arrive at this conclusion by noting that $ 2^x>0\\forall x$ and $ \\sqrt{y}\\geq0\\forall y$, so their product is $ \\geq0$.", "Solution_6": "I meant that it does not converge to 0 :wink:", "Solution_7": "The answer depends on whether the first square root outside the 3 and/or the final square root are considered as part of the $ x$ square roots.\r\n\r\n[hide=\"you can use\"]\n\\[ 2\\cos \\frac{\\theta}{2}\\equal{}\\sqrt{2\\plus{}2\\cos \\theta}\\]\n[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be real numbers such that \\[2 \\leq a_i/b_i \\leq 3\\textrm{ for }1 \\leq i \\leq n.\\]\r\nDetermine the maximal value of \\[\\frac{\\left(\\sum a_i^2\\right) \\left(\\sum b_i^2\\right)}{\\left(\\sum a_ib_i\\right)^2}.\\]", "Solution_1": "Let's see if this works.\r\n\r\nWe have [tex](a_i-2b_i)(a_i-3b_i)\\le0[/tex], i.e. [tex]a_i^2+6b_i^2\\le5a_ib_i[/tex]. Sum this to get [tex]\\sum a_i^2+6\\sum b_i^2\\le5\\sum a_ib_i[/tex]. Next by AM-GM,\r\n\r\n[tex]2\\sqrt{6\\sum a_i^2\\sum b_i^2}\\le\\sum a_i^2+6\\sum b_i^2\\le5\\sum a_ib_i[/tex], i.e.\r\n\r\n[tex]\\displaystyle \\frac{(\\sum a_i^2)(\\sum b_i^2)}{(\\sum a_ib_i)^2}\\le\\frac{25}{24}[/tex].\r\n\r\nI'm not sure if we actually can get 25/24, though.\r\n\r\n(sorry, its 2 in the morning... i'll take a second look when i get up)", "Solution_2": "Ok, it's 3 in the morning, I've showered, and I now know that we can get 25/24. Set\r\n\r\n[tex]a_1=\\sqrt{6(n-1)}[/tex],\r\n[tex]a_2=\\cdots=a_n=3[/tex],\r\n[tex]b_1=\\sqrt{3(n-1)/2}[/tex], and\r\n[tex]b_2=\\cdots=b_n=1[/tex].\r\n\r\nOf course this only works if [tex]n\\ge2[/tex]. If n=1 then the whole thing equals 1 anyway, so I suppose I should be more specific and say that the maximum value of 25/24 is attainable iff n>1.", "Solution_3": "Moreover we can prove the following generalisation:\r\n\r\nLet $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be real numbers such that \\[0 < m \\leq a_i/b_i \\leq M\\textrm{ for }1 \\leq i \\leq n.\\]\r\nThen\r\n\r\n \\[1 \\leq \\frac{\\left(\\sum a_i^2\\right) \\left(\\sum b_i^2\\right)}{\\left(\\sum a_ib_i\\right)^2} \\leq \\frac{(M+ m)^2}{4Mm}.\\]\r\n\r\nWe could for example prove that the trinomial \\[ f(x) = \\left(m \\sum a_i^2\\right) x^2 - \\left( (m + M) \\sum a_ib_i \\right) x + M \\sum b_i^2\\] has at least one zero, because then its discriminant is $\\geq 0$.\r\n(In fact, $f(1/m) \\leq 0$ and $f(1/M) \\geq 0$.)" } { "Tag": [ "algebra", "polynomial" ], "Problem": "$120^4 + 272^4 + 315^4 + n^4 = 353^4$\r\nFind n.\r\n\r\nAnswer:\r\n[hide]\nI got the answer 900 using Chinese Remainder theorem. But are there any better methods? Mine was kinda tedious.\n[/hide]", "Solution_1": "Do you mean $n^{2}=900\\implies n=30$?", "Solution_2": "$120^4+315^4+n^4=(353^2+272^2)(353+272)(353-272)=(353)(15)^4$\r\n\r\n$8^4+21^4+(\\frac{n}{15})^4=353^2+272^2$\r\n\r\n$64^2 + 441^2 + (\\frac{n}{15})^4 = 353^2 + 272^2$\r\n\r\n$(\\frac{n}{15})^4 + 441^2 - 353^2=272^2 - 64^2$\r\n\r\n$(\\frac{n}{15})^4 = (208)(336) - (794)(88)$\r\n\r\n$(\\frac{n}{30})^4 = (51)(84) - (397)(11) = 1$\r\n\r\n$n=30$", "Solution_3": "[quote=\"amcavoy\"]Do you mean $n^{2}=900\\implies n=30$?[/quote]\r\n\r\noops, sorry, yup. typo. :blush:", "Solution_4": "Well, $n=-30$ is also a solution.\r\n\r\nMasoud Zargar", "Solution_5": "So is $n=30i,-30i$ :)", "Solution_6": "Well in that case, there are infinite solutions since we could have $\\pm 30i^k$, where $k\\in\\mathbb{N}$. :P\r\n\r\nMasoud Zargar", "Solution_7": "[quote=\"boxedexe\"]Well in that case, there are infinite solutions since we could have $\\pm 30i^k$, where $k\\in\\mathbb{N}$. :P\n\nMasoud Zargar[/quote]\r\n\r\nthose are not all distinct solutions...consider the peridicity of powers of $i$, also this argument is like\r\n\r\nx=1 has infinite solutions since\r\n\r\n$1,1^2,1^3,1^4,...$ are all solutions", "Solution_8": "Solutions are only $\\pm 30$ I think.... Why do you say that there are more??", "Solution_9": "As a fourth-degree polynomial in $n$ there exist(s):\r\n\r\nOne solution $\\in \\mathbb{N}$,\r\nTwo solutions $\\in \\mathbb{Z}$,\r\nTwo solutions $\\in \\mathbb{R}$,\r\nFour solutions $\\in \\mathbb{C}$,\r\n\r\nOkay? :D", "Solution_10": "[quote=\"t0rajir0u\"]As a fourth-degree polynomial in $n$ there exist(s):\n\nOne solution $\\in \\mathbb{N}$,\nTwo solutions $\\in \\mathbb{Z}$,\nTwo solutions $\\in \\mathbb{R}$,\nFour solutions $\\in \\mathbb{C}$,\n\nOkay? :D[/quote]\r\n\r\nYes, but they can be the same sometimes", "Solution_11": "[quote=\"Jos\u00e9\"][quote=\"t0rajir0u\"]As a fourth-degree polynomial in $n$ there exist(s):\n\nOne solution $\\in \\mathbb{N}$,\nTwo solutions $\\in \\mathbb{Z}$,\nTwo solutions $\\in \\mathbb{R}$,\nFour solutions $\\in \\mathbb{C}$,\n\nOkay? :D[/quote]\n\nYes, but they can be the same sometimes[/quote]\r\n\r\nFor example, I don't know, $x^4-1=0$", "Solution_12": "$x^4-1=0$ has four distinct solutions...... right?\r\n\r\n$1$\r\n$-1$\r\n$i$, $-i$\r\n\r\nAn example of where there are not four [u]distinct[/u] solutions is:\r\n\r\n$x^4-4x^3+6x^2-4x+1=0$" } { "Tag": [ "geometry", "rhombus", "perimeter", "rotation", "3D geometry", "AMC", "AIME" ], "Problem": "EXPLAIN HOW YOU SOLVED THE PROBLEM PLEASE.\r\n\r\n1.In a plane, 10 lines intersect such that no 3 lines meet at the same point. What is the maximum number of regions created by the lines?\r\n\r\n2.The lengths of the diagonals of a rhombus are 6 inches and 8 inches. If a circle is inscribed in the rhombus, how many inches are in its circumference? Use 3.14 as an approximation for pi, and express your answer as a decimal to the nearest tenths.\r\n\r\n5.The areas of two adjacent squares are 256 square inches and 16 square inches, respectively, and their bases lie on the same line. What is the number of inches in the length of the segment that joins the centers of the two inscribed circles? Express your answer as a decimal to the nearest tenth.\r\n\r\n6.A sequence of numbers a1, a2, a3... is defined by a1=7, a2=-6 and a, =an-1-an-2 for n>2. What is the sum of the first 2000 terms of the sequence?\r\n\r\n7. Two runners began running in the same direction o a quarter-mile track. The first runner began five feet ahead of the second runner. The first runner's pace is 6 minutes per mile, while the second runner's pace is 8 minutes per mile. Both runners continue at their respective paces for 119 minutes. How many times will the first runner pass the second runner?\r\n\r\n9. Regular hexagons are placed side-by-side in a continuous pattern. What is the maximum number of congruent hexagons that can be placed sid-by-side such that the perimeter of the resulting figure is less than 100cm? Side to side is 3cm\r\n\r\n10. Right AABC has legsmeasuring 8cm and 15cm. The triangle is rotated about one of its legs. What is the number of cubic centimeters in the maximum possible volume of the resulting solid? Express your answer in terms of pi.", "Solution_1": "#1: It's one more than a 10th triangular number.\r\n#2: Rhombus is orthodiagnal. So find the right triangle.\r\n#5: Draw a triangle connecting both of the radii.\r\n\r\nRest, I am not sure of..", "Solution_2": "9th triangluar number. take 1 for example. it obviously have no intersection point", "Solution_3": "what year is this and what round (chapter, state, nats)? YOu might be able to find some solution on the internet..", "Solution_4": "I figured out 1. need help on others", "Solution_5": "For number 10, the formula for a cone is \u03c0r\u00b2h/3, where r is the radius of the base and h is the height of the cone. So find a way to maximize the radius of the base.", "Solution_6": "5.\r\n\r\nThere are two squares, 256 sq. units and 16 sq. units\r\n\r\nIt says they are adjacent, so they are touching. Also, their bases lie on the same line.\r\n\r\nKinda like this.\r\n\r\n|---------|\r\n| |_____\r\n| | |\r\n|_______|____|\r\nLet's call the 256 sq. unit square A and the 16 unit square B\r\nso find the side length of each square. square a is 16 units and square B is 4 units\r\n\r\nSo to get the center of the inscribed circle, find the center of the square.\r\n\r\nSo the center of square B is half of 4 off of each side, or 2.\r\n\r\nThe center of square A is half of each side, or 8.\r\n\r\nThen make a right triangle.\r\nThe center of square A is 10 to the left of the center of square B\r\n\r\nThe center of square A is 6 above the center of square B\r\n\r\n10^2 + 6^2 = 136 using a calculator to sq. root it you get 11.7 when u round it", "Solution_7": "how is a cone related", "Solution_8": "[quote=\"abacadaea\"]9th triangluar number. take 1 for example. it obviously have no intersection point[/quote]True, but go beyond it. It always gives 1 more than the number of line you can draw. But I don't think it's 9th one.\n\n[quote=\"darit28\"]how is a cone related[/quote]\r\nThe question said: \"The triangle is rotated about one of its legs.\"\r\nI think the radius becomes 15, because we know that 8,15,17 is a pythagorean triple, but we were told that the leg was rotated..\r\n\r\nOh, I think someone mentioned this before, but the area of a cone is just $ \\frac{\\pi r^2 \\cdot h}{3}$\r\n\r\nFor #9,I will assume that the side of a hexagon is 3.\r\n I think it's going to be $ 7$.\r\n\r\nThe first and last hexagons will have 5 sides out and one in between have 4.\r\nSo I think the equation is $ (3)5\\plus{}(4\\cdot3)n\\plus{}(3)5<100$\r\n$ n\\equal{}5$ and we count the first and last to get $ \\boxed{7}$", "Solution_9": "[quote=\"darit28\"]how is a cone related[/quote]\r\nAfter rotating, you get a cone. I seem to remember an AIME problem similar to this. Anyway, you just find the area." } { "Tag": [ "geometry", "incenter", "angle bisector" ], "Problem": "I'm incredibly stuck on two proofs:\r\n\r\n[i]1. Prove that the altitudes of any triangle are concurrent.\n\n2. Prove that the angle bisectors of any triangle are concurrent.[/i]\r\n\r\nDoes anyone know how to do these?", "Solution_1": "[hide=\"Hint on 2\"] Draw the incircle. [/hide]", "Solution_2": "[hide=\"Hint on #1\"]Connect the feet of the altitudes and prove that the orthocenter is actually the incenter of the smaller triangle.[/hide]", "Solution_3": "I've been able to prove #1 but am still very stuck on #2 (I tried the hint)", "Solution_4": "The angle bisector from $A$ and the angle bisector from $B$ meet at a point $I$, such that ....\r\nCan you see how this tells you that the bisector from $C$ goes through this point?\r\n\r\n(Remember, an angle bisector of an angle happens to be the locus of points that are [b]equidistant from the two lines that make up the angle[/b])", "Solution_5": "I understand that this is true. And I believe that it works. But I can't for the life of me come up with a rigorous proof.", "Solution_6": "A rigourous proof: (?)\r\n\r\nLet the angle bisector from the vertex $A$ and the angle bisector from the vertex $B$ meet at a point $I$. (They cannot be parallel, just by considering that the sum of the angles in a triangle is $180^{o}$).\r\n\r\n$I$ lies on the bisector of $\\angle A$ and hence the perpendicular distance from $I$ to the sides $AB$ and $AC$ are equal.\r\nSimilarly, since it lies on the bisector of $\\angle B$, the perpendicular distance from $I$ to the sides $AB$ and $BC$ are equal.\r\n\r\nFrom these two, the perpendicular distance from $I$ to the sides $AC$ and $BC$ are equal- hence $I$ lies on the bisector from $\\angle C$. We have shown that the angle bisectors from $\\angle A,\\angle B$ and $\\angle C$ all pass thorugh a point $I$: they are concurrent.", "Solution_7": "[hide=\"2\"]Use bysector theorem and then Ceva reciprocal :)[/hide]" } { "Tag": [ "geometry", "circumcircle", "cyclic quadrilateral", "geometry proposed" ], "Problem": "Let $ A_1A_2...A_n $ be a n-equilateral gons inscribed in (O) . $ M \\in $ the arc $ A_nA_1 $ (M is different from $ A_1 , A_n $) . Prove that :\r\n a) $ \\frac{1}{MA_1.MA_2} + \\frac{1}{MA_2.MA_3} + ... + \\frac{1}{MA_{n-1}.MA_n} = \\frac{1}{MA_n.MA_1} $ \r\n b) If n is an odd number then : $ MA_1 + MA_3 + ... + MA_n = MA_2 + MA_4 + ... + MA_{n-1} $", "Solution_1": "This is problem 253 from the last chapter of Geometric Transformations by I. M. Yaglom (Circular Transformations), which has never been translated to English. The first three chapters has been published by MAA as Geometric Transformations I (Isometries), Geometric Transformations II (Similarities) and Geometric Transformations III (Projectivities).\r\n\r\nDenote $a = A_1A_2 = A_2A_3 = ... = A_nA_1$ the side of the regular n-gon $A_1A_2...A_n$. Let $d_1 = MA_1$, $d_2 = MA_2$, ... $d_n = MA_n$ be the distances of the point $M$ on the arc $A_nA_1$ of its circumcircle $(O)$ from the vertices $A_1, A_2, ..., A_n$. We have to show that\r\n\r\n$\\frac{1}{d_1d_2} + \\frac{1}{d_2d_3} + ... + \\frac{1}{d_{n-1}d_n} = \\frac{1}{d_nd_1}$ [color=white]...[/color] (for any n)\r\n\r\n$d_1 + d_3 + ... + d_{n-2} + d_n = d_2 + d_4 + ... + d_{n-1}$ [color=white]...[/color] (for an odd n)\r\n\r\nInvert the n_gon in a circle centered at the point $M$ and an arbitrary radius $r$. The circumcircle $(O)$ is carried into a line $o$ and the vertices $A_1, A_2, ..., A_n$ into points $A_1', A_2', ..., A_n'$ on this line. The distance of the images $A_i', A_j'$ of points $A_i, A_j$ in an inversion with center $M$ and power $k = r^2$ is\r\n\r\n$A_i'A_j' = A_iA_j \\frac{r^2}{MA_i \\cdot MA_j}$\r\n\r\nHence, the distances $a_{12} = A_1'A_2'$, $a_{23} = A_2'A_3'$, ..., $a_{n1} = A_n'A_1'$ are equal to\r\n\r\n$a_{12} = a \\frac{r^2}{d_1d_2}$, $a_{23} = a \\frac{r^2}{d_2d_3}$, ..., $a_{n1} = a \\frac{r^2}{d_nd_1}$\r\n\r\nSince the inverted vertices $A_1', A_2', ..., A_n'$ are collinear and arranged on the line $o$ in this order, it is clear that\r\n\r\n$a_{12} + a_{23} + ... + a_{n-1,n} = a_{n1}$\r\n\r\nSubstituting the above expressions for $a_{12}, a_{23}, ..., a_{n-1,n}, a_{n1}$ and reducing by $ar^2$, we get the desired result:\r\n\r\n$\\frac{1}{d_1d_2} + \\frac{1}{d_2d_3} + ... + \\frac{1}{d_{n-1}d_n} = \\frac{1}{d_nd_1}$\r\n\r\nLet $b = A_1A_3 = A_2A_4 = ... = A_{n-1}A_1 = A_nA_2$ be the shortest diagonal of the regular n-gon $A_1A_2...A_n$, where n is odd.\r\n\r\nThe distances $b_{13} = A_1'A_3'$, $b_{24} = A_2'A_4'$, ..., $b_{n-1,1} = A_{n-1}'A_1'$, $b_{n2} = A_n'A_2'$ are equal to\r\n\r\n$b_{13} = b \\frac{r^2}{d_1d_3}$, $b_{24} = b \\frac{r^2}{d_2d_4}$, ..., $b_{n-1,1} = b \\frac{r^2}{d_{n-1}d_1}$, $b_{n2} = b \\frac{r^2}{d_nd_2}$\r\n\r\nSince the inverted vertices $A_1', A_2', ..., A_n'$ are collinear and arranged on the line $o$ in this order, it is clear that\r\n\r\n$a_{12} + a_{23} = b_{13}$\r\n\r\n$a_{23} + a_{34} = b_{24}$\r\n...\r\n\r\n$a_{n-2,n-1} + a_{n-1,n} = b_{n-2,n}$\r\n\r\nSince the inversion center $M$ is carried into the point at infinity, the last 2 equations are different:\r\n \r\n$a_{n-1,n} + b_{n-1,1} = a_{n1}$\r\n\r\n$b_{n2} + a_{12} = a_{n1}$\r\n\r\nSubstituting the above expressions for $a_{12}, a_{23}, ..., a_{n-1,n}, a_{n1}$, $b_{13}, b_{24}, ..., b_{n-1,1}, b_{n2}$ and reducing by $r^2$, we get:\r\n\r\n$a\\frac{1}{d_1d_2} + a\\frac{1}{d_2d_3} = b\\frac{1}{d_1d_3}$\r\n\r\n$a\\frac{1}{d_2d_3} + a\\frac{1}{d_3d_4} = b\\frac{1}{d_2d_4}$\r\n...\r\n\r\n$a\\frac{1}{d_{n-2}d_{n-1}} + a\\frac{1}{d_{n-1}d_n} = b\\frac{1}{d_{n-2}d_n}$\r\n\r\n$a\\frac{1}{d_{n-1}d_n} + b\\frac{1}{d_{n-1}d_1} = a\\frac{1}{d_nd_1}$\r\n\r\n$b\\frac{1}{d_nd_2} + a\\frac{1}{d_1d_2} = a\\frac{1}{d_nd_1}$\r\n\r\nRemoving the fractions, rearranging and indexing the equations by $bd_i$,\r\n\r\n[color=blue][b](1)[/b][/color] [color=white].....[/color] $bd_1 = ad_2 - ad_n$\r\n\r\n[color=blue][b](2)[/b][/color] [color=white].....[/color] $bd_2 = ad_3 + ad_1$\r\n\r\n[color=blue][b](3)[/b][/color] [color=white].....[/color] $bd_3 = ad_4 + ad_2$\r\n...\r\n\r\n[color=blue][b](n-2)[/b][/color] [color=white]..[/color] $bd_{n-2} = ad_{n-1} + ad_{n-3}$\r\n\r\n[color=blue][b](n-1)[/b][/color] [color=white]..[/color] $bd_{n-1} = ad_n + ad_{n-2}$\r\n\r\n[color=blue][b](n)[/b][/color] [color=white].....[/color] $bd_n = ad_{n-1} - ad_1$\r\n\r\nAdding separately the odd equations (1), (3), ... (n-2), (n) and the even equations (2), (4), ... (n-1), we get\r\n\r\n$b(d_1 + d_3 + ... + d_{n-2} + d_n) = 2a(d_2 + d_4 + ... + d_{n-1}) - a(d_n + d_1)$\r\n\r\n$b(d_2 + d_4 + ... + d_{n-1}) = a(d_1 + d_n) + 2a(d_3 + d_5 + ... + d_{n-2})$\r\n\r\nSubtracting the 2nd equation from the 1st one, rearranging the result and reducing by $2a + b$, we get the desired result\r\n\r\n$(2a + b)(d_1 + d_3 + ... + d_{n-2} + d_n) = (2a + b)(d_2 + d_4 + ... + d_{n-1})$\r\n\r\n$d_1 + d_3 + ... + d_{n-2} + d_n = d_2 + d_4 + ... + d_{n-1}$\r\n\r\nFor n = 3 (equilateral triangle $\\triangle A_1A_2A_3$), both results reduce to\r\n\r\n$d_1 + d_3 = d_2$\r\n\r\nwhere $d_1, d_2, d_3$ are the distances of the vertices $A_1, A_2, A_3$ from a point $M$ on the arc $A_3A_1$ of its circumcircle. This directly follows from Ptolemy's theorem for the cyclic quadrilateral $A_1A_2A_3M$:\r\n\r\n$A_1A_2 \\cdot MA_3 + A_2A_3 \\cdot MA_1 = A_3A_1 \\cdot MA_2$ \r\n\r\nSince $A_1A_2 = A_2A_3 = A_3A_1 = a$, \r\n\r\n$a d_3 + a d_1 = a d_2$\r\n\r\n$d_3 + d_1 = d_2$" } { "Tag": [ "modular arithmetic" ], "Problem": "Find the smallest positive integer $ n$ such that $ 107n$ has the same last two digits as $ n$.", "Solution_1": "[hide=\"Solution\"]We are looking for the smallest positive integer $ n$ such that $ 107n\\equiv n\\pmod{100}$.\n\n$ 107n\\equiv 7n\\pmod{100}\\quad\\Longrightarrow\\quad7n\\equiv n\\pmod{100}$\n\nSubtracting $ n$ from both sides, we get $ 6n\\equiv 0\\pmod{100}$.\n\nThe smallest positive integer $ n$ such that $ 100\\mid 6n$ is $ \\boxed{50}$.[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "I am going to post a short tutorial on combinations and permutations. \r\n\r\nFirst I will go through an example before I define these terms:\r\n\r\nHow many ways can you choose a shortstop, a second baseman, and a catcher from a team of 9 players?\r\n\r\nFirst we consider how many choices we have for the shortstop... there are 9. Now that we have a shortstop how many choices do we have for the second baseban... we have 8 choices(any of the 8 remaining players). Finally how many choices do we have for the catcher... 7. We multiply these numbers together to find the total number of ways we can pick these three positions 9*8*7=504. \r\n(note: it doesn't matter which position we pick first, each way will give us the same answer)\r\n\r\nNow let's say that we wanted to find out how many ways we can choose two of these players to play catch. \r\n\r\nUsing the same logic as last time we find 9*8=72. But wait we have overcounted... Let's say that two players are A and B. If we just do 9*8 we will count A playing catch with B as different than B playing catch with A. So we need to find out how many ways we can arrange A and B or any of the two players. We can arrange them in 2!=2 ways. Since we have overcounted by a factor of two we must divide 72 by 2 to get 36 different ways.\r\n\r\nIf we wanted to find out how many ways to pick groups of three we would do 9*8*7/(3*2*1)=84. \r\n\r\nOk the first example demonstrated the idea of permutations. A permutation is an arrangement of some objects when order DOES matter. It is often denoted by $P(n,r)$ or $_nP_r$ where you have a large group of n things and you want to find how many ways you can arrange r of them(so that order does matter). \r\n\r\nThe second example was of combinations. A combination is an arrangement of some objects where the order of them DOES NOT matter. There are a few ways it is denoted: $_nC_r= ^nC_r=C^n_r=C(n,r)={n\\choose r}$ I personally like the last one but the second to last one is easiest to use if you don't have $\\LaTeX$ ;) .\r\n\r\nProblems:\r\n(please leave these problems to those who are new to this)\r\n\r\n1. How many arrangements are there of the word \"math\"?\r\n\r\n2. How many arrangements are there of the word \"Mississippi\"?\r\n\r\n3. How many 5 letter words are there( a word is any arrangement of letter i.e. it doesn't have to be an actual \"word\")?\r\n\r\n4. How many 5 letter words are there if you can only use a letter once?\r\n\r\n5. How many diagonals does an n-sided polygon have?\r\n\r\n6. How many ways can six people sit around a table?\r\n\r\n7. How many ways can six people sit around a table if Ben and Jamie wont sit next to each other?\r\n\r\n8. How many distinct arrangements of 1001011110 are there? \r\n\r\nFeel free to post more problems if you want.", "Solution_1": "In how many ways can you split up a group of n persons into $k \\leq n$ teams (a team must have at least 1 person)?", "Solution_2": "1. [hide]\n\nWe have 4 diffrent letters and positions so we have 4!=24 diffrent possipilitys [/hide]\n\n3.[hide]\n\nWe have 5 positions 26 letters thus for every position are 26 diffrent possibilitys that means 26*26*26*26*26=26^5=11881376 diffrent ways\n\n[/hide]\n\n4.[hide]\n\nIt's like problem 3 ecept we have 26letters for position 1 25letters for position 2 etc. so there are 26*25*24*23*22=7893600 possibilitys \n\n[/hide]\n\n\n\nIs that correct so far?", "Solution_3": "For joml's #1, and other similar to it, don't you mean \"How many arangements are there when putting the letter in the work 'math' together\"? Looking at 1, I would say 1, the word \"math\". It isn't really that clear to me what I'm supossed to be solving (my questions isn't much better though).", "Solution_4": "By arrangements I mean like tham, and athm...", "Solution_5": "6.\n\n[hide]The persons ABCDEF are sitting around the table. A can change 5 times his position with one of the others to get a new order. B has already changed possition with A so he can still change with CDEF. Then C has changed with AB so there remain DEF etc. Thus we can use the sum formula\n\nn(n-1)/2=6*5/2=15[/hide]", "Solution_6": "i have another problem. how many ways can you draw n-2 triangles on an n-gon?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "I haven't proved the problem :\r\n \"Proved that have no k,n,m,a >=2 in N satisfy:\r\n (m-k+1)....m=a^n.\r\nwith k1)$\r\n\r\n :)", "Solution_1": "$ t=\\tan \\frac{x}{2}\\Longrightarrow \\cos x=\\frac{1-t^{2}}{1+t^{2}}$.", "Solution_2": "[quote=\"kunny\"]$ t=\\tan \\frac{x}{2}\\Longrightarrow \\cos x=\\frac{1-t^{2}}{1+t^{2}}$.[/quote]\r\n\r\n$ I=\\int\\frac{dx}{1+a\\cos x}$\r\n\r\n$ Let\\;t = \\tan{\\frac{x}{2}}$\r\n$ \\Rightarrow\\;dt = \\frac{1}{2}\\sec^{2}{\\frac{x}{2}}dx$\r\n$ \\Rightarrow\\;I=\\int\\frac{2dt}{(1+t^{2})(1+a\\frac{1-t^{2}}{1+t^{2}})}$\r\n$ \\Rightarrow\\;I=-\\frac{2}{a+1}\\int\\frac{dt}{\\frac{a-1}{a+1}t^{2}-1}$\r\n\r\n$ Let\\;u =\\sqrt{\\frac{a-1}{a+1}}t$\r\n$ \\Rightarrow\\;du = \\sqrt{\\frac{a-1}{a+1}}dt$\r\n$ \\Rightarrow\\;I=-\\frac{2}{a+1}*\\sqrt{\\frac{a+1}{a-1}}\\int\\frac{du}{u^{2}-1}$\r\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}\\int\\frac{du}{1-u^{2}}$\r\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;u+C$\r\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;(\\sqrt{\\frac{a-1}{a+1}}\\tan{\\frac{x}{2}})+C$", "Solution_3": "[quote=\"IrinaSharova\"]$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}\\int\\frac{du}{1-u^{2}}$\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;u+C$\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;(\\sqrt{\\frac{a-1}{a+1}}\\tan{\\frac{x}{2}})+C$[/quote]\r\n\r\nOr $ I=\\frac{2}{\\sqrt{a^{2}-1}}\\int\\frac{du}{1-u^{2}}$ which gives $ I = \\frac{2}{\\sqrt{a^{2}-1}}\\int \\frac{du}{(1+u)(1-u)}$ and use partial fractions:\r\n\r\n$ I = \\frac{1}{\\sqrt{a^{2}-1}}\\int (\\frac{1}{1-u}+\\frac{1}{1+u}).du$\r\n\r\nSo: $ I = \\frac{1}{\\sqrt{a^{2}-1}}ln | \\frac{1+u}{1-u}|$ and then re-substitute the variables.", "Solution_4": "[quote=\"BanishedTraitor\"][quote=\"IrinaSharova\"]$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}\\int\\frac{du}{1-u^{2}}$\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;u+C$\n$ \\Rightarrow\\;I=\\frac{2}{\\sqrt{a^{2}-1}}arctanh\\;(\\sqrt{\\frac{a-1}{a+1}}\\tan{\\frac{x}{2}})+C$[/quote]\n\nOr $ I=\\frac{2}{\\sqrt{a^{2}-1}}\\int\\frac{du}{1-u^{2}}$ which gives $ I = \\frac{2}{\\sqrt{a^{2}-1}}\\int \\frac{du}{(1+u)(1-u)}$ and use partial fractions:\n\n$ I = \\frac{1}{\\sqrt{a^{2}-1}}\\int (\\frac{1}{1-u}+\\frac{1}{1+u}).du$\n\nSo: $ I = \\frac{1}{\\sqrt{a^{2}-1}}ln | \\frac{1+u}{1-u}|$ and then re-substitute the variables.[/quote]\r\n\r\nYes, either way will do, :) they are equivalent." } { "Tag": [], "Problem": "How many of the 4 digit integers ( from 1,000 to 9,999) have 4 distinct digits that are either increasing or decreasing?", "Solution_1": "Do they have to be strictly increasing/decreasing? e.i. Does the number $ 2344$ count? What about $ 2222$ ?\r\n\r\nSame question for \"Help 2 :( \" .", "Solution_2": "[quote=\"marcopolo\"]How many of the 4 digit integers ( from 1,000 to 9,999) have 4 distinct digits that are either increasing or decreasing?[/quote]\r\n\r\nJust saying, if you want a better response, you may want to try to name your threads better. We understand that you need help, say something like: increasing/decreasing digits combo\r\nNot something like: combinations (there's thousands of problems involving combinations) or help!" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "How to prof that sin( sqrt(x) ) < sqrt ( sin (x) ), 0 < x < Pi/2 ???", "Solution_1": "ARO 2006 : http://www.mathlinks.ro/Forum/viewtopic.php?t=86557", "Solution_2": "Thank you very very much!!!! It's not very difficult problem, i don't know why i couldn't to solve it." } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "A function $f(x)$ satisfies $f(x)=f\\left(\\frac{c}{x}\\right)$ for some real number $c(>1)$ and all positive number $x$. \r\n\r\nIf $\\int_1^{\\sqrt{c}} \\frac{f(x)}{x} dx=3$, evaluate $\\int_1^c \\frac{f(x)}{x} dx$", "Solution_1": "[quote=\"kunny\"]A function $f(x)$ satisfies $f(x)=f\\left(\\frac{c}{x}\\right)$ for some real number $c(>1)$ and all positive number $x$. \n\nIf $\\int_1^{\\sqrt{c}} \\frac{f(x)}{x} dx=3$, evaluate $\\int_1^c \\frac{f(x)}{x} dx$[/quote]\r\nI guess the answer is $6$ :P", "Solution_2": "Let $u=\\frac{c}{x}$, so $du=\\frac{-c}{x^{2}}dx$, so $\\int_1^{\\sqrt{c}} \\frac{f(x)}{x} dx= \\int_c^{\\sqrt{c}}\\frac{u f(u)}{c}(\\frac{-x^2}{c})du=\\int_{\\sqrt{c}}^c \\frac{f(u)}{u} du$.\r\n\r\nTherefore, $\\int_1^c \\frac{f(x)}{x} dx=\\int_1^{\\sqrt{c}} \\frac{f(x)}{x} dx+\\int_{\\sqrt{c}}^c \\frac{f(u)}{u} du=3+3=6$, $\\mathbb{QED}.$", "Solution_3": "Your answers are correct, two guys. :)" } { "Tag": [ "number theory", "greatest common divisor" ], "Problem": "Express 7 as a linear combination of 15 and 4, or prove that it is impossible.\r\n\r\nthere is something wrong with the questions above. It was given by my teacher for revision. He says there's an error inside and u got to show it why it is wrong. \r\n\r\nI so far know that gcd (15,4) = 1 hence it contradicts to 7??? \r\n\r\nI got lost from there as if I don't know to show that there's an error ( or I could just fail to explain correctly that there's an error)\r\n\r\nMy apologies cuz this is my first time trying such questions as well as first time studying elementary no. theory.", "Solution_1": "I would guess that the question is wrong since it didn't specify whether the combinations of 15 or 7 could be negative? (Which is required for the problem to be solved correctly...)\r\n\r\nBut I didn't see any other errors." } { "Tag": [ "inequalities", "LaTeX", "inequalities unsolved" ], "Problem": "Show that", "Solution_1": "much easier to read it latex: $\\sqrt{e^e\\cdot\\pi^{\\pi}} > e^{\\pi}$." } { "Tag": [ "geometry", "perimeter", "geometry proposed" ], "Problem": "Let $ ABC$ be a triangle with angles $ \\pi/4 < \\alpha, \\beta, \\gamma < \\pi/2$. \r\n\r\na) Prove that it is possible to put, inside this triangle, three squares with this four properties:\r\n\r\ni) the three squares have the same area\r\n\r\nii) they all have a common vertex $ K$\r\n\r\niii) $ K$ is the only point which belongs to two different squares\r\n\r\niv) each square has two [b]opposite[/b] vertex on the perimeter of $ ABC$.\r\n\r\n\r\n\r\nb) Consider the square (external to $ ABC$) whose side is $ BC$, and let O be its centre. Let $ r_A$ be the symmetric of the line $ AK$ with respect to the bisector of $ \\alpha$. Prove that $ O \\in r_A$.", "Solution_1": "The problem should not be so difficult. Drawing equal antiparallel MN, PQ and RS to BC, CA and AB respectively, we know the hexagon MNQPSR is cyclic, let's K being its' excircle's center. All we need, basically, is to find the lenght of MN such as the distance from K to it is half of its lenght, then to show K is the isogonal conjugate of intersection of OA and similar ones from B and C w.r.t. ABC.\r\nHowever I do not have time for such a search.\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [], "Problem": "How many different ways can the letters of the world FOUR be scrambled so that the first letter is a vowel and the last letter is a consonant?", "Solution_1": "There are 2 vowels and 2 consonants.\r\nThe first letter is a vowel, so there are $ 2$ choices.\r\nThe last letter is a consonant, so there are again $ 2$ choices.\r\nThe 2 middle letters are 1 consonant and 1 vowel, in any order.\r\nThere is only 1 vowel left and 1 consonant left, but there are $ 2$ possible orders\r\nSo the answer is $ 2*2*2 \\equal{} \\boxed{8}$" } { "Tag": [ "function", "calculus", "derivative", "analytic geometry", "graphing lines", "slope", "inequalities" ], "Problem": "find if this function is increasing or decreasing\r\n$ f(x)\\equal{}ln(x^2\\plus{}1)\\minus{}e^{\\minus{}x}\\plus{}1$", "Solution_1": "EDIT : oops, sry. its $ f(x_1)$ and $ f(x_2)$, not $ f'(x_1)$ and $ f'(x_2)$.\r\n\r\nwhen $ x_1 < x_2$, if $ f(x_1)\\le f(x_2)$,a function is increasing.\r\n\r\nas you said, $ f(x) = ln(x)$ is an increasing function in $ (0,\\infty)$\r\n\r\ntherefore, x can't be negative.\r\n\r\n$ f(0) = 0$\r\n$ f(\\infty) = \\infty$\r\n\r\nbut it could possibly be decreasing at some point.\r\n\r\nwe should find the derivative of the function.\r\n\r\n$ f'(x) = \\frac {2x}{x^2 + 1} + e^x$\r\n\r\nin a graph, if a slope of a point is 0, that point is the graph's maximum or minimum.\r\n\r\nso we are looking for a value of x that $ \\frac {2x}{x^2 + 1} + e^x = 0$.\r\n\r\nif $ 0\\lex$, the value of expression is larger than 0.\r\n\r\nand x can't be negative.\r\n\r\ntherefore, there are no maximum or minimum in this function.\r\n\r\nso the function is continuously increasing.", "Solution_2": "\"When $ x_1\\frac{\\minus{}2x}{x^2\\plus{}1}$\n\nSince $ x^2\\plus{}1$ is always positive\n\n$ \\frac{x^2\\plus{}1}{e^x}>\\minus{}2x$\n\n\nAdding $ 2x$ to both sides,\n\n$ \\frac{x^2\\plus{}1\\plus{}2xe^x}{e^x}>0$\n\nSince e^x is positive,\n\n$ x^2\\plus{}1\\plus{}2xe^x>0$\n\nI can't figure out where to go from there\n[/hide]", "Solution_5": "Little error [b]mewto55555[/b]\r\n[hide=\"Correction\"]\nIn the case considering, $ f' \\left( \\minus{} x \\right) \\equal{} e^{x} \\minus{} \\frac {2x}{x^{2} \\plus{} 1}$.\nFollowing the inequality with the correction reaches the conclusion.\n$ e^{x} > \\frac { \\minus{} 2x}{x^{2} \\plus{} 1} \\implies f' \\left( x \\right) > 0 \\ \\forall \\ x \\Leftrightarrow f \\left( x \\right)$ is an increasing function.\n$ \\left( e^{x} \\equal{} \\sum_{n \\equal{} 0}^{\\infty} \\frac {x^{n}}{n!} \\right)$.\n[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a,b,c \\ge 0, a^2+b^2+c^2=3$, then:\r\n\\[ a+b+c \\ge (ab)^2+(bc)^2+(ca)^2 \\]", "Solution_1": "I think it is here :\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=82512" } { "Tag": [ "geometry", "percent", "AMC", "USA(J)MO", "USAMO", "trigonometry", "area of a triangle" ], "Problem": "Is it generally easy to change courses once you get there? Has anyone had any experience doing so?\r\n\r\nAre the two geometry courses [i]Computational Geometry[/i] and [i]Geometric Proofs[/i] independent of each other? Does one require the knowledge of the other course? (I ask this because I can hardly recognize any of the terms under the description of \"Computational Geometry\", but I feel that I understand the description of \"Geometric Proofs\".) How difficult is \"Geometric Proofs\"?\r\n\r\nI think the most appropriate courses for me in each topic would be Alg. 2.5, Combinatorial Arguments, Computational Geometry, and Number Theory. However, these are all afternoon session classes :( .\r\n\r\nI am also very worried that I have overestimated my own skill...", "Solution_1": "In my experience, it's not that hard to change courses even when you've already started. The teachers didn't take roll last year, so it was possible to just ditch one class and go to another.\r\n\r\nAlso, don't worry about overestimating your own skill. You are the person who knows best what's good for you, and if you did overestimate, then you will end up learning more. Don't worry about it!", "Solution_2": "[quote=\"Meggy-Meg\"]In my experience, it's not that hard to change courses even when you've already started. The teachers didn't take roll last year, so it was possible to just ditch one class and go to another.\n\nAlso, don't worry about overestimating your own skill. You are the person who knows best what's good for you, and if you did overestimate, then you will end up learning more. Don't worry about it![/quote]\r\n\r\nSo if we underestimate our skill, we can change classes?", "Solution_3": "Would it be bad if you took Geo Proofs but didn't understand the material of Computational Geometry? I need to find a course for the morning....", "Solution_4": "I don't think that's a very good idea though...since you'll probably just fall behind\r\n\r\nFor me, I'm thinking about Geo Proofs and Alg 2.5, since I understand basically everything in Computational Geo, and I need an intermediate afternoon course.\r\n\r\nBut the problem is that Geo and Alg are probably my best 2 subjects, so I'm thinking I might want to take Combo or NT, since I'm pretty bad at those.\r\n\r\nSo should I do my first choice, or should I do Computational Geo and Modular Arithmetic/Counting Strategies, since I don't want to take any beginner classes.", "Solution_5": "Yeah, I have a similar problem as Carbon57's.... I don't want to take any beginner classes, and I think Combs 2 and NumThe 2 might be too easy... but then that leaves all my classes in the afternoon session...\r\n\r\nI was really disappointed when I got the course selection sheet for two reasons:\r\n-I thought we were taking 4 courses\r\n-Certain courses have conflicting schedules\r\n\r\n :( Oh well I'm sure it'll be fun anyways", "Solution_6": "Last year I wasn't sure about the whole idea of taking only two courses, but having been at the camp, I think that it is best. Four classes is too much information to absorb in such a short time. There is a difference between learning a technique, and absorbing it into your repertoire of techniques that you are comfortable with.\r\n\r\nI think that each class has a pretty wide range of difficulty. I remember in the problem sets, there were some people who solved very few problems alone, some that could do say half, I could do like 95 percent of them and pretty much nobody could solve everything. Also you can always get more difficult problems from your teachers if you think its too easy. But if that doesn't help, I don't think that it should be that hard to switch classes.\r\n\r\nAnd I don't think that the counting strategies class should be that easy. Topics like recursion or bijections can be easy in some contexts, but they can be very difficult in other contexts. Apparently this class also covers some entry level USAMO problems...", "Solution_7": "[quote=\"leoxnlin\"] Are the two geometry courses [i]Computational Geometry[/i] and [i]Geometric Proofs[/i] independent of each other? Does one require the knowledge of the other course? (I ask this because I can hardly recognize any of the terms under the description of \"Computational Geometry\", but I feel that I understand the description of \"Geometric Proofs\".) How difficult is \"Geometric Proofs\"?[/quote]\r\n\r\nHello, I will be part of the team teaching the \"Geometric Proofs\" class. I can give you my thoughts. Of course we plan to teach the class at a level after assessing the nature of the students in the class. But you should probably review those theorems (It shouldn't take more than 5 minutes to look up say Ceva or Menelaus so you know the statement at least). The goal of the class is to teach how to use these techniques more than to introduce them. I plan to prove a number of those theorems (like ptolemy). The law of sines is useful. Stewart's, law of cosines, heron's formula are unlikely to come up because they are mainly computational.", "Solution_8": "Altheman, in your opinion is it good to take both olympiad geometry from AOPS and Geometric Proofs from AMSP?", "Solution_9": "leoxnlin. dont wry bout ur skill. ur rly talented :) u'll be fine in the classes u mentioned. \r\ncomp. geo was a cool class [ i liked it --- even tho it was prettyy like 'wth am i doing' for me..] but u'll be fine XD", "Solution_10": "Check the topic post date, shelly32494. :wink:", "Solution_11": "I am fairly certain that the courses change from year to year.\r\nI am desperately hoping for combinatorics and number theory." } { "Tag": [ "linear algebra", "matrix", "integration", "Putnam", "superior algebra", "superior algebra solved" ], "Problem": "We have the matrices A,B in M_n(R) which verify the relations: A^(2002)=A^(2003) and A+B=In. Prove that In-AB is invertible!\r\n\r\nCheers! :D :D", "Solution_1": "are you sure this is the right question? maybe A2002= B2003 ?!", "Solution_2": "This problem was given in GM 3/2003 and the text is like I gave it .. perhaps it's a typing error there?! but I have found no erata! \r\n\r\ncheers! :D :D", "Solution_3": "in that case from (1-x)2002(1-1+x) = 0 => (1-x)2002x=0 which means that the proper values of B can only be 1 and 0 (1). \r\n\r\nbut In - AB = In-B+B2 = (B+ eIn)(B + e2In) where e3=1, e <> 1 and we obviously cannot have e or e2 as proper values of B (because of (1)), so det In -AB >0 (!).", "Solution_4": "[quote=\"Lagrangia\"]A,B in M_n(R) which verify the relations: A^(2002)=A^(2003) and A+B=In. Prove that In-AB is invertible!\n[/quote]\r\n\r\nI-AB=I-(A-A)\r\n\r\nFirst case if A=A projection matrix, I-AB=I is invertible\r\n\r\nSecond case if A-A=/=0 \r\n\r\nlemma 1: (A-A)^2002=0\r\nprof: (A-A)^2002=A^(2002)(A-I)(A-I)^2001=(A^2003-A^2002)(A-I)^2001=0\r\n\r\nlemma 2: If M \\in Mn(R) such that there exist q \\in N \\{0;1}, M^q=0 then \r\nI-M is invertible, its inverse is I+M+...+M^(q-1)\r\n\r\nprof: (I-M)(I+M+...+M^(q-1))=\r\nI+M+...+M^(q-1)-M-M-...-M^(q-1) - M^q = I \r\n---------------------------------------------------------------------\r\nlemma 2 with M=A-A gives the inverse of I-AB=I-(A-A) is \r\nI+(A-A)+....+(A-A)^2001\r\n\r\n\r\nRemark: If M is nilpotent then det(I-M)=1 \r\nIndeed nilpotent matrix M is similar to triangular superior with diagonal \r\n(0,...,0), under the diagonal only 0\r\nsi I-M is similar to \r\n\r\n(1 * * .........................................* *)\r\n( :D 1 *......................................* *)\r\n(....................................................*)\r\n(....................................................*)\r\n( :D :D :D ........................... :D 1) \r\nthe determinant of this matrix is 1\r\n\r\nLet me guess this is 12th grade problem in GM, isn't it ?\r\nProblems in G :D M Gazeta :D Matematica :D is very nice :D \r\n :D \\cup \\beta \\int n :D :D \\lambda", "Solution_5": "well ... that's about what I proved :) nice soln though mo \\cup bi :)", "Solution_6": "This one is actually an 11th grade problem,moubi! :D :D\r\n\r\nThis is alos another 11th grade problem:\r\n\r\nConsider the matrices A,B of order n with real entries such that A^3=B^3 and A^2B=B^2A.\r\nProve that at least one of the matrices A^2+B^2, A-B, A+B is invertible..\r\n\r\nthis shouldn't be too hard! :D \r\n\r\ncheers! :D :D", "Solution_7": "Your second problem proves to be false if A=B=O_n. What's the correct text?", "Solution_8": "maybe both A and B are different from 0n ?!", "Solution_9": "Like the other problem, I took it from the GM, and the text is correct how I posted it! At least it is the same like in the gazzette. I also think that A,B <>On!\r\n\r\ncheers! :D :D", "Solution_10": "Well then just take A=B such that det(A)=0. :D", "Solution_11": "Hmm.. I see what you mean andrei.. :D :D \r\nLets try to solve the problem if A and be are both different than On!\r\n\r\ncheers! :D :D", "Solution_12": "This is an older Putnam problem or something, from 1991. It's in the book with olympiads for grades 11-12 , 2003 from Gil Publishing House at the \"Chosen Problems in Algebra\" section. \r\n\r\nCheers ;)", "Solution_13": "http://www.kalva.demon.co.uk/putnam/psoln/psol912.html", "Solution_14": "There's another thing : A<>B. The soln: (A^2+B^2)(A-B)=On (calculate it) so if A^2+B^2 is invertible then A-B=On, which is false.", "Solution_15": "Okay! I'll move this post to solved problems because the last problem is totally flawed. (If A=B and detA=0 then ...)" } { "Tag": [], "Problem": "These are not my equations, they are h's equations.\r\n\r\nh = e^2 * z0 / (2 alpha)\r\nh = [(1.6021765(31) x 10^-19 A-s)^2] * (3.767303134... x 10^2 kg-m^2/A^2-s^3) / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)]\r\nh = (2.5669696(36) x 10^-38 A^2-s^2) * (3.767303134... x 10^2 kg-m^2/A^2-s^3) / (1.4594705(14) x 10^-2 rad)\r\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\r\n\r\nh = e^2 / (2 alpha) * e0 * c\r\nh = [(1.6021765(31) x 10^-19 A-s)^2] / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)] * (8.854187817... x 10^-12 A^2-s^4/kg-m^3) * (2.99792458 x 10^8 m/s)\r\nh = (2.5669696(36) x 10^-38 A^2-s^2) / (1.4594705(14) x 10^-2 rad) * (8.854187817... x 10^-12 A^2-s^4/kg-m^3) * (2.99792458 x 10^8 m/s)\r\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\r\n\r\nh = e^2 * u0 * c / (2 alpha)\r\nh = [(1.6021765(31) x 10^-19 A-s)^2] * (1.256637061... x 10^-6 kg-m/A^2-s^2) * (2.99792458 x 10^8 m/s) / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)]\r\nh = (2.5669696(36) x 10^-38 A^2-s^2) * (1.256637061... x 10^-6 kg-m/A^2-s^2) * (2.99792458 x 10^8 m/s) / (1.4594705(14) x 10^-2 rad)\r\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\r\n\r\nAre h's equations right, I cannot find h anywhere.\r\n\r\nThanks!", "Solution_1": "[quote=\"Garry Denke\"]These are not my equations, they are h's equations.\n\nh = e^2 * z0 / (2 alpha)\nh = [(1.6021765(31) x 10^-19 A-s)^2] * (3.767303134... x 10^2 kg-m^2/A^2-s^3) / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)]\nh = (2.5669696(36) x 10^-38 A^2-s^2) * (3.767303134... x 10^2 kg-m^2/A^2-s^3) / (1.4594705(14) x 10^-2 rad)\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\n\nh = e^2 / (2 alpha) * e0 * c\nh = [(1.6021765(31) x 10^-19 A-s)^2] / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)] * (8.854187817... x 10^-12 A^2-s^4/kg-m^3) * (2.99792458 x 10^8 m/s)\nh = (2.5669696(36) x 10^-38 A^2-s^2) / (1.4594705(14) x 10^-2 rad) * (8.854187817... x 10^-12 A^2-s^4/kg-m^3) * (2.99792458 x 10^8 m/s)\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\n\nh = e^2 * u0 * c / (2 alpha)\nh = [(1.6021765(31) x 10^-19 A-s)^2] * (1.256637061... x 10^-6 kg-m/A^2-s^2) * (2.99792458 x 10^8 m/s) / [(2.00000000 x 10^0 rad/sr) * (7.2973525(68) x 10^-3 sr)]\nh = (2.5669696(36) x 10^-38 A^2-s^2) * (1.256637061... x 10^-6 kg-m/A^2-s^2) * (2.99792458 x 10^8 m/s) / (1.4594705(14) x 10^-2 rad)\nh = 6.6260693(11) x 10^-34 kg-m^2/s-rad\n\nAre h's equations right, I cannot find h anywhere.\n\nThanks![/quote]\r\n\r\n\r\nwho or what is h . if it refers to $plancks$ $constant$ then what do u want?", "Solution_2": "Where are h's equations published?\r\n\r\n[b]h = e^2 * z0 / (2 alpha)\nh = e^2 / (2 alpha) * e0 * c\nh = e^2 * u0 * c / (2 alpha)[/b]\r\n\r\nI can't find h anywhere.\r\n\r\nThanks!", "Solution_3": "[quote=\"Garry Denke\"]Where are h's equations published?\n\n[b]h = e^2 * z0 / (2 alpha)\nh = e^2 / (2 alpha) * e0 * c\nh = e^2 * u0 * c / (2 alpha)[/b]\n\nI can't find h anywhere.\n\nThanks![/quote]\r\n\r\ni really could not understad by \"h's equations \" and please specify what ur variables are :rotfl:", "Solution_4": "$\\alpha$ is the fine structure constant, $h$ is Planck's constant, $e_{0}$ (rather $\\varepsilon_{0}$) is the vacuum electric permeability, $c$ is the speed of light, $e$ is the electron charge... The middle equation is correct, I don't know what $z_{0}$ and $u_{0}$ mean. Can you clarify your question? Do you want to confirm that these equations are correct?", "Solution_5": "In sci.physics Sue found the second,\r\n\r\nhttp://physics.nist.gov/cuu/Images/alphaeq.gif \r\nhttp://physics.nist.gov/cuu/Constants/alpha.html\r\n\r\nbut the first and the last I can't find.\r\n\r\nhttp://physics.nist.gov/cgi-bin/cuu/Value?z0\r\nhttp://physics.nist.gov/cgi-bin/cuu/Value?mu0\r\n\r\nYes, thanks!" } { "Tag": [ "logarithms", "calculus", "calculus computations" ], "Problem": "Find the sum:\r\n\r\n$ \\sum\\limits_{n\\equal{}1}^{\\infty}\\left(n\\ln \\left(\\frac{2n\\plus{}1}{2n\\minus{}1}\\right)\\minus{}1\\right)$.", "Solution_1": "It can be shown using a series expansion for ln (1+x) that the series converges. the sum? well, mathematica gives it as ___. :roll: \r\n\r\nMaybe someone can find where the alternating harmonic series is.", "Solution_2": "$ \\sum_{n\\equal{}1}^m n\\ln(2n\\plus{}1)\\minus{}\\sum_{n\\equal{}1}^m n \\ln(2n\\minus{}1)\\minus{}m$\r\n\r\n$ \\equal{}\\sum_{n\\equal{}0}^m n\\ln(2n\\plus{}1)\\minus{}\\sum_{n\\equal{}0}^m (n\\plus{}1) \\ln(2n\\plus{}1)\\plus{}(m\\plus{}1)\\ln(2m\\plus{}1)\\minus{}m$\r\n\r\n$ \\equal{}\\underbrace{\\minus{}\\sum_{n\\equal{}0}^m \\ln(2n\\plus{}1)}_{\\ln\\frac{m!\\, 2^m}{(2m\\plus{}1)!}}\\plus{}(m\\plus{}1)\\ln(2m\\plus{}1)\\minus{}m$\r\n\r\nFor $ m\\to\\infty$ use the stirling approximation and you get $ \\frac{1\\minus{}\\ln 2}{2}$ ." } { "Tag": [ "function", "modular arithmetic", "floor function", "number theory", "relatively prime", "number theory unsolved" ], "Problem": "Let $ N$ be a positive integer. How many non-negative integers $ n \\le N$ are there that have an integer multiple, that only uses the digits $ 2$ and $ 6$ in decimal representation?", "Solution_1": "[quote=\"FelixD\"]Let $ N$ be a positive integer. How many non-negative integers $ n \\le N$ are there that have an integer multiple, that only uses the digits $ 2$ and $ 6$ in decimal representation?[/quote]\r\nFor any $ p$ which is relative prime to 10 , we will show that there exist an integer $ k$ such that it divisible by $ p$ .Let's consider number\r\n$ k\\equal{}\\frac{6.10^n(10^m\\minus{}1)}{9}\\plus{}\\frac{2.(10^n\\minus{}1)}{9}\\equal{}\\frac{6.10^{m\\plus{}n}\\minus{}4.10^n\\minus{}2}{9}$\r\nOnly need to chose $ m,n$ such that both of $ m\\plus{}n$ and $ n$ are multiple of $ \\varphi(9p)$ ,by using Euler's function we can easy to show that $ k$ is a multiple of $ p$\r\nConsider a number $ k$ only use the digit $ 2,6$ then it is not hard to show that $ 5$ is not a divisor of \r\n$ k$ and $ \\frac{k}{2}$ only uses the digit $ 1,3$ , it means that $ 2\\parallel{}k$\r\nTherefore n satisfy condition if and only if $ n$ is not a multiple of $ 4$ or $ 5$ . The number of positive integer satisfy condition is : \r\n\\[ N\\minus{}[\\frac{N}{4}]\\minus{}[\\frac{N}{5}]\\plus{}[\\frac{N}{20}]\r\n\\]", "Solution_2": "It is necessary that $ n$ is not divisible by both $ 4$ and $ 5$,because $ 22,26,62,66$ are all not divisible by $ 4$,and units digit of multiple of $ 5$ never be $ 2$ or $ 6$.\r\nIf even $ n$ satisfies the condition,then so does $ \\frac {n}{2}$,\r\ntherefore we prove there exists such multiple of $ n$ that satisfies the condition when $ n$ is even and not divisible by $ 4$ or $ 5$.\r\n\r\nLet $ n \\equal{} 2m$,where $ m\\in \\mathbb{N}$ is relatively prime to $ 10$.\r\nwe know there exists a positive integer $ a$ such that\r\n$ 10^a\\equiv 1\\pmod{m}$.\r\nthen $ ma$-digit integer $ b \\equal{} \\underbrace{111\\cdots 11}_{ma}\\equiv \\underbrace{\\underbrace{11\\cdots 1}_{a} \\plus{} \\underbrace{11\\cdots 1}_{a} \\plus{} \\cdots \\underbrace{11\\cdots 1}_{a}}_{m}\\equiv 0\\pmod{m}$,\r\n\r\nwhich means $ 2b\\equal{}\\underbrace{222\\cdots 22}_{ma}$ is a multiple of $ n$,and we get the answer $ N \\minus{} \\left\\lfloor \\frac {N}{4}\\right\\rfloor \\minus{} \\left\\lfloor \\frac {N}{5}\\right\\rfloor \\plus{} \\left\\lfloor \\frac {N}{20}\\right\\rfloor$." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra theorems" ], "Problem": "I want to be able to create a multiplication table of Zmod2[x]/. I just don't even know what to google. Can someone please provide the name of the topic that would cover this technique of polynomial multiplication over a polynomial modulus? Thanks!", "Solution_1": "The Chinese remainder theorem for rings may be most useful here, as it reduces your case to a very simple one.\r\nThe topic you'd be looking for here would be ideals and modulo in rings. In this case, chinese remainder theorem for rings.\r\n=Uncool-" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $a,b,c \\geq 0$\r\nif\r\n$b+c \\leq 1+a$ , \r\n$a+c \\leq 1+b$,\r\n$a+b \\leq 1+c$.\r\nProve that\r\n$\\sum_{cyc}a^{2}$ $\\leq$ $2abc+1$", "Solution_1": "we know $\\prod_{cyc}(a+b-c) \\leq abc$\r\nso the given inequality is $\\equiv$ to proving \r\n$\\sum_{cyc}a^{2}\\leq 2\\prod_{cyc}(a+b-c)+1$ ------------*\r\n\r\nset,\r\n$a+b-c=u$\r\n$b+c-a=v$\r\n$c+a-b=w$\r\nso,\r\nthe inequality * $\\equiv$ $8uvw+4$ $\\geq$ $\\sum_{cyc}(u+v)^{2}$\r\nwhere $0 \\leq u,v,w \\leq 1$\r\nset ,$f(x)=(x+u)^{2}+(x+v)^{2}+(u+v)^{2}-8uvx-4$\r\n$f^{''}(x)=4>0$\r\nso $f$ attains maximum in it's boundary .so we have to check just $4$ values now\r\nnamely $(d,e,f)$ where $d,e,f$ are either 0 or 1.\r\nso we have by checking $max[f(u,v,w)]=f(1,1,1)=0$\r\nand we are done.", "Solution_2": "Hm.. If $f''(x) > 0$, then I think $f$ attains its minimum at its boundary.. :maybe:", "Solution_3": "[quote=\"RDeepMath91\"]If $f''(x) > 0$, then I think $f$ attains its minimum at its boundary[/quote]\r\n\r\nYes that's true . since the variables are in [0,1] , (1,1,1) is a boundary :wink:", "Solution_4": "[quote=\"RDeepMath91\"]Hm.. If $f''(x) > 0$, then I think $f$ attains its minimum at its boundary.. :maybe:[/quote]\r\nno that's not true it attains it's maximum in the boundary .", "Solution_5": "[quote=\"SOUMYASHANT NAYAK\"]let $a,b,c \\geq 0$\nif\n$b+c \\leq 1+a$ , \n$a+c \\leq 1+b$,\n$a+b \\leq 1+c$.\nProve that\n$\\sum_{cyc}a^{2}$ $\\leq$ $2abc+1$[/quote]\r\n\r\nlet be\r\n$A=1-a$\r\n$B=1-b$\r\n$C=1-c$\r\n\r\nthe conditions become\r\n$A,B,C\\leq 1$\r\n$A+B\\geq C$\r\n$B+C\\geq A$\r\n$C+A\\geq B$\r\n\r\nadding the first two we have $B\\geq 0$ and the same for $A$ and $C$, so $A,B$ and $C$ are sides of a triangle, and exist $x,y,z\\geq 0$ such that\r\n$x+y=C$\r\n$y+z=A$\r\n$x+z=B$\r\n\r\n(we have $x,y,z\\leq 1$ too)\r\n\r\nthe inequality becomes\r\n$ABC+A^{2}+B^{2}+C^{2}\\leq2AB+2BC+2CA$\r\n\r\n$4xy+4yz+4zx\\geq x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)+2xyz$\r\n\r\nbut we have\r\n\r\n$4xy+4yz+4zx=2xy+2yz+2zx+x(y+z)+y(z+x)+z(x+y)\\geq 2xyz+x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)$\r\nfor $1\\geq x,y,z \\geq 0$.", "Solution_6": "This is actually an old problem, and I have such kind of idea.\r\n\r\nAssume that $a \\leqslant b \\leqslant c$, with\r\n$b+c \\leqslant 1+a \\leqslant 1+b$\r\nwe have $c \\leqslant 1$.\r\n\r\n1) If $a \\leqslant bc$, we get\r\n$(a^{2}+b^{2}+c^{2})-(2abc+1)\\\\ =a^{2}+(b+c)^{2}-2(1+a)bc-1\\\\ \\leqslant a^{2}+(1+a)^{2}-2(1+a)bc-1\\\\ \\leqslant 2(1+a)(a-bc) \\leqslant 0$\r\n\r\n2) If $a>bc$, we get\r\n$(a^{2}+b^{2}+c^{2})-(2abc+1)\\\\ =(a-bc)^{2}-(1-b^{2})(1-c^{2})\\\\ \\leqslant (b-bc)^{2}-(1-b^{2})(1-c^{2})\\\\ =(1-c)[b^{2}(1-c)-(1-b^{2})(1+c)]\\\\ =(1-c)(2b^{2}-c-1)\\\\ \\leqslant(1-c)(2b-b-b)=0$", "Solution_7": "[quote=\"shalex\"]This is actually an old problem, and I have such kind of idea.\n[/quote]\r\ndo you know the source of this problem", "Solution_8": "I'm sorry I don't know the exact source it comes from,\r\nbut I saw this problem in a document before this competition was held.", "Solution_9": "[quote=\"shalex\"]I'm sorry I don't know the exact source it comes from,\nbut I saw this problem in a document before this competition was held.[/quote]\r\nthey have copied almost all the problems i suppose.... :mad:", "Solution_10": "Let $b\\geq c$ then $b+c-1\\leq a\\leq c-b+1\\leq b-c+1$\r\nSo $1\\geq b\\geq c$\r\n$f(a)=a^{2}-2bca+b^{2}+c^{2}-1$\r\nWe only have to prove that $f(b+c-1)\\leq 0,f(c-b+1)\\leq 0$\r\n$f(b+c-1)=-2(b+c)(1-b)(1-c)\\leq 0$\r\n$f(c-b+1)=-2(b-c)(1-b)(1+c)\\leq 0$\r\nSo we done." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that r(6,3)=18.", "Solution_1": "I could say nothing but to confess it's to hard :D" } { "Tag": [ "algebra", "polynomial", "analytic geometry", "function", "domain", "conics", "parabola" ], "Problem": "when solving polynomials, for example $ P(x) \\equal{} x^2 \\minus{} 3x \\plus{} 2$ \r\nwe say $ p(x) \\equal{} 0$ then when we find the roots right?\r\nwe say $ p(x) \\equal{} 0$ because the roots of that polynomial is when it the polynomial crosses the x-axis, the $ y$ coordinate equals to $ 0$ therefore $ P(x) \\equal{} 0$ right ?\r\n\r\nwhen it comes to imaginary roots we still do the same thing, eventhough they dont pass the x-axis at all, why? \r\nfor example we say $ x^2 \\plus{} 1 \\equal{} 0$ then we solve it. and it gives $ \\pm{i}$ i am confused. because it doesnt pass the x-axis, why do we asume $ P(x) \\equal{} 0$,and we then solve, it doesnt pass the x-axis so $ p(x)\\neq0$\r\n\r\ncan someone explain this please.", "Solution_1": "I think it crosses the \"x-axis\" on the imaginary plane.", "Solution_2": "I might be being really dumb right now, but I'm pretty sure that's not true, 7h3.D3m0n.117.\r\n\r\nWhen you talk about crossing the x-axis on the coordinate plane, you're only considering a domain of real numbers, not complex numbers. So not passing the x-axis does not imply that there are no roots (real or non-real) to the equation.", "Solution_3": "[quote=\"azjps\"]I might be being really dumb right now, but I'm pretty sure that's not true, 7h3.D3m0n.117.[/quote]\r\n\r\nYeah, my illogical way of thinking got in the way of my reasoning; even I'm not sure anymore. \r\n\r\nBut wouldn't it make sense in the sense of the complex plane though?", "Solution_4": "[quote] you're only considering a domain of real numbers, not complex numbers.[/quote]\r\n\r\nhow can you say the complex numbers domain exists , when the first equation that found the imaginary number $ x^2 \\plus{} 1$ doesnt have imaginary root, (when no imaginary number exists how can you say that imaginary domain exists, any proof?)", "Solution_5": "Once we start using words like \"complex plane\" in the context of solving polynomials, we have to be very careful to sort out what we mean by \"plane.\"\r\n\r\nImaginary roots cannot be depicted on a [b]real[/b] Cartesian plane at all. You are correct that $ x^2 \\plus{} 1 \\equal{} 0$ has no real solutions - in other words, there is no real point at which the graph crosses the [b]real[/b] $ x$-axis. \r\n\r\nIt's possible to graph the \"complete\" graph of a parabola by letting both $ x$ and $ y$ take complex values. Then the graph of $ y \\equal{} x^2 \\plus{} 1$ will \"cross\" the [b]complex[/b] $ x$-axis at $ (\\pm i, 0)$. (In terms of real dimensions, the parabola $ y \\equal{} x^2 \\plus{} 1$ over the complex numbers is actually a surface (and the complex $ x$-axis is a plane), so our notion of \"crossing\" must be modified somewhat.) \r\n\r\nThe problem with visualizing this graph is that $ x$ now has two (real) dimensions and so does $ y$, so the resulting graph is [b]four-dimensional.[/b]", "Solution_6": "when the existence of imaginary number is not proven, how can we say that imaginary or complex axis exists? any proof? or is it just an assumption?", "Solution_7": "Like all numbers, it's just an abstract term defining something observed. Why can't we picture a 4th dimension based on 3 dimensional space?", "Solution_8": "Complex numbers are not \"proven\" to exist. They are [b]defined[/b] in a logically consistent way that we want them to be.\r\n\r\nEDIT: Henry beat me...", "Solution_9": "Post conflict like 3x; Hmm, that's what I thought :) , but I didn't realize that the complex x-axis wasn't referring to a line; thanks.\r\n\r\nI was going to say that complex numbers are defined etc but it seems that a bunch of people have beaten me to it.", "Solution_10": "[quote=\"binomial_4eva\"]when the existence of imaginary number is not proven, how can we say that imaginary or complex axis exists? any proof? or is it just an assumption?[/quote]\r\n\r\nHow can we say that the real axis exists? There are actually some mathematicians who think that the infinite precision of a real number has no sensible meaning.", "Solution_11": "so its just an assumption that its there? i thought in mathematics things are proven first then accepted", "Solution_12": "Remember even though it's mathematics, math is still a science. Science can be overwhelmingly and [seemingly] true, but all it takes is one counterexample to shatter what we think is true now and send scientists back to the drawing boards.", "Solution_13": "[quote]How can we say that the real axis exists? [/quote]\r\nthis is a good question. thanks for reminding me i wanted to ask, i am looking for an answer for that my self", "Solution_14": "Mathematics is based on an axiomatic system, which basically means that there are some things that we just have to accept as true based upon nothing but observation.", "Solution_15": "[quote]based up on nothing but observation[/quote]\r\nwhen have we observed the complex axis, have you seen the complex axis?", "Solution_16": "No. Have you seen the real axis?", "Solution_17": "[quote=\"Temperal\"]Mathematics is based on an axiomatic system, which basically means that there are some things that we just have to accept as true based upon nothing but observation.[/quote]\r\n\r\nThat's not what an axiomatic system is at all. Mathematics is a tool we use to study patterns that we observe. To formalize those patterns, we [b]define[/b] the terms we use - line, point, real number - and write out [b]axioms[/b] regarding those terms - statements that we hold to be true - and then prove things from those definitions and axioms. The resulting patterns may or may not describe the observations we make, but the point is that [b]from our definitions[/b], the results are irrevocably true.\r\n\r\nThe real axis exists because we have axiomatically defined the real numbers (for example, using [url=http://mathworld.wolfram.com/DedekindCut.html]Dedekind cuts[/url]). The complex numbers exist because we can axiomatically define operations on them (or we can define the notion of a field extension or an algebraic closure and work from there). We use those systems because they are [b]useful[/b].", "Solution_18": "no but you said [quote]based up on nothing but observation[/quote] so i am asking when have u observed the complex-axis, how does it look like ?", "Solution_19": "[quote=\"binomial_4eva\"]so i am asking when have u observed the complex-axis, how does it look like ?[/quote]\r\n\r\nKeyword is [i]when[/i]. When mankind develops a sensible visual for the 4th dimension, the microscopic world, and the imaginary axis, you'll be the first to know.", "Solution_20": "The complex $ x$-axis is the set of points $ (x, 0)$ where $ x \\in \\mathbb{C}$. As such it, it looks like an [url=http://mathworld.wolfram.com/ArgandDiagram.html]Argand plane[/url] (which I prefer to the use of the term \"complex plane,\" since when we do not distinguish between real and complex dimensions the term is ambiguous). \r\n\r\n\"Observation\" is a tricky word when it comes to mathematics. Observations are the basis for our axiomatic systems, but they do not comprise them. Mathematics is fundamentally different from other branches of science. The other branches of science [b]use mathematics as a tool[/b] (and I do not know how often I can say this) to analyze patterns in physical observations. Mathematics itself is fundamentally different. Mathematical objects are [b]formal[/b] - they exist because we say they exist. Whether we can visualize four-dimensional space is [b]irrelevant[/b] to whether we can define it mathematically.", "Solution_21": "Sorry, perhaps not on observation, then. But the basis of the post is that some things are not proven in an axiomatic system.", "Solution_22": "Not since the 1800s. What are you talking about?", "Solution_23": "What? What about the 1800s?", "Solution_24": "Starting (I believe) in the late 1800s, mathematics underwent a crisis of formality. Mathematicians realized that they did not actually know what their intuitive notions (say, of a set) meant and sought to rigorize all of mathematics (see for example [url=http://en.wikipedia.org/wiki/Principia_Mathematica]Principia Mathematica[/url]). All of modern mathematics now takes place in the context of axioms. \r\n\r\nSo when you say that some things (by which I take it you mean \"mathematical statements\") aren't proven in an axiomatic system, what things are you referring to?", "Solution_25": "Aren't axioms unproven?", "Solution_26": "[url=http://mathworld.wolfram.com/Axiom.html]Axioms[/url] don't need to be proven. The way in which you've worded your statements has behind it the implicit assumption that there is a possibility that a given axiom might be wrong. \r\n\r\nThe actual word we want to discuss here is [url=http://mathworld.wolfram.com/Consistency.html]consistency[/url]. Interestingly enough, it's not actually possible (in some sense) to prove that the axioms of mathematics are consistent. See [url=http://mathworld.wolfram.com/GoedelsIncompletenessTheorem.html]Godel's incompleteness theorem[/url], and look up a few things about the philosophy of mathematics if you really want to get into this kind of stuff :)" } { "Tag": [ "group theory", "abstract algebra" ], "Problem": "Daca (G; *) grup necomutativ infinit, atunci rezulta ca exista H, subgrup propriu, necomutaiv si infinit ? \r\n\r\nEnglish version : \r\nIf (G; *) is an infinite nonabelian group, implies the existence of an infinite, nonabelian proper subgroup H ?", "Solution_1": "Cautam pe net chestii care nu prea aveau legatura cu asta, si am gasit din greseala definitia [url=http://planetmath.org/encyclopedia/InfiniteDihedralGroup.html]grupului diedral generalizat[/url]. Dupa aia a fost usor :).\r\n\r\nSe pare ca raspunsul e \"nu\". Adica exista grupuri infinite neabeliene care nu au subgrupuri proprii infinite neabeliene. Fie $N=\\mathbb Z_{p^\\infty}$ grupul radacinilor unitatii de ordin putere a lui $p$, unde $p$ e un numar prim. Asta e un exemplu binecunoscut de grup abelian infinit care are numai subgrupuri proprii finite. Acum fie $G$ produsul semidirect $N\\rtimes\\mathbb Z_2$, unde elementul netrivial din $\\mathbb Z_2$ actioneaza pe $N$ prin conjugare trimitand fiecare element in inversul sau. \r\n\r\nFie acum $H\\le G$ un subgrup. $H\\cap N$ e subgrup normal in $H$, de indice $1$ sau $2$. Daca e $1$, atunci $H\\le N$, si $H$ e abelian. Daca e $2$, atunci $H$ e, la randul sau, izomorf cu produsul semidirect $(H\\cap N)\\rtimes\\mathbb Z_2$. Daca $H\\cap N$ e infinit, atunci $H\\cap N=N$, si deci $H=G$. Am aratat deci ca avem una din trei variante: $H$ e ori abelian, ori finit, ori egal cu $G$.\r\n\r\nSper ca nu am gresit pe undeva." } { "Tag": [ "inequalities", "induction" ], "Problem": "Let $F_{i}$ represent the ith fibonacci number. Let $F_{a}$, $F_{b}$, $F_{c}$, and $F_{d}$ be the sides of a convex (nondegenerate) quadrilateral, with $ad\n\nWe prove that $\\frac32 \\cdot F_{i}\\leq F_{i+1}\\leq 2F_{i}$\n\nClearly $F_{i-1}\\leq F_{i}$, so ${F_{i+1}= F_{i}+F_{i-1}\\leq 2F_{i}}$\nThis establishes the second part of the inequality.\n\n$F_{i}\\leq 2 F_{i-1}$ so $F_{i+1}= F_{i}+F_{i-1}\\geq \\frac32 F_{i}$\nwhich establishes the first part of the inequality.\n\nNow assume that cd\n\nWe prove that $\\frac32 \\cdot F_{i}\\leq F_{i+1}\\leq 2F_{i}$\n\nClearly $F_{i-1}\\leq F_{i}$, so ${F_{i+1}= F_{i}+F_{i-1}\\leq 2F_{i}}$\nThis establishes the second part of the inequality.\n\n$F_{i}\\leq 2 F_{i-1}$ so $F_{i+1}= F_{i}+F_{i-1}\\geq \\frac32 F_{i}$\nwhich establishes the first part of the inequality.\n\nNow assume that c0 or <0 ),for exemple if z>0 and Im(n)<0 the norme increase ...\r\n(b) B=E/c , ||||=(1/2\u00b5c)E\u00b2 .\r\n(c)i dont understand the statement \r\ni'm i reight ?" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Find all the functions $f: R \\to R$ satisfy: \r\n $(f(x))'.\\int{f(x)dx}=f(x)$", "Solution_1": "[quote=\"Lovasz\"]Find all the functions $f: R \\to R$ satisfy: \n $(f(x))'.\\int{f(x)dx}=f(x)$[/quote]\r\nWhat do you mean $\\int{f(x)dx}$???\r\nIs it a set of $\\{\\int{f(x)dx}+C\\}$? Or this is concrete integral?", "Solution_2": "For that matter, $(f(x))'$ is terrible notation; $f(x)$ is a real number that can't be differentiated, unless there's another variable somewhere.\r\n\r\nIf $\\int$ indicates an antiderivative, that's a lot of trouble too, because it doesn't have a unique value anywhere." } { "Tag": [ "modular arithmetic", "induction", "number theory unsolved", "number theory" ], "Problem": "This one I think should be an easy one, but I can't solve it.\r\n\r\nShow that if $ a$ is an odd integer, then for any $ n \\ge 1$\r\n\r\n\\[ a^{2^n} \\equiv 1 \\pmod{2^{n\\plus{}2}}\\]", "Solution_1": "[b]Lemma:[/b] The greatest power of $ 2$ dividing $ {2^n \\choose k}$ is $ n$ minus the greatest power of $ 2$ dividing $ k$. \r\n\r\n[i]Proof.[/i] Apply [url=http://planetmath.org/encyclopedia/KummersTheorem.html]Kummer's theorem[/url]. \r\n\r\nNow write $ a \\equal{} 2k \\plus{} 1$. Then binomial expansion gives\r\n\r\n$ a^{2^n} \\equiv 1 \\plus{} 2^{n \\plus{} 1} k \\plus{} {2^n \\choose 2} (2k)^2 \\bmod 2^{n \\plus{} 2}$\r\n\r\nwhere the other terms vanish $ \\bmod 2^{n \\plus{} 2}$ by the lemma. This is congruent to $ 1 \\plus{} 2^{n \\plus{} 1} \\left( k \\minus{} k^2 \\right) \\bmod 2^{n \\plus{} 2}$ and $ k \\minus{} k^2$ is even.\r\n\r\nThere's also a totally straightforward proof by induction, but where's the fun in that :P The binomial argument lets you prove something stronger (as I recall): namely, that $ 5$ has maximal order." } { "Tag": [ "vector", "linear algebra", "matrix", "geometry", "linear algebra unsolved" ], "Problem": "Non-singular symetric bilinear form $ g$ acts on $ E \\times E$, where $ E$ is a finite dimentional vector space. \r\n\r\nDefine a symetric bilinear form $ g_{r}$ on $ E^{r}\\times E^{r}$ by:\r\n$ )$.\r\n\r\nIf $ g$ is positive definite, is $ g_{r}$ also positive definite?", "Solution_1": "This is essentially the [url=http://en.wikipedia.org/wiki/Gramian_matrix]Gram matrix[/url] of a system of vectors, except that the inner product is not the standard one. Let $ A$ be the matrix of $ g$ (in your notation, $ \\langle g, x\\times y\\rangle =y^{*}A x$). Let $ B_{h}$ be the matrix with $ r$ columns, where the columns are vectors $ h_{1},\\dots , h_{r}$. Define $ B_{k}$ similarly. You should be able to check that the value of $ g_{r}$ is $ \\det (B_{k}^{*}AB_{h})$. When $ k=h$, the matrix $ B_{h}^{*}AB_{h}$ is positive semidefinite, because $ A$ is [url=http://en.wikipedia.org/wiki/Positive-definite_matrix]positive definite[/url]. It is positive definite if and only if the matrix $ B_{h}$ has rank $ r$. For instance, if $ h_{1}=\\dots =h_{r}$, then $ B_{h}$ and $ B_{h}^{*}AB_{h}$ have rank $ 1$, and therefore $ g_{r}$ vanishes. So, the form $ g_{r}$ is not positive definite.", "Solution_2": "[quote=\"mlok\"]This is essentially the [url=http://en.wikipedia.org/wiki/Gramian_matrix]Gram matrix[/url] of a system of vectors, except that the inner product is not the standard one. Let $ A$ be the matrix of $ g$ (in your notation, $ \\langle g, x\\times y\\rangle =y^{*}A x$). Let $ B_{h}$ be the matrix with $ r$ columns, where the columns are vectors $ h_{1},\\dots , h_{r}$. Define $ B_{k}$ similarly. You should be able to check that the value of $ g_{r}$ is $ \\det (B_{k}^{*}AB_{h})$. When $ k=h$, the matrix $ B_{h}^{*}AB_{h}$ is positive semidefinite, because $ A$ is [url=http://en.wikipedia.org/wiki/Positive-definite_matrix]positive definite[/url]. It is positive definite if and only if the matrix $ B_{h}$ has rank $ r$. For instance, if $ h_{1}=\\dots =h_{r}$, then $ B_{h}$ and $ B_{h}^{*}AB_{h}$ have rank $ 1$, and therefore $ g_{r}$ vanishes. So, the form $ g_{r}$ is not positive definite.[/quote]\r\n\r\nThanks, now I totaly understand.\r\n\r\nAlso the question was about r-dimentional area in a Riemann structure on a pure manifold. The correct defination of $ g_{r}$ should be:\r\n\r\nDefine a symetric bilinear form $ g_{r}$ on $ \\bigwedge^{r}E \\bigotimes \\bigwedge^{r}E$ by:\r\n$ }=det()$.\r\n\r\nSo $ B_{h}$ in your post is either rank $ r$ or $ h_{1}\\wedge ... \\wedge h_{r}=0$ by defination of \r\nexterior product. So $ g_{r}$ is positive definite." } { "Tag": [ "trigonometry", "Euler", "Pythagorean Theorem", "geometry", "algebra", "binomial theorem" ], "Problem": "Solve the following equation: \r\n$\\cos^{n}x-\\sin^{n}x = 1$, \r\nwhere n is a positive integer and x is a real number.", "Solution_1": "${\\cos^{n}(x)-\\sin^{n}(x)=1}$\r\n\r\nBy Euler\r\n\r\n${\\cos (x)=\\frac{1}{2}\\left(e^{-i x}+e^{i x}\\right)}$\r\n${\\sin (x)=\\frac{1}{2}i \\left(e^{-i x}-e^{i x}\\right)}$\r\n\r\nSubstituing\r\n\r\n${2^{-n}\\left(e^{-i x}+e^{i x}\\right)^{n}-2^{-n}\\left(i \\left(e^{-i x}-e^{i x}\\right)\\right)^{n}=1}$\r\n\r\nSeting\r\n\r\n${e^{i x}=a+b i}$\r\n${e^{-i x}=a-b i}$\r\n\r\nSubstituing\r\n\r\n${a^{n}-b^{n}=1}$\r\n${a^{n}=b^{n}+1}$\r\n\r\n${\\{a,b\\}\\in \\mathbb{R}}$\r\n${e^{i x}=\\cos (x)+i \\sin (x)}$\r\n${a=\\cos (x)}$\r\n${b=\\sin (x)}$\r\n\r\n${a^{n}=b^{n}+1}$\r\n${a^{2}+b^{2}=1}$\r\n\r\nJust Need to solve this system", "Solution_2": "nice ! that's exactly the question :P \r\nthe second equation u wrote says only it exists x such that a = cos x and b = sin x, then replace it into the first one...\r\nwhat did you try with the complex forms ??", "Solution_3": "[quote=\"Thales418\"]${\\cos^{n}(x)-\\sin^{n}(x)=1}$\n\n\n${a=\\cos (x)}$\n${b=\\sin (x)}$\n\n${a^{n}=b^{n}+1}$\n${a^{2}+b^{2}=1}$\n\nJust Need to solve this system[/quote]\r\n\r\nYou could have said this immediately, without all that intermediate work.", "Solution_4": "Yea, i don't think either that u need that Euler-stuff:\r\nGiven a right triangle a, b, 1 with 1 as the hypotenuse,\r\nand x as the angle opposed to side b, we have:\r\n$cos(x) = a; sin(x) = b$\r\nSubstituting&simplifying:\r\n$I: a^{n}-b^{n}= 1$\r\nAnd because of the Pythagorean Theorem:\r\n$II: a^{2}+b^{2}= 1$\r\n\r\nNow we can square the first equation and take the 2nd one to the power of n and set them equal in order to get:\r\n\r\n$(a^{n}-b^{n})^{2}= (a^{2}+b^{2})^{n}$\r\n\r\nwhich is, using the Binomial Theorem:\r\n\r\n$a^{2n}-2a^{n}b^{n}+b^{2n}= a^{2n}+\\sum_{k=1}^{n-1}{n-1 \\choose k}a^{2(n-1-k)}b^{2k}+b^{2n}$\r\n\r\n(note that the sum must always be positive for $a,b > 0$)\r\n\r\n$0 = \\sum_{k=1}^{n-1}{n-1 \\choose k}a^{2(n-1-k)}b^{2k}+2a^{n}b^{n}$\r\n\r\nAs the sum and $2a^{n}b^{n}$ are both positive for $a, b > 0$, it is obvious that there is no 2-tupel $(a, b)$ for $a, b > 0$ solving this equation.\r\n\r\nBecause $a, b$ are the legs of the right-angled triangle, we now know that there exists no such triangle such that,\r\n$a^{n}-b^{n}= 1$, for $n \\in \\mathbb{N}$.\r\nSubstituting back, this yields that there is also no real x such that, $cos^{n}(x)-sin^{n}(x) = 1$.\r\n\r\nLol, this must be wrong, it was too easy. sry if it is, im a beginner!\r\n\r\n\r\nEDIT:\r\nlol gawd im stupid^^ ya its pretty ovious what i proved and it doesn't bring us any further^^ i should have looked at the problem more closely", "Solution_5": "i know that dont need euler..\r\n\r\nbut when i was doing i thougth that if i put this it will simplicate, but not. but i didnt want to erase the \"euler\" so....\r\n\r\n\r\nand TearOfMercury you did not find a explicit formula for x", "Solution_6": "[quote=\"TearOfMercury\"]Lol, this must be wrong, it was too easy. sry if it is, im a beginner![/quote]\r\nYou correctly showed there are no solutions in the case that $\\cos x$ and $\\sin x$ are the legs of a non-degenerate right triangle with hypotenuse 1, i.e. when $0 < \\cos x, \\sin x < 1$. (For these situations, you also did far more work than was necessary.) The tricky part is what happens when we are outside of this range, particularly when $\\cos x$ is still positive but $\\sin x$ is negative and $n$ is odd. Those solutions will be ugly, I imagine. (There are also trivial solutions when $\\sin x = 0$.)", "Solution_7": "$\\boxed{n\\ge 2\\Longrightarrow 1=\\cos^{n}x-\\sin^{n}x\\le |\\cos x |^{n}+|\\sin x |^{n}\\le \\cos^{2}x+\\sin^{2}x=1}$\r\n\r\na.s.o.", "Solution_8": "[quote]$\\boxed{n\\ge 2\\Longrightarrow 1=\\cos^{n}x-\\sin^{n}x\\le |\\cos x |^{n}+|\\sin x |^{n}\\le \\cos^{2}x+\\sin^{2}x=1}$[/quote]\r\nso is this supposed to be the solution? :huh: if it is could s.o. explain it to me? otherwise would s.o. please post a solution :roll:", "Solution_9": "$a\\le b\\le a\\Longrightarrow a=b\\ .$\r\nWhen $a-b=|a|+|b|$ ? \r\nWhen for $0\\le a\\le 1$, $0\\le b\\le 1$, $n\\ge 2$ we have $a^{n}+b^{n}=a^{2}+b^{2}$ ?" } { "Tag": [], "Problem": "Well,\r\nYes, now lemme formally introduce myself,\r\nI am Raghavendra from Bangalore.\r\nDa same one who pissed shre off... \r\nSame class as Abhimanyu (alter ego)\r\n\r\nWell how s the board mugging goin on all...\r\nFor me its not started as yet. I really like da last minute slogging. Under all pressure...\r\nYou know the thrill one can obtain from dat is itself 'thani'\r\nLOADSSS of thrill :D", "Solution_1": "I had guessed who you are from your PM... :P", "Solution_2": "I had guessed who you are when pardesi told me...", "Solution_3": "[quote=\"Euclidean Geometer\"]Well,\nYes, now lemme formally introduce myself,\nI am Raghavendra from Bangalore.\nDa same one who pissed shre off... \nSame class as Abhimanyu (alter ego)\n\nWell how s the board mugging goin on all...\nFor me its not started as yet. I really like da last minute slogging. Under all pressure...\nYou know the thrill one can obtain from dat is itself 'thani'\nLOADSSS of thrill :D[/quote] \r\nhello raghavendra, and \r\n :welcome: welcome to this forum :welcomeani:", "Solution_4": "hi da...chumma formality ke... :D ....but anyways it has been real fun knowing u :)" } { "Tag": [], "Problem": "show that for n greater than or equal to 2, (n^3+(n+2)^3)/4 is a composite integer.", "Solution_1": "[quote=\"mdk\"]show that for n greater than or equal to 2, (n^3+(n+2)^3)/4 is a composite integer.[/quote]\r\n[hide=\"Hint\"]Consider parity.\nIf $ n$ is even,\n$ \\frac{n^{3}+(n+2)^{3}}{4}= (n+1)\\cdot \\frac{3(n(n+2))+4}{4}$\nThe second factor must be an integer (since n(n+2) has 2 factors of 2). \nI'll leave the odd case to you.\n[/hide]", "Solution_2": "$ n^{3}+(n+2)^{3}=8(n+1)+2(n)(n+1)(n+2)=8(n+1)+12\\binom{n+2}{3}$ which is obviously divisible by $ 4$." } { "Tag": [ "linear algebra", "matrix", "limit", "linear algebra unsolved" ], "Problem": "Let $A_{n}=\\left(\\begin{matrix}\\frac{1}{2}& 1 & 1\\\\ 0 & \\frac{1}{3}& 1\\\\ 0 & 0 & \\frac{1}{5}\\\\ \\end{matrix}\\right)^{n}$. Find $\\lim_{n\\to+\\infty}A_{n}$.", "Solution_1": "It is readily checked that the eigenvalues of $A_{1}$ are ${\\frac{1}{2},\\frac{1}{3},\\frac{1}{5}}$.\r\nUsing these, the matrix $A_{1}$ is diagonizable because they form an basis for $R^{3}$. Hence $A_{1}=P^{-1}BP$ where $B$ har the eigenvalues on the diagonal.\r\nWhen $n-\\to infinity$, $B$ will turn towards the nullmatrix, leaving your answer the nullmatrix." } { "Tag": [ "USAMTS", "geometry", "AMC", "AIME", "LaTeX", "analytic geometry", "graphing lines" ], "Problem": "[size=150][b]DO NOT[/b][/size] POST ANSWERS, SOLUTIONS, COMMENTS, ETC. IN THIS THREAD UNTIL WEDNESDAY OCTOBER 8!!!\r\n\r\nTHIS THREAD IS ONLY HERE TO ELIMINATE THE NEED FOR 500 THREADS ON ROUND 1, AND TO SERVE AS A DISCLAIMER AGAINST POSTING [b]ANYTHING[/b] ON ROUND 1 UNTIL OCTOBER 8!!", "Solution_1": "Mmm, good idea. :) \r\nI'll just link to the [url=http://www.nsa.gov/programs/mepp/usamts.html]USAMTS website[/url] and the [url=http://www.nsa.gov/programs/mepp/usamts/problems/problems15R1.pdf]problem[/url].\r\n\r\nP.S. We can start discussing these problems tomorrow since according to the official website's [url=http://www.nsa.gov/programs/mepp/usamts/mail.html]page about deadlines[/url] \"Solutions need to be postmarked (or faxed) [i]before [/i]the deadline date of the round.\"", "Solution_2": "Also, you can't mail anything out on a Sunday. But, actually, you can fax in answers I believe. (Or am I just making that up?) So we should wait until after Sunday.", "Solution_3": "without discussing particular problems, how did you guys do?", "Solution_4": "My justification to one of the problems was about half a solution from which the answer came out of nowhere, but I think the rest were OK... at any rate, I'm 99% sure the numerical answers were right...", "Solution_5": "Finding the answers themselves was not hard; proving them was. I found #3 the easiest and #4 the hardest.\r\n\r\n#2, I believe, is impossible to \"prove\", since prime numbers have no patterns whatsoever.", "Solution_6": "All of them can be proved. I found #1 the hardest.", "Solution_7": "Question: how soon do you get your results back. Is it like they say, 2 months, or do you USUALLY get it back before the next round starts.", "Solution_8": "You usually get the results after the next round has started but before the deadline, if I remember correctly.", "Solution_9": "It's a free public service. If you want them to go faster, I'm sure they'd accept payment for additional graders.", "Solution_10": "Or stop giving so much money away for tax-exemption (USAMTS is sponcered by the government) :wink:", "Solution_11": "Which problem do you guys think is the easiest? Don't explain how, just which....", "Solution_12": "The third one I guess.", "Solution_13": "Number three was, then 5, 4, 2, and 1. i did not find a solution to one :cry: .", "Solution_14": "definately 3 is easiest, then 2, then 1 , 4 , 5. seems like a lot of people thought 1 was hard.... it thought it was ok.", "Solution_15": "Right, but a lot more work than that must be shown to get any credit. Particularly for #1. Some insightful discovery (although patterns work, I suppose) is probably preferred.\r\n\r\nThe [url=http://www.nsa.gov/programs/mepp/usamts/problems/credits15R1.pdf]credits and quick answers[/url] are posted at http://www.nsa.gov/programs/mepp/usamts/problems/credits15R1.pdf", "Solution_16": "got scores for the first round today.", "Solution_17": "yep, 20 for 20. i thought that the graders would be tougher...some of mine were a bit verbose.", "Solution_18": "I thought it was out of 25 . . .", "Solution_19": ":lol:, you're absolutely right, 5 a piece, 5 problems, 4 times. 100 points total.", "Solution_20": "am i the only one who hasn't gotten miine yet?", "Solution_21": "Bah. No. I haven't gotten mine yet, either. It's probably because I live significantly farther from the NSA headquarters than some of the other people on this forum...but I guess that means there's a lesser chance of NSA hitmen gunning me down. :D", "Solution_22": "Phht, they'll get you anywhere on Earth even if you were hiding in a bunker in A********** (which the government successfully hushed the media because of political reasons)", "Solution_23": "Got my results in the mail today...yay. 24/25.\r\n\r\nI too thought they would be much harsher in grading...especially with my sorta-sticky-did-at-the-last-minute #1, which was fairly ugly, or my let's-work-with-50-pairs-of-similar-triangles #5, which ended up with a commendation. But I don't mind being graded fairly loosely, now that I think about it.", "Solution_24": "Just how good is a commendation?\r\n\r\nI got 25 and two commendations, both of which are extremely unexpected.\r\n\r\nThe first problem was a sort of put together some BS last second proof that made no sense, yet I got it commended.\r\n\r\nIn the fifth problem, I only did superimposition onto coordinate plane and got commended.\r\n\r\nDid any of you get commendations that were unexpected?", "Solution_25": "cool!! you asked them to post the answers on my birthday! wow! cool!", "Solution_26": "Well, my proof for #5 was even simpler, probably - I just constructed some similar triangles - and I got commended on that...\r\n\r\nMaybe it's the best x% of the proofs for such-and-such a problem...or no, maybe not.", "Solution_27": "No, it's totally subjective -- whenever a grader likes it for whatever reason, it gets commended.", "Solution_28": "I got a 20, with 1 commendation. I completely skipped 4.\r\n\r\nAnyone else find 4 difficult?", "Solution_29": "Hm. #4 was the second one I got (after #3). I found #1 the hardest, just because the gaps were kinda abstract." } { "Tag": [ "inequalities", "induction", "combinatorics solved", "combinatorics", "graph theory" ], "Problem": "In a group of kn persons, each person knows more than (k - 1)n others.\n(Here k and n are positive integers.)\nProve that one can choose k + 1 persons from this group,\nsuch that every chosen person knows all the others chosen.\n\n(A knows B if and only if B knows A, as usual)", "Solution_1": "It is a simple induction.", "Solution_2": "Myth, I'm wondering ... Is this really that easy for you ?", "Solution_3": "Yes.\r\n\r\nDid you solve this problem?", "Solution_4": "indeed is not such a simple induction. This is actually a particular case of [b]Zarankiewcz's[/b] theorem:\r\n\r\n[i] If a graph G does not contain any k-complete sub-graphs, then the set of its vertices' degrees verifies the following inequality\n\nmin { d(x) | x\\in G } \\leq [ n(k-2) / (k-1) ] \n\n\nwhere by [x] we have denoted the integer part of the real x.[/i]\r\n\r\nOne can see that the problem above is an application of this theorem for n -> kn , k -> k+1.\r\n\r\n\r\nThe solution I know is based on using the cardinal set inclusion and exclusion theorem. \r\n\r\nIs it the same with yours, Myth?", "Solution_5": "Nope I didn't solve it ... :( I'm not so strong you know :(\r\n\r\nAnd I almost don't know what a graph is ...\r\nYeah of course I heard of it but I never used it.\r\nAnd we didn't learn about graphs in our IMO preparation session.\r\n\r\nSo Myth, what is your simple induction ?", "Solution_6": "If each persons knows greater than (k-1)n+1 other persons then we delete some acquaintances.\r\n\r\nLet P be person with (k-1)n+1 acquaintances. Let M be a set of all friends of P, |M|=(k-1)n+1. Each person in M has in M at least ((k-1)n+1)-(n-1)=(k-2)n+2 friends. Now we delete arbitrary element of M ==> |M|=(k-1)n and each person in M has at least (k-2)n+1 friends in M ==> there are k persons in M who know each other ==> these k persons + person P know each other.\r\n\r\nAm I right?" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "find all $ x,y \\in R$ such that:\r\n$ (x^2\\minus{}x)(x^2\\minus{}2x\\plus{}2)\\equal{}y^2\\minus{}1$", "Solution_1": "I thin we should search for solution ins $ \\mathbb{N}$ or $ \\mathbb{Z}$ but not in $ \\mathbb{R}$. However wherever you look either in $ \\mathbb{R}$ or in $ \\mathbb{N}$ or $ \\mathbb{Z}$ solving towards $ x$ we get $ x \\equal{} \\{\\frac {1 \\minus{} \\sqrt {4y \\plus{} 5}}{2},\\frac {1 \\plus{} \\sqrt {4y \\plus{} 5}}{2},\\frac {1 \\minus{} \\sqrt {5 \\minus{} 4y}}{2},\\frac {1 \\plus{} \\sqrt {5 \\minus{} 4y}}{2}\\}$. From these solutiom we can make infinitely many solutions in every set.", "Solution_2": "why $ x\\equal{}{\\frac{1\\minus{}sqrt{4y\\plus{}5}}{2},...}$?\r\nCould you discribe?" } { "Tag": [ "trigonometry", "calculus", "integration", "algebra", "polynomial", "AMC" ], "Problem": "Is there anyway to \"simplify\" $\\sin(x) \\sin(y)$.\r\nTo splash D, I realized that moments after posting so I changed it.", "Solution_1": "[quote=\"bpms\"]Is there anyway to \"simplify\" the following:\n$\\sin^{-1}\\cos(a)$\n$\\cos^{-1}\\sin(a)$[/quote]\r\n\r\nwell $\\cos a=\\sin \\frac{\\pi}{2}-a$, so your expression both equal to $\\frac{\\pi}{2}-a$", "Solution_2": "[quote=\"bpms\"]Is there anyway to \"simplify\" $\\sin(x) \\sin(y)$.\nTo splash D, I realized that moments after posting so I changed it.[/quote]\r\n$\\sin(x) \\sin(y)=\\frac{\\cos(x-y)-\\cos(x+y)}{2}$\r\n$\\cos(x)\\cos(y)=\\frac{\\cos(x-y)+\\cos(x+y)}{2}$\r\n$\\sin(x)\\cos(y)=\\frac{\\sin(x+y)+\\sin(x-y)}{2}$", "Solution_3": "How about $\\sin(\\cos^{-1}x)$? I got this far.\r\n[hide]$\\cos(90-\\cos^{-1}x)$\n$\\sin(90)\\sin(\\cos^{-1}x)+x\\cos{90}$\n$\\sin(\\cos^{-1}x)$\n$\\sin(90-\\cos^{-1}x)$\n$\\sin(90)x-\\cos(90)\\sin(\\cos^{-1}x)$\n$x$[/hide]\r\nWhich is obviously not true... I made a mistake somewhere... Could someone point it out...", "Solution_4": "[quote=\"bpms\"]How about $\\sin(\\cos^{-1}x)$? I got this far.\n[hide]$\\cos(90-\\cos^{-1}x)$\n$\\sin(90)\\sin(\\cos^{-1}x)+x\\cos{90}$\n$\\sin(\\cos^{-1}x)$\n$\\sin(90-\\cos^{-1}x)$\n$\\sin(90)x-\\cos(90)\\sin(\\cos^{-1}x)$\n$x$[/hide]\nWhich is obviously not true... I made a mistake somewhere... Could someone point it out...[/quote]\r\n\r\ndude draw a triangle", "Solution_5": "[quote=\"bpms\"]How about $\\sin(\\cos^{-1}x)$? I got this far.\n[hide]$\\cos(90-\\cos^{-1}x)$\n$\\sin(90)\\sin(\\cos^{-1}x)+x\\cos{90}$\n$\\sin(\\cos^{-1}x)$\n$\\sin(90-\\cos^{-1}x)$\n$\\sin(90)x-\\cos(90)\\sin(\\cos^{-1}x)$\n$x$[/hide]\nWhich is obviously not true... I made a mistake somewhere... Could someone point it out...[/quote]\nYou just said $\\sin(\\cos^{-1}x)=\\cos(90-\\cos^{-1}x)$ and then $\\sin(\\cos^{-1}x)=\\sin(90-\\cos^{-1}x)$. The first one is correct, but that doesn't help.\nFrom Iversonfan's hint: [hide] $\\sin(\\cos^{-1}x)=\\sqrt{1-x^{2}}$ [/hide]", "Solution_6": "What about $\\sin(xy)$?", "Solution_7": "[quote=\"bpms\"]What about $\\sin(xy)$?[/quote]\r\nThat's unlikely to have a simpler form. Consider the trouble of getting a formula for integral $x$: $\\sin(xy)=\\sqrt{S_{x}(\\sin^{2}y)}$, where $S_{x}$ is the x-th degree spread polynomial described by the explicit formula $S_{x}(n) = n\\sum_{k=0}^{x-1}{x \\over x-k}{2x-1-k \\choose k}(-4n)^{x-1-k}$.\r\n(all found on Wikipedia :D)", "Solution_8": "you could use complex numbers for sin(xy)", "Solution_9": "More specifically, $\\sin(xy) = \\Im{(\\cos x+i \\sin x)^{y}}.$\r\n\r\nThen do some complicated work involving non-integer factorials." } { "Tag": [ "geometry", "parallelogram", "ratio", "cyclic quadrilateral" ], "Problem": "When doing certain geometry problems, mainly the olympiad ones, what tactics do you guys use?\r\n\r\nWhen you think you have to join lines up (as you usually do with olympiad geometry) to prove and relate certain things, how do you know where to join, produce or connects lines?\r\n\r\nWhen you start an olympiad geometry problem, do you just look at and think about possible ways? Or do you just consider a way to do it and rush into it? Do you think about what lines you could draw to help before you start?\r\n\r\n--I really want to know how you guys think as i want to be able to do that. When i start problems, its usually train wreaks, i will randomly connect lines and hopes it works. I also think of one way to start it and just stick with it. Leonardo Da Vinci believed that the first way he looked at a problem was to bias. \r\n\r\nSo how can i think like the way yous guys think? What did yous do?\r\n\r\nI would really appreciate any input!\r\n-Kurt", "Solution_1": "This is someone that's probably as equally stunted in geometry as you are, but I have heard some things from people that are very good at geometry (sometimes in a failed attempt to help me get better :D ), so hopefully this will help.\r\n\r\nMost of the time when these extra lines or points are constructed, it's not a random process. There's a goal in mind; maybe making that extra line gives you a parallelogram, or that extra point creates a cyclic quadrilateral, or you want a right angle.\r\n\r\nIn the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247506#1247506]only geometry problem I ever solved in contest time[/url], I recall needing to prove something about ratios of sides as my first step. It seemed intuitively obvious to me from all of the parallel lines, but the only way I found to rigorously do it was to extend those parallel lines to form several parallelograms and use a whole bunch of equal sides. From this experience, my guess would be that people good at geometry have an idea of what they're trying to do when they make these extra constructions, the same way I knew exactly what I was going for.\r\n\r\nThe way you would get to the point of noticing what extra things can be made by extending lines of stuff, I presume, is by doing lots of problems and thus finding what patterns and situations give an opportunity for constructing them.", "Solution_2": "Thanks, thats really good info. \r\n\r\nAnyone else?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $p$ be an odd prime congruent to 2 modulo 3. Prove that at most $p-1$ members of the set $\\{m^2 - n^3 - 1 \\mid 0 < m,\\ n < p\\}$ are divisible by $p$.", "Solution_1": "The equation $x^3 \\equiv a \\mod p$ has exactly one solution for given $a$.\r\nProof: Take some $s,t$ with $3s-(p-1)t=1$, since this shows $(x^3)^s \\equiv x \\cdot x^{t\\cdot (p-1)} \\equiv x \\mod p$, thus $x \\to x^3$ is injective, thus surjective.\r\nNow for any $m$, there is exactly one $n$ when $n=0$ is also allowed, thus at most $p-1$ solutions (indeed there are $p-3$ of them for $p \\geq 5$: $m \\equiv \\pm 1 \\mod p \\iff n\\equiv 0 \\mod p$ fall out, too).", "Solution_2": "Sorry for reviving the old topic , \n\nI feel ,this problem can be generalized for $m^{j} - n^{3} - k $ where $k$ and $j$ are fixed constants . :maybe: , instead of just $m^{2} - n^{3} - 1 $ :maybe:", "Solution_3": "Isn't it true that there are exactly $p-1$ members of the set?", "Solution_4": "Actually, this is weird. Given that this is an old problem, and (supposedly) known to Eastern European community, I would not expect it, but it seems, there is a typo, and it is asking to prove that such a set has at most $\\mathbf{p}$, rather than $\\mathbf{p-1}$ elements, and $S=\\{m^2-n^3-1:0\\leq m,n\\leq p-1\\}$. (In fact, Andreescu's contest problems book shows the same modification, too).\n\nAssuming this is the problem statement, here is a proof. The key here is that if $p\\equiv 2 \\pmod{3}$ prime, then $-3$ is not a quadratic residue modulo $p$. Show first that, $x^3+1 \\equiv x'^3 +1 \\pmod{p} \\iff x\\equiv x' \\pmod{p}$. This is indeed true, since,\n$$\np\\mid (x'-x)(x'^2 + x'x+x^2) \\quad \\text{and if } p\\mid x'^2 + x'x+x^2\\implies p\\mid (2x'+x)^2+3x^2 \\implies ((2x'+x)x^{-1})^2 \\equiv -3 \\pmod{p},\n$$\na contradiction. This further proves that, $\\{1^3,2^3,\\dots,(p-1)^3\\}\\equiv \\{1,2,\\dots,p-1\\}\\pmod{p}$. Therefore, for any given $y$, there is a unique $x$ for which, $x^3 \\equiv y^2-1 \\pmod{p}$, showing that at most $p$ pairs of $(x,y)$ can exist.", "Solution_5": "Alternative way to show that $x^3$ forms a CRS $\\pmod{p}$.$x^{p}\\equiv x\\pmod{p}$ and $x^{p-1}\\equiv 1\\pmod{p}$ multiplying the two yields $x^{2p-1}\\equiv x\\pmod{p}$ since $p\\equiv 2\\pmod{3}$, $3|2p-1$ so it must form a CRS.", "Solution_6": "Notice that third powers are surjective modulo $p$, thus for every $y \\in \\{0, 1, 2, \\dots, p-1\\}$, there is exactly one $x$ under modulo $p$ that works. Thus, there are at most $p$ pairs of solutions.", "Solution_7": " Actually,if p is congrunt to1 modulo 3,{1\u00b3,2\u00b3,...,\uff08p-1\uff09\u00b3}has only\uff08p-1\uff09/3values\uff08modp\uff09", "Solution_8": "The problem comes down to noticing that $\\frac{p-2}{3}$ is integer implies\n\\[x^3 \\equiv y^3 \\iff x^{p-2} \\equiv y^{p-2} \\iff x^{-1} \\equiv y^{-1} \\iff x \\equiv y \\pmod p.\\]\n\nTherefore $n^3 \\pmod p$ is both injective and surjective, so for every value of $m$, there is exactly one corresponding value of $n$ to make $m^2-n^3-1 \\equiv 0 \\pmod p$. Therefore we have exactly $p$ multiples of $p$." } { "Tag": [ "algebra", "polynomial", "LaTeX", "linear algebra", "matrix", "linear algebra theorems" ], "Problem": "Let T:V-->V be any linear map.Suppose ch(X)=(X-c_1)^m_1...(X-c_k)^m_k is the characteristic polynomial such that c_i not equal to c_j, for all i not equal to j.Put W_i =Ker(T-ci.Identity)^m_i, with 1<=i<=k..then prove that dim(W_i)=m_i.", "Solution_1": "Please learn to type in $ \\text{\\LaTeX}$, otherwise reading your posts will be a nightmare: in other words (and symbols), if $ \\lambda_1, \\lambda_2, ..., \\lambda_s$ are the $ s$ disctint eigenvalues of any matrix $ A$ that represents $ T$, whose (algebraic) multiplicities are $ m_1, m_2, ..., m_s$, prove that $ \\dim N(A - \\lambda_iI)^{m_i} = m_i$.\r\n\r\nObserve that $ A$ is similar to a Jordan canonical form $ J$: therefore, $ \\dim N(A - \\lambda_iI)^{m_i} = \\dim N(J - \\lambda_iI)^{m_i}$. But $ J$ has on its main diagonal the eigenvalue $ \\lambda_i$ exactly $ m_i$ times. So, $ J-\\lambda_iI$ has exactly $ m_i$ zero elements on the main diagonal, the same happening with $ (J - \\lambda_iI)^{m_i}$ (and in fact for $ (J - \\lambda_iI)^k, \\, \\forall \\, k \\in \\mathbb{N}$). $ \\Diamond$" } { "Tag": [], "Problem": "I know this is really random, but what's your favorite word? mine's holyschmanitzenwiggle :D which is totally Jeffbot5000's word but i love it", "Solution_1": "Definitions please. :maybe: My pick: [b]flam\u00b7boy\u00b7ant[/b]\r\n1: characterized by waving curves suggesting flames \r\n2: marked by or given to strikingly elaborate or colorful display or behavior", "Solution_2": "holyschmanitzenwiggle: an exclamation of surprise", "Solution_3": "how about blubber?", "Solution_4": "PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (also spelled PNEUMONOULTRAMICROSCOPICSILICOVOLCANOKONIOSIS) = a lung disease caused by breathing in particles of siliceous volcanic dust.", "Solution_5": "chthonic\r\n\r\nlike a demon", "Solution_6": "how do you pronounce that?!? :?:", "Solution_7": "sesquipedalian(person who uses long words)", "Solution_8": "[quote=\"infinity4ever\"]how do you pronounce that?!? :?:[/quote]\r\n\r\ndon't say the ch\r\n\r\nthen its easy", "Solution_9": "zyzzyva\r\n\r\nIt's some sort of weevil", "Solution_10": "crwth is another one i like\r\n\r\n-a Celtic instrument", "Solution_11": "[b]buffalo[/b]\r\n\r\nIt's an amazing word.\r\n- The animal\r\n- to bully, confuse, or intimidate\r\n\r\nPlus, \"Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.\" is a grammatically correct sentence :)", "Solution_12": "mine is ACETYL\u00adSERYL\u00adTYROSYL\u00adSERYL\u00adISO\u00adLEUCYL\u00adTHREONYL\u00adSERYL\u00adPROLYL\u00adSERYL\u00adGLUTAMINYL\u00adPHENYL\u00adALANYL\u00adVALYL\u00adPHENYL\u00adALANYL\u00adLEUCYL\u00adSERYL\u00adSERYL\u00adVALYL\u00adTRYPTOPHYL\u00adALANYL\u00adASPARTYL\u00adPROLYL\u00adISOLEUCYL\u00adGLUTAMYL\u00adLEUCYL\u00adLEUCYL\u00adASPARAGINYL\u00adVALYL\u00adCYSTEINYL\u00adTHREONYL\u00adSERYL\u00adSERYL\u00adLEUCYL\u00adGLYCYL\u00adASPARAGINYL\u00adGLUTAMINYL\u00adPHENYL\u00adALANYL\u00adGLUTAMINYL\u00adTHREONYL\u00adGLUTAMINYL\u00adGLUTAMINYL\u00adALANYL\u00adARGINYL\u00adTHREONYL\u00adTHREONYL\u00adGLUTAMINYL\u00adVALYL\u00adGLUTAMINYL\u00adGLUTAMINYL\u00adPHENYL\u00adALANYL\u00adSERYL\u00adGLUTAMINYL\u00adVALYL\u00adTRYPTOPHYL\u00adLYSYL\u00adPROLYL\u00adPHENYL\u00adALANYL\u00adPROLYL\u00adGLUTAMINYL\u00adSERYL\u00adTHREONYL\u00adVALYL\u00adARGINYL\u00adPHENYL\u00adALANYL\u00adPROLYL\u00adGLYCYL\u00adASPARTYL\u00adVALYL\u00adTYROSYL\u00adLYSYL\u00adVALYL\u00adTYROSYL\u00adARGINYL\u00adTYROSYL\u00adASPARAGINYL\u00adALANYL\u00adVALYL\u00adLEUCYL\u00adASPARTYL\u00adPROLYL\u00adLEUCYL\u00adISOLEUCYL\u00adTHREONYL\u00adALANYL\u00adLEUCYL\u00adLEUCYL\u00adGLYCYL\u00adTHREONYL\u00adPHENYL\u00adALANYL\u00adASPARTYL\u00adTHREONYL\u00adARGINYL\u00adASPARAGINYL\u00adARGINYL\u00adISOLEUCYL\u00adISOLEUCYL\u00adGLUTAMYL\u00adVALYL\u00adGLUTAMYL\u00adASPARAGINYL\u00adGLUTAMINYL\u00adGLUTAMINYL\u00adSERYL\u00adPROLYL\u00adTHREONYL\u00adTHREONYL\u00adALANYL\u00adGLUTAMYL\u00adTHREONYL\u00adLEUCYL\u00adASPARTYL\u00adALANYL\u00adTHREONYL\u00adARGINYL\u00adARGINYL\u00adVALYL\u00adASPARTYL\u00adASPARTYL\u00adALANYL\u00adTHREONYL\u00adVALYL\u00adALANYL\u00adISOLEUCYL\u00adARGINYL\u00adSERYL\u00adALANYL\u00adASPARAGINYL\u00adISOLEUCYL\u00adASPARAGINYL\u00adLEUCYL\u00adVALYL\u00adASPARAGINYL\u00adGLUTAMYL\u00adLEUCYL\u00adVALYL\u00adARGINYL\u00adGLYCYL\u00adTHREONYL\u00adGLYCYL\u00adLEUCYL\u00adTYROSYL\u00adASPARAGINYL\u00adGLUTAMINYL\u00adASPARAGINYL\u00adTHREONYL\u00adPHENYL\u00adALANYL\u00adGLUTAMYL\u00adSERYL\u00adMETHIONYL\u00adSERYL\u00adGLYCYL\u00adLEUCYL\u00adVALYL\u00adTRYPTOPHYL\u00adTHREONYL\u00adSERYL\u00adALANYL\u00adPROLYL\u00adALANYL\u00adSERINE \r\n\r\nIt means tobacco or mosaic disease.", "Solution_13": "bubka\r\nit is a real amazing word for reasons which have presently escaped my mind", "Solution_14": "first of all, if its your favorite word, YOU NEED TO KNO HOW TO SPELL IT", "Solution_15": "I like all words... no favorite :D\r\n\r\nbut i do have a favorite number", "Solution_16": "whats your favorite number? mine are 7,12 and infinity :D", "Solution_17": "I've liked 4 ever since i've had a favorite number ...\r\n the others are 22/7, 42, and 2400... or $ \\sqrt{5760000}$ :wink:", "Solution_18": "Ever since I've learned them, I've particularly liked fustigate and spifflicate.", "Solution_19": "#s? ...\r\n\r\n13, 130, 1300, 13000, etc...\r\n\r\n& $ \\pi$", "Solution_20": "permutation, lamentation, jubilant, luminescent, phosphorescent, galvanizing, morbid, sequential, impetuous, insipid, frivolous, convivial, temperamental, all fun words. Try using the word deplorable in a sentence with a bunch of lowlife idiots and see how they respond :D", "Solution_21": "Bubbling. Just say it to yourself. Bubbling. Bubbling.", "Solution_22": "aperture = opening that admits light, in cameras for example", "Solution_23": "Hippopotomonstrosesquippedaliophobia- Fear of long words", "Solution_24": "transcendentalism...it was one of the last words at a local spelling bee...", "Solution_25": "aardvark\r\n\r\nFirst word in the dictionary and it's named after some kind of animal.", "Solution_26": "[quote=\"1=2\"]PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (also spelled PNEUMONOULTRAMICROSCOPICSILICOVOLCANOKONIOSIS) = a lung disease caused by breathing in particles of siliceous volcanic dust.[/quote]\r\n\r\nthat's a made up word\r\n\r\nmy favorite word\r\n\r\nis\r\n\r\nI\r\n\r\nreferring to the first person subject pronoun", "Solution_27": "no, thats a real word 1337whatevercomesnext", "Solution_28": "ouch. It's the greatest word of all tim. Say it over and over again:\r\n\r\n\"ouch ouch ouch ouch ouch ouch ouch ouch chow chow chow mmm... chow is good.\"", "Solution_29": "favorite word- bucket\r\n\r\ndon't know why", "Solution_30": "rhetorical. reason: one of the only times that r is next to h." } { "Tag": [ "calculus", "algebra unsolved", "algebra" ], "Problem": "$ P(x) \\equal{} x^{2} \\plus{} ax \\plus{} b,(a,b\\in \\mathbb{Z})$\r\nProve that,there is such an $ m\\in\\mathbb{Z}$ for $ \\forall n\\in \\mathbb{Z}$ \r\n$ P(n).P(n \\plus{} 1) \\equal{} P(m)$ sorry for my bad english", "Solution_1": "[quote=\"cnyd\"]$ P(x) \\equal{} x^{2} \\plus{} ax \\plus{} b,(a,b\\in \\mathbb{Z})$\nProve that,there is such an $ m\\in\\mathbb{Z}$ for $ \\forall n\\in \\mathbb{Z}$ \n$ P(n).P(n \\plus{} 1) \\equal{} P(m)$ sorry for my bad english[/quote]\r\n\r\n$ \\forall m,n$ :\u00a8$ P(n)P(n\\plus{}1)\\equal{}P(n^2\\plus{}(a\\plus{}1)n\\plus{}b)$", "Solution_2": "How?Please show.", "Solution_3": "[quote=\"pankajsinha\"]How?Please show.[/quote]\r\n\r\nAre you joking ? That's just basic verification calculus :\r\n\r\n$ P(n)\\equal{}n^2\\plus{}an\\plus{}b$\r\n$ P(n\\plus{}1)\\equal{}(n\\plus{}1)^2\\plus{}a(n\\plus{}1)\\plus{}b\\equal{}n^2\\plus{}(a\\plus{}2)n\\plus{}a\\plus{}b\\plus{}1$\r\n\r\n$ P(n)P(n\\plus{}1)\\equal{}(n^2\\plus{}an\\plus{}b)(n^2\\plus{}(a\\plus{}2)n\\plus{}a\\plus{}b\\plus{}1)$ $ \\equal{}(n^4\\plus{}(a\\plus{}2)n^3\\plus{}(a\\plus{}b\\plus{}1)n^2)$ $ \\plus{}(an^3\\plus{}(a^2\\plus{}2a)n^2\\plus{}(a^2\\plus{}ab\\plus{}a)n)$ $ \\plus{}(bn^2\\plus{}(ab\\plus{}2b)n\\plus{}ab\\plus{}b^2\\plus{}b)$\r\nand so $ P(n)P(n\\plus{}1)\\equal{}n^4\\plus{}(2a\\plus{}2)n^3\\plus{}(a^2\\plus{}3a\\plus{}2b\\plus{}1)n^2$ $ \\plus{}(a^2\\plus{}2ab\\plus{}a\\plus{}2b)n$ $ \\plus{}ab\\plus{}b^2\\plus{}b$\r\n\r\n$ P(n^2\\plus{}(a\\plus{}1)n\\plus{}b)\\equal{}(n^2\\plus{}(a\\plus{}1)n\\plus{}b)^2\\plus{}a(n^2\\plus{}(a\\plus{}1)n\\plus{}b)\\plus{}b$ $ \\equal{}n^4\\plus{}(a\\plus{}1)^2n^2\\plus{}b^2\\plus{}2(a\\plus{}1)n^3\\plus{}2bn^2\\plus{}2(ab\\plus{}b)n$ $ \\plus{}an^2\\plus{}(a^2\\plus{}a)n\\plus{}ab\\plus{}b$\r\nand so $ P(n^2\\plus{}(a\\plus{}1)n\\plus{}b)\\equal{}$ $ n^4\\plus{}(2a\\plus{}2)n^3\\plus{}(a^2\\plus{}3a\\plus{}2b\\plus{}1)n^2\\plus{}(a^2\\plus{}a\\plus{}2ab\\plus{}2b)n\\plus{}b^2\\plus{}ab\\plus{}b$\r\n\r\nHence the claim.\r\n\r\nIs there a step you want to be explained a little bit more ?", "Solution_4": "Thank You and Sorry for making you work hard .I thought there would be some method without getting into calculations" } { "Tag": [], "Problem": "In how many ways can a committee of 4 people be chosen from five married couples if no committee is to include a husband-and-wife pair?\r\nA: 105 B:54 C:10 D:80\r\n\r\nI'm really bad with any kind of combinatorics or probability... I almost skipped over this problem on the test. But I tried it out.\r\n\r\nI found that first we find ${10} \\choose 1$ ways to choose the first person. Then his/her spouse can't be chosen, so there are $8$ people left. Then do $8 \\choose 1$ for the second, and so on. I realized of course that my answer would be a lot bigger than the possibilities. Now this was a problem I felt confident on... I never know when to use combinations or permutations, and I felt I had picked the right time here - obviousy not so. What am I doing wrong here?", "Solution_1": "Hi!\r\n\r\nI think a differrent way\r\n\r\nWe have 5 couples, 1,2,3,4, and 5\r\n\r\nAll 4 members must belong into different couples. Hence, there is just one couple which will not be choosen\r\nWe have 5 ways to choose this couple.\r\n\r\nAfter that, we have to choose the Man or Woman from any other couples\r\n\r\ne.g. we take out the couple number 5\r\nThen for every of the couples 1,2,3,4 we choose M or W\r\n\r\n2 possible choices for every couple, total ways=5*2*2*2*2=80", "Solution_2": "deej, that's not a bad idea, but you have to check two things:\r\n\r\n(1) Did you count every possibility?\r\n(2) Did you count any possibility more than once?\r\n\r\nIn this case, you did (1) fine -- any possible committee can be found by your process. The problem is in (2) -- how many ways did you count each committee?" } { "Tag": [ "search" ], "Problem": "I'll just post the required steps, please give your comments about the validity of the apparent fallacy, I'm not really sure.\r\n\r\nFirst we derive $ K_{eq}$ for $ A \\plus{} B \\minus{}\\minus{}> C \\plus{} D$\r\n\r\n$ R_f \\equal{} K_f[A][B]$\r\n$ R_b \\equal{} K_b[C][D]$\r\n\r\nAt Equilibrium, $ R_f \\equal{} R_b$\r\n$ \\frac{K_f}{K_b} \\equal{} K_{eq} \\equal{} \\frac{[C][D]}{[A][B]}$\r\n\r\nAnd so we get a formula for Keq. \r\n\r\nLooking at the 1st step we have written rate of the reaction is proportional to concentration of its reactants raised to the power of its stoichiometric co-efficients. However we know that this is not valid, in Kinetics the order of the reaction is not always equal to the sum of co-efficents/stoi and it is only a experimentally determined quantity.\r\n\r\nSo is the formula for $ K_{eq}$ valid in case ALL the reactions. Or perhaps are all the reactions we consider under equilibrium only elementary reactions with obvious orders or in the worst case Pseudo first order reactions?", "Solution_1": "i had already posted this [url=http://www.mathlinks.ro/viewtopic.php?search_id=150332017&t=159089]here[/url]", "Solution_2": "I don't think you got an answer because Carcul didn't understand your question. So let me wait.", "Solution_3": "[quote]\n$ \\Delta G \\equal{} \\Delta G^{\\circ} \\plus{} RT \\ln{Q}$ where $ Q \\equal{} \\prod^{N}_{i \\equal{} 1} (\\frac {f_i}{P^{\\circ}})^{v_i}$[/quote]\r\n\r\nJust elaborating on this. Say the gas is not ideal. By definition $ P^{\\circ} \\equal{} 1$. We define the activity for gases as $ \\frac {f_i}{f^{\\circ}_i} \\equal{} a_i$. Since we defined $ f^{\\circ} \\equal{} 1$ hence $ f_i \\equal{} a_i$.\r\n\r\nSo $ K \\equal{} \\prod^{N}_{i \\equal{} 1} a_i^{v_i}$ defined in terms of activity.\r\n\r\nEdit: Oh [censored] I just killed my own post. Anyway, it wasn't a hard derivation. Did everyone see it before I stuffed it up?" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "\\[ \\int {\\sqrt {\\frac{{\\sin (x \\minus{} a)}}{{\\sin (x \\plus{} a)}}} } dx\r\n\\]", "Solution_1": "[quote=\"Xaenir\"]\n\\[ \\int {\\sqrt {\\frac {{\\sin (x \\minus{} a)}}{{\\sin (x \\plus{} a)}}} } dx\n\\]\n[/quote]\r\n\r\nthanks Xaenir\r\nnice integral this my solution", "Solution_2": "hello, alternatively you can set $ t=\\sqrt{\\frac{\\sin(x-a)}{\\sin(x+a)}}$ then you will get $ x=-a+\\arctan\\left(\\frac{\\sin(2a)}{-t^2+\\cos(2a)}\\right)$ and we have $ dx=-\\frac{2t\\sin(2a)}{-t^4+2t^2\\cos(2a)-1}\\,dt$\r\nand our integral is $ \\int\\frac{2t^2\\sin(2a)}{t^4-2t^2\\cos(2a)+1}\\,dt$.\r\nSonnhard." } { "Tag": [ "trigonometry", "geometry", "3D geometry" ], "Problem": "Welcome to the 2nd Pytheagle Mathematical Competition (PMC)!\r\n\r\n[b]Purpose:[/b]\r\nThis competition was created by me, Pytheagle, because of boredom. However, I spent hours of thought on the problems, so I hope there will be nothing wrong there. \r\n\r\n[b]Rules:[/b]\r\n1) Calculators are allowed but for only adding, subtracting, multiplying, dividing, and for exponential work. Acceptable uses of calculators include $ 2^{12}, 134224/433, 342\\cdot33,$ etc. Unacceptable uses include graphing, $ \\sin x$, and any other \"non-basic\" operation.\r\n\r\n2) Solutions are required; if only an answer is given, no more than 1 point shall be awarded.\r\n\r\n3) I grade each solution out of 5 points based on Clarity, Completeness, and Correctness. NO GRADING VOLUNTEERS.\r\n\r\n4) NO CHEATING (asking others for help, asking other's answers).\r\n\r\n5) Submissions are due by September 12, 2009, however, turning it in by September 9, 2009 will get you 2 more points. \r\n\r\n=====\r\n\r\nThe last PMC ended horribly because of no one submitting, inactive graders, and inexperience from my part. I hope this will be better.\r\n\r\n=====\r\n\r\nIf you are participating, post that in this thread. ONLY PM ME TO SEND YOUR SOLUTIONS OR IF YOU HAVE ANY QUESTION OR PROBLEM.\r\n\r\n=====\r\n\r\nAward: The winner will get X posts of his rated 6 cubes (X will be atleast 1). X is the difference between the top score and the 2nd top score.\r\n\r\n\t\r\nDOWNLOAD PMC HERE!!!", "Solution_1": "1. ABCAK!\r\nSo hopefully this will be rated 6 CUBE!\r\nFirst post!\r\n :!: \r\n\r\nOk will there be a finals round?", "Solution_2": "I am in :)", "Solution_3": "Sure, I'll join.", "Solution_4": "[quote=\"pytheage\"]\nAward: The winner will get X posts of his rated 6 cubes (X will be atleast 1). X is the difference between the top score and the 2nd top score.\n[/quote]\r\n\r\nYou realize this defies AoPS rules?\r\n\r\n--\r\n\r\nI might join, depending on how much homework I get this week.", "Solution_5": "It is?\r\nCmon, cause I really need that :P:P\r\nAnd how hard will this be?", "Solution_6": "Yes it does.", "Solution_7": "[quote=\"abcak\"]And how hard will this be?[/quote]\r\nLook at the problems.\r\n\r\nUnlimited time? Internet resources allowed?\r\n\r\nBTW, I might be in for the same reasons izzy stated.", "Solution_8": "I will join.", "Solution_9": "No it's not against the rules; I asked Mr. Rusczyk and he said it's fine.\r\n\r\nAnd yes, you have unlimited time, but you MAY NOT USE ANY RESOURCES OTHER THAN YOURSELF.", "Solution_10": "It still defeats the purpose of post rating, but whatever. :D", "Solution_11": "Oh, yourself counts your brain right...? :D", "Solution_12": "I got only ONE submittion. You hear that? ONE submission.\r\n\r\nThat's why PMC is canceled and thunder365 gets 1 post rated 6 cubes.\r\n\r\nWhat do I need to do to make more people participate? Make the problems easier?", "Solution_13": "No, make them harder...it's a waste of my time to type up 6 solutions to 6 (relatively) easy problems." } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "An aluminum sphere with an irrational value as its radius can be melted down to form two new spheres with radii of integer length in two different ways. What is the smallest possible volume of that sphere?", "Solution_1": "[hide]Well, the smallest number is \n\n$1729=12^{3}+1^{3}=10^{3}+9^{3}$.\n\nSo, the radius is $\\sqrt[3]{1729}$[/hide]" } { "Tag": [ "logarithms" ], "Problem": "If log base 4 of 6 = p, express log base 3 of 12 in terms of p.", "Solution_1": "$ p \\equal{} \\log_46 \\equal{} \\log_4(2\\cdot3) \\equal{} \\frac12 \\plus{} \\log_43\\implies p \\minus{} \\frac12 \\equal{} \\log_43$\r\n\r\n$ \\implies \\frac2{2p \\minus{} 1} \\equal{} \\log_34\\implies \\frac1{2p \\minus{} 1} \\equal{} \\log_32$\r\n\r\n$ \\log_3{12} \\equal{} \\log_3\\left(2^2\\cdot3\\right) \\equal{} 2\\log_32 \\plus{} 1$\r\n\r\n$ \\equal{} \\frac2{2p \\minus{} 1} \\plus{} 1 \\equal{} \\boxed{\\frac {2p \\plus{} 1}{2p \\minus{} 1}}$" } { "Tag": [ "logarithms", "function", "inequalities" ], "Problem": "Team #3\r\n\r\nSolve for $x$: $\\log_2 (x+2) + 5 = 8 + \\log_2 x$.", "Solution_1": "[hide]$\\log_2 (x+2) = \\log_2 8 + \\log_2 x = \\log _2 8x$\n\n$8x = x+2$\n\n$x = \\frac 27$[/hide]", "Solution_2": "[hide]$\\log_2(x+2)+5=8+\\log_2x$\n$\\log_2(x+2)-\\log_2x=3$\n$\\log_2\\frac{x+2}{x}=3$\n$\\frac{x+2}{x}=8$\n\nSolving for $x$, $x=\\frac{2}{7}$.[/hide]", "Solution_3": "I remember this problem, because I actually participated in NCIML this year. DOes anyone else here participate in it?", "Solution_4": "I'm going to give you a detailed explanation even if you can't read it :diablo: \r\n\r\nokay:\r\n\r\nlog2 of (x+2) + 5= 8 + log2 of (x)\r\n\r\nlog2 of (x+2)-log2 of (x)=8-5\r\n\r\nlog2 of (x+2)/x=3\r\n\r\nThen exponentiate to get (x+2)/x=2^3\r\n\r\nFrom there, just cross multiply and get [b]x=2/7[/b]", "Solution_5": "Everyone, You all forgot the condition for anti-logarithm. :D \r\n\r\nkunny", "Solution_6": "[quote=\"kunny\"]Everyone, You all forgot the condition for anti-logarithm. :D \n\nkunny[/quote]\r\n\r\nHm? Can you expand on that statement? Never heard of that before.", "Solution_7": "[quote=\"kunny\"]Everyone, You all forgot the condition for anti-logarithm. :D \n\nkunny[/quote]\r\n\r\nUm.. Not sure what you mean because that's what I got and it's what the site had too. ;)", "Solution_8": "Here's what I found out what AntiLogarithm is:\r\n\r\nIt's the inverse function of a log defined as:\r\n\r\nlog b of (antilog b of z) = z = antilog b of (log b of z)\r\n\r\nThe antilog of base b of z is therefore $b^z$\r\n\r\nHowever, I don't see how we needed this, I think the solutions posted above should be allright :sleep2: :sleeping:", "Solution_9": "Can anyone solve the following logarithmic equation and inequality?\r\n\r\n[1] $\\log_2(x-2)+\\log_2(7-x)=2$\r\n\r\n[2] $\\log_2(x-2)+\\log_2(7-x)\\leq2$", "Solution_10": "[hide]\n$(x-2)(7-x) = 4$\n\n$(x-2)(7-x) \\le 4$\n\nThen solve.[/hide]", "Solution_11": "Then what's your answer,tarquin.", "Solution_12": "I don't get it...for the first one isn't the answer just $6$ or $3$... :?:", "Solution_13": "Yep, at least that's what I got.", "Solution_14": "second one is:\r\n$x<3 \\cup x>6$", "Solution_15": "[quote=\"amirhtlusa\"]second one is:\n$x<3 \\cup x>6$ :( [/quote]", "Solution_16": "Second one: $-x^2+9x-18\\leq 4$\r\n\r\nSo $x\\in(-\\infty,3]\\cup[6,\\infty)$", "Solution_17": "Everyone,you forgot [color=blue]the condition for anti-logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)", "Solution_18": "$x>2$ :?", "Solution_19": "[quote=\"Slizzel\"]$x>2$ :?[/quote]\r\n\r\n :(", "Solution_20": "[quote=\"kunny\"]Everyone,you forgot [color=blue]the condition for anti=logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)[/quote]\r\nBut it still doesn't affect Silverfalcon's problem, does it?", "Solution_21": "Yes, you are right, but in case of logarithmic inequality, you can't skimp. ;)", "Solution_22": "So the condition is $2< x< 7$, right?", "Solution_23": "Yes. :)", "Solution_24": "errr...\r\nyup kunny is right i forgot the defenition of anit logaritm here is the answer:\r\n$2\\le x\\le3 \\cup 6\\le x\\le 7$", "Solution_25": "[quote=\"amirhtlusa\"]errr...\nyup kunny is right i forgot the defenition of anit logaritm here is the answer:\n$2\\le x\\le3 \\cup 6\\le x\\le 7$ :? :( [/quote]", "Solution_26": "It doesn't work for $x=2$ and $7$", "Solution_27": "yes it's right its open interval\r\n$x\\in (2,3]\\cup [6,7)$", "Solution_28": "That's right. :)", "Solution_29": "Yeah.. thanks kunny for reminding us about anti-logarithm", "Solution_30": "Don't mention it. :)", "Solution_31": "So everytime we deal with logarithms, we have to think of the anti-logarithms too? \r\n\r\nIs that the same with the restriction? \r\n\r\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :(", "Solution_32": "[quote=\"YooSam\"]So everytime we deal with logarithms, we have to think of the anti-logarithms too? \n\nIs that the same with the restriction? \n\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :([/quote]\r\n\r\nYes. We should take into consideration the condition for which the logarithm can be defined, that is to say, \r\n\r\n[color=blue]anti-logarithm is always positive real number[/color] and moreover [color=red]the base is positive real number and not equal 1[/color].\r\n\r\nkunny", "Solution_33": "[quote=\"kunny\"][quote=\"YooSam\"]So everytime we deal with logarithms, we have to think of the anti-logarithms too? \n\nIs that the same with the restriction? \n\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :([/quote]\n\nYes. We should take into consideration the condition for which the logarithm can be defined, that is to say, \n\n[color=blue]anti-logarithm is always positive real number[/color] and moreover [color=red]the base is positive real number and not equal 1[/color].\n\nkunny[/quote]\r\n\r\nOh, that is what we call restriction :D . I never knew that people call it anti-logarithms :( . Thanks kunny." } { "Tag": [ "inequalities", "geometry", "circumcircle", "percent", "email", "national olympiad" ], "Problem": "I'd like to know who are the members of mathlinks who are going to partecipate at the 8th Junior Balkan in Novi Sad.It would be very nice to meet there. :) \r\n\r\nCheers!", "Solution_1": "I am going!!!", "Solution_2": "Nice to hear that. :)", "Solution_3": "there are also some romanian members going to jbmo 2004, but I assume that they do not visit the website that often ... :maybe:", "Solution_4": "And you? Have you ever been at the JBMO?If yes could you please tell me something about it. :) How was that like?The level of the problems, the organisation, everything...\r\n\r\nIris", "Solution_5": "I have participated in the 2nd JBMO in 1998 (first one at which romania went) and got 39p :( out of 40p. Gold medal and special award of the jury for the best solution of problem 4. The JBMO was my first official contest where I represented Romania, and I just loved it. Plus it was in Athens, and I got to see a lot of stuff there, which I hope to revisit now.", "Solution_6": "Wow,congratulations :) 39 out of 40!I guess that's the best score ever achieved at the JBMO.I also guess that that gold medal has been the first of a long serie of one.Your solution of problem 4 was the best one.Congratulations for that too,because I've noticed that it was the hardest one there.(it was proposed by Bulgaria,wasn't it)I can't solve it and i've posted it here at mathlinks,but no acceptable solution came :( .It would be very nice of you to send a detailed solution. ;) I've been to Athens too,but not for a competition cause i'm just a freshmen.It's a lovely city.You said you hope to visit it again this year.Are you going at the IMO?", "Solution_7": "well a lot of people got 40p afterwards, for example Claudiu Raicu, Andrei Negut, Kappa :D so it's not such a great performance :P\r\n\r\nand yes I do plan to visit at IMO. \r\n\r\nPS in the future you may want to use space \" \" character after the punctuation \",\" and \".\" :D", "Solution_8": "BTW last year I got 39p. :)", "Solution_9": "[quote=\"Sasha\"]BTW last year I got 39p. :)[/quote] we all know that 39p is the best score :P:P;)", "Solution_10": "Well, I have participated at JBMO and I am from the roumanian team.I have got 40. :D :D ;)", "Solution_11": "iris, ku jeton ne shqiperi?? une jam shqiptare po jetoj ne amerike. nice to hear from a fellow albanian who is interested in math. keep in touch.", "Solution_12": "[quote=\"Persu Madalina\"]Well, I have participated at JBMO and I am from the roumanian team.I have got 40. :D :D ;)[/quote]Hi there ... congrats I must admit that I did not expect such a nice result from the Romanian team (4 maximum scores, one 39 and one 38 :)). \r\n\r\nMore about Madalina, she is 13, so she's got another 2 more years of potentially making a full-score at the Jbmo. Good luck in that.\r\n\r\nSo, tell us how was the trip there?", "Solution_13": "Congrats, Madalina! Bravo!\r\nP.S. Meet you next year in Moldova??? or Greece!!", "Solution_14": "I was wondering, what did you do at the jbmo, iris_aliaj?", "Solution_15": "Hi Madalina!\r\nI had the pleasure to get to know you there at the JBMO and I congratulated you on your perfect score, but I didnt know you were 13.Thats amazing. Good job! Wish you the best in all the competitions youre going to participate in the future. :)", "Solution_16": "[quote=\"Valentin Vornicu\"]I was wondering, what did you do at the jbmo, iris_aliaj?[/quote]\r\n\r\nWell, I got bronze medal and i'm so happy but i wanna thank mathlinks and those active members like: Darij, Pierre, Lagrangia, Grobber, Arne, Zabelman, Mindflyer, Maja, Dante, Myth, Valiowk for all their help and supply during the period i was training for it. :) \r\nAnd thanks to you Valentin for the solution of that second JBMO problem. :) \r\n\r\nThanks a lot.\r\n\r\n\r\nP.S: It's Iris Aliaj and not \"iris_aliaj\" :D", "Solution_17": "Hi, Iris Aliaj!\r\nI also participated at JBMO and also got bronze. How many points did you get? (I got 24)", "Solution_18": "strange jbmo ... too easy maybe ... so many full-scores, getting 39 / 40 would mean silver :maybe: jeez ... \r\n\r\nanyway, can you guys post the questions?", "Solution_19": "Hi freemind!\r\nCongrats on your bronze medal.\r\nI got 25 points :D (of course 1 point doesn't show anything :) )\r\nProblem 1 10p\r\nProblem 2 5p\r\nproblem 3 10p\r\nProblem 4 0p :maybe: \r\n\r\nHi Valentin!\r\nwell maybe it was easy for the strong teams like Romania and Bulgaria since there were 11 gold medals.\r\nI think determining sth as easy or as difficult is never completely true or false cause it depends on the training you have had and on the level of problems you have most dealed with(e.g i've never dealed with any combinatoric problem though i'm gonna start now).\r\nI'll see to posting the problems here as soon as possible. :) \r\n\r\nCheers! :D", "Solution_20": "Hi Iris Aliaj!\r\nActually, i did like you, just at the 2nd problem I got 4 points.\r\nBy the way, how old are you? I'm 13.\r\nKeep in touch.\r\n\r\nHi, V.V. I already have the problems. Here you are:\r\n 8th JUNIOR BALKAN MATHEMATICAL OLYMPIAD\r\n NOVI SAD, Serbia & Montenegro, 26.06.2004\r\n\r\nPROBLEM 1\r\n\r\nProve that the inequality \r\n\t\t\t(x+y) / (x\u00b2 \u2013 xy + y\u00b2) \u2264 2\u221a2 / \u221a(x\u00b2 + y\u00b2)\r\nHolds for all real numbers x and y, not both equal to 0.\r\n\r\nPROBLEM 2\r\n\r\nLet ABC be an isosceles triangle with AC=BC, M the midpoint of AC, and Z be the line through C perpendicular to AB. The circle through B, C, M intersects the line Z at points at C and Q. Find the radius of the circumcircle of ABC in terms of m = CQ.\r\n\r\nPROBLEM 3\r\n\r\nIf the positive integers x and y are such that 3x + 4y and 4x + 3y are both perfect squares, prove that both x and y are both divisible with 7.\r\n\r\nPROBLEM 4\r\n\r\nConsider a convex polygon having n vertices, n\u22654. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.\r\n\r\nProve that there are two more black triangles that white ones.\r\n\r\n \r\n\t\t\t\t\tWorking time: 4\u00bd hours\r\n\r\n[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/resources.php?c=1&cid=21&year=2004 .][/color]", "Solution_21": "Hi!!! I am back again!!! :lol: All the problemes were easy for the roumanian team!!!I mean first we all said we did all the problems. Freemind what's up? I hope to see you soon... maibe at our ONM. What's the first strong contest in Roumania?Vranceanu?I am so bored in this holiday... :?", "Solution_22": "[quote=\"Persu Madalina\"]Hi!!! I am back again!!! :lol: [...] I am so bored in this holiday... :?[/quote] nice of you to join our \"little\" gathering here on the site. if you are bored, you can always try to solve problems from the site ;) :D", "Solution_23": "Hello to everyone :D . I have oalso participate on JBMO 2004 and got bronze medal with 25 points.I am defintly late to post in this topic but i am posting with 2 reasons : 1.Does anybody know in which city in Moldova will be held next JBMO?\r\n And the second is to show my solution of problem 2(unfortunatly i did not do it on competention pm\u010dy 4 or 5 month after :)))\r\n[solution reposted at http://www.mathlinks.ro/Forum/viewtopic.php?t=14344 ]\r\n\r\n P.S. i almost forgot to ask from which to which date will JBMO 2005 be held", "Solution_24": "These details are not yet known I'm afraid ... probably in the city of Chisinau (the capital of Moldova)...", "Solution_25": "Hi..\r\nI thought jbmo will be in greece this year \r\nand in moldova in 2006\r\nand all contestants from moldova, adn the teachers say it will be in greece this year!", "Solution_26": "I am 99,99 percent sure that the competition is in Moldova but Valentin knows the best :D doesnt he ? :)", "Solution_27": "[quote=\"vedran6\"]I am 99,99 percent sure that the competition is in Moldova but Valentin knows the best :D doesnt he ? :)[/quote]Well I didn't say that it will be in Moldova. All I said was that if the olympiad is held there, then it probably will be in the Capital city :)", "Solution_28": "[quote=\"Valentin Vornicu\"][quote=\"vedran6\"]I am 99,99 percent sure that the competition is in Moldova but Valentin knows the best :D doesnt he ? :)[/quote]Well I didn't say that it will be in Moldova. All I said was that if the olympiad is held there, then it probably will be in the Capital city :)[/quote]\r\n Than we will have to wait until we got the information about the host of JBMO 2005 :).I would that it will be Greece because it is more atractive than Moldova :lol:", "Solution_29": "I went to Greece on my JBMO in 1998 and it was wonderfull :) Also the IMO last year was great, so I assume it'll be great if the JBMO is held again in Greece ...", "Solution_30": "I hope it will be in moldova", "Solution_31": "[quote=\"ruza\"]I hope it will be in moldova[/quote]\r\n I hope i will be in the first six on BiH Junior Mathematical Olimpiad this week :D", "Solution_32": "why does everyone want jbmo to be in moldova?\r\nit would be much nicer to be in greece\r\n:)))", "Solution_33": "Well if I get into BiH JBMO team(hopefully I will :) ) I think it would be my only chance to visit Moldova in my life.Beside I don`t know much about Moldova and this would be perfect opportunity to learn more about it :lol: .", "Solution_34": "Moldova is a small and i'd say that it's a poor country compared with romania, bulgary or greece. of course becuse i'm a contestant i don't want jbmo to be in moldova. but anyway if it is not this year then it will be next year and i will also participate (if i'll qualify :))) )\r\ngood luck tu ruza and vedran6 at their tst", "Solution_35": "I found out that JBMO this year will be in Greece, between 20 and 26 June! I am going to participate,too! :D See you there and good luck at your tst!", "Solution_36": "[quote=\"Persu Madalina\"]I found out that JBMO this year will be in Greece, between 20 and 26 June! I am going to participate,too! :D See you there and good luck at your tst![/quote]Great! Good luck in Greece Madalina. You have the chance of making for the first time two consecutive maximum scores at the jBMO (and even 3 right? :) ).", "Solution_37": "Not really...because I miss my third JBMO with only about 2 month :( :mad: Anyway, thanks for your wishes! :D", "Solution_38": "Just to mention that ruza and I have qualifaid for JBMO which will be held in city of Varlea in Greece :).I was first with 100 points :lol: (i am very lucky man :lol:).and ruza was fourthSo see you in Greece", "Solution_39": "It`s not Varlea it`s Veria :lol:", "Solution_40": "[quote=\"ruza\"]It`s not Varlea it`s Veria :lol:[/quote]\r\n There is also problem number two on the map of Greece there isnt neither Varlea or Veria there is only something like Vorua", "Solution_41": "Iris Aliaj, give me you're email please?!" } { "Tag": [ "geometry", "3D geometry", "prism", "Pythagorean Theorem" ], "Problem": "Does anyone know a way to find the space diagonal of a hexagonal prism?", "Solution_1": "[hide=\"Assuming the bases are regular:\"]\nLet the side length be $ s$, and the height be $ h$. The long diagonal of the base is $ 2s$, and thus, by the Pythagorean Theorem, the space diagonal is $ \\sqrt{h^2\\plus{}4s^2}$.\n[/hide]\r\n\r\nOkay, that was not a very good explanation, but hopefully it will suffice." } { "Tag": [ "geometry", "rectangle", "inradius", "incenter", "ratio", "trigonometry", "perimeter" ], "Problem": "Let $ ABCD$ a cuadrilateral inscribed in a circumference. Suppose that there is a semicircle with its center on $ AB$, that\r\nis tangent to the other three sides of the cuadrilateral. \r\n\r\n(i) Show that $ AB \\equal{} AD \\plus{} BC$. \r\n\r\n(ii) Calculate, in term of $ x \\equal{} AB$ and $ y \\equal{} CD$, the maximal area that can be reached for such quadrilateral.", "Solution_1": "If $AD \\parallel BC,$ then $AD, BC \\perp AB$ and ABCD is a rectangle with $AD = BC = \\frac{AB}{2}.$ Assume that AC, BC are not parallel and they meet at a point P. Since ABCD is cyclic, $\\angle PCD = \\angle PAB,\\ \\angle PDC = \\angle PBA$ and the triangles $\\triangle ABP \\sim \\triangle CDP$ are similar. Let $E \\in AB$ be the center of the semicircle tangent to AD, BC, CD. (E) is the excircle of the triangle $\\triangle CDP$ against the vertex P with exradius $r_p.$ Let (I) be the incircle of the triangle $\\triangle ABC$ with inradius r. Denote $p = CD,\\ c = DP,\\ d = PC$ the sides and $s = \\frac{c+ d + p}{2}$ the semiperimeter of the triangle $\\triangle CDP.$ Let the bisector $PI \\equiv PE$ of the angle $\\angle P$ meet the opposite side CD at a point K. The similarity coefficientb of the triangles $\\triangle ABC \\sim \\triangle CDP$ is $\\eta = \\frac{PE}{PK}.$ The incenter I cuts the angle bisector PK in the ratio $\\frac{PI}{IK} = \\frac{c + d}{p},$ hence $\\frac{PI}{PK} = \\frac{c + d}{c + d + p}.$ Also, $\\frac{PE}{PI} = \\frac{r_p}{r} = \\frac{s}{s - p} = \\frac{c + d + p}{c + d - p}.$ Thus the similarity coefficient is equal to\r\n\r\n$\\eta = \\frac{PE}{PK} = \\frac{PE}{PI} \\cdot \\frac{PI}{PK} = \\frac{c + d + p}{c + d - p} \\cdot \\frac{c + d}{c + d + p} = \\frac{c + d}{c + d - p}$\r\n\r\nConsequently,\r\n\r\n$AB = \\eta CD = \\frac{(c + d)p}{c + d - p}$\r\n\r\n$AD = PA - PD = \\eta PC - PD = \\eta d - c$\r\n\r\n$BC= PB - PC = \\eta PD - PC = \\eta c - d$\r\n\r\n$AD + BC = (\\eta - 1)(c + d) = \\frac{p(c + d)}{c + d - p} = AB$", "Solution_2": "a) Call $ O$ the center of the circle $ \\Gamma$ tangent to $ BC$, $ CD$, $ DA$, and $ P$, $ Q$, $ R$ the respective points where $ \\Gamma$ touches $ BC$, $ CD$, $ DA$. Call $ 2\\alpha=\\angle DAB$, $ 2\\beta=\\angle CBA$. Triangles $ OPC$ and $ OQC$ are obviously equal and rectangle at $ P$ and $ Q$ respectively, or $ \\angle POC=\\angle QOC=\\frac{\\BAD}{2}=\\alpha$, Similarly, $ \\angle QOD=\\angle ROD=\\beta$. Calling $ r$ the radius of $ \\Gamma$, $ AR=\\frac{r}{\\tan(2\\alpha)}$ and $ CP=r\\tan(\\alpha)$. Therefore, since $ \\frac{1}{\\tan(2\\alpha)}+\\tan(\\alpha)=\\frac{\\cos(2\\alpha)+2\\sin^{2}\\alpha}{\\sin(2\\alpha)}=\\frac{1}{\\sin(2\\alpha)}=\\frac{OA}{r}$, then $ AR+CP=OA$, and similarly $ BP+DR=OB$, or $ AD+BC=AB$, qed.\r\n\r\n\r\nb) If $ BC$ and $ AD$ are parallel, so are $ OP$ and $ OR$, or $ BC$ and $ AD$ are perpendicular to $ AB$. The area of $ ABCD$ is then $ \\frac{AB(BC+DA)}{2}=\\frac{x^{2}}{2}$ (incidentally, $ ABCD$ is then a rectangle or it could not be cyclic, and $ AB=CD=2BC=2DA$).\r\n\r\nIf $ BC$ and $ AD$ are not parallel, let $ E$ be the point where they meet. Since $ ABCD$ is cyclic, $ ABE$ and $ DCE$ are similar, with proportionality constant $ \\frac{AB}{CD}=\\frac{x}{y}$, or the area of $ ABCD$ is equal to $ 1-\\frac{y^{2}}{x^{2}}$ times the area of $ ABE$. Let us now show that the perimeter of $ ABE$ is determined by $ x$ and $ y$: it is obvious that $ DE+CE=\\frac{y}{x}(AE+BE)=AE+BE-(AD+BC)=AE+BE-x$, or $ AE+BE+AB=\\frac{x^{2}}{x-y}+x$. Since $ AB$ is given, the area will reach a maximum when $ ABE$ is isosceles in $ E$, ie, when $ ABCD$ is an isosceles trapezoid with $ AB$ parallel to $ CD$, and $ BC=AD=\\frac{x}{2}$. Then, the altitude from $ C$ or $ D$ onto $ AB$ is given by $ \\sqrt{AD^{2}-\\left(\\frac{x-y}{2}\\right)^{2}}=\\frac{1}{2}\\sqrt{2xy-y^{2}}$. Multiply this by $ \\frac{x+y}{2}$ and the maximum area of $ ABCD$ is found.", "Solution_3": "Part a) is IMO 1985 :wink: I know one very nice proof of this part(without using trigonometry) :) If someone want i could post it.", "Solution_4": "Let $ O$ be the center of the circle centered on $ AB$ and tangent to $ AD,DC,CB.$ Rays $ OC,OB$ bisect $ \\angle BCD \\equal{} \\theta$ and $ \\angle CDA \\equal{} \\varphi.$ Pick the point $ M$ on $ AB$ such that $ AD \\equal{} AM.$ Then the triangle $ \\triangle ADM$ is isosceles, where \n\n$ \\angle DMA \\equal{} \\frac {\\pi \\minus{} \\angle DAB}{2} \\equal{} \\frac {\\pi}{2} \\minus{} \\frac {\\pi \\minus{} \\theta}{2} \\equal{} \\frac {\\theta}{2}$\n\n$ \\angle DCO \\equal{} \\frac {\\theta}{2}$ $ \\Longrightarrow$ $ DCOM$ is cyclic $\\Longrightarrow$ $ \\angle CMB \\equal{} \\angle ODC \\equal{} \\frac {\\varphi}{2}$ \n\nOn the other hand, in the $ \\triangle CMB,$ we have \n\n $ \\angle MCB \\equal{} \\pi \\minus{} \\frac {\\varphi}{2} \\minus{} (\\pi \\minus{} \\varphi) \\equal{} \\frac {\\varphi}{2} \\Longrightarrow \\angle CMB \\equal{} \\angle MCB \\equal{} \\frac {\\varphi}{2}$ \n\n$ \\triangle CMB$ is then isosceles with apex $ B$ $ \\Longrightarrow$ $ AB \\equal{} AM \\plus{} BM \\equal{} AD \\plus{} BC.$", "Solution_5": "[quote=Erken]Part a) is IMO 1985 :wink: I know one very nice proof of this part(without using trigonometry) :) If someone want i could post it.[/quote]\n\nYes it's the circle centered at A with the length AD then we show that the second triangle is isosceles by some angle chasing and we are done ...", "Solution_6": "[quote=Fouad-Almouine][quote=Erken]Part a) is IMO 1985 :wink: I know one very nice proof of this part(without using trigonometry) :) If someone want i could post it.[/quote]\n\nYes it's the circle centered at A with the length AD then we show that the second triangle is isosceles by some angle chasing and we are done ...[/quote]\nYes, I like this solution... \nFor who wants to see this solution here (first solution) :\nhttps://artofproblemsolving.com/community/c6h60782p366584\n", "Solution_7": "Could someone explain why in part b of daniel73 solution it is needed to show that the perimeter of $ ABE$ is determined by $ x$ and $ y$?\nThanks" } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "Pint $ M$ is taken in interior of quadrilateral $ ABCD$ such that $ ABMD$ is parallelogram. If $ \\angle CBM \\equal{} \\angle CDM$ prove that $ \\angle ACD \\equal{} \\angle BCM$.", "Solution_1": "This is my proof.Nice problem! :) \r\nPoint N is taken such that DMCN is a parallelogram.\r\nSo ABCN is also a parallelogram.\r\n$ \\angle ADC \\equal{} \\angle ADM \\plus{} \\angle MDC \\equal{} \\angle ABM \\plus{} \\angle MBC \\equal{} \\angle ABC \\equal{} \\angle ANC$\r\n$ \\Rightarrow \\angle ADC \\equal{} \\angle ANC$\r\n$ \\Rightarrow \\angle ACD \\equal{} \\angle AND$\r\nWe have $ \\angle BCM \\equal{} \\angle AND$\r\nThus $ \\angle BCM \\equal{} \\angle ACD$ :lol:" } { "Tag": [ "modular arithmetic", "IMC", "college contests" ], "Problem": "Find the number of positive integers x satisfying the following two conditions:\r\n1. $x<10^{2006}$\r\n2. $x^{2}-x$ is divisible by $10^{2006}$", "Solution_1": "Funny how I guessed this property from long before the IMC... I never thought there would be an easy proof though. :)", "Solution_2": "Really easy for IMC :D !!!\r\nThere are 3 answer $x=1$ and there exist exactly one $00$if $a\\in [0,1]$ prove that:\r\n$(1+x_1+x_2+...+x_n)^a\\leq 1+x_1^a+\\frac{1}{2}(2x_2)^a+\\frac{1}{3}(3x_3)^a+...+\\frac{1}{n}(nx_n)^a$", "Solution_1": "Isn't this trivial? I mean we just observe that $ i\\cdot x_i\\leq 1+x_1+...+x_n $ and then write that $ (\\frac{i\\cdot x_i}{1+x_1+...+x_n})^a\\geq\\frac{i\\cdot x_i}{1+...+x_n} $ and then multiply by $\\frac{1}{i} $ and then add up." } { "Tag": [], "Problem": "I have 85000 words in my book. HOw many words would I have left over on the last page if I organized the pages so that 300 words were on each page?", "Solution_1": "[hide]I think 100 words. $85000/300= 283 R 100$[/hide]", "Solution_2": "Yes, correct :)", "Solution_3": "[hide]$\\frac{85000}{300} = 283 \\text{remainder} 100$ \nso the answer is 100\n[/hide]", "Solution_4": "Plug it into a calculator, and you get something that ends in .333333333333333333333, and .333333333333333333333333 is about 1/3, and 1/3 of 300 = 100. Yay.", "Solution_5": "$\\frac{85000}{300} = 283 r 100$\r\n\r\n$100$" } { "Tag": [ "quadratics", "modular arithmetic" ], "Problem": "Find all possible solutions to $x^2+y^2+z^2=1996$, where x, y and z are positive integers.", "Solution_1": "[hide=\"hint\"]\nmod 4\n[/hide]", "Solution_2": "[hide]\n$1996\\equiv 0 (mod 4)$. The unic quadratic residual $mod 4$ are $0$ and $1$. So it must be $x^2\\equiv y^2 \\equiv z^2 \\equiv 0 (mod 4)$\n\n$x=4k$\n$y=4m$\n$z=4n$\n\nThe equation becomes $4k^2+4m^2+4n^2=499$ The left-hand side is $\\equiv 0 (mod 4)$ $\\forall k,m,n \\in \\mathbb{N}$, the right-hand is $\\equiv 1 (mod 4)$. So there are no solutions.[/hide]", "Solution_3": "I think there is at least one solution, here is an example:\r\n[hide]\n(x,y,z) = (6, 14, 42 ) :!:\n[/hide]", "Solution_4": "[quote=\"hydro\"]So it must be $x^2\\equiv y^2 \\equiv z^2 \\equiv 0 (mod 4)$\n\n$x=4k$\n$y=4m$\n$z=4n$[/quote]\r\n\r\n$x^2\\equiv 0 \\pmod {4}\\not \\implies 4|x$.", "Solution_5": "[hide=\"Partial Solution\"]So $4|x^{2}\\implies 2|x$. This means $x=2m$, $y=2n$, and $z=2k$. So then $m^{2}+n^{2}+k^{2}=499$. The right side $499\\equiv 3\\pmod{4}$ so $m^{2},n^{2},k^{2}\\equiv 1\\pmod{4}$.\n\nWe have $\\left(m-1\\right)|4$ or $\\left(m+1\\right)|4$ (the same holds true for $n$ and $k$). This shows why $\\left(6,14,42\\right)$ works.\n\nSomeone else will have to finish it from there, I have to go...[/hide]", "Solution_6": "once we reduce it to \r\n\r\n$m^2+n^2+p^2=499$ , by looking at mod 4 , we find that they are all odd number . Also , every odd square number ends with 1,5,9 . So the combination to have a $9$ behind is $[1,9,9]$ or $[5,5,9]$ \r\n\r\nFor the case $[5,5,9]$ by trial and error ( only need to try $5^2,15^2$ ) and we find only $(15,15,7)$ satisfy it .\r\n\r\nFor the case $[1,9,9]$ , those end with $1$ are $[1,81,121,361,441]$ while those end with $9$ are $[9,49,169,289]$ . By some calculation , only $(21,7,3)$ satisfy it . Hence the two solution is\r\n\r\n$(30,30,14)$ and $(42,14,6)$ and their ordered pair satisfy the condition", "Solution_7": "Congratulations :!:\r\nMy solution is almost the same as yours except in the stage of \"trial and error\".", "Solution_8": "maybe we can also consider $\\mod 3$ and deduce that two of them are divisible by 3. Then taking $\\mod 9$ to deduce that the third one must be $7 (\\mod 9)$ ( since every square mod 9 can only be 0,1,4,7).\r\n\r\nFrom this , it is obvious that one of them is $7$ ( or else it will be 7+9x2=25 which is too big) .\r\n\r\nSo now the equation is reduce to\r\n\r\n$m^2+n^2=499-49=450=2\\cdot 5^2\\cdot 3^2$\r\n\r\nSo we can have $(5^2+5^2)(3^2+0^2)=15^2+15^2$ or $(6^2+3^2)(3^2+1^2)=21^2+3^2$" } { "Tag": [ "function", "induction", "percent", "algebra proposed", "algebra" ], "Problem": "Find all functions $f: \\mathbb R \\to \\mathbb R$ which satisfy \\[f( x + f ( xy ) )= f( x ) + x f( y ) \\] for all reals $x$ and $y$.", "Solution_1": "[quote=\"kenan aze\"]f:R - R/ x, y are real numbers, Find all functions such that \nf[ x + f ( xy ) ] = f( x ) + x f( y ) :)[/quote]\r\n\r\nI found this problem rather hard. Thank you for giving us a simpler solution if you got one.\r\n\r\nLet $ P(x,y)$ be the assertion $ f(x+f(xy))=f(x)+xf(y)$\r\n\r\n$ f(x)=0$ $ \\forall x$ is a trivial solution. So we'll now consider non all-zero solutions.\r\n\r\n1) $ f(x)=0$ $ \\iff$ $ x=0$\r\n========================\r\n\r\nSuppose that $ \\exists u\\neq 0$ such that $ f(u)=0$. then, for $ x\\neq 0$, $ P(\\frac ux,x)$ $ \\implies$ $ \\frac uxf(x)=0$ and so :\r\n\r\n$ f(x)=0$ $ \\forall x\\neq 0$ and then $ P(0,0)$ $ \\implies$ $ f(f(0))=f(0)$ and so $ f(0)=0$ (else we would have a non zero real $ f(0)$ whose image would be non zero.\r\n\r\nSo $ f(x)=0$ $ \\implies$ $ x=0$.\r\nThen $ P(-1,-1)$ $ \\implies$ $ f(f(1)-1)=0$ and so $ f(1)=1$ and $ f(0)=0$\r\nQ.E.D.\r\n\r\n2) $ f(n)=n$ $ \\forall n\\in\\mathbb Z$\r\n==================================\r\n\r\n$ P(1,x)$ $ \\implies$ $ f(f(x)+1)=f(x)+1$\r\n$ P(1,f(x)+1)$ $ \\implies$ $ f(f(x)+2)=f(x)+2$\r\nAn immediate induction gives $ f(f(x)+n)=f(x)+n$ $ \\forall x$, $ \\forall n>0\\in\\mathbb N$\r\n\r\nThen $ P(-n,-1)$ $ \\implies$ $ f(-n)=nf(-1)$\r\n\r\nThen $ P(-1,-2)$ $ \\implies$ $ f(1)=f(-1)-f(-2)=-f(-1)$ and so $ f(-1)=-1$ and $ f(n)=n$ $ \\forall n\\in\\mathbb Z$\r\nQ.E.D.\r\n\r\n3) $ f(x)=1$ $ \\iff$ $ x=1$\r\n=========================\r\n\r\nWe already know that $ f(1)=1$. Suppose now $ f(a)=1$, then : \r\n\r\n$ P(\\frac 1x,x)$ $ \\implies$ $ f(\\frac 1x+1)=f(\\frac 1x)+\\frac{f(x)}x$\r\n\r\n$ P(\\frac 1x,ax)$ $ \\implies$ $ f(\\frac 1x+1)=f(\\frac 1x)+\\frac{f(ax)}x$\r\n\r\nAnd so $ f(ax)=f(x)$ $ \\forall x\\neq 0$ and so $ f(ax)=f(x)$ $ \\forall x$\r\n\r\nSo $ f(-a)=-1$ and then $ P(a,-1)$ $ \\implies$ $ f(a-1)=1-a$\r\nSo $ f(a(a-1))=f(a-1)=1-a$ and then $ P(a,a-1)$ $ \\implies$ $ f(a+f(a(a-1)))=f(a)+af(a-1)$ and so $ 1=1+a(1-a)$ and so $ a=1$\r\nQ.E.D\r\n\r\n4) $ f(x)$ is injective\r\n=======================\r\n\r\nConsider now $ f(y_1)=f(y_2)$ with $ y_1\\neq y_2$. Obviously $ y_1,y_2\\neq 0$ (since $ f(x)=0$ implies $ x=0$) and let then $ a=\\frac{y_1}{y_2}$ :\r\nComparing $ P(y_2,a)$ and $ P(y_2,1)$, we get $ f(a)=1$, so $ a=1$ (point 3 above) and $ y_1=y_2$ and $ f(x)$ is injective.\r\nQ.E.D.\r\n\r\n5) $ f(x)=x$ $ \\forall x$\r\n=======================\r\n\r\nLet $ A=f(\\mathbb R)$\r\n\r\n$ P(1,x)$ $ \\implies$ $ f(f(x)+1)=f(x)+1$ and so $ x\\in A$ $ \\implies$ $ x+1\\in A$\r\n$ P(-1,x)$ $ \\implies$ $ f(f(-x)-1)=-1-f(x)$ and so $ x\\in A$ $ \\implies$ $ -1-x\\in A$ and so, using line above, $ -x\\in A$\r\nSo, since $ x\\in A$ implies $ x+1\\in A$ and $ -x\\in A$ we get $ -x+1\\in A$ and $ x-1\\in A$\r\n\r\nThen for a given $ x$, and since $ f(x)-1\\in A$, let $ z$ such that $ f(z)=f(x)-1$ :\r\n$ P(1,z)$ $ \\implies$ $ f(f(z)+1)=f(z)+1$ and so $ f(f(x))=f(x)$ $ \\forall x$ and, since $ f(x)$ is injective, $ f(x)=x$ $ \\forall \nx$\r\n\r\nAnd it's immediate to check back that this indeed is a solution.\r\n\r\n6) synthesis of solutions\r\n==========================\r\n\r\nHence the two solutions of this equation :\r\n$ f(x)=0$ $ \\forall x$\r\n$ f(x)=x$ $ \\forall x$", "Solution_2": "congrats you. i think your solution is very good. but i prove it by finding f(1) and f(-1). i have no time to write solution. i will send my solution in a week when school holiday will begin. again thank you for your solution. :)", "Solution_3": "f(x+f(xy))= f(x) + xf(y) \r\n\r\n1) let find f(0)=0\r\nlet f(0) gets other value than 0. x=y=o ff(0)=f(0)\r\nx=f(0),y=1 then x=f(0),y=0. then we get f(0)=f(1)\r\nx=y=1 then x=1, y= 1 + f(0) then x= 1 + f(1) , y =0 we will get f(0)f(1)=0 . which means f(0)=0\r\n\r\n2)let find f(1)=1, \r\nif f(1) + 1 = A = 0 then x = y = 1 then we get f(1) = 0 condraction A is not zero\r\nx= 1 => f( 1 + f ( y )) = f(1) + f ( y ) is B\r\nat B y= 1 + f( y ) => f( A + f(y) ) = f( A ) + f (y)\r\nx=A y= y/A f ( A + f(y) ) = f( A ) + A f ( y/A )\r\nthen we get f(y) = A f ( y / A )\r\nat B y=A then we get f(1) ( f(1) - 1 ) = 0 \r\nif f(1) = 0 x= 1/m y= m then we get f(m) = 0 and f(0) = 0 which means f is obviously zero\r\nthen f(1) = 1 \r\n\r\n3)let prove that if n is N then f(yn)=nf(y)\r\nit is easy to prove by helping of B by induction if n is NATURAL then C=> f( n + f(y) ) = n + f(y) \r\nat C y = 1 then xis natural f( x ) = x\r\nat C y = yn f( n + f( yn ) ) = n + f( yn ) \r\nx= n, n is N, f ( n + f( yn ) ) = f (n) + nf(y) = n + f (yn) we get if n is N then f(yn)=nf(y)\r\nn=2 y= -1/2 then f( -1/2 ) = - 1/2\r\n\r\n4)let find f( -1 ) = -1\r\nif f(k) = 0 x=k , y= 1 then we get k = 0\r\nx= -1 y= 1 then f( -1 ) = 1 + f ( -1 + f( -1 )) then ff( -1 ) = f(1 + f ( -1 + f( -1 ))) = 1 + f( -1 + f( -1 )) = 1 + f ( -1) -1 we get ff(-1) = f ( -1)\r\nx=1 y= -1 then f(-1) = -1 + f ( 1 + f ( -1 )) then ff( -1 ) = f ( -1 + f ( 1 + f ( -1 ))) = f(-1) + f(-1 - f( -1 )) then f(-1 - f( -1 ))= 0 => f(-1) = -1\r\n\r\n5) let finish problem\r\nat B y = -1 + f( -y) f( 1 + f( -1 + f(-y)))= 1 + f ( -1 + f(-y) ) = -f(y) then we get D=>f( - f(y)) = -f(y)\r\nat D let give y = -f(y) we get ff(y) = f(y)\r\ny= -1 f(x - f(-x)) = f(x) - x then f( f(x - f(-x)))= f(f(x) - x) = f ( x + f( -x)) = x + f( -x) = f(x) - x then we get 2x + f(-x) = f(x)\r\nthen we get 2x + f(-x) = f(x) , then f(2x + f(-x)) = ff(x) = f(x) = f(2x) + 2xf( -1/2 ) = 2f(x) - x which means f(x) = x\r\n\r\n6) last part is checking\r\n............................................ is obviously true :D \r\n\r\n\r\n=========================================================================================================================\r\nthe land of fire AZERBAIJAN", "Solution_4": "[quote=\"kenan aze\"]f(x+f(xy))= f(x) + xf(y) \n\n1) let find f(0)=0\nlet f(0) gets other value than 0. x=y=o ff(0)=f(0)\nx=f(0),y=1 then x=f(0),y=0. then we gets f(0)=f(1)\nx=y=1 then x=1, y= 1 + f(0) then x= 1 + f(1) , y =0 we will get f(0)f(1)=0 . which means f(0)=0\n[/quote]\r\n\r\n$ x \\equal{} y \\equal{} 1$ $ \\implies$ $ f(1 \\plus{} f(1)) \\equal{} f(1) \\plus{} f(1)$\r\n$ x \\equal{} 1, y \\equal{} 1 \\plus{} f(0)$ $ \\implies$ $ f(1 \\plus{} f(1 \\plus{} f(0))) \\equal{} f(1) \\plus{} f(1 \\plus{} f(0))$\r\n\r\nAnd I dont understand how you conclude from these two lines that $ f(0)f(1) \\equal{} 0$\r\n :blush:", "Solution_5": "f(x+f(xy))= f(x) + xf(y) \r\n\r\n1) let find f(0)=0 \r\nlet f(0) gets other value than 0. x=y=o ff(0)=f(0) \r\nx=f(0),y=1 then x=f(0),y=0. then we gets f(0)=f(1) \r\nx=y=1 f(1+f(1))=2f(1)\r\nx=1, y= 1 + f(0) f(1 + f(1+f(0)))=f(1) + f( 1 + f(0))= f(1) + f(1 + f(1)) = f(1) + 2f(1)= 3 f(1)\r\nx= 1 + f(1) , y =0 f(1 + f(1) + f(0))= f( 1 + f(1)) + (1 + f(1))f(0) = 2f(1) + f(0) + f(1)f(0) = 3f(1) + f(1)f(0)\r\nfrom last two equations we get that f(0)f(1)=0 . which means f(0)=0\r\n\r\ni think there is no dificculty to calculate some simple operations.... :D \r\n\r\n====================================================================================================================================\r\nLAND OF FIRE AZERBAIJAN", "Solution_6": "[quote=\"kenan aze\"] x= 1 + f(1) , y =0 f(1 + f(1) + f(0))= f( 1 + f(1)) + (1 + f(1))f(0) = 2f(1) + f(0) + f(1)f(0) = 3f(1) + f(1)f(0)\n\ni think there is no dificculty to calculate some simple operations.... :D \n[/quote]\r\n\r\nSure, but you did not give the last step and since you gave the two first, I thought there was no missing step.\r\n\r\nAnd I dont worry finding $ f(0)$. I gave an earlier proof of this problem and was just reading yours in order to help you to improve.\r\n\r\nI'm sorry for your reaction and will no longer read your proofs.\r\n\r\nEnjoy mathlinks and maths.", "Solution_7": "i am sorry if you feel worse yourself for my reply. i use it randomly. big percent you are elder than me. i dont want to make you unhappy. maybe i make somethink wrongly", "Solution_8": "If we just assume that $f(x)$ is continuous and differentiable then we can proceed as follows. \n\nIf, $r$ be a root of $f(x)$, then we take $(x,y)=(x,\\frac{r}{x})$. The given equation then implies, $xf(\\frac{r}{x})=0 \\rightarrow f(t)=0 \\forall t \\in \\mathbb R$. So,a non trivial solution would imply that there are no non-zero root....$(1)$.\n\nNow, let $f(0)=a > 0$ without loss of generality. So, $f(x+a)=f(x)+ax$, which contains a special periodicity, namely $f'(x+a)=f'(x)+a$. Now, for $4x \\in [0,a)$, let $\\text{min}\\{f(x)\\}=-g; \\text{max}\\{f(x)\\}=G$ ($g>0$ as $f(0)=f(a)$). Hence, for all $x>\\lceil \\frac{g}{a} \\rceil ,f'(x)>0$ and for all $x<\\lfloor \\frac{G}{a} \\rfloor, f'(x)<0$. Hence, $\\exists x_0 <0$ such that $f(x_0)=0$. But that is not possible as $(1)$ implies. Hence, $f(0)=0$.\n\nNow, let $y=k \\to 0, f(xk)=h \\to 0$, the given equation gives $\\lim \\limits_{k \\to 0} \\frac{f(x+h)-f(x)}{h}=\\frac{xf(k)}{f(xk)}=f'(x)$. Using L'Hospital rule in the middle expression of $\\frac{0}{0}$ for we get $f'(x)=\\lim \\limits_{k \\to 0} \\frac{f'(k)}{f'(xk)}$. this implies that $f'(0) \\neq 0$. Because using L,Hospital rule repeatedly if there exists some $k >1$ s,t $f^{(k)}(0) \\neq 0$, then as $x \\to 0$, the middle equation blows up even though $f'(x)=0$ (what is just assumed). So, either $f^{(k)}(0)=0$ $\\forall k \\in \\mathbb N \\rightarrow f(x)=0 \\forall x \\in \\mathbb R$ or $f'(0) \\neq 0 \\rightarrow f'(x)=1 \\rightarrow f(x)=x$" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "[b]Thank you reply soon![/b]\r\n\r\nLet circle (O) with diameter AB. Let line $ \\Delta$ is perpendicular with AB at I. Let a point M change on circle (O). MA, MB meet $ \\Delta$ order at P and Q. QA meet (O) at M. Prove that MN pass a fixed point.\r\n\r\n--------------------------------------------\r\nI love you. Because you a good friend.", "Solution_1": "[quote=\"thanhnam2902\"][b]Thank you reply soon![/b]\r\n\r\nLet circle (O) with diameter AB. Let line $ \\Delta$ is perpendicular with AB at I. Let a point M change on circle (O). MA, MB meet $ \\Delta$ order at P and Q. QA meet (O) at M. Prove that MN pass a fixed point.\r\n\r\n--------------------------------------------[quote]\r\nI thinh QA meet (O) at N\r\n$ MN\\cup (AB) \\equal{} K$\r\nThen $ (KIAB) \\equal{}\\minus{}1$ so K is the fix point." } { "Tag": [ "search", "algorithm", "advanced fields", "advanced fields unsolved" ], "Problem": "Simple question. My mind isn't working right now. Consider that there is a graph G = (V,E) which is an undirected, connected graph, and let sEV. The breadth-first search obviously constructs a breadth-first tree rooted at s that, for each vEV, represents a shortest path from s to v in G. Of course, there may be many shortest paths from s to v. All such shortest paths are represented in the breadth-first directed acyclic graph (BFDAG) G^D = (V,D), where\r\n\r\nD = {(u,v) | d(s,v) = d(s,u) + 1} ----> d represents a lower-case delta symbol.\r\n\r\nBasically, I need to modify the BFS algorithm to also compute D. My problem is that I can't seem to break down the definition of D into english terms. If someone could help me to sort out the meaning of the above equation for D then I can easily finish this problem. Like I said, my mind just isn't working right now.\r\n\r\nThanks.", "Solution_1": "I'm not sure, but I hope $\\delta (s,u)$ represents the distance from $s$ to $u$ in the tree built by the DFS algorithm. Let $T$ be this tree.\r\n\r\nThis $\\delta$ allows us to see $T$ as structured in levels.\r\n$s$ is the only vertex on the $0$ level.\r\nOn the first level lie all the vertices that are connected to $s$ by an edge.\r\nOn the $i+1^{th}$ level lie all the vertices that are connected to a vertex on the $i^{th}$ level and than don't appear in one of the previous levels. This gives the DFS algorithm.\r\n\r\nNow that you have this tree, link all the vertices on the level $i$ with all the vertices on the level $i+1$, for all $i$. Like in a complete bipartite graph. This gives the set of edges $D$. You say this is a directed set of edges, but I don't see the reason why we can't see it as undirected. \r\n$D$ is a set of edges that doesn't have much to do with $E$, the edges of $G$.\r\nThe idea is that if we have an edge in $(E\\cap D)\\setminus T$ then we have a new shortest path from $s$ to the highest leveled vertex of this edge. This is easily seen if you draw the picture, but if you need help I can solve it.\r\n\r\nIf you make the definition of $D$ like this:\r\n$D=\\{(u,v)|\\delta(s,v)=\\delta(s,u)+1\\mbox{ and } \\{u,v\\}\\in E\\}$, then the set of directed paths from $s$ to a vertex $v$ is precisely the set of shortest paths from $s$ to $v$." } { "Tag": [], "Problem": "Evaluate the following sum\r\n 1/2^2 + 1/2^3 + 1/2^4 + ...\r\n+ 1/3^2 + 1/3^3 + 1/3^4 +...\r\n+1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 + ...\r\n+ 1/5^2 + 1/5^3 + ...\r\n+ ...", "Solution_1": "The n'th row, starting at $ n\\equal{}2$, has value $ \\dfrac{\\frac{1}{n^2}}{1\\minus{}\\frac{1}{n}}\\equal{}\\frac{1}{n^2\\minus{}n}\\equal{}\\frac{1}{n\\minus{}1}\\minus{}\\frac{1}{n}$. This clearly telescopes to 1." } { "Tag": [ "inequalities", "algebra", "polynomial", "inequalities solved", "4-variable inequality", "Symmetric inequality", "AM-GM" ], "Problem": "Let $a,b,c,d$ be positve numbers. Prove that\r\n\\[\\sum_{\\rm cyclic}\\frac{a^{3}}{(a+b)(a+c)(a+d)}\\geq\\frac12.\\]", "Solution_1": "Today, I not delete any post :D . Please post at least one solution, That link is from diendantoanhoc.net and NOT English. :maybe:\r\n@pvthuan: You are wrong when post this problem in Inequalities Solved Problems.", "Solution_2": "It is a solved problem I received from one of my friends and I put it here in hope of a new solution. Of course the author sent me the solution in diendantoanhoc.", "Solution_3": "See here ''The problems that were discussed and solved in the topics above are moved here.'' You are wrong, please don't delete my post and someone's in this topic :P", "Solution_4": "By AM-GM,\r\n\\[(a+b)(a+c)(a+d) \\le \\left( \\frac{a+b+a+c+a+d}{3}\\right)^{3}= \\left( a+\\frac{b+c+d}{3}\\right)^{3},\\]\r\nso\r\n\\[\\frac{a^{3}}{(a+b)(a+c)(a+d)}\\ge \\frac{a^{3}}{\\left( a+\\frac{b+c+d}{3}\\right)^{3}}.\\]\r\n\r\nWOLOG, assume that $a+b+c+d = 4$. Then\r\n\\[\\frac{a^{3}}{\\left( a+\\frac{b+c+d}{3}\\right)^{3}}= \\left( \\frac{a}{a+\\frac{4-a}{3}}\\right)^{3}= \\left( \\frac{3a}{2a+4}\\right)^{3}= \\frac{27}{8}\\left( \\frac{a}{a+2}\\right)^{3},\\]\r\nso the inequality becomes\r\n\\[\\sum_{\\text{cyc}}\\left( \\frac{a}{a+2}\\right)^{3}\\ge \\frac{4}{27}.\\]\r\n\r\nWe claim that\r\n\\[\\left( \\frac{x}{x+2}\\right)^{3}\\ge \\frac{2x-1}{27}\\]\r\nfor all $x \\in [0,4]$.\r\n\r\nThis is because\r\n\\[\\left( \\frac{x}{x+2}\\right)^{3}-\\frac{2x-1}{27}= \\frac{2(x-1)^{2}(-x^{2}+6x+4)}{27(x+2)^{3}},\\]\r\nand $-x^{2}+6x+4 \\ge 0$ for $3-\\sqrt{13}\\le x \\le 3+\\sqrt{13}$, which includes the interval $0 \\le x \\le 4$.\r\n\r\nTherefore,\r\n\\[\\sum_{\\text{cyc}}\\left( \\frac{a}{a+2}\\right)^{3}\\ge \\frac{2(a+b+c+d)-4}{27}= \\frac{4}{27}.\\]", "Solution_5": "Wonderful! That is what I am looking for. You will be credited for this solution in my book, Naoki Sato (on Crux editorial).\r\n\r\nAs for the claim, we may write\r\n\\[\\left(\\frac x{x+2}\\right)^{3}\\geq \\frac1{27}+k(x-1). \\]\r\nWe come up with this idea by taking into account the fact of equality case. We need to find $k$ such that the claim is valid. Write it in the form\r\n\\[(x-1)\\left[\\frac{2x^{2}}{(x+2)^{3}}-k\\right]\\geq0. \\]\r\nWe wish to find $k$ such that the polynomial in the square bracket has $x-1$ as an factor. Thus, we can plug $x=1$ into that bracket to get $k=\\tfrac2{27}$. Alternatively, we can check the $g'(1)=0$, $g(x)$ be the polynomial in the bracket. Then we can verify the claim as what nsato did.", "Solution_6": "[quote=\"pvthuan\"]Wonderful! That is what I am looking for. You will be credited for this solution in my book, Naoki Sato (on Crux editorial).\n[/quote]\r\n\r\nThank you! These days, I need all the credit I can get, although I don't work for Crux anymore.\r\n\r\nIncidentally, the corresponding inequality for five variables is false.", "Solution_7": "The following is easy \r\n\\[\\sum_{\\rm cyclic}\\frac{a^{4}}{(a+b)(a+c)(a+d)}\\geq\\frac{a+b+c+d}8.\\]\r\nI guess that\r\n\\[\\sum_{\\rm cyclic}\\frac{a^{4}}{(a+b)(a+c)(a+d)}\\geq\\frac{\\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}}4.\\]" } { "Tag": [ "AMC 10", "AMC" ], "Problem": "AoPS will host another AMC contest [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Math Jam[/url] at 7 PM ET (4 PM PT) on Friday, February 17. We will work through solutions for each of the last 5 problems on each the AMC 10 and AMC 12 and then take requests for discussing solutions to other problems on the exams.", "Solution_1": "sorry to revive an old post, but is there an AMC 10A or 10B math jam this year?", "Solution_2": "You can find these on the math jam schedule:\r\n[url]http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php[/url]" } { "Tag": [ "inequalities", "Gauss", "quadratics" ], "Problem": "If two real numbers $ x$ and $ y$ satisfy the equation $ \\frac{x}{y} \\equal{} x \\minus{} y$, then:\r\n$ \\textbf{(A)}\\ {x \\ge 4}\\text{ and }{x \\le 0}\\qquad \\\\\r\n\\textbf{(B)}\\ {y}\\text{ can equal }{1}\\qquad \\\\\r\n\\textbf{(C)}\\ \\text{both }{x}\\text{ and }{y}\\text{ must be irrational}\\qquad \\\\\r\n\\textbf{(D)}\\ {x}\\text{ and }{y}\\text{ cannot both be integers}\\qquad \\\\\r\n\\textbf{(E)}\\ \\text{both }{x}\\text{ and }{y}\\text{ must be rational}$", "Solution_1": "$ x=xy-y^2$\r\n$ x-xy=-y^2$\r\n$ x(1-y)=-y^2$\r\n$ x=\\frac{y^2}{y-1}$\r\n\r\nNote that $ y^2$ is always nonnegative. $ x$ is always negative if $ y<1$. If we want to find a lower bound for $ x$, set $ y^2%Error. \"geky\" is a bad command.\n-k$, where $ k$ is a real constant. $ y^2-ky+k\\ge0$, which means the LHS of that inequality must be a perfect square trinomial. So $ k=4$, which means $ x\\ge4$ as well.\r\n\r\nAnswer: A\r\n\r\nI'm not sure if I'm right or if my solution's legit though... :maybe:", "Solution_2": "Your solution would be what one should do at the actual contest, but the complete solution to the problem is provided as follows:\r\n\r\n[hide=\"My Solution\"]\nContinuing from where gauss left off and solving for $ y$, we obtain\n\n$ x \\equal{} \\frac{y^2}{y \\minus{} 1}\\implies xy \\minus{} x \\equal{} y^2\\implies y^2 \\minus{} xy \\plus{} x \\equal{} 0.$\n\nThe quadratic in $ y$ yields roots of $ \\frac{x \\pm \\sqrt{x^2 \\minus{} 4x}}{2}.$ Clearly, the quantity under the square root must be nonnegative. Therefore, $ x^2 \\minus{} 4x \\ge 0\\implies x(x \\minus{} 4)\\ge 0\\implies x \\in (\\minus{}\\infty, 0] \\cup [4, \\infty) \\implies \\fbox{(A)}.$\n[/hide]", "Solution_3": "[quote=\"sunehra\"]Your solution would be what one should do at the actual contest[/quote]\r\n\r\nBy that, do you mean in terms of the general gist of the idea one should be thinking, instead of writing/noting the whole solution?", "Solution_4": "Yes, because a contest is about speed. Especially in a large contest like this one. I'm not really sure how much time they had when taking the AHSME back then." } { "Tag": [ "LaTeX", "limit", "calculus", "calculus computations" ], "Problem": "Sorry for my bad latex but.. :)\r\n\r\nHere is the problem\r\nFind lim(x->0)[1/ln(x+sqrt(1+x^2)) - 1/ln(1+x)]\r\n\r\nI think its a L'hopital problem but not sure :)", "Solution_1": "c'mon guys.. my exam is in 3 hours.. i need this solution", "Solution_2": "$= \\lim_{x \\rightarrow 0}\\frac{\\ln(1+x)-\\ln(x+\\sqrt{1+x^{2}})}{\\ln(x+\\sqrt{1+x^{2}})\\ \\ln(1+x)}$\r\nThen 2 times l'H\u00f4pital.", "Solution_3": "Let's not use L'H\u00f4pital's rule. Starting from the simplification made by Petruspan we have:\r\n\r\n$ln(1+x) = x-\\frac{x^{2}}{2}+\\frac{x^{3}}{3}$\r\n$ln(x+\\sqrt{1+x^{2}}) = argsinhx = x-\\frac{x^{3}}{6}$\r\n\r\nNow substitute these expansions in your limit (which I will designate by L) and we get:\r\n\r\n$L = \\frac{\\frac{x^{3}}{2}-\\frac{x^{2}}{2}}{x^{2}-\\frac{x^{3}}{2}}= \\frac{x-1}{2-x}=-\\frac{1}{2}$" } { "Tag": [ "geometry", "geometry solved" ], "Problem": "This problem was communicated to me and I really love it.\r\n\r\nGiven a circle X and three points A, B, C inside it, let us draw three circles :\r\n\r\n1) a circle $c_1$ tangent to the circle X, to the line AB, and to the line BC.\r\n2) a circle $c_2$ tangent to the circle X, to the line BC, and to the line CA.\r\n3) a circle $c_3$ tangent to the circle X, to the line CA, and to the line AB.\r\n\r\nLet us denote $C_1$, $C_2$, $C_3$ the points of tangency of the circles $c_1$, $c_2$, $c_3$ with X.\r\n\r\nProve that the lines $BC_1$, $CC_2$, $AC_3$ are concurrent.", "Solution_1": "Let N be the external center of similitude of the circle X and the incircle I of triangle ABC. We are going to prove that the lines $BC_1$, $CC_2$ and $AC_3$ pass through N. (This will, of course, establish the concurrency of these lines.)\r\n\r\nThe Monge theorem states that the pairwise external centers of similitude of three circles lie on one line. Applying this to the circles X, I and $c_1$, we obtain that the external center of similitude of the circles X and I, the external center of similitude of the circles I and $c_1$, and the external center of similitude of the circles $c_1$ and X lie on one line. In other words, the points N, B and $C_1$ lie one one line (in fact, the point B is the external center of similitude of the circles $C_1$ and I because B lies on the two external common tangents AB and BC of these circles, and the point $C_1$ is the external center of similitude of the circles $c_1$ and X because it is the point of tangency of these two circles). This means that the line $BC_1$ passes through N. Similarly we can prove the same for $CC_2$ and $AC_3$, and we are done.\r\n\r\n Darij", "Solution_2": "Let K be that intersection. You can also prove I , O, K are collinear. :cool: \r\n The solution uses ditaliation. I 'll give you later.", "Solution_3": "I intend to give you my solution next time, but I have time now.\r\n You can easily prove that the detalitation centers of (c1) and (c2), (c2) and (c3), (c3) and (c1) such that the product of the three propotions is positive are collinear (remember there are two detalitation centers of (c1) and (c2)).\r\n Now, what you have to do now is to apply it. ( Sorry, I 'm not very good at English geometry languange :( )." } { "Tag": [ "Stanford", "college" ], "Problem": "I took acting fundamentals as a freshmen. I need a total of 2 credits in art to graduate. (so far I have 1/2 credits) acting was alright, i'm deciding weather i should take actingII as a sophmore or should I take Introduction to pyschology(after taking the class you get get to be a leader/mentor in camps my school organizes, n i've heard it's a fun class) instead. How much do art(meaning acting,band, ect) classes count for colleges? I could fullfill my art requirement in my junior n senior year, but I don't want to have a whole bunch of classes on top of college apps. advice anyone?\r\n\r\nThanks :)", "Solution_1": "It depends what colleges you are applying to, and what you like to do. If you are applying to University of California campuses, it's a good idea to meet the \"a-g\" requirements for admission in California \r\n\r\nhttp://www.ucop.edu/a-gGuide/ag/welcome.html \r\n\r\nand those include art classes of specified kinds. But study what you like for your electives. PLEASE don't try to microanalyze what a college wants you to do--believe me, within broad limits college admission officers want you to do what YOU want to do. Do something challenging and interesting, and it will all work out.", "Solution_2": "Do you like it? If you like it, do it. If you love it, do it more.", "Solution_3": "I heard you need a full year of art to get into some colleges in the west coast. Is this only for university of California or is it for all colleges on the west coast? How about stanford? Acting was alright, it wasn't great but it wasn't horrible either. I enjoy participating in the Speech and Debate club more than taking the class.", "Solution_4": "[quote]I heard you need a full year of art to get into some colleges in the west coast. Is this only for university of California or is it for all colleges on the west coast?[/quote]\r\nWe in the CSU system had that as a requirement before the UC did. But as far as I know, it's just a UC-CSU requirement, that is, public universities in California. I can't speak for any private universities or for other states. And it's \"visual or performing arts,\" so yes, for all you violin players out there, your orchestra class does count." } { "Tag": [ "\\/closed" ], "Problem": "I can't figure out how to order the Aops books online. Is ordering something online the same as ordering by fax?", "Solution_1": "Go to \"bookstore\"\r\nAdd a book to your cart\r\nClick \"view cart\"\r\nThen click \"complete order\"\r\n\r\nI'm not exactly sure about these steps...but I think they're correct.." } { "Tag": [], "Problem": "Point $B$ is in the exterior of the regular $n$-sided polygon $A_1A_2\\cdots A_n,$ and $A_1A_2B$ is an equilateral triangle. What is the largest value of $n$ for which $A_n, A_1,$ and $B$ are consecutive vertices of a regular polygon?", "Solution_1": "[hide=\"Simon-Favorite Factoring Trick\"] We have $m\\angle A_nA_1A_2=\\frac{(180^\\circ)(n-2)}{n}$ and $m\\angle A_1A_2B=60^\\circ$. Hence, $m\\angle A_nA_1B=360^\\circ-\\frac{(180^\\circ)(n-2)}{n}-60^\\circ$. For this angle to be one of a regular polygon, it must be equivalent to $\\frac{180^\\circ(m-2)}{m}$ for some positive integer $m$. \\[ 360^\\circ-\\frac{(180^\\circ)(n-2)}{n}-60^\\circ=\\frac{180^\\circ(m-2)}{m}, \\] \\[ 2-\\frac{n-2}{n}-\\frac{1}{3}=\\frac{m-2}{m}, \\] \\[ \\frac{6n-3(n-2)-n}{3n}=\\frac{m-2}{m}, \\] \\[ \\frac{2n+6}{3n}=\\frac{m-2}{m}, \\] \\[ (2n+6)(m)=(3n)(m-2), \\] \\[ 0=nm-6m-6n, \\] \\[ 36=nm-6m-6n+36, \\] \\[ 36=(m-6)(n-6). \\] $n$ must be an integer, so it is clearly maximized when $n-6=36$ and $m-6=1$. Therefore the desired answer is $\\boxed{042}$. Go figure. :roll: [/hide]", "Solution_2": "Shouldn't the order in the problem be $ A_n$, $ A_1$, and then $ B$? The current order doesn't seem to make sense...", "Solution_3": "You're right. \r\n\r\n(well, except for that #1 in your sig .. ) :P" } { "Tag": [ "function", "quadratics", "real analysis", "real analysis unsolved" ], "Problem": "Study if there is a derivable function $ f: R->R$ with:\r\n\r\n$ f'(x+f(y))=y+f'(x) \\forall x,y \\in R$", "Solution_1": "Integrate to $ x$:\r\n$ f(x \\plus{} f(y)) \\equal{} yx \\plus{} C \\plus{} f(x)$\r\nDifferentiate to $ y$:\r\n$ f'(x \\plus{} f(y))\\cdot f'(y) \\equal{} x$\r\nIntegrate with respect to $ x$:\r\n$ (f(x \\plus{} f(y)) \\minus{} f(f(y)))\\cdot f'(y) \\equal{} \\frac {x^2}{2}$\r\nSo if $ f$ exists, then it is a quadratic function.\r\nSubstitution of a quadratic function shows that there is no solution.", "Solution_2": "Yes\r\nAnother proof:if we take x=0 it results $ f'(f(y))\\equal{}y\\plus{}a$ (1)\r\nwhere we put $ a\\equal{}f'(0)$\r\nFrom this it is clear that $ f$ is an injection,and because it is also continuous(it is differentiable) it results that $ f$ is strictly monotonous.\r\nThis is obvious a contadiction,because $ f'$ must have the same sign on $ R$ but we can see from (1) $ f'(f(y\\minus{}a))\\equal{}y$,so $ f'(R)\\equal{}R$.", "Solution_3": "$ x \\mapsto f(y)f'(x) \\minus{} yx$ is periodic with period $ f(y)$.\r\nThen (differentiate) $ x \\mapsto f'(y)f'(x) \\minus{} x$ is periodic with period $ f(y)$,\r\nwhich means $ yf'(y) \\equal{} f(y)$, or $ f$ is linear, which is not possible." } { "Tag": [ "inequalities", "calculus", "puzzles" ], "Problem": "15^(1/2) + 15^(1/3) + 15^(1/4) + 15^(1/5) + 15^(1/6) + 15^(1/7) + \r\n15^(1/8) versus 15", "Solution_1": "Unlike [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169876]previous[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164380]examples[/url], this one doesn't require extremely high precision, and is accessible using relatively simple inequalities that can be done by hand.\r\nI still don't see the appeal, but some might like this.\r\n\r\n[I used a spreadsheet to test my suspicions, and won't say any more about the answer]", "Solution_2": "[hide]\n15 is close to 16 (actually, 16-1).\n16^1/2=4, 16^1/4=2, 2^1/3=16^1/12=1.26.\n16^1/3=2^4/3=2*1.26=2.52\n16^1/6=2^2/3=1.26*1.26=1.59\n16^1/8=2^1/2=1.414\nWe use elementary calculus:\n15^(1/2) is about $ 4 \\minus{} \\dfrac{1}{2*4} \\equal{} 4 \\minus{} 1/8 \\equal{} 3.875$.\n15^(1/3) is about $ 2.52 \\minus{} \\dfrac{2.52}{3*16} \\equal{} 2.52 \\minus{} 0.84/16 \\equal{} 2.52 \\minus{} 0.21/4 \\equal{} 2.47$\n15^(1/4) is about $ 2 \\minus{} \\dfrac{2}{4*16} \\equal{} 2 \\minus{} 1/32 \\equal{} 1.97$\n15^(1/6) is about $ 1.59 \\minus{} \\dfrac{1.59}{6*16} \\equal{} 1.59 \\minus{} .159 \\equal{} 1.57$\n15^(1/8) is about $ 1.41 \\minus{} \\dfrac{1.41}{8*16} \\equal{} 1.41 \\minus{} .01 \\equal{} 1.4$\n15^1/5 is between 15^1/4 and 15^1/6 = 1.77\n15^1/7 is between 15^1/6 and 15^1/8 = 1.48\nsumming these, we get 14.54 is less than 15.\n(hopefully the errors aren't larger than .5)\n\n[/hide]\r\n\r\nEDIT: oh wow, the error was like -.07, that's surprisingly good with all the rounding i did", "Solution_3": "If I use 1.4142 for sqrt(2), 1.26 for cbrt(2), then the cbrt(4) is about 1.5874.\r\nWhere did I get these? They're not to have come from a calculator.\r\n\r\n\r\n15^(1/2) < 16^(1/2) = 4\r\n15^(1/3) < 16^(1/3) = (2^4)^(1/3) = 2^(4/3) = 2*2^(1/3) ~ 2.5198\r\n15^(1/4) < 16^(1/4) = 2\r\n15^(1/5) < 16^(1/4) = 2\r\n15^(1/6) < 16^(1/6) = (2^4)^(1/6) = 2^(4/6) = 2^(2/3) ~ 1.5874\r\n15^(1/7) < 16^(1/6) = (2^4)^(1/6) = 2^(4/6) = 2^(2/3) ~ 1.5874\r\n15^(1/8) < 16^(1/8) = (2^4)^(1/8) = 2^(4/8) = 2^(1/2) ~ 1.4142\r\n\r\nThe numbers on the right of the equals sign add up to about 15.1088,\r\nso it means that 15^(1/2) + ... + 15^(1/8) < 15.1088, and my \r\nmethod gives an inconclusive result.", "Solution_4": "I suggest truncating the problem, meaning actually alter the problem to a close one, but slightly shorter. (Get rid of the 15^(1/8) term; it doesn't fit in with inequalities like the ones below, because it crowds the problem.)\r\n\r\nThe following does not use calculus or decimal approximations of roots. It does use inequalities involving roots of perfect powers.\r\n\r\n\r\n15^(1/2) < 16^(1/2)......= 4\r\n15^(1/3) < 27^(1/3)......= 3\r\n15^(1/4) < 16^(1/4)......= 2\r\n15^(1/5) < 32^(1/5)......= 2\r\n15^(1/6) < 64^(1/6)......= 2\r\n15^(1/7) < 128^(1/7)....= 2\r\n------------ . . . . . . . . . . . . -----\r\n15^(1/2) +...+ 15^(1/7) < 15" } { "Tag": [], "Problem": "This is actually an easy problem but I found it hard to show algebrically and I want to know other solutions. Please help. Thanks in advance.\r\n\r\nA, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?", "Solution_1": "[hide=\"hint\"]\nThink mods.\n[hide=\"further hint\"]\n$3x \\equiv 6x \\equiv 18x \\mod{750}$\n[/hide][/hide]", "Solution_2": "[hide]They meet again for the first time only when A finishes one lap. Proof: Clearly, B will have to have lapped A exactly once when B meets A again. B meets A again at the starting point, when B has run 2 laps, and A has run 1 lap. Well, the first time that happens, C is there as well, having run 6 laps...[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "$ x,y$ integer number . $ y^{3}\\equal{}x^{3}\\plus{}8x^{2}\\minus{}6x\\plus{}8$ find $ x,y\\equal{}?$", "Solution_1": "[hide=\"hint\"] we use the well-known idea,we put the expression between two cubes(consecutive one) :wink: [/hide]" } { "Tag": [ "geometry", "perimeter", "rectangle" ], "Problem": "The perimeter of a retangle is 100 and diagonal mesure is x. What's the area of the retangle ?\r\n\r\nA) 625 \u2013 x\u00b2\t\t\t\t\r\nB) 625 \u2013 $ \\frac{x^{2}}{2}$\r\nC) 1250 \u2013 $ \\frac{x^{2}}{2}$\r\nD) 250 - $ \\frac{x^{2}}{2}$\r\nE) 2500 - $ \\frac{x^{2}}{2}$", "Solution_1": "[hide=\"Solution\"]Let the length of the rectangle by $ y$, and the width $ 50\\minus{}y$. $ x^2 \\equal{} y^2 \\plus{} (50\\minus{}y)^2 \\equal{} y^2 \\plus{} 2500 \\minus{} 100y \\plus{} y^2 \\equal{} 2y^2 \\minus{} 100y \\plus{} 2500$. We want to compute $ A \\equal{} y(50\\minus{}y) \\equal{} 50y \\minus{} y^2$. $ x^2 \\equal{} \\minus{}2A \\plus{} 2500$. $ 2A \\equal{} 2500\\minus{}x^2$. $ A \\equal{} 1250 \\minus{} \\displaystyle\\frac{x^2}{2}$. The answer is $ \\boxed{C}$.\n\n[/hide]", "Solution_2": "[quote=\"darkdieuguerre\"][hide=\"Solution\"]Let the length of the rectangle by $ y$, and the width $ 50 \\minus{} y$. $ x^2 \\equal{} y^2 \\plus{} (50 \\minus{} y)^2 \\equal{} y^2 \\plus{} 2500 \\minus{} 100y \\plus{} y^2 \\equal{} 2y^2 \\minus{} 100y \\plus{} 2500$. We want to compute $ A \\equal{} y(50 \\minus{} y) \\equal{} 50y \\minus{} y^2$. $ x^2 \\equal{} \\minus{} 2A \\plus{} 2500$. $ 2A \\equal{} 2500 \\minus{} x^2$. $ A \\equal{} 1250 \\minus{} \\displaystyle\\frac {x^2}{2}$. The answer is $ \\boxed{C}$.\n\n[/hide][/quote]\r\n\r\nThanks, but I don't understand this line: \r\n\r\n$ x^2 \\equal{} \\minus{} 2A \\plus{} 2500$. $ 2A \\equal{} 2500 \\minus{} x^2$. $ A \\equal{} 1250 \\minus{} \\displaystyle\\frac {x^2}{2}$. The answer is $ \\boxed{C}$.\r\n\r\nHow do you find that x\u00b2 = -2A +2500 ?\r\nHow to do this ?", "Solution_3": "\\begin{align*}\r\n\\left\\{\\begin{array}{l}\r\n2(a+b)=100 \\\\\r\na^2+b^2=x^2\r\n\\end{array}\\right.\r\n&\\iff \r\n\\left\\{\\begin{array}{l}\r\na+b=50 \\\\\r\na^2+b^2=x^2\r\n\\end{array}\\right. \\\\\r\n&\\iff \r\n\\left\\{\\begin{array}{l}\r\na+b=50 \\\\\r\nab=\\frac{(a+b)^2-(a^2+b^2)}{2}=\\frac{50^2-x^2}{2}=1250-\\frac{x^2}{2}\r\n\\end{array}\\right. \\\\\r\n\\end{align*}\r\n\r\nThus, $ \\boxed{C}$.", "Solution_4": "Thanks Nakagawa, good method, and now I understand. Thanks ;)" } { "Tag": [ "geometry", "inequalities", "vector" ], "Problem": "What books would anyone recomend for olympiad level geometry? Im so terrible at it, i need it for the IMO...", "Solution_1": "For inequalities I recomend \"Old and new inequalities\" from Gil Publishing House.The authors are T.Andreescu,G.Dospinescu and others.", "Solution_2": "And for geometry??? :lol:", "Solution_3": "I recomend two books:\r\n \r\n 1) Virgil Nicula-\"Geometrie sintetica,vectoriala si analitica\" GIL Publishing House\r\n 2) Marian Dinca & Marcel Chirita-\"Numere complexe in geometrie\"", "Solution_4": "Errr, i only speak english, are those books in english?", "Solution_5": "No they are not. But, I think you understand what I have written. Am I right?", "Solution_6": "Something to do with complex numbers?", "Solution_7": "I'm sorry. I find the above conversation hilarious. :laugh:", "Solution_8": "Vector, Analytical, and Classical??? Geometry, and Geometry with Complex Numbers? LOL :rotfl:", "Solution_9": "Youve never done geometry with complex numbers??? I had more classical geometry in mind...", "Solution_10": "Books: FJM 1,2,3 and Titeica and Lalescu." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Find the maximum and minimum of $ f(a,b,c) \\equal{} \\frac {a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3}{(a^2 \\plus{} b^2 \\plus{} c^2)^3}$ for $ a,b,c\\in\\mathbb{R}$.", "Solution_1": "@bakerbakura: Please check your problems after posting them. $ d$ is missing. :wink:", "Solution_2": "[quote=\"bakerbakura\"]Find the maximum of $ f(a,b,c,d) \\equal{} \\frac {a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3}{(a^2 \\plus{} b^2 \\plus{} c^2)^3}$ for $ a,b,c,d\\in\\mathbb{R}$.[/quote]\r\n$ \\frac{1}{8}.$", "Solution_3": "Sorry about the mistake! Also notice that I have included the minimum as well...", "Solution_4": "[quote=\"bakerbakura\"]Find the maximum and minimum of $ f(a,b,c) = \\frac {a^3b^3 + b^3c^3 + c^3a^3}{(a^2 + b^2 + c^2)^3}$ for $ a,b,c\\in\\mathbb{R}$.[/quote]\r\n\r\nWe have\r\n\\[ f(1, - 1,0) = - \\frac {1}{8},f(1,1,0) = \\frac {1}{8},\\]\r\nand we can prove\r\n\r\n $ f^2(a,b,c)\\le{\\frac {1}{8^2}}$holds $ \\forall(a,b,c)\\in{R^3},$\r\n\r\nso\r\n\\[ minimum = f(1, - 1,0) = - \\frac {1}{8},\nmaxmum = f(1,1,0) = \\frac {1}{8}.\\]", "Solution_5": "[quote=\"bakerbakura\"]Find the maximum and minimum of $ f(a,b,c) \\equal{} \\frac {a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3}{(a^2 \\plus{} b^2 \\plus{} c^2)^3}$ for $ a,b,c\\in\\mathbb{R}$.[/quote]\r\n\r\n$ \\frac18 \\minus{} \\frac {a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3}{(a^2 \\plus{} b^2 \\plus{} c^2)^3}$\r\n\r\n$ \\equal{} \\frac {4\\sum{b^2c^2(b \\minus{} c)^2} \\plus{} \\sum{a^2(a^2 \\minus{} b^2)(a^2 \\minus{} c^2)} \\plus{} 3a^2b^2c^2}{8(a^2 \\plus{} b^2 \\plus{} c^2)^3}$\r\n\r\n$ \\ge 0,$\r\n\r\nwe get $ f(a,b,c) \\le \\frac18$,\r\n\r\nwhen $ (a,b,c) \\equal{} (0,1,1)$ get \"$ \\equal{}$\".\r\n\r\nand\r\n\r\n$ \\frac18 \\plus{} \\frac {a^3b^3 \\plus{} b^3c^3 \\plus{} c^3a^3}{(a^2 \\plus{} b^2 \\plus{} c^2)^3}$\r\n\r\n$ \\equal{} \\frac {4\\sum{b^2c^2(b \\plus{} c)^2} \\plus{} \\sum{a^2(a^2 \\minus{} b^2)(a^2 \\minus{} c^2)} \\plus{} 3a^2b^2c^2}{8(a^2 \\plus{} b^2 \\plus{} c^2)^3}$\r\n\r\n$ \\ge 0,$\r\n\r\nwe get $ f(a,b,c) \\ge \\minus{}\\frac18$,\r\n\r\nwhen $ (a,b,c) \\equal{} (0,\\minus{}1,1)$ get \"$ \\equal{}$\".\r\n\r\nso..." } { "Tag": [ "quadratics", "geometry", "3D geometry", "algebra", "quadratic formula", "number theory solved", "number theory" ], "Problem": "Find all integer solutions of:\r\n\r\nx^2+1=y^3\r\n\r\n\r\nBye", "Solution_1": "if i remember correctly it is from one of mathlinks contest and if my memory doesnt betray it has been solved by Z[i].", "Solution_2": "Ok but do you know the smallest solution?", "Solution_3": "Wouldn't the smallest solution be x=0 and y=1?\r\n\r\nBecuase 0^a and 1^b equal 0 and 1 respectively for all numbers a and b. Since 0 is 1 less than1, and both 2 and 3 are numbers, 0^2+1 = 1^2", "Solution_4": "A try\r\nx^2=(y-1)(y^2+y+1)\r\nSince y^2+y+1 is between y^2 and (y+1)^2, so y^2+y+1 can never be a perfect square, so (y-1) and (y^2+y+1) must have common factors>1.\r\nthen (y-1)(y+2)=y^2+y-2 and y^2+y+1 must have common divisors. The only possible common divisor they have must be 3, thus y=1(mod 3).\r\ny-1=3a^2, y^2+y+1=3b^2. Use quadratic formula $y= \\frac{-1+\\sqrt{3(4b^2-1)}}{2}$, so 4b^2-1=3k^2 where k is integer. (2b)^2-3k^2=1. Apply Pell's equation, solutions for (b, k) are (1, 1), (13, 15), (181, 209)....\r\nBut y=3a^2+1, stuck there....", "Solution_5": "the soln that i mentioned before was from valentin vornicu.and it is realy nice .u can easily conclude that x is even number.and then in Z[i] u can also conclude that x+i and x-i are coprime so they should be product of unit and cube in Z[i] .so we can say x+i=(a+ib)^3 .and the remain is trivial." } { "Tag": [ "geometry", "calculus" ], "Problem": "Can we use \"let alone\" in affirmative sentence?\r\n\r\nThanks in advance", "Solution_1": "Do you mean something like \"It is not true that (weak statement), let alone (stronger statement).\" E.g., \"I can't even picture higher dimensions, let alone do geometry in them.\" It's perfectly good English, but is not terribly common. The only other use I can think of is for \"leave alone\", as in \"let that man alone!\".\r\n\r\nYou can look at the free dictionary's idioms section: [url]http://idioms.thefreedictionary.com/let+alone[/url]", "Solution_2": "Thank you for your reply.\r\n\r\nIs the following sentence correct in grammaticaly?\r\n\r\nShe speaks French let alone English.", "Solution_3": "[quote=\"kunny\"]Thank you for your reply.\n\nIs the following sentence correct in grammaticaly?\n\nShe speaks French let alone English.[/quote]\r\nI'm not an expert on this but this sentence doesn't sound right to me. EDIT: I see what you mean here, I guess you want to use 'let alone' as an inclusion, like 'She speaks French and English', as if that's not enough language! ;)\r\n\r\nThe affirmative usage would be something like '(general statement), let alone (less general statement)'\r\n\r\nFor example: kunny is an avid calculus problem solver, let alone doing 'Today's Calculation of Integral'!!! :)", "Solution_4": "In that sentence construction, \"and\" is definitely the right word to use. If you want the emphasis and connotations of \"let alone\", modify the second part with something like \"as well\".\r\n\r\nFor something that sounds right: \"I can't believe she speaks French, let alone English!\"", "Solution_5": "I am pretty sure the first clause has to be negative to use \"let alone\".", "Solution_6": "[quote=\"kunny\"]Thank you for your reply.\n\nIs the following sentence correct in grammaticaly?\n\nShe speaks French let alone English.[/quote] I am not sure but it looks wrong" } { "Tag": [ "quadratics", "algebra", "quadratic formula", "algebra unsolved" ], "Problem": "1) $p,q \\in Z$ satisfy the equation:\r\n$x^{2}+px+q=0$\r\nDetermine the roots of equation.\r\n2) Let $m,n \\in N$ satisfy the equation\r\n$x^{2}-m(n+1).x+m+n+1=0$ which has all roots are positive integers. Prove that $mn \\leq 4$", "Solution_1": "Oh, are they very difficult?\r\nWho can solve them? :D", "Solution_2": "I think people just aren't interested.\r\n[hide=\"1)\"]\n$x^{2}+px+q=0=\\left(x+\\frac{p}2\\right)^{2}+q-\\frac{p}4$, so we can easily solve for the two roots which are the positive or negative square root (yes that was a derivation of the quadratic formula)\n[/hide]\r\nSame Idea for other one" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find a $ n$ such that the number $ n^2 \\minus{} 1$ have $ 10$ positive divisor.\r\nFind a $ n$ such that the number $ n^2 \\minus{} 2$ have $ 10$ positive divisor.\r\nFind a $ n$ such that the number $ n^2 \\minus{} 3$ have $ 10$ positive divisor.\r\n\r\n\r\n\r\n\r\n\r\n> I edited <", "Solution_1": "[quote=\"mayhem\"]Find a $ n$ such that the number $ n^2 \\minus{} 1$ have $ 10$ positive divisor.\nFind a $ n$ such that the number $ n^2 \\minus{} 2$ have $ 10$ positive divisor.\nFind a $ n$ such that the number $ n^2 \\minus{} 3$ have $ 10$ positive divisor.\n\n\n\n\n\n> I edited <[/quote]\r\nLet $ d(n)$ be the number of divisors of $ n$.\r\n\r\n$ n^2 \\minus{} 1 \\equal{} (n \\minus{} 1)(n \\plus{} 1)$.\r\nCase $ n$ is even:\r\nThen $ d(n^2 \\minus{} 1) \\equal{} d(n \\minus{} 1) \\cdot d(n \\plus{} 1)$. $ n \\equal{} 2$ gives $ n^2 \\minus{} 1 \\equal{} 3$ which doesn't have $ 10$ divisors. So assume $ n > 2$. Obviously $ d(n) > 1 \\forall n > 1$ so either $ d(n \\minus{} 1) \\equal{} 2$ and $ d(n \\plus{} 1) \\equal{} 5$ or $ d(n \\minus{} 1) \\equal{} 5$ and $ d(n \\plus{} 1) \\equal{} 2$.\r\nCase $ d(n \\minus{} 1) \\equal{} 2$ and $ d(n \\plus{} 1) \\equal{} 5$:\r\nThen $ n \\minus{} 1 \\equal{} p$ and $ n \\plus{} 1 \\equal{} q^4$ where $ p$ and $ q$ are primes. So $ q^4 \\minus{} 2 \\equal{} p$. $ q \\equal{} 3$ gives $ p \\equal{} 79$. $ q \\equal{} 7$ gives $ p \\equal{} 2399$. $ q \\equal{} 11$ gives $ p \\equal{} 14639$. $ q \\equal{} 13$ gives $ p \\equal{} 28559$. $ q \\equal{} 29$ gives $ p \\equal{} 707279$. $ q \\equal{} 41$ gives $ p \\equal{} 2825759$. usw...\r\nCase $ d(n \\minus{} 1) \\equal{} 5$ and $ d(n \\plus{} 1) \\equal{} 2$:\r\nThen $ n \\minus{} 1 \\equal{} q^4$ and $ n \\plus{} 1 \\equal{} p$ where $ p$ and $ q$ are primes. So $ q^4 \\plus{} 2 \\equal{} p$. If $ q \\neq 3$ then $ 3 \\mid q^4 \\plus{} 2$ so $ q \\equal{} 3$. Giving $ p \\equal{} 83$. So $ n \\equal{} 82$ is a solution. $ 82^2 \\minus{} 1 \\equal{} 6723 \\equal{} 3^4 \\cdot 83$.\r\nCase $ n$ is odd:\r\nThen $ 8 \\mid n^2 \\minus{} 1$. So since $ n^2 \\minus{} 1 \\equal{} pq^4$ for primes $ p,q$. we get $ q \\equal{} 2$. So $ n^2 \\minus{} 1 \\equal{} 16p$. So $ (n \\minus{} 1)(n \\plus{} 1) \\equal{} 16p$.\r\nCase $ n \\minus{} 1 \\equal{} 8$, $ n \\plus{} 1 \\equal{} 2p$:\r\nThen $ 2p \\equal{} 10$ or $ p \\equal{} 5$. So $ n \\equal{} 9$ is a solution.\r\nCase $ n \\minus{} 1 \\equal{} 2p$, $ n \\plus{} 1 \\equal{} 8$:\r\nThen $ 6 \\equal{} 2p$ so $ p \\equal{} 3$, so $ n \\equal{} 7$ is a solution.\r\n\r\n\r\nSo a smallest solutions are $ n \\in \\{7, 9, 80,82,2398,14638,27557,707278,2825758,...\\}$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find positive integers $ a,b$ such that the following equation satisfies: \\[ (a\\minus{}b)^5\\equal{}(a\\plus{}b)^3\\minus{}a^2b^2\\]\r\n :)", "Solution_1": "No solutions!\r\nA very obvious one is $ (a,b) \\equal{} (5,2)$\r\n :) \r\nI am not sure whether any more solution exists.... :maybe:", "Solution_2": "If $ a\\equal{}b$,\r\n$ (a\\minus{}a)^5 \\equal{} (a\\plus{}a)^3 \\minus{} a^2a^2 \\iff 8a^3\\equal{}a^4 \\iff a\\equal{}0, 8$\r\nSo, $ ((a, b)\\equal{}(0, 0), (8, 8))$\r\n\r\nIf $ b\\equal{}0$,\r\n$ (a\\minus{}0)^5 \\equal{} (a\\plus{}0)^3 \\minus{} a^20^2 \\iff a^5\\equal{}a^3 \\iff a\\equal{}0, \\pm 1$\r\nSo, $ (a, b) \\equal{} (0, 0), (1, 0)$\r\n\r\nI don't try the other possibility.\r\nBut, In the range of $ 0 \\le a \\le 500$, $ 0 \\le b \\le 500$, , a solution is only $ (a, b)\\equal{}(0, 0), (1, 0), (5, 2), (8, 8)$.\r\n(by Mathematica)", "Solution_3": "In my problem, I mentioned $ a,b>0$, see.\r\nSo, only two solutions within $ 500$ is $ (5,2) \\& (8,8)$! That's amazing indeed! :lol: \r\nI wonder what surprises lie beyond 500!", "Solution_4": "Mathematica\r\n[code]\nTiming[For[a=1, a<=10000, a++{For[b=1, b<=10000, b++,{If[(a-b)^5 == (a+b)^3 - a^2 b^2, Print[{a, b}]]}]}]]\n[/code]\r\n\r\n$ \\{5, 2\\}$\r\n$ \\{8, 8\\}$\r\n\r\n$ \\{3455.41 \\ \\text{Second}, \\text{Null}\\}$", "Solution_5": "???????\r\n\r\nAre you writing Java? I don't know Java.\r\nWhat does $ \\{3455.41 \\ \\text{Second}, \\text{Null}\\}$ mean?", "Solution_6": "[quote=\"Sunkern_sunflora\"]???????\n\nAre you writing Java? I don't know Java.\nWhat does $ \\{3455.41 \\ \\text{Second}, \\text{Null}\\}$ mean?[/quote]\r\nNo, That's Mathematica programing. The computer was instructed to find all possible solutions with $ a,b\\le 10000$, and it returned the two solutions and displayed that no other solution found and that total time taken= $ 3455.41$ seconds.", "Solution_7": "[quote=\"Sunkern_sunflora\"]???????\n\nAre you writing Java? I don't know Java.\nWhat does $ \\{3455.41 \\ \\text{Second}, \\text{Null}\\}$ mean?[/quote]\r\n\r\ncf. [url=http://reference.wolfram.com/mathematica/ref/Timing.html]Timing[/url] - Wolfram Mathematica", "Solution_8": "But is there a soln? Can this be mathematically deduced?", "Solution_9": "[hide=\"does it help?\"]\nLet $ d\\equal{}a\\minus{}b$ then $ (a \\minus{} b)^5 \\equal{} (a \\plus{} b)^3 \\minus{} a^2b^2$ becomes\n$ d^5\\equal{}(2b\\plus{}d)^3\\minus{}(b\\plus{}d)^2b^2$\ntrivial cases for $ d\\equal{}0,1,2,3$\n\n[/hide]" } { "Tag": [ "function", "analytic geometry", "graphing lines", "slope", "calculus", "calculus computations" ], "Problem": "If the line tangent to the graph of the function f at the point (1,7) passes through the point (-2,2) then f'(1) = \r\nCan someone please explain this to me? Thanks", "Solution_1": "The line passing through those 2 points is defined by $y = 5/3*x+16/3$\r\n$f'(1)$ means: what is the slope of the tangent line at that point and from the equation of the line we know that the slope is $5/3$ so $f'(1)=5/3$", "Solution_2": "could it possibly be another answer because 5/3 dosen't match any of the choices that i have?\r\nthe choices are\r\na. 1\r\nb. 3\r\nc. -5\r\nd. 7\r\ne. undefined", "Solution_3": "This is simple. $f'(1)=\\frac{7-2}{1-(-2)}=3.$ thus the ansawer is $b.$\r\n\r\nLast edit: $f'(1)=\\frac{7-2}{1-(-2)}=\\frac{5}{3}.$ thus the answer is $e.$ \r\n\r\nkunny", "Solution_4": "You may want to check your arithmetic.... Maybe the second point is really supposed to be $(-2, -2)$? Then kunny is right :D\r\n\r\nedit: it's been fixed. Yay!", "Solution_5": "Yeh, oops. :blush: Thanks, Xevarion.\r\n\r\nkunny" } { "Tag": [ "geometry", "perimeter" ], "Problem": "The perimeter of a square lot is lined with trees, and there are three yards between the centers of adjacent trees. There are eight trees on a side, and a tree is at each corner. What is the number of yards in the perimeter of the lot?", "Solution_1": "Since there are 8 trees on a side, there are 7 spaces between trees.\r\nEach space is 3 yards, so the length of each side is $ 7*3 \\equal{} 21$ yards.\r\nThe perimeter is therefore $ 21 * 4 \\equal{} 84$ yd." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "let O is the center of circumcircle of ABC triangle.M is a point on AB arc.K & N are 2 point on AB such that :\r\n$ MK\\bot OA and MN\\bot OB$\r\n(MK,AC)=L and (MN,BC)=P. \r\nwhat is j$, so $ \\forall_{k\\equal{}1}^n (i\\minus{}j)\\alpha_k < \\frac{1}{N}$. [/hide]\r\nNow for $ P \\equal{} AB$, we have $ | 1 \\plus{} x_i^P | \\ge 1$ with equality when $ x_i \\equal{} 0$. \r\n\r\nNote that this solution only used the condition $ 1 \\equal{} \\prod | 1 \\plus{} x_i^p |$.", "Solution_6": "Difficult problem ! Another proof ?!" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "[color=darkblue]Let $ O$ be the center of the circumscribe circle of the triangle $ ABC$.The lines $ AO,BO,CO$ are meeting the oposite sides in points $ D,E,F$.Prove that :\n\\[ \\frac{1}{AD}\\plus{}\\frac{1}{BE}\\plus{}\\frac{1}{CF}\\equal{}\\frac{2}{R}\\][/color]", "Solution_1": "By relation of Gergonne we obtain:\r\n$ \\frac{OD}{AD}\\plus{}\\frac{OE}{BE}\\plus{}\\frac{OF}{CF}\\equal{}1$\r\n$ \\frac{AD\\minus{}R}{AD}\\plus{}\\frac{BE\\minus{}R}{BE}\\plus{}\\frac{CF\\minus{}R}{CF}\\equal{}1$\r\n$ 3\\minus{}(\\frac{R}{AD}\\plus{}\\frac{R}{BE}\\plus{}\\frac{R}{CF})\\equal{}1$\r\n$ R(\\frac{1}{AD}\\plus{}\\frac{1}{BE}\\plus{}\\frac{1}{CF})\\equal{}2$\r\n$ \\frac{1}{AD}\\plus{}\\frac{1}{BE}\\plus{}\\frac{1}{CF}\\equal{}\\frac{2}{R}$, q.e.d.", "Solution_2": "This is the simple and the nicest way to solve this problem.\r\nHere is a Trigonometric proof, if you don;t know Gergonne's relation.\r\n[hide]Let $ AA'$ be the alttude from $ A$.Then with the sin law:\n$ AA'%Error. \"ADcos\" is a bad command.\n(C-B)=ACsinC=2RsinBsinC$ therefore $ \\frac{2R}{AD}=\\frac{cos(C-B)}{sinBsinC}$\nFor $ BE$ and $ CF$ is the same and by suming we obtain \\[ 2R(\\frac{1}{AD}+\\frac{1}{BE}+\\frac{1}{CF})=\\frac{cos(C-B)}{sinBsinC}+\\frac{cos(A-C)}{sicCsinA}+\\frac{cos(B-A)}{sinBsinA}\\]\nTherefore we have $ 2RsinAsinBsinC(\\frac{1}{AD}+\\frac{1}{BE}+\\frac{1}{CF})=4sinAsinBsinC$[/hide]", "Solution_3": "This is a repeated post... Problem 6 of 1985 Iberoamerican Olympiad :) \r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=473507#473507[/url]" } { "Tag": [ "inequalities", "algebra", "IMO Shortlist" ], "Problem": "Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality\r\n\\[ \\frac {(a \\minus{} b)(a \\minus{} c)}{a \\plus{} b \\plus{} c} \\plus{} \\frac {(b \\minus{} c)(b \\minus{} d)}{b \\plus{} c \\plus{} d} \\plus{} \\frac {(c \\minus{} d)(c \\minus{} a)}{c \\plus{} d \\plus{} a} \\plus{} \\frac {(d \\minus{} a)(d \\minus{} b)}{d \\plus{} a \\plus{} b}\\ge 0\\]\r\nholds. Determine all cases of equality.\r\n\r\n[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i]", "Solution_1": "Without loss of generality, we can assume that $ a\\geq c$, $ b\\geq d$. Then\r\n$ \\frac{(a\\minus{}b)(a\\minus{}c)}{a\\plus{}b\\plus{}c}\\plus{}\\frac{(b\\minus{}c)(b\\minus{}d)}{b\\plus{}c\\plus{}d}\\plus{}\\frac{(c\\minus{}d)(c\\minus{}a)}{c\\plus{}d\\plus{}a}\\plus{}\\frac{(d\\minus{}a)(d\\minus{}b)}{d\\plus{}a\\plus{}b}$\r\n$ \\equal{}\\frac{(a\\minus{}c)^2}{a\\plus{}c\\plus{}d}\\plus{}\\frac{(b\\minus{}d)^2}{a\\plus{}b\\plus{}d}\\minus{}\\frac{(a\\minus{}b)(a\\minus{}c)(b\\minus{}d)(c^2\\plus{}d^2\\plus{}2ac\\plus{}ad\\plus{}bc\\plus{}2bd}{(a\\plus{}b\\plus{}c)(a\\plus{}b\\plus{}d)(a\\plus{}c\\plus{}d)(b\\plus{}c\\plus{}d)}$.\r\nIf $ a\\leq b$, then $ (a\\minus{}b)(a\\minus{}c)(b\\minus{}d)\\leq 0$, and if $ a\\geq c$, $ b\\geq c$, then $ (b\\minus{}c)(b\\minus{}d)(c\\minus{}a)\\leq 0$. Therefore, in these cases, we proved the inequality.\r\nWe only need prove the inequality when $ a\\geq c\\geq b\\geq d$. Let $ x\\equal{}a\\minus{}c$, $ y\\equal{}c\\minus{}b$, $ z\\equal{}b\\minus{}d$. Then the inequality is equal to\r\n$ \\frac{x^2}{a\\plus{}b\\plus{}c}\\plus{}\\frac{z^2}{a\\plus{}b\\plus{}d}\\plus{}xyz\\left(\\frac{1}{(a\\plus{}c\\plus{}d)(a\\plus{}b\\plus{}d)}\\minus{}\\frac{1}{(a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}d)}\\minus{}\\frac{1}{(a\\plus{}b\\plus{}d)(b\\plus{}c\\plus{}d)}\\right)$ $ \\geq 0$.\r\nWe know that\r\n$ \\frac{1}{a\\plus{}b\\plus{}c}\\geq \\frac{1}{b\\plus{}c\\plus{}d}\\cdot \\frac{y\\plus{}2z}{x\\plus{}2y\\plus{}3z}\\geq \\frac{1}{b\\plus{}c\\plus{}d}\\cdot \\frac{y}{x\\plus{}2y\\plus{}3z}$.\r\nTherefore,\r\n$ \\frac{x^2}{a\\plus{}b\\plus{}c}\\plus{}\\frac{z^2}{a\\plus{}b\\plus{}d}\\geq \\frac{x^2(x\\plus{}y\\plus{}2z)\\plus{}z^2(x\\plus{}2y\\plus{}3z)}{(a\\plus{}b\\plus{}c)(a\\plus{}b\\plus{}d)}$\r\n$ \\geq\\frac{(x^2\\plus{}z^2)(x\\plus{}2y\\plus{}3z)}{2(a\\plus{}b\\plus{}c)(a\\plus{}b\\plus{}d)}\\geq \\frac{xyz}{(a\\plus{}b\\plus{}d)(b\\plus{}c\\plus{}d)}$.\r\nAlso, because $ (a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}d)\\geq (a\\plus{}c\\plus{}d)(a\\plus{}b\\plus{}d)$,\r\n$ xyz\\left(\\frac{1}{(a\\plus{}c\\plus{}d)(a\\plus{}b\\plus{}d)}\\minus{}\\frac{1}{(a\\plus{}b\\plus{}c)(a\\plus{}c\\plus{}d)}\\right)\\geq 0$.\r\nTherefore, we proved the inequality when $ a\\geq c\\geq b\\geq d$.\r\nIt is easy to see that the equality holds if and only if $ a\\equal{}c$ and $ b\\equal{}d$.", "Solution_2": "I will write just the main steps because I'm really tired since it's 3am\r\n\r\nWLOG $ a \\equal{} max(a,b,c,d)$\r\n\r\nnote that the inequality is equivalent to :\r\n\r\n$ \\frac {(a \\minus{} d)(b \\minus{} c)(a \\plus{} 2d)}{(d \\plus{} a \\plus{} b)(c \\plus{} a \\plus{} d)} \\plus{} \\frac {(a \\minus{} d)(b \\minus{} c)(2b \\plus{} c)}{(a \\plus{} b \\plus{} c)(b \\plus{} c \\plus{} d)} \\plus{} \\frac {(c \\minus{} d)^2}{c \\plus{} d \\plus{} a} \\plus{} \\frac {(a \\minus{} b)^2}{a \\plus{} b \\plus{} c}$\r\n\r\nhence it remain to prove the inequality in the case $ c \\geq b$\r\n\r\nif $ a\\geq d \\geq c \\geq b$\r\nwe have :\r\n$ LHS \\geq \\frac {(a \\minus{} b)(a \\minus{} c)}{a \\plus{} b \\plus{} c} \\plus{} \\frac {(d \\minus{} a)(d \\minus{} b)}{d \\plus{} a \\plus{} b} \\geq (d \\minus{} b) ( \\frac {a \\minus{} c}{a \\plus{} b \\plus{} c} \\plus{} \\frac {d \\minus{} a}{b \\plus{} a \\plus{} d} ) \\geq \\frac { (d \\minus{} b)(d \\minus{} c)}{b \\plus{} a \\plus{} d} \\geq 0$\r\n\r\nif $ a\\geq c \\geq d \\geq b$\r\n\r\n case 1 : $ a \\minus{} c \\leq d \\minus{} b$\r\n\r\nhence : $ a \\minus{} d \\leq c \\minus{} b$\r\nthus $ LHS \\geq (a \\minus{} c)( \\frac {a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac { c \\minus{} d}{c \\plus{} d \\plus{} a} ) \\plus{} \\frac { (a \\minus{} c)(d \\minus{} b)(c \\minus{} b)}{(b \\plus{} c \\plus{} d)(d \\plus{} a \\plus{} b)} \\geq 0$\r\ncase 2 : $ a \\minus{} c \\geq d \\minus{} b$\r\n\r\nthen the inequality is equivalent to :\r\n$ (a \\minus{} c)( \\frac { a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} ) \\plus{} (d \\minus{} b)(\\frac {d \\minus{} a}{d \\plus{} a \\plus{} b} \\plus{} \\frac {c \\minus{} b}{b \\plus{} c \\plus{} d} ) \\geq 0$\r\nthus $ LHS \\geq (d \\minus{} b)( \\frac { a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} \\plus{} \\frac {d \\minus{} a}{d \\plus{} a \\plus{} b} \\plus{} \\frac {c \\minus{} b}{b \\plus{} c \\plus{} d} ) \\geq 0$ easy to prove !! \r\n\r\nif $ a\\geq c \\geq b \\geq d$\r\n\r\nnote that the inequality is equivalent to :\r\n$ (a \\minus{} c)( \\frac { a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} ) \\plus{} (b \\minus{} d) (\\frac {a \\minus{} d}{d \\plus{} a \\plus{} b} \\plus{} \\frac {b \\minus{} c}{b \\plus{} c \\plus{} d} ) \\geq 0$\r\nnow $ \\frac {a \\minus{} d}{d \\plus{} a \\plus{} b} \\plus{} \\frac {b \\minus{} c}{b \\plus{} c \\plus{} d} \\geq 0$ just let : $ a \\equal{} d \\plus{} x \\plus{} y \\plus{} z ,c \\equal{} d \\plus{} x \\plus{} y....$\r\n\r\nhence we have 2 cases :\r\n\r\nif $ b \\minus{} d \\geq a \\minus{} c$ as the previous method the inequality is easy to prove\r\nif $ a \\minus{} c \\geq b \\minus{} d$ \r\nnote that : $ \\frac { a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} \\geq 0$ is not true so we will use the trivial identity :\r\n$ \\frac { (a \\minus{} c)(b \\minus{} d)}{a \\plus{} b \\plus{} c} \\plus{} \\frac { (b \\minus{} d)(c \\minus{} a)}{a \\plus{} b \\plus{} c} \\geq 0$\r\n\r\n hence the inequality is equivalent to :\r\n$ (a \\minus{} c)( \\frac { a \\minus{} b}{a \\plus{} b \\plus{} c} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} \\plus{} \\frac {b \\minus{} d}{a \\plus{} b \\plus{} c} ) \\plus{} (b \\minus{} d)(\\frac {a \\minus{} d}{d \\plus{} a \\plus{} b} \\plus{} \\frac {b \\minus{} c}{b \\plus{} c \\plus{} d} \\plus{} \\frac {c \\minus{} a}{a \\plus{} b \\plus{} c} ) \\geq 0$\r\nit is easy to prove that :$ \\frac{ a\\minus{}b}{a\\plus{}b\\plus{}c} \\minus{}\\frac{c\\minus{}d}{c\\plus{}d\\plus{}a} \\plus{} \\frac{b\\minus{}d}{a\\plus{}b\\plus{}c} \\geq 0$with the condition : $ a \\minus{} c \\geq b \\minus{} d$\r\nthus : $ LHS \\geq (b \\minus{} d)( \\frac {a \\minus{} d}{b \\plus{} c \\plus{} a} \\minus{} \\frac {c \\minus{} d}{c \\plus{} d \\plus{} a} \\plus{} \\frac {a \\minus{} d}{d \\plus{} a \\plus{} b} \\plus{} \\frac {b \\minus{} c}{b \\plus{} c \\plus{} d} \\plus{} \\frac {c \\minus{} a}{a \\plus{} b \\plus{} c}) \\geq 0$ the last inequality can be handeled with easy manipulations , hence the inequality is proved :) omg what an ugly proof :(", "Solution_3": "[quote=\"April\"]Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality\n\\[ \\frac {(a \\minus{} b)(a \\minus{} c)}{a \\plus{} b \\plus{} c} \\plus{} \\frac {(b \\minus{} c)(b \\minus{} d)}{b \\plus{} c \\plus{} d} \\plus{} \\frac {(c \\minus{} d)(c \\minus{} a)}{c \\plus{} d \\plus{} a} \\plus{} \\frac {(d \\minus{} a)(d \\minus{} b)}{d \\plus{} a \\plus{} b}\\ge 0\\]\nholds. Determine all cases of equality.\n\n[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i][/quote]\r\nDear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23).", "Solution_4": "My solution(very sorry if repeated):\r\nFirstly it is equivalent to \r\n\r\n$ \\frac{(a-c)^2}{c+d+a}+\\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\\frac{2b+d}{(a+b+d)(b+c+d)}-\\frac{2a+c}{(a+b+c)(c+d+a)}\\ge 0$\r\n\r\nWe only need to prove:$ (\\frac{2b+d}{(a+b+d)(b+c+d)}-\\frac{2a+c}{(a+b+c)(c+d+a)})^2\\ge \\frac{4}{(a+c+d)(b+d+a)}$\r\n\r\nWe are sufficient to show both $ (\\frac{2b+d}{(a+b+d)(b+c+d)})^2$ and $ (\\frac{2a+c}{(a+b+c)(c+d+a)})^2$ are not bigger than $ \\frac{4}{(a+c+d)\n(b+d+a)}$\r\n\r\nThey are almost trivial if we notice $ (2b+d)^2 \\le 4(b+d)^2$ and some easy calculation :P", "Solution_5": "[quote=\"can_hang2007\"][quote=\"April\"]Prove that for any four positive real numbers $ a$, $ b$, $ c$, \n[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i][/quote]\nDear friend, please take a look at: http://mathvn.org/forum/viewthread.php?thread_id=1254 (problem O23, page 23).[/quote]\r\n\r\nDear can_hang2007 I canot open the file .pdf , please can you put here", "Solution_6": "My solution is more computational than the ones provided above, but in my opinion it is also a bit more motivated (though perhaps this could probably just be attributed to my lack of experience with inequalities.) \n\n[hide]We shall show that if $a,b,c,d$ are any nonnegative real numbers such that $a \\neq c$ or $b \\neq d$\n\n[b]Lemma 1[/b]: $x^4 - x^3 - x^2 + 2x + 1 \\geq 0$ for all nonnegative $x$.\n[i]Proof:[/i] We will use [url=http://en.wikipedia.org/wiki/Sturm%27s_theorem]Sturm's theorem.[/url]. Let $P_0(x) = x^4 - x^3 - x^2 + 2x + 1$. We compute the Sturm chain:\n\\begin{align*}\nP_0(x) &= x^4 - x^3 - x^2 + 2x + 1\\\\\nP_1(x) &= P_0'(x) &= 4x^3 - 3x^2 - 2x+2 \\\\\nP_2(x) &= \\frac{x}{4}(4x^3 - 3x^2 - 2x+2) - (x^4 - x^3 - x^2 + 2x + 1) &=& \\frac{x^3}{4} + \\frac{x^2}{2} - 2x + 1 \\\\\nP_3(x) &= 16(\\frac{x^3}{4} + \\frac{x^2}{2} - 2x + 1) - (4x^3 - 3x^2 - 2x+2) &= 11x^2 - 30x + 14 \\\\\nP_4(x) &= (\\frac{x}{44}+\\frac{13}{121})(11x^2 - 30x + 14) - (\\frac{x^3}{4} + \\frac{x^2}{2} - 2x + 1) &=\\frac{-192x}{121} - \\frac{61}{121} \\\\\nP_5(x) &= (-\\frac{192x}{121} - \\frac{61}{121})(\\frac{-1331x}{192} + \\frac{778151}{36864}) - (11x^2 - 30x + 14) &= \\frac{-908387}{36864}.\n\\end{align*}\n(I could not get this to format properly; just right click on the image and click \"view image.\")\n\nNote that $P_0(0) > 0$. By Sturm's theorem, the number of distinct roots in the interval $(0, \\infty)$ is the difference in the number of sign changes of the leading coefficient of each polynomial in the Sturm chain and the number of sign changes in the constant term of each polynomial in the Sturm chain, which can easily be seen to be 0, so $P_0$ has no nonnegative real roots, as desired. $\\blacksquare$\n\n------------------------------------------------------------------------------------------\n\n(I shall rename $a,b,c,d$ as $w,x,y,z$, and call another set of variables $a,b,c,d$. This is due to my writing the latter portion, which is too difficult to change now, first.)\n\nIt is easy to see that the inequality is true when any three variables are equal, so we may suppose that there are not 3 equal variables among $w,x,y,z$. Without loss of generality, let $z$ be the smallest of $w,x,y,z$, and let $a = w-z$, $b = y-z$, and $c = w-z$. Define $f(k) = \\frac{(a-b)(a-c)}{a+b+c+k} + \\frac{(b-c)b}{b+c+k} + \\frac{c(c-a)}{c+a+k} + \\frac{ab}{a+b+k}$ (division by zero is not a problem here since no two of $a, b, c$ can equal 0, by our assumption that there are not 3 equal variables among $a,b,c,d$.) We wish to show that $f(z) \\geq 0$.\n\nThe sign of $f$ is given by \n$P(k) = (b+c+k)(a+b+c+k)(c+a+k)(a+b+k)f(k) \\\\ \n= (a^4 b + 2a^3 b^2 + 2a^2 b^3 + ab^4 + a^4 c - a^3 bc + 2ab^3 c + b^4 c - a^3 c^2 - 3a^2 bc^2 + 2b^3 c^2 - a^2 c^3 + 2abc^3 + 2b^2 c^3 + ac^4 + bc^4) + k(a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4) + k^2(2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3) + k^3((a-c)^2 + b^2).$\nWe claim that all coefficients of $P$ are nonnegative, which will finish this problem.\n\n$a^4 b + 2a^3 b^2 + 2a^2 b^3 + ab^4 + a^4 c - a^3 bc + 2ab^3 c + b^4 c - a^3 c^2 - 3a^2 bc^2 + 2b^3 c^2 - a^2 c^3 + 2abc^3 + 2b^2 c^3 + ac^4 + bc^4 \\geq 0$ is equivalent to $ac(a+c)(a-c)^2 + b(a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2) + b^2(2a^3 + 2a^2 b + ab^2 + 2abc + b^2 c + 2bc^2 + 2c^3).$ If $c = 0$, $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \\geq 0.$ Otherwise, substituting $w = \\frac{a}{c}$ into lemma 1 gives $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \\geq 0$, so the constant coefficient of $P$ is nonnegative.\n\n$a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4 \\geq 0$ is equivalent to $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4ab + b^2 + 3bc + 4c^2) \\geq 0.$ The inequality is obviously true if $c=0$, so suppose for now that $c \\neq 0$. Note that it is sufficient to prove that $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4c^2) \\geq 0.$ When viewed as a quadratic in $b$, the discriminant of the polynomial is $-a(a-c)^2(7a^3 + 60a^2 c + 52 ac^2 + 32c^3)$, which is clearly nonpositive. Hence, the coefficient of $k$ in $P$ is nonnegative.\n\n$2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3 \\geq 0$ is equivalent to $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a + 2b + 2c) \\geq 0$, so it is sufficient to show that $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a+2c) \\geq 0$. As before, we view the inequality as a quadratic in $b$. The discriminant of $b$ is $-(a-c)^2(20a^2 + 52ac + 7c^2) \\leq 0$, so the coefficient of $k^2$ in $P$ is nonnegative.\n\nBecause the coefficient of $k^3$ in $P$ is clearly nonnegative, all coefficients of $P$ are nonnegative, as desired. It can easily be seen from the above inequalities that equality holds exactly when $a=c$ and $b=d$. [/hide]", "Solution_7": "math154 has pointed out that there is an error in this solution. $a^4 + 2ac^3 + c^4 - a^3 c - 3a^2 c^2 \\geq 0$ is false (and is completely irrelevant to my lemma :(.) Here is a revised proof, hopefully with no mistakes this time: \n\n\n[hide]We shall show that if $a,b,c,d$ are any nonnegative real numbers such that $a \\neq c$ or $b \\neq d$\n\n[b]Lemma 1[/b]: If $x \\geq 0$ and $(x-1)(x^3 - 3x - 1) < 0$, then $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 \\leq 0$.\n[i]Proof:[/i] Let $P(x) = x^3-3x-1$. $P(2) > 0$, $P(1) < 0$, $P(-1) > 0$, and $P(-2) < 0$, so by the intermediate value theorem, the three roots of $P$ lie in the intervals $(-2, -1)$, $(-1, 0)$, and $(1, 2)$. Since $P(0), P(1) < 0$, $P$ must be negative over $[0,1]$, so $(x-1)(P(x) \\geq 0$ for $x \\in [0,1]$. It follows that if $x \\geq 0$ and $(x-1)(x^3 - 3x - 1) < 0$, we must have $x \\in (1, 2)$. \n\nWe shall show that $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 < 0$ for $x \\in (1,2)$. The polynomial $3x^2 - 18x + 2$ has a minimum at $x=3$, so it is monotonic over $(1,2)$. Since it is negative at both $1$ and $2$, $3x^2 - 18x + 2 < 0$ for $x \\in (1,2)$. Hence, $(1+4x)(3x^2 - 18x + 2) = 12x^3 - 69x^2 - 10x - 2 < 0$ for $x \\in (1,2)$. Since $3x^4 - 23x^3 - 5x^2 + 2x - 1$ has derivative $12x^3 - 69x^2 - 10x - 2 < 0$ for $x \\in (1,2)$ and is negative at $x=1$, $3x^4 - 23x^3 - 5x^2 + 2x - 1 < 0$ for $x \\in (1,2)$. Hence, $2(1+x)(3x^4 - 23x^3 - 5x^2 + 2x - 1) = 6x^5 - 40x^4 - 56x^3 - 6x^2 + 2x - 2 < 0$ for $x \\in (1,2)$. Because $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1$ is negative at $x=1$ and has derivative $6x^5 - 40x^4 - 56x^3 - 6x^2 + 2x - 2 < 0$ for $x \\in (1,2)$, we have that $x^6 - 8x^5 - 14x^4 - 2x^3 + x^2 - 2x + 1 < 0$ for all $x \\in (1,2)$, as desired. $\\blacksquare$\n\n------------------------------------------------------------------------------------------\n\n(I shall rename $a,b,c,d$ as $w,x,y,z$, and call another set of variables $a,b,c,d$. This is due to my writing the latter portion, which is too difficult to change now, first.)\n\nIt is easy to see that the inequality is true when any three variables are equal, so we may suppose that there are not 3 equal variables among $w,x,y,z$. Without loss of generality, let $z$ be the smallest of $w,x,y,z$, and let $a = w-z$, $b = y-z$, and $c = w-z$. Define $f(k) = \\frac{(a-b)(a-c)}{a+b+c+k} + \\frac{(b-c)b}{b+c+k} + \\frac{c(c-a)}{c+a+k} + \\frac{ab}{a+b+k}$ (division by zero is not a problem here since no two of $a, b, c$ can equal 0, by our assumption that there are not 3 equal variables among $a,b,c,d$.) We wish to show that $f(z) \\geq 0$.\n\nThe sign of $f$ is given by \n$P(k) = (b+c+k)(a+b+c+k)(c+a+k)(a+b+k)f(k) \\\\ \n= (a^4 b + 2 a^3 b^2 + 2 a^2 b^3 + a b^4 + a^4 c - a^3 b c + 2 a b^3 c + \n b^4 c - a^3 c^2 - 3 a^2 b c^2 + 2 b^3 c^2 - a^2 c^3 + 2 a b c^3 + \n 2 b^2 c^3 + a c^4 + b c^4) + k(a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4) + k^2(2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3) + k^3((a-c)^2 + b^2).$\nWe claim that all coefficients of $P$ are nonnegative, which will finish this problem.\n\n$a^4 b + 2 a^3 b^2 + 2 a^2 b^3 + a b^4 + a^4 c - a^3 b c + 2 a b^3 c + \n b^4 c - a^3 c^2 - 3 a^2 b c^2 + 2 b^3 c^2 - a^2 c^3 + 2 a b c^3 + \n 2 b^2 c^3 + a c^4 + b c^4 \\geq 0$ is equivalent to $ac(a+c)(a-c)^2 + b(a-c)(a^3 - 3ac^2 + c^3) + b^2(2a^3 + 2a^2 b + ab^2 + 2abc + b^2 c + 2bc^2 + 2c^3) \\geq 0$, so it is sufficient to show that $ac(a+c)(a-c)^2 + b(a-c)(a^3 - 3ac^2 + c^3) + b^2(2a^3 + 2c^3) \\geq 0$. Since this inequality is homogeneous and obvious for $c=0$, we may assume without loss of generality that $c=1$, so we wish to show that $a(a+1)(a-1)^2 + b(a-1)(a^3 - 3a - 1) + b^2(2a^3 + 2) \\geq 0$. The result is obvious if $(a-1)(a^3-3a-1) \\geq 0$, so suppose that $(a-1)(a^3 - 3a - 1) < 0$. Viewing this inequality as a quadratic in $b$, we see that it is sufficient to have $(a-1)^2(a^3 - 3a - 1)^2 - 4a(a+1)(2a^3 + 2)(a-1)^2 = (a-1)^2(a^6 - 8a^5 - 14a^4 - 2a^3 + a^2 - 2a + 1) \\leq 0$, which is true by lemma 1. Hence, the constant coefficient of the polynomial is nonnegative. \n\n$a^4+3a^3b + 4a^2 b^2 + 4ab^3 + b^4 + a^3 c - 5a^2 bc + 3b^3 c - 4a^2 c^2 - 2abc^2 + 4b^2 c^2 + ac^3 + 4bc^3 + c^4 \\geq 0$ is equivalent to $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4ab + b^2 + 3bc + 4c^2) \\geq 0.$ Note that it is sufficient to prove that $(a-c)^2(a^2 + 3ac + c^2) + b(a-c)(3a^2-2ac-4c^2) + b^2(4a^2 + 4c^2) \\geq 0.$ We view the inequality as a quadratic in $b$. The discriminant of the quadratic is $-a(a-c)^2(7a^3 + 60a^2 c + 52 ac^2 + 32c^3)$, which is clearly nonpositive. Hence, the coefficient of $k$ in $P(k)$ is nonnegative.\n\n$2a^3 + 2a^2 b + 3ab^2 + 2b^3 - 2a^2 c - 5abc + 2b^2 c - 2ac^2 + 3bc^2 + 2c^3 \\geq 0$ is equivalent to $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a + 2b + 2c) \\geq 0$, so it is sufficient to show that $2(a-c)^2(a+c) + b(a-c)(2a-3c) + b^2(3a+2c) \\geq 0$. As before, we view the inequality as a quadratic in $b$. The discriminant of the quadratic is $-(a-c)^2(20a^2 + 52ac + 7c^2) \\leq 0$, so the coefficient of $k^2$ in $P(k)$ is nonnegative.\n\nBecause the coefficient of $k^3$ in $P(k)$ is clearly nonnegative, all coefficients of $P$ are nonnegative, as desired. It can easily be seen from the above inequalities that equality holds exactly when $a=c$ and $b=d$. [/hide]", "Solution_8": "A7 in 2008 seems to be easy.\nFrom $ a,b,c,d \\in\\mathbb{R}^+ $, we yield\n \\[ \\frac{(a-b)(a-c)}{a+b+c}\\geq\\frac{(a-b)(a-c)}{a+b+c+d}=\\frac{a^2-ab-ac+bc}{a+b+c+d} \\] Similarly, we got \\[ \\frac{(b-c)(b-d)}{b+c+d}\\geq\\frac{b^2-bc-bd+cd}{a+b+c+d} \\] \\[ \\frac{(c-d)(c-a)}{c+d+a}\\geq\\frac{c^2-cd-ca+ad}{a+b+c+d} \\] \\[ \\frac{(d-a)(d-b)}{d+a+b}\\geq\\frac{d^2-ad-bd+ab}{a+b+c+d} \\] Sum all of the above inequalities together, we obtain: \\[ \\frac{(a-b)(a-c)}{a+b+c}+\\frac{(b-c)(b-d)}{b+c+d}+\\frac{(c-d)(c-a)}{c+d+a}+\\frac{(d-a)(d-b)}{d+a+b}\\geq\\frac{a^2-ab-ac+bc+b^2-bc-bc+cd+c^2-cd-ca+ad+d^2-da-db+ab}{a+b+c+d}=\\frac{a^2+b^2+c^2+d^2}{a+b+c+d}\\geq 0 \\] Equality occurs iff $ a=c $ or $ b=d $", "Solution_9": "[quote=quangminhltv99]A7 in 2008 seems to be easy.\nFrom $ a,b,c,d \\in\\mathbb{R}^+ $, we yield\n \\[ \\frac{(a-b)(a-c)}{a+b+c}\\geq\\frac{(a-b)(a-c)}{a+b+c+d}=\\frac{a^2-ab-ac+bc}{a+b+c+d} \\] [/quote]\n\nWell, no. That holds only if the fractions are nonnegative. For example if $c>a>b$, then it doesn't hold.", "Solution_10": "[quote=hxy09]My solution(very sorry if repeated):\nFirstly it is equivalent to \n\n$ \\frac{(a-c)^2}{c+d+a}+\\frac{(b-d)^2}{d+a+b}+(a-c)(b-d)(\\frac{2b+d}{(a+b+d)(b+c+d)}-\\frac{2a+c}{(a+b+c)(c+d+a)}\\ge 0$\n\n[/quote]\n\nWhat would the motivation for splitting the expression in this way be? Thanks!\n\n", "Solution_11": "Just a query; after clearing the denominators, isn't it pretty much do-able by Lagrange Multipliers(of course, after normalizing)", "Solution_12": "[quote=April]Prove that for any four positive real numbers $ a$, $ b$, $ c$, $ d$ the inequality\n\\[ \\frac {(a \\minus{} b)(a \\minus{} c)}{a \\plus{} b \\plus{} c} \\plus{} \\frac {(b \\minus{} c)(b \\minus{} d)}{b \\plus{} c \\plus{} d} \\plus{} \\frac {(c \\minus{} d)(c \\minus{} a)}{c \\plus{} d \\plus{} a} \\plus{} \\frac {(d \\minus{} a)(d \\minus{} b)}{d \\plus{} a \\plus{} b}\\ge 0\\]\nholds. Determine all cases of equality.\n\n[i]Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany[/i][/quote]\n\nWe have\n\\[LHS=\\frac{1}{660}\\frac{(4a^2b+297a^2c+20ab^2+490abc+297ac^2+490acd+144ad^2+144b^2c+77b^2d+77bd^2+4c^2d+20cd^2)(a-2b+2d-c)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\]\n\\[+\\frac{1}{660}\\frac{(144a^2b+77a^2c+20a^2d+490abd+77ac^2+4ad^2+4b^2c+297b^2d+20bc^2+490bcd+297bd^2+144c^2d)(2a+b-2c-d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\]\n\\[+\\frac{1}{132}\\frac{(11a^2c+116a^2d+232abd+11ac^2+16ad^2+16b^2c+116bc^2+232bcd)(a-b-c+d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\]\n\\[+\\frac{1}{132}\\frac{(16a^2b+116ab^2+232abc+232acd+11b^2d+11bd^2+16c^2d+116cd^2)(a+b-c-d)^2}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\]\n\\[\\ge{0}\\]\n", "Solution_13": "well, hxy09...\n[quote=hxy09]\nWe only need to prove$ (\\frac{2b+d}{(a+b+d)(b+c+d)}-\\frac{2a+c}{(a+b+c)(c+d+a)})^2\\ge \\frac{4}{(a+c+d)(b+d+a)}$[/quote]\nBut, this does not satisfy when $a=b=c=d$!!\n.....isn't it!? :wallbash_red: \n", "Solution_14": "[b]Note.[/b] The term ``LHS'' will always refer to the expression $\\frac {(a - b)(a - c)}{a + b + c} + \\frac {(b - c)(b - d)}{b + c + d} + \\frac {(c - d)(c - a)}{c + d + a} + \\frac {(d - a)(d - b)}{d + a + b}$.\n\nWe will rewrite the LHS as \\[ (a-c)(\\frac{a-b}{a+b+c} + \\frac{d-c}{c+d+a}) + (b-d)(\\frac{b-c}{b+c+d} + \\frac{a-d}{d+a+b}). \\] Let us focus on the first term $(a-c)(\\frac{a-b}{a+b+c} + \\frac{d-c}{c+d+a})$. We will try to express this in terms of $(a-c)$ and $(b-d)$. Thus we write \\[(a-c)(\\frac{a-b}{a+b+c} + \\frac{d-c}{c+d+a}) = (a-c)(\\frac{(ac+ad+a^2-bc-bd-ba) + (da+db+dc-ca-cb-c^2)}{(a+b+c)(c+d+a)}) \\] \\[ = (a-c)(\\frac{2ad - 2bc - ab + cd + a^2 - c^2}{(a+b+c)(c+d+a)}) \\] \\[ = (a-c)(\\frac{(a-c)(a+c+\\frac{1}{2}b+\\frac{1}{2}d) + (b-d)(-\\frac{3}{2}a-\\frac{3}{2}c)}{(a+b+c)(c+d+a)}) \\] \\[ = (a-c)^2\\frac{a+c+\\frac{1}{2}b+\\frac{1}{2}d}{(a+b+c)(c+d+a)} + (a-c)(b-d)\\frac{-\\frac{3}{2}a-\\frac{3}{2}c}{(a+b+c)(c+d+a)}\\]\n\nSimilarly, the second term $(b-d)(\\frac{b-c}{b+c+d} + \\frac{a-d}{d+a+b})$ equals \\[ (b-d)^2\\frac{b+d+\\frac{1}{2}a+\\frac{1}{2}c}{(b+c+d)(d+a+b)} + (a-c)(b-d)\\frac{\\frac{3}{2}b+\\frac{3}{2}d}{(b+c+d)(d+a+b)} \\]\n\nThus, the entire LHS equals \\[(a-c)^2X + (b-d)^2Z + (a-c)(b-d)Y,\\] where $X = \\frac{a+c+\\frac{1}{2}b+\\frac{1}{2}d}{(a+b+c)(c+d+a)}$, $Y = \\frac{3}{2}(\\frac{b+d}{(b+c+d)(d+a+b)} - \\frac{a+c}{(a+b+c)(c+d+a)})$, and $Z = \\frac{b+d+\\frac{1}{2}a+\\frac{1}{2}c}{(b+c+d)(d+a+b)}$. Let's simplify $Y$ a little bit. Miraculously, we can write \\[Y = \\frac{3}{2}(\\frac{(b+d)(a+b+c)(c+d+a) - (a+c)(b+c+d)(d+a+b)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\\] \\[ = \\frac{3}{2}(\\frac{(b+d)(bd+(b+d)(a+c)+(a+c)^2) - (a+c)(ac+(a+c)(b+d) + (b+d)^2)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)})\\] \\[ = \\frac{3}{2}(\\frac{bd(b+d)-ac(a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}).\\] Note how most of the terms in the numerator cancel. \n\nLet $u = a-c$, $v = b-d$. Then the LHS is \\[u^2X + v^2Z + uvY,\\] which is quadratic in $u,v$. I claim that $Y^2 - 4XZ < 0$, or equivalently:\n\n[b]Lemma.[/b] $Y^2 < 4XZ$\n\n[i]Proof.[/i] To prove this, we first note that \\[4XZ = \\frac{(2a+2c+b+d)(2b+2d+a+c)}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\] and \\[Y^2 = \\frac{9}{4}(\\frac{(bd(b+d)-ac(a+c))^2}{((a+b+c)(b+c+d)(c+d+a)(d+a+b))^2}).\\] Thus $Y^2 < 4XZ$ is equivalent to \\[\\frac{9}{4}(bd(b+d)-ac(a+c))^2 < (2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b)\\]\n\nNote that, if $x,y > 0$, $(x-y)^2 < x^2$ if $x \\ge y$ and $(x-y)^2 < y^2$ if $y \\ge x$. Let's let $x = ac(a+c)$, $y = bd(b+d)$. Then we can assume WLOG that $x \\ge y$. We have \\[\\frac{9}{4}(bd(b+d)-ac(a+c))^2 < \\frac{9}{4}(ac(a+c))^2\\]\n\nBut we also have \\[(2a+2c+b+d)(2b+2d+a+c)(a+b+c)(b+c+d)(c+d+a)(d+a+b) > (2a+2c)(a+c)(a+c)(c)(a+c)(a) = 2ac(a+c)^4\\]\n\nThus, if we can show that \\[2ac(a+c)^4 > \\frac{9}{4}(ac(a+c))^2,\\] we would be done with the proof of this Lemma. But this is equivalent to \\[(a+c)^2 > \\frac{9}{8}ac,\\] which of course is true. $\\blacksquare$\n\nNow that we have $Y^2 - 4XZ < 0$, let's show that the LHS, which is $u^2X + v^2Z + uvY$, is always nonnegative. Note that $X,Z > 0$. First suppose that $u = v = 0$; then obviously the LHS equals $0$, which is nonnegative. Now suppose that $u = 0$ but $v \\not= 0$. Then the LHS is $v^2Z$, which is positive. Similarly, if $u \\not= 0 $ and $v = 0$, then the LHS is $u^2X$, which is positive. \n\nNow suppose that $u,v \\not= 0 $. We can write the LHS as \\[v^2(X(u/v)^2 + Y(u/v) + Z)\\] Since $Y^2 - 4XZ < 0$, we know that $Xr^2 + Yr + Z > 0$ for all real numbers $r$. This means that $X(u/v)^2 + Y(u/v) + Z > 0$. Since $v^2 > 0$, we know that the LHS is positive.\n\nThus the LHS is always nonnegative, with equality occurring if and only if $u = v = 0$, which is equivalent to $a=c$ and $b=d$ holding simultaneously. $\\blacksquare$", "Solution_15": "[b][color=red]Claim:[/color][/b] \t[The main rewrite] \tThe given inequality may be rewritten as \t\\begin{align*} \t\t(a-c)^2 & \\left( \\frac{1}{a+b+c} + \\frac{1}{c+d+a} \\right) \t\t+ (b-d)^2 \\left( \\frac{1}{b+c+d} + \\frac{1}{d+a+b} \\right) \\\\ \t\t&\\ge 3(a-c)(b-d) \t\t\\left[ \\frac{a+c}{(a+b+c)(c+d+a)} \t\t\t- \\frac{b+d}{(b+c+d)(d+a+b)} \\right]. \t\\end{align*}\n\n[hide=\"Proof\"] Using the SOS identity $2(a-b)(a-c) = (a-b)^2 + (a-c)^2 - (b-c)^2$, we get \\[ 2 \\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{a+b+c} \t = \\sum_{\\text{cyc}} \\frac{(a-c)^2}{a+b+c} \t + \\sum_{\\text{cyc}} \\frac{(a-b)^2 - (b-c)^2}{a+b+c} \\] We observe that the second sum should be divisible by $(a-c)(b-d)$. After \\[ \\sum_{\\text{cyc}} \\frac{(a-c)(a-2b+c)}{a+b+c} \t= \\sum_{\\text{cyc}} \\left[ \\frac{(a-c)(a-2b+c)}{a+b+c} + 2(a-c) \\right] \t= \\sum_{\\text{cyc}} \\frac{3(a-c)(a+c)}{a+b+c} \\] we can pair up two of the terms via \\begin{align*} \t\\frac{3(a-c)(a+c)}{a+b+c} \t+ \\frac{3(c-a)(c+a)}{c+d+a} \t&= 3(a-c)(a+c) \\left[ \\frac{1}{(a+c)+b} - \\frac{1}{(a+c)+d} \\right] \\\\ \t&= -3(a-c)(b-d) \\cdot \\frac{a+c}{(a+b+c)(c+d+a)}. \\end{align*} Doing the same calculation on the other half finishes. $\\blacksquare$[/hide]\n\nIn light of this, taking the absolute value of right-hand side, it suffices to prove \\begin{align*} \t(a-c)^2 & \\left( \\frac{1}{a+b+c} + \\frac{1}{c+d+a} \\right) \t+ (b-d)^2 \\left( \\frac{1}{b+c+d} + \\frac{1}{d+a+b} \\right) \\\\ \t&\\ge 3 |a-c| \\cdot |b-d| \\cdot \t\\max \\left\\{ \\frac{a+c}{(a+b+c)(c+d+a)}, \t\\frac{b+d}{(b+c+d)(d+a+b)} \\right\\}. \\end{align*} We show only the first one (other one ditto). In fact, a crude AM-GM suffices:\n\n[b][color=red]Claim:[/color][/b] \tWe have \t\\[ 4\\sqrt[4]{\\prod_{\\text{cyc}} \\frac{1}{a+b+c}} \t\t> \\frac{3(a+c)}{(a+b+c)(c+d+a)}. \\]\n\n[hide=\"Proof\"]It's equivalent to prove that \t\\[ 256 (a+b+c)^3 (c+d+a)^3 > 81 (b+c+d) (d+a+b) (a+c)^4. \\] \tWe have $(b+c+d)(d+a+b) \\le (b+d+\\frac{a+c}{2})^2$ \tby Jensen on $x \\mapsto \\log(b+d+x)$. \tSo if we let $x = \\frac{b}{a+c}$ and $y = \\frac{d}{a+c}$, \tthen it is sufficient to prove \t\\[ 256(x+1)^3 (y+1)^3 > 81(x+y+\\tfrac12)^2. \\] \tHowever $(x+1)(y+1) = xy + x + y + 1 > x+y+\\tfrac12$ \tso this is obvious.[/hide]\n\nAnd thus we are done. Equality holds iff $|a-c| = |b-d| = 0$, which is $a=c$ and $b=d$.", "Solution_16": "\\[ \\frac{(a-b)(a-c)}{a+b+c}+\\frac{(b-c)(b-d)}{b+c+d}+\\frac{(c-d)(c-a)}{c+d+a}+\\frac{(d-a)(d-b)}{d+a+b}\\geq 0 \\]\n\\[ \\frac{2(a-b)(a-c)}{a+b+c}+\\frac{2(b-c)(b-d)}{b+c+d}+\\frac{2(c-d)(c-a)}{c+d+a}+\\frac{2(d-a)(d-b)}{d+a+b}\\geq 0 \\]\n\\[(\\frac{2(a-b)(a-c)}{a+b+c}+\\frac{(2b-c-a)(a-c)}{a+b+c})+(\\frac{2(b-c)(b-d)}{b+c+d}+\\frac{(2c-d-b)(b-d)}{b+c+d})+(\\frac{2(c-d)(c-a)}{c+d+a}+\\frac{(2d-a-c)(c-a)}{c+d+a})+(\\frac{2(d-a)(d-b)}{d+a+b}+\\frac{(2a-b-d)(d-b)}{d+a+b})\\geq \\] \\[\\geq\\frac{(2b-c-a)(a-c)}{a+b+c}+\\frac{(2c-d-b)(b-d)}{b+c+d}+\\frac{(2d-a-c)(c-a)}{c+d+a}+\\frac{(2a-b-d)(d-b)}{d+a+b} \\]\n\\[\\frac{(a-c)^2}{a+b+c}+\\frac{(b-d)^2}{b+c+d}+\\frac{(c-a)^2}{c+d+a}+\\frac{(d-b)^2}{d+a+b}\\geq (a-c)(\\frac{2b-c-a}{a+b+c}-\\frac{2d-a-c}{c+d+a})+(b-d)(\\frac{2c-d-b}{b+c+d}-\\frac{2a-b-d}{d+a+b})\\]\nTransform the right side of the inequality:\n\\[(a-c)(\\frac{2b-c-a}{a+b+c}-\\frac{2d-a-c}{c+d+a})+(b-d)(\\frac{2c-d-b}{b+c+d}-\\frac{2a-b-d}{d+a+b})=(a-c)(\\frac{3b}{a+b+c}-\\frac{3d}{c+d+a})+(b-d)(\\frac{3c}{b+c+d}-\\frac{3a}{d+a+b})= \\]\n\\[=\\frac{3(a-c)(b(c+d+a)-d(a+b+c))}{(a+b+c)(c+d+a)}+\\frac{3(b-d)(c(d+a+b)-a(b+c+d))}{(b+c+d)(d+a+b)}=\\frac{3(a-c)(b-d)(a+c)}{(a+b+c)(c+d+a)}+\\frac{3(b-d)(c-a)(b+d)}{(b+c+d)(d+a+b)}=\\]\n\\[=\\frac{3(a-c)(b-d)((a+c)(b+c+d)(d+a+b)-(b+d)(a+b+c)(c+d+a))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}=\\]\n\\[=\\frac{3(a-c)(b-d)((a+c)(b+d)^2+(a+c)^2(b+d)+ac(a+c)-(b+d)(a+c)^2-(b+d)^2(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}=\\]\n\\[=\\frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \\]\nWe need to prove that:\n\\[\\frac{(a-c)^2}{a+b+c}+\\frac{(b-d)^2}{b+c+d}+\\frac{(c-a)^2}{c+d+a}+\\frac{(d-b)^2}{d+a+b}\\geq \\frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \\]\nWe write a set of such inequalities:\n\\[a*c*(a+c)<(a+b+c)*(b+c+d)*(c+d+a)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{d+a+b} \\]\n\\[a*c*(a+c)<(d+a+b)*(b+c+d)*(a+b+c)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{c+d+a} \\]\n\\[a*c*(a+c)<(d+a+b)*(a+b+c)*(c+d+a)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{b+c+d} \\]\n\\[a*c*(a+c)<(d+a+b)*(b+c+d)*(c+d+a)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{a+b+c} \\]\nWe get:\n\\[ac(a+c)<\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\\{a+b+c;b+c+d;c+d+a;d+a+b\\}}\\]\nSimilarly:\n\\[b*d*(b+d)<(a+b+c)*(c+d+a)*(b+c+d)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{d+a+b} \\]\n\\[b*d*(b+d)<(a+b+c)*(d+a+b)*(b+c+d)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{c+d+a} \\]\n\\[b*d*(b+d)<(a+b+c)*(c+d+a)*(d+a+b)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{b+c+d} \\]\n\\[b*d*(b+d)<(d+a+b)*(c+d+a)*(b+c+d)=\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{a+b+c} \\]\nWe get:\n\\[bd(b+d)<\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\\{a+b+c;b+c+d;c+d+a;d+a+b\\}}\\]\nCombine 2 inequalities obtained:\n\\[max\\{ac(a+c);bd(b+d)\\}<\\frac{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}{max\\{a+b+c;b+c+d;c+d+a;d+a+b\\}} \\]\nWe still need the following inequality, which we prove right away:\n\\[2(a-c)^2+2(b-d)^2\\geq 3*|a-c|*|b-d| \\]\n\\[\\frac{1}{2}(a-c)^2+\\frac{3}{2}(|a-c|-|b-d|)^2+\\frac{1}{2}(b-d)^2\\geq 0 \\]\nNow we can prove the main inequality:\n\\[ \\frac{3(a-c)(b-d)(ac(a+c)-bd(b+d))}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\leq \\frac{3*|a-c|*|b-d|*|ac(a+c)-bd(b+d)|}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)}\\leq \\frac{3*|a-c|*|b-d|*(max\\{ac(a+c);bd(b+d)\\}}{(a+b+c)(b+c+d)(c+d+a)(d+a+b)} \\leq \\]\n\\[\\leq \\frac{3*|a-c|*|b-d|}{max\\{a+b+c;b+c+d;c+d+a;d+a+b\\}}\\leq \\frac{2(a-c)^2+2(b-d)^2}{max\\{a+b+c;b+c+d;c+d+a;d+a+b\\}} \\leq \\frac{(a-c)^2}{a+b+c}+\\frac{(b-d)^2}{b+c+d}+\\frac{(c-a)^2}{c+d+a}+\\frac{(d-b)^2}{d+a+b} \\]\nEquality condition: a=c;b=d" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "Volume", "geometric inequality", "IMO", "IMO 1967" ], "Problem": "Prove that a tetrahedron with just one edge length greater than $1$ has volume at most $ \\frac{1}{8}.$", "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": "[hide]Assume $ CD>1$ and let $ AB\\equal{}x$. Let $ P,Q,R$ be the feet of perpendicular from $ C$ to $ AB$ and $ \\triangle ABD$ and from $ D$ to $ AB$, respectively.\n\nSuppose $ BP>PA$. We have that $ CP\\equal{}\\sqrt{CB^2\\minus{}BT^2}\\le\\sqrt{1\\minus{}\\frac{x^2}4}$, $ CQ\\le CP\\le\\sqrt{1\\minus{}\\frac{x^2}4}$. We also have $ DQ^2\\le\\sqrt{1\\minus{}\\frac{x^2}4}$. So the volume of the tetrahedron is $ \\frac13\\left(\\frac12\\cdot AB\\cdot DR\\right)CQ\\le\\frac{x}6\\left(1\\minus{}\\frac{x^2}4\\right)$.\n\nWe want to prove that this value is at most $ \\frac18$, which is equivalent to $ (1\\minus{}x)(3\\minus{}x\\minus{}x^2)\\ge0$. This is true because $ 00$ so that $x^2+2\\equiv 0\\pmod{p}$ (by mutiple the inverse of $d$ in modulo $p$.) Now consider all pair $(u,v)$ which $0\\leq u,v\\leq [\\sqrt{p}]=k$ we have $(k+1)^2>p$ pair so have more than $p$ number in form $u+xv$ so we can find two distince pair $(u_i,v_i)$ which $i=1,2$ so that congurent to each other modulo $p$. Which gives us \r\n\\[ u_1-u_2\\equiv x(v_2-x_1)\\pmod{p} \\]\r\nChose $c=|u_1-u_2|$ and $d=|v_1-v_2$ then $c,d\\neq0$ and \r\n\\[ c^2\\equiv x^2d^2\\equiv -2d^2\\pmod{p} \\]\r\nSo $p\\mid c^2+2d^2$. Now note that $0<2+2d^2 p^2-a^2=2b^2\n \n=>(p-a)(p+a)=2b^2\n \nand we know that p^2=1(mod 4)\n => b^2=0(mod 4) =>b=0(mod 2)\nand (p-a, p+a)=2 \n\n=> p-a=2(x^2) and p+a=y^2 where x=1(mod 2) and y=0(mod 2)\nor \np-a=x^2 and p+a=2(y^2) where x=0(mod 2) and y=1(mod2)\n\nin each of these two conditions we have:\n2p=2w^2+z^2 which w=1(mod 2)and z=0(mod 2)(=>z^2= (mod 4))\n\n=>p=w^2+(z^2)/2 and we had z^2=0(mod 4)\nso we could show that p is a member of A...", "Solution_5": "[hide=HINT] note that there exists infinite integers (a,b) with their gcd = p .. now according to Bezout Lemma : there exists integers (x,y) such that p=ax+by now choose x=az^2 and y=2q^2b and try showing that from this p^{2} exists in A " } { "Tag": [], "Problem": "How many units apart is any pair of parallel sides of a regular hexagon with side of 6 units? Express your answer in simplest radical form.", "Solution_1": "[geogebra]d5f5a04b8ef85aef78fb20ba0d8c52a2aa8de571[/geogebra] \r\n\r\nNotice that the triangle is a 30-60-90, which means it has the property that its hypotenuse is $ 2$ times the length of the shorter leg, while the longer leg is $ \\sqrt{3}$ times the shorter leg. The length of the longer leg is thus $ \\frac{6}{2}\\sqrt{3}\\equal{}3\\sqrt{3}$. However, we want [i]twice[/i] the length of the longer leg, because that's the length between the two sides.\r\nThe answer is thus $ 3\\sqrt{3}\\times2\\equal{}\\boxed{6\\sqrt{3}}$" } { "Tag": [ "trigonometry" ], "Problem": "Express $ \\sin {4t} \\plus{} \\sqrt{3} \\cos {4t}$ in the form $ R\\sin {(4t\\plus{} \\alpha)}$. Thanks", "Solution_1": "For $ a\\neq 0,\\ b\\neq 0,$ rewrite the expression in the form of \r\n\r\n$ a\\sin \\theta \\plus{}b\\cos \\theta \\equal{}\\sqrt{a^2\\plus{}b^2}\\left(\\sin \\theta \\cdot \\frac{a}{\\sqrt{a^2\\plus{}b^2}}\\plus{}\\cos \\theta \\cdot \\frac{b}{\\sqrt{a^2\\plus{}b^2}}\\right).$", "Solution_2": "[hide=\"Answer\"]$ \\sin 4t \\plus{} \\sqrt{3} \\cos 4t \\equal{} R \\sin(4t \\plus{} \\alpha)$\n$ \\equal{}> \\sin 4t \\plus{} \\sqrt{3} \\cos 4t \\equal{} R (\\sin 4t \\cos \\alpha \\plus{} \\cos 4t \\sin \\alpha)$\n$ \\equal{}> \\sin 4t \\plus{} \\sqrt{3} \\cos 4t \\equal{} (R \\cos \\alpha) \\sin 4t \\plus{} (R \\sin \\alpha) \\cos 4t$\n\nand so we must have\n\n$ R \\sin \\alpha \\equal{} \\sqrt{3}$\n$ R \\cos \\alpha \\equal{} 1$.\n\nDividing both these equations we get $ \\tan \\alpha \\equal{} \\sqrt{3} \\equal{}> \\alpha \\equal{} \\arctan(\\sqrt{3}) \\equal{} \\frac{\\pi}{3}$. So, $ R \\equal{} \\frac{1}{\\cos \\alpha} \\equal{} \\frac{1}{\\cos(\\pi/3)} \\equal{} 2$. Therefore,\n\n$ \\sin 4t \\plus{} \\sqrt{3} \\cos 4t \\equal{} 2 \\sin\\left(4t \\plus{} \\frac{\\pi}{3}\\right)$.[/hide]", "Solution_3": "@kunny, thanks... I'll be remembering that formula :D\r\n\r\n@carcul, awesome solution, thanks :D" } { "Tag": [ "function", "search", "algebra proposed", "algebra" ], "Problem": "Find all strictly increasing functions f from the naturals to the naturals such that:\r\n\r\nf(f(n))=3n, for all naturals n.\r\n\r\n\r\nBomb", "Solution_1": "Posted before. Try to play around with some small values and try to generalise the things you notice. Look at where $3^k$ and $2 \\cdot 3^k$ go. Or use the search button :p" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c>0$ and $ a^3\\plus{}b^3\\plus{}c^3 \\ge 3$ then:\r\n\r\n$ \\frac{a^6}{a^2\\plus{}b\\plus{}1}\\plus{}\\frac{b^6}{b^2\\plus{}c\\plus{}1}\\plus{}\\frac{c^6}{c^2\\plus{}a\\plus{}1}\\ge 1$", "Solution_1": "By Cauchy-Schwartz in Engel Form,\r\n\r\n$ \\frac{a^6}{a^2\\plus{}b\\plus{}1}\\plus{}\\frac{b^6}{b^2\\plus{}c\\plus{}1}\\plus{}\\frac{c^6}{c^2\\plus{}a\\plus{}1}\\ge \\frac{(a^3\\plus{}b^3\\plus{}c^3)^3}{a^2\\plus{}a\\plus{}1\\plus{}b^2\\plus{}b\\plus{}1\\plus{}c^2\\plus{}c\\plus{}1}$\r\n\r\n$ \\equal{}\\frac{9}{a^2\\plus{}a\\plus{}1\\plus{}b^2\\plus{}b\\plus{}1\\plus{}c^2\\plus{}c\\plus{}1}$\r\n\r\n$ \\equal{}\\frac{3\\plus{}\\sum_{cyc}((\\frac{2}{3}a^3\\plus{}\\frac{1}{3})\\plus{}(\\frac{1}{3}a^3\\plus{}\\frac{2}{3}))}{a^2\\plus{}a\\plus{}1\\plus{}b^2\\plus{}b\\plus{}1\\plus{}c^2\\plus{}c\\plus{}1}$\r\n\r\n$ \\ge\\frac{3\\plus{}\\sum_{cyc}(a^2\\plus{}a)}{a^2\\plus{}a\\plus{}1\\plus{}b^2\\plus{}b\\plus{}1\\plus{}c^2\\plus{}c\\plus{}1}\\equal{}1$", "Solution_2": "almost the same...\r\n[hide]\n\\[ \\sum{\\dfrac{a^6}{a^2 \\plus{} b \\plus{} 1}}\\ge\n\\]\n\n\\[ \\sum{\\dfrac{a^6}{\\dfrac{a^3 \\plus{} a^3 \\plus{} 1}{3} \\plus{} \\dfrac{b^3 \\plus{} 1 \\plus{} 1}{3} \\plus{} 1}} \\equal{}\n\\]\n\n\\[ 3\\sum{\\dfrac{a^6}{2a^3 \\plus{} b^2 \\plus{} 6}}\\ge\n\\]\n\n\\[ \\dfrac{(a^3 \\plus{} b^3 \\plus{} c^3)^2}{(a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6)}\\ge\n\\]\n\n\\[ \\dfrac{3(a^3 \\plus{} b^3c^3)}{a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6}\\ge\n\\]\n\n\\[ \\dfrac{a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 6}{a^3 \\plus{} b^3 \\plus{} c^2 \\plus{} 6} \\equal{}\n\\]\n\n\\[ 1\n\\]\n[/hide]\r\n:)", "Solution_3": "My solution uses [b] Cauchy-Schwartz [/b] and the fact that: $ a^3\\plus{}1 \\ge a^2 \\plus{} a$." } { "Tag": [ "induction" ], "Problem": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1292501#1292501\r\n\r\nThis is borderline on spam, but... :wink: \r\n\r\nProblem:\r\nProve the AM-GM inequality. (No intervention from other states, please).", "Solution_1": "Yes and AM>GM", "Solution_2": "[quote=\"dragon96\"]I from another state (sorry). Anyways, mewto was lying once again. AM-GM stands for Arithmetic mean-Geometric mean.[/quote]\r\n\r\n :rotfl: :rotfl: :rotfl: \r\n\r\nHe was pretending to be stupid...\r\n\r\nLet X=The average MPG of the cars and let Y=the price of gas (in dollars)\r\n\r\nBY the definitions of Asian and General Motors, $ AM \\equal{} \\frac {x \\plus{} y}{2}$ and $ GM \\equal{} \\sqrt {xy}$\r\n\r\nBy the Automobile Theorem,\r\n\r\n$ (x \\minus{} y)^2\\geq{0}$\r\n\r\n\r\n$ x^2 \\minus{} 2xy \\plus{} y^2\\geq{0}$\r\n\r\n$ x^2 \\plus{} 2xy \\plus{} y^2\\geq{4xy}$\r\n\r\n$ (x \\plus{} y)^2\\geq{4xy}$\r\n\r\n$ (x \\plus{} y)\\geq{2\\sqrt {xy}}$\r\n\r\n$ \\frac {(x \\plus{} y)}{2}\\geq{\\sqrt {xy}}$\r\n\r\n$ AM\\geq{GM}$\r\n\r\nEquality holds true iff MPG=1 and Price of Gas= $ \\$1$", "Solution_3": "Correct! Now, prove it in general. What if there are several more dimensions, such as z=the number of Asians allowed in the car?", "Solution_4": "I would use hybrid induction to prove for $ 2^k$ asians and then prove it for $ k\\minus{}1$ asians.", "Solution_5": "[quote=\"tinytim\"]I would use hybrid induction to prove for $ 2^k$ asians and then prove it for $ k \\minus{} 1$ asians.[/quote]\r\n\r\nYup.\r\nIf it is true for $ a_1,a_2,\\ldots,a_{n \\plus{} 1}$, then we have\r\n\r\n$ \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_{n \\plus{} 1}}{n \\plus{} 1}\\geq\\sqrt [n \\plus{} 1]{a_1a_2\\cdots a_{n \\plus{} 1}}$.\r\n\r\nSubstituting $ a_{n \\plus{} 1} \\equal{} \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n}n$, we get\r\n\r\n$ \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n}n\\geq\\sqrt [n \\plus{} 1]{a_1a_2\\cdots a_{n \\plus{} 1}}$\r\n\r\n$ \\frac {(a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^{n \\plus{} 1}}{n^{n \\plus{} 1}}\\geq a_1a_2\\cdots a_n\\cdot a_{n \\plus{} 1}$\r\n\r\n$ \\frac {(a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n)^n}{n^n}\\geq a_1a_2\\cdots a_n$\r\n\r\nTaking the $ n$th root, we have\r\n\r\n$ \\frac {a_1 \\plus{} a_2 \\plus{} \\cdots \\plus{} a_n}n\\geq\\sqrt [n]{a_1a_2\\cdots a_n}$.\r\n\r\nBlah.\r\n\r\nNew Problem:\r\nProve the QM-AM-GM-HM Inequality. :wink:", "Solution_6": "Oh dang did this thread get cleaned out? MOD ALERT! (siren) MOD ALEERT!", "Solution_7": "Hmm, kinky." } { "Tag": [ "geometry", "trigonometry", "calculus", "geometry proposed" ], "Problem": "Let $ ABCD$ be a quadrilateral which has an incircle centered at $ O$. Prove that\r\n\\[ OA\\cdot OC\\plus{}OB\\cdot OD\\equal{}\\sqrt{AB\\cdot BC\\cdot CD\\cdot DA}\\]", "Solution_1": "Beautiful identity!\r\n\r\nI have just uploaded a note, \"Circumscribed quadrilaterals revisited\", on my website ( http://www.cip.ifi.lmu.de/~grinberg/ ), where this identity is Theorem 10 and a synthetic proof is given. Hope you will like it (although you will probably find most of the stuff quite well-known).\r\n\r\n Darij", "Solution_2": "[b]Very nice identity ![/b] Thanks,[i] Mekrazywong[/i] !\r\nIndeed, this problem is totally trivial by trigonometry.\r\nIt is equivalently with the following conditioned identity:\r\n\r\n$x+y+z+t=\\pi\\Longrightarrow\\cos x\\cdot \\cos z +\\cos y\\cdot \\cos t =\\sin (x+y)\\cdot \\sin (y+z)\\ \\ (1)$.\r\n\r\nReally it isn't worth proving syntetically, at the most\r\na syntetical prove for the conditioned identity (1) !\r\n\r\n[u]Remark.[/u] In general,\r\n$\\cos x\\cdot \\cos z+\\cos y\\cdot \\cos t =\\sin (x+y)\\cdot\\sin (y+z)\\Longleftrightarrow$$\\cos \\frac{x+y+z+t}{2}\\cdot \\cos y\\cdot \\cos \\frac{x+y+z-t}{2}=0$.\r\n[i]Mekrazywong[/i] and [i]Darij[/i], can give you a geometrical interpretation\r\n(for a quadrilateral, without circumscribed) of this equivalence ?", "Solution_3": "First of all, I would like to clarify here this result was [i]not[/i] due to me. Actually this is a problem from Chinese TST 2003, proposed by a Chinese professor. By the way, there is a really simple solution via spiral similarity.", "Solution_4": "Denote by $ a\\equal{}AB$, $ b\\equal{}BC$, $ c\\equal{}CD$, $ d\\equal{}DA$ the lengths of the sides of $ ABCD$.\r\nLet $ r$ be the ray of the inscribed circle. Then $ OA \\equal{}\\frac{r}{\\sin\\frac{A}{2}},OB \\equal{}\\frac{r}{\\sin\\frac{B}{2}},OC \\equal{}\\frac{r}{\\sin\\frac{C}{2}},OD \\equal{}\\frac{r}{\\sin\\frac{C}{2}}$.\r\nLet $ M$ be the point where $ AB$ touches the circle. Then $ AM \\equal{} rctg\\frac{A}{2},\\ MB \\equal{} rctg\\frac{B}{2}\\Rightarrow a \\equal{} AM\\plus{}MB \\equal{} r\\cdot\\frac{\\sin\\frac{A\\plus{}B}{2}}{\\sin\\frac{A}{2}\\sin\\frac{B}{2}}$,\r\n$ b \\equal{} r\\cdot\\frac{\\sin\\frac{B\\plus{}C}{2}}{\\sin\\frac{B}{2}\\sin\\frac{C}{2}},c \\equal{} r\\cdot\\frac{\\sin\\frac{C\\plus{}D}{2}}{\\sin\\frac{C}{2}\\sin\\frac{D}{2}},d \\equal{} r\\cdot\\frac{\\sin\\frac{D\\plus{}A}{2}}{\\sin\\frac{D}{2}\\sin\\frac{A}{2}}$.\r\nReplacing $ OA,OB,OC,OD,a,b,c,d$, making some calculus and using the equalities $ \\sin\\frac{A\\plus{}B}{2}\\equal{}\\sin\\frac{C\\plus{}D}{2}, sin\\frac{B\\plus{}C}{2}\\equal{}\\sin\\frac{A\\plus{}D}{2}$, the relation becomes\r\n$ \\sin\\frac{A}{2}\\sin\\frac{C}{2}\\plus{}\\sin\\frac{B}{2}\\sin\\frac{D}{2}\\equal{}\\sin\\frac{A\\plus{}B}{2}\\sin\\frac{B\\plus{}C}{2}$.\r\nBut $ \\sin\\frac{A}{2}\\sin\\frac{C}{2}\\plus{}\\sin\\frac{B}{2}\\sin\\frac{D}{2}\\equal{}\\frac{1}{2}\\left[\\cos\\frac{A\\minus{}C}{2}\\minus{}\\cos\\frac{A\\plus{}C}{2}\\plus{}\\cos\\frac{B\\minus{}D}{2}\\minus{}\\cos\\frac{B\\plus{}D}{2}\\right] \\equal{}$\r\n$ \\equal{}\\frac{1}{2}\\left(\\cos\\frac{A\\minus{}C}{2}\\plus{}\\cos\\frac{B\\minus{}D}{2}\\right) \\equal{}\\sin\\frac{A\\plus{}B}{2}\\sin\\frac{B\\plus{}C}{2}$.", "Solution_5": "use this lemma :\r\nwe have that $ AB.BC \\equal{} OB^{2} \\plus{} \\frac {OA.OB.OC}{OD}$", "Solution_6": "My solution:\n\nLet $ E $ be a point satisfy $ \\triangle CDE \\sim \\triangle BAO $ .\n\nSince $ C, D, E, O $ are concyclic ,\nso from Ptolemy theorem we get $ DE \\cdot OC+CE\\cdot OD=CD\\cdot OE $ . ... $ (1) $\n\nSince $ \\frac{DE}{OA}=\\frac{CE}{OB}=\\frac{CD}{AB} $ ,\n\nso from $ (1) $ we get $ OA\\cdot OC+OB\\cdot OD=AB\\cdot OE $ . ... $ (2) $ \n\nSince $ \\angle EOD=\\angle ECD=\\angle OBA=\\angle CBO , \\angle DEO=\\angle DCO=\\angle OCB $ ,\n\nso we get $ \\triangle DOE \\sim \\triangle OBC $ and $ \\frac{OE}{BC}=\\frac{OD}{OB} $ . ... $ (3) $\n\nSimilarly, we can prove $ \\triangle OCE \\sim \\triangle AOD $ and $ \\frac{OE}{DA}=\\frac{CE}{OD} $ . ... $ (4) $\n\nFrom $ (3), (4) $ we get $ \\frac{OE^2}{BC\\cdot DA}=\\frac{CE}{OB}=\\frac{CD}{AB} $ . ... $ (5) $\n\nFrom $ (2), (5) $ we get $ (OA\\cdot OC+OB\\cdot OD)^2=AB^2\\cdot \\frac{BC\\cdot CD\\cdot DA}{AB}=AB\\cdot BC\\cdot CD\\cdot DA $ .\n\nQ.E.D", "Solution_7": "Another proof :", "Solution_8": "Dear Mathlinkers,\n\nhttp://jl.ayme.pagesperso-orange.fr/Docs/Quadrilatere%20circonscriptible.pdf p. 7...\n\nSincerely \nJean-Louis \t\n", "Solution_9": "Just solving this for no reason as I just came for this page for a different problem but saw this and wanted to gain some fake satisfaction before sleep of solving an [hide]easy[/hide] chinese tst.\nJust consider the point of tangencies as $KEK'W$.Let $F$ be $KK' \\cap EW$\nThen consider a projective transformation that fixes the incircle and sends $F$ to the center of the incircle.\nNow ,we just have to prove the problem for a rhombus which is a mere Pythagorean .", "Solution_10": "[b]Solved with Max Lu. The work here is almost entirely copy-pasted from our work on USAMO 2004/6 :maybe: . [/b]\n\nLet the incircle have radius $r$, and let the lengths of the tangents from $A,B,C,D$ be $a,b,c,d$. Then, note that \n\\[(w+ri)(x+ri)(y+ri)(z+ri) \\text{is a negative real number}\\]\nsince the argument of it is $\\frac{A}{2}+\\frac{B}{2}+\\frac{C}{2}+\\frac{D}{2}=180$. Thus, we have $r\\cdot \\sum_{cyc} wxy - r^3 \\sum_{cyc} a = 0\\Longrightarrow r^2 = \\frac{\\sum_{cyc} wxy}{\\sum_{cyc} w}$.\n\nNow, note that this gives\n\\[a^2+r^2 = a^2 + \\frac{\\sum abc}{\\sum a} = \\frac{a^2\\sum a + \\sum abc}{\\sum a} = \\frac{(a+b)(a+c)(a+d)}{\\sum a}\\]\n\nUsing this, we get that \n\\[OA\\cdot OC + OB\\cdot OD = \\sqrt{\\frac{(a+b)(a+c)(a+d)}{a+b+c+d}} \\cdot \\sqrt{\\frac{(a+c)(b+c)(c+d)}{a+b+c+d}}+ \\sqrt{\\frac{(a+b)(b+c)(b+d)}{a+b+c+d}} \\cdot \\sqrt{\\frac{(a+d)(b+d)(c+d)}{a+b+c+d}} \\]\\[= \\frac{((a+c)+(b+d)) \\sqrt{(a+b)(a+d)(b+c)(c+d)}}{a+b+c+d} = \\sqrt{AB\\cdot BC\\cdot CD\\cdot DA}\\]\nand we're done!", "Solution_11": "Let $X,Y,Z,W$ be touch points of the incircle of $ABCD$ and sides $DA,AB,BC,CA$, respectively. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$ of an inversion around $(XYZW)$, respectively. Note that $A^*,B^*,C^*,D^*$ are the midpoints of $XY,YZ,ZW,WX$, respectively.\n[asy]\nimport olympiad;import geometry;\nsize(8cm);defaultpen(fontsize(10pt));\n\npair A,B,C,D,X,Y,Z,W,O,A0,B0,C0,D0;\nX=dir(110);Y=dir(205);Z=dir(280);W=dir(10);O=(0,0);\nA=intersectionpoint(perpendicular(X, line(X,O) ),perpendicular(Y, line(Y,O) ));\nB=intersectionpoint(perpendicular(Y, line(Y,O) ),perpendicular(Z, line(Z,O) ));\nC=intersectionpoint(perpendicular(Z, line(Z,O) ),perpendicular(W, line(W,O) ));\nD=intersectionpoint(perpendicular(W, line(W,O) ),perpendicular(X, line(X,O) ));\nA0=midpoint(X--Y);B0=midpoint(Y--Z);C0=midpoint(Z--W);D0=midpoint(W--X);\n\ndraw(circumcircle(X,Y,Z),royalblue);draw(A--B--C--D--cycle,heavyred+1);draw(A0--B0--C0--D0--cycle, fuchsia);draw(X--Y--Z--W--cycle, red+0.5);\ndraw(A--O--B,palered+1);draw(C--O--D,palered+1);\n\ndot(\"$O$\",O,dir(90));\ndot(\"$X$\",X,dir(X));\ndot(\"$Y$\",Y,dir(Y));\ndot(\"$Z$\",Z,dir(Z));\ndot(\"$W$\",W,dir(W));\ndot(\"$A$\",A,dir(A));\ndot(\"$B$\",B,dir(B));\ndot(\"$C$\",C,dir(C));\ndot(\"$D$\",D,dir(D));\ndot(\"$A^*$\",A0,dir(A0));\ndot(\"$B^*$\",B0,dir(B0));\ndot(\"$C^*$\",C0,dir(C0));\ndot(\"$D^*$\",D0,dir(D0));\n[/asy]\n\\begin{align*}\nOA\\cdot OC+OB\\cdot OD&=\\sqrt{AB\\cdot BC\\cdot CD\\cdot DA}\\Longleftrightarrow\\\\\n\\frac{r^2}{OA^*}\\cdot\\frac{r^2}{OC^*}+\\frac{r^2}{OB^*}\\cdot \\frac{r^2}{OD^*}&= \\sqrt{\\frac{r^2\\cdot A^*B^*}{OA^*\\cdot OB^*}\\cdot \\frac{r^2\\cdot B^*C^*}{OB^*\\cdot OC^*}\\cdot \\frac{r^2\\cdot C^*D^*}{OC^*\\cdot OD^*}\\cdot \\frac{r^2\\cdot D^*A^*}{OD^*\\cdot OA^*}}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\sqrt{A^*B^*\\cdot B^*C^*\\cdot C^*D^*\\cdot D^*A^*}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\frac{XZ\\cdot YW}{4}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\frac{XY\\cdot ZW+YZ\\cdot WX}{4}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=XA^*\\cdot C^*Z+B^*Y\\cdot D^*W\\Longleftrightarrow\\\\\nr^2(\\sin{\\angle YXO}\\cdot \\sin{\\angle WZO}+\\sin{\\angle ZYO}\\cdot \\sin{\\angle XWO})&=r^2(\\cos{\\angle YXO}\\cdot \\cos{\\angle WZO}+\\cos{\\angle ZYO}\\cdot \\cos{\\angle XWO})\\Longleftrightarrow\\\\\n\\cos{(\\angle YXO+\\angle WZO)}&=-\\cos{(\\angle ZYO+\\angle XWO)},\n\\end{align*}which is true as $\\angle YXO+\\angle WZO=\\frac{2\\pi-\\angle XOY-\\angle ZOW}{2}=\\frac{\\angle YOZ+\\angle WOX}{2}=\\pi-\\angle ZYO-\\angle XWO$. $\\blacksquare$\n", "Solution_12": "[quote=rafaello]Let $X,Y,Z,W$ be touch points of the incircle of $ABCD$ and sides $DA,AB,BC,CA$, respectively. Let $A^*,B^*,C^*,D^*$ denote the inverses of $A,B,C,D$ of an inversion around $(XYZW)$, respectively. Note that $A^*,B^*,C^*,D^*$ are the midpoints of $XY,YZ,ZW,WX$, respectively.\n[asy]\nimport olympiad;import geometry;\nsize(8cm);defaultpen(fontsize(10pt));\n\npair A,B,C,D,X,Y,Z,W,O,A0,B0,C0,D0;\nX=dir(110);Y=dir(205);Z=dir(280);W=dir(10);O=(0,0);\nA=intersectionpoint(perpendicular(X, line(X,O) ),perpendicular(Y, line(Y,O) ));\nB=intersectionpoint(perpendicular(Y, line(Y,O) ),perpendicular(Z, line(Z,O) ));\nC=intersectionpoint(perpendicular(Z, line(Z,O) ),perpendicular(W, line(W,O) ));\nD=intersectionpoint(perpendicular(W, line(W,O) ),perpendicular(X, line(X,O) ));\nA0=midpoint(X--Y);B0=midpoint(Y--Z);C0=midpoint(Z--W);D0=midpoint(W--X);\n\ndraw(circumcircle(X,Y,Z),royalblue);draw(A--B--C--D--cycle,heavyred+1);draw(A0--B0--C0--D0--cycle, fuchsia);draw(X--Y--Z--W--cycle, red+0.5);\ndraw(A--O--B,palered+1);draw(C--O--D,palered+1);\n\ndot(\"$O$\",O,dir(90));\ndot(\"$X$\",X,dir(X));\ndot(\"$Y$\",Y,dir(Y));\ndot(\"$Z$\",Z,dir(Z));\ndot(\"$W$\",W,dir(W));\ndot(\"$A$\",A,dir(A));\ndot(\"$B$\",B,dir(B));\ndot(\"$C$\",C,dir(C));\ndot(\"$D$\",D,dir(D));\ndot(\"$A^*$\",A0,dir(A0));\ndot(\"$B^*$\",B0,dir(B0));\ndot(\"$C^*$\",C0,dir(C0));\ndot(\"$D^*$\",D0,dir(D0));\n[/asy]\n\\begin{align*}\nOA\\cdot OC+OB\\cdot OD&=\\sqrt{AB\\cdot BC\\cdot CD\\cdot DA}\\Longleftrightarrow\\\\\n\\frac{r^2}{OA^*}\\cdot\\frac{r^2}{OC^*}+\\frac{r^2}{OB^*}\\cdot \\frac{r^2}{OD^*}&= \\sqrt{\\frac{r^2\\cdot A^*B^*}{OA^*\\cdot OB^*}\\cdot \\frac{r^2\\cdot B^*C^*}{OB^*\\cdot OC^*}\\cdot \\frac{r^2\\cdot C^*D^*}{OC^*\\cdot OD^*}\\cdot \\frac{r^2\\cdot D^*A^*}{OD^*\\cdot OA^*}}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\sqrt{A^*B^*\\cdot B^*C^*\\cdot C^*D^*\\cdot D^*A^*}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\frac{XZ\\cdot YW}{4}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=\\frac{XY\\cdot ZW+YZ\\cdot WX}{4}\\Longleftrightarrow\\\\\nOA^*\\cdot OC^*+OB^*\\cdot OD^*&=XA^*\\cdot C^*Z+B^*Y\\cdot D^*W\\Longleftrightarrow\\\\\nr^2(\\sin{\\angle YXO}\\cdot \\sin{\\angle WZO}+\\sin{\\angle ZYO}\\cdot \\sin{\\angle XWO})&=r^2(\\cos{\\angle YXO}\\cdot \\cos{\\angle WZO}+\\cos{\\angle ZYO}\\cdot \\cos{\\angle XWO})\\Longleftrightarrow\\\\\n\\cos{(\\angle YXO+\\angle WZO)}&=-\\cos{(\\angle ZYO+\\angle XWO)},\n\\end{align*}which is true as $\\angle YXO+\\angle WZO=\\frac{2\\pi-\\angle XOY-\\angle ZOW}{2}=\\frac{\\angle YOZ+\\angle WOX}{2}=\\pi-\\angle ZYO-\\angle XWO$. $\\blacksquare$[/quote]\n\nMOST BEAUTIFUL IDEE!(AND DIAGRAME!)" } { "Tag": [ "geometry" ], "Problem": "What is the force exerted by the atmosphere on one side of a two feet by three feet board? (Answer choices are given in lb)", "Solution_1": "the pressure times the area, so atmospheric pressure times 6..\r\nsilly imperial units!" } { "Tag": [ "geometry", "algebra", "polynomial", "Vieta", "area of a triangle", "Heron\\u0027s formula", "algebra unsolved" ], "Problem": "The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.", "Solution_1": "hayyyyyyy,,no one :? it's very beatiful ;)", "Solution_2": "ehsan2004\r\n\r\nplease!\r\ncould you send the solution??\r\nThis was also in greek IMO Tst 2005\r\n(problem 1)", "Solution_3": "Suppose that $a,b,c$ are the roots of the polynomial\r\n\\[f(x)=x^3-mx^2+nx-p,\r\n\\]\r\nwith $m,n,p\\in \\mathbb{Q}$ and let $S$ denote the triangle's area.\r\nHeron's formula and standard manipulations with Viete's formulas yield\r\n\\[S^2=-m^4+4m^2n-8mp,\r\n\\]\r\nhence $S=\\sqrt{A}$, for some $A\\in \\mathbb{Q}.$\r\nWe have $h_a=\\frac{2S}{a}$, hence $a=\\frac{2\\sqrt{A}}{h_a}$. Write that $f(a)=0$, group the terms containing $\\sqrt{A}$ and square again. After clearing the denominators, we find that $h_a$ is the root of a 6th degree rational polynomial.", "Solution_4": "I have a good solution.\r\nBecause the coefficients are rational we write the equation as\r\n$ x^3-Ax^2+Bx-C=0 $\r\nNow let a,b,c the lengths of the triangle.\r\na+b+c=2t=A\r\nab+bc+ca=B\r\nabc=C\r\nWe get all these from Vieta.\r\nAnd from Iron we have that $ E^2=t(t-a)(t-b)(t-c)=t(t^3+At^2+Bt-C)$ which is rational.\r\nAlso we have that \r\n$\\upsilon_a=\\frac{2E}{a}$\r\n$\\upsilon_b=\\frac{2E}{b}$\r\n$\\upsilon_c=\\frac{2E}{c}$\r\nare roots of $ (x^2-\\frac{4E^2}{a^2})(x^2-\\frac{4E^2}{b^2})(x^2-\\frac{4E^2}{c^2})=0 $\r\nThis equation is 6th degree with rational coefficients.\r\nThere are rational because $ \\frac{4E^2}{b^2}+\\frac{4E^2}{c^2}+\\frac{4E^2}{a^2}=\\frac{4E^2}{C^2}(B^2-2AC) $ is rational\r\nAlso ${ \\frac{16E^4}{b^2a^2}+\\frac{16E^4}{b^2c^2}+\\frac{16E^4}{c^2a^2}=\r\n\\frac{16(E^2)^2}{C^2}(A^2-2B)}$\r\nThe last is obviously rational so all are rational and we are done.", "Solution_5": "[hide=\"Solution\"]\nIf $a, b, c$ are the sides, then the polynomial $(x-a)(x-b)(x-c)$ has rational coefficients. Then the altitudes are $\\dfrac{2\\sqrt{s(s-a)(s-b)(s-c)}}{a, b, c}$. By Vieta $s$ is rational, so $\\sqrt{s(s-a)(s-b)(s-c)}$ is rational too. For convenience, call this $q$.\n\nThen $\\Pi (x-\\dfrac{2\\sqrt q}{a})(x+\\dfrac{2\\sqrt q}}{a})=\\Pi (x^2-\\dfrac{4q}{a^2})$, and this clearly has rational coefficients.\n[/hide]", "Solution_6": "This problem is not as daunting as it may seem. \n\nSuppose $r_1, r_2, r_3$ are roots of some cubic $ax^3 + bx^2 + cx + d$. Then we have that by Vieta, $r_1 + r_2 + r_3 = \\frac{-b}{a} \\in \\mathbb{Q}$, and $r_1r_2 + r_2r_3 + r_3r_1 = \\frac{c}{a} \\in \\mathbb{Q}$, and finally $r_1r_2r_3 = -d/a \\in \\mathbb{Q}$. Hence each $r_i, 1 \\leq i \\leq 3$ is rational also. \n\nNote that the altitudes are given by $2[\\text{area}]/r_i$. Now the area is simply $\\sqrt{s(s - r_1)(s - r_2)(s - r_3)}$, where $s$ is the semiperimeter (this is due to Heron's formula). Now for simplicity, let $X = \\sqrt{s(s - r_1)(s - r_2)(s - r_3)}$. Consider the polynomial \\[\\left(x^2 - \\frac{4X^2}{r_1^2}\\right) \\left(x^2 - \\frac{4X^2}{r_2^2} \\right) \\left(x^2 - \\frac{4X^2}{r_3^2} \\right). \\] Here each $\\frac{4X^2}{r_i^2}$ is clearly rational. Hence this new polynomial will have rational coefficients, and also have degree $6$. $\\blacksquare$" } { "Tag": [ "Divisor Functions" ], "Problem": "Determine all positive integers $n$ such that $n={d(n)}^2$.", "Solution_1": "Obviously so $ n$ is odd (a square has an odd number of divisors). Let $ n\\equal{}p_{1}^{2k_{1}}\\cdots p_{m}^{2k_{m}}$, then we have to solve \\[ (2k_{1}\\plus{}1)\\cdots (2k_{m}\\plus{}1)\\equal{}p_{1}^{k_{1}}\\cdots p_{m}^{k_{m}},\\]\r\nand since $ 2x\\plus{}1<3^{x}$ for all $ x>1$ we have $ p_{1}\\le 3$ and $ k_{1}\\le 1$, while all other primes cannot appear. So $ n\\in\\{1,9\\}$.", "Solution_2": "[hide=J12]Note that $n$ is a perfect square. So we know that $n=(p_1^{e_1}\\cdot p_2^{e_2}...p_k^{e_k})^2$. We know that $\\sqrt{n}=p_1^{e_1}\\cdot p_2^{e_2}...p_k^{e_k}=d(n)=(2e_1+1)(2e_2+1)...(2e_k+1)$ so it follows that since $3^x>2x+1$ for $x>1$ that if $k_1>1$ we cannot have $p_1\\ge3$. If $k_1=1$ then we have $p_1=3$ which is a solution. So $n=9$ is one solution. If $k_1>1$ and $p_1=2$ this won't work because we know that $d(n)$ is odd implying that $n$ is odd. The only case we need to check now is $n=1$ which is indeed a solution. So $n=1,9$ are our solutions. [/hide]" } { "Tag": [ "linear algebra", "matrix", "MATHCOUNTS", "floor function", "function", "vector", "trigonometry" ], "Problem": "What does having numbers stacked together in tall parentheses denote? :blush: [/code]", "Solution_1": "You mean like $\\left|\\begin{matrix}1&2\\\\3&4\\end{matrix}\\right|$?\r\n\r\nThose are called matrices, and are above Mathcounts level.", "Solution_2": "I think he means combinations. \r\n\r\nLook it up in the AoPS Wiki. It's \\binom{top number}{bottom number}, in dollar signs, of course.", "Solution_3": "you mean like $\\binom{a}{b}$?\r\n\r\nit just means the number of ways a [i]b[/i] number of things can be chosen from an [i]a[/i] number of things.\r\n\r\nits' just $\\frac{a(a-1)(a-2)...(a-b+1)}{b!}$", "Solution_4": "Or more like $\\binom{n}{r}=\\frac{n!}{(n-r)!r!}$", "Solution_5": "$\\binom{n}{r}$ is equivalent to nCr.", "Solution_6": "$\\lfloor$ $\\rfloor$ What do those symbols mean?", "Solution_7": "It is the floor function/greatest integer function. $\\lfloor n \\rfloor$ denotes the greatest integer less than or equal to $n$.\r\n\r\nFor example:\r\n\r\n$\\lfloor 3.7 \\rfloor = 3$\r\n$\\lfloor-12.32 \\rfloor =-13.$\r\n\r\n[url]http://en.wikipedia.org/wiki/Greatest_integer_function[/url]", "Solution_8": "[quote=\"i_like_pie\"]You mean like $\\left|\\begin{matrix}1&2\\\\3&4\\end{matrix}\\right|$?\n\nThose are called matrices, and are above Mathcounts level.[/quote]\r\n\r\nHOW ARE MATRICES ABOVE MATHCOUNTS LEVEL? I did matrices this year, and I'm in Algebra. How matrces are too advanced for a forum that makes my head hurt day after day is beyond me.", "Solution_9": "I think you mean vectors.\r\n\r\nMaybe you are allowed to just use basic matrix and vector stuff in MC forum(like addition, and that sort of stuff)\r\n\r\nBTW i like pie, that i think means the determinant of that matrix \r\n\r\n[hide]\nsince i am really dumb take the negation of every statement but this one and that is probably the truth[/hide]", "Solution_10": "[quote=\"rd5493\"][quote=\"i_like_pie\"]You mean like $\\left|\\begin{matrix}1&2\\\\3&4\\end{matrix}\\right|$?\n\nThose are called matrices, and are above Mathcounts level.[/quote]\n\nHOW ARE MATRICES ABOVE MATHCOUNTS LEVEL? I did matrices this year, and I'm in Algebra. How matrces are too advanced for a forum that makes my head hurt day after day is beyond me.[/quote]\r\n\r\nI think he means using matrices to transform vectors with trig and that kind of higher stuff. Adding and subtacting them isn't above Mathcounts, buts it's useless for competitions.", "Solution_11": "Matrices dealing with linear algebra are above MC level. I'd be very angry if they asked a question about them." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Show (synthetically) that distance between vertices X and Y of two triangles \r\n XBC and YBC is greater than distance between their incenters.\r\n\r\n\r\n\r\n Lt. Pestich", "Solution_1": "At least this has been discussed here: http://www.mathlinks.ro/Forum/viewtopic.php?t=15277 .\r\n\r\n darij" } { "Tag": [], "Problem": "Find the coefficient of $ x^{3}y^{3}z^{2}$ in the expansion of $ (x\\plus{}y\\plus{}z)^{8}$. \r\n \r\n\r\nI know you are supposed to use the Binomial Thereom but when I tried so i got the wrong answer.please :help: me", "Solution_1": "[hide=\"Solution\"]We can see that, before expanding, the coefficient of the variables are $ 1$, so we don't need to worry about that ($ 1^3$ and $ 1^2$ are all $ 1$, which changes nothing when multiplying). First, we want to find out how many ways we can choose three $ x$ terms. We can get that by doing $ \\binom{8}{3}\\equal{}56$. Then, we choose our three $ y$ terms. We can do that by $ \\binom{8\\minus{}3}{3}\\equal{}\\binom{5}{3}\\equal{}10$. We did $ 8\\minus{}3$ because after we chose our $ x$ terms, we are left with only $ 5$ terms. Therefore, our coefficient is $ 56\\cdot10\\equal{}\\boxed{560}$[/hide]" } { "Tag": [], "Problem": "I probably forgot a few... :lol: \r\n\r\nI know, for many of these there isn't a lot of recordings out there to listen to (like Ivan Galamian and Pinchas Zukerman) but I thought it wouldn't do them justice not to include them.", "Solution_1": "Jascha Heifetz is the KING OF THEM ALL. His technique is flawless and I feel so ashamed watching him play. :wallbash_red: \r\n\r\nIvan Galamian was pretty good too. He my teacher was one of his students. :D \r\n\r\nActually, EVERYONE up there have made it into history books.", "Solution_2": "Yeah, I didn't vote for anyone because they're all beast. \r\n\r\nFunny...Ivan Galamian was once my teacher's teacher too!", "Solution_3": "Scary... who's your teacher? Are you seriously good at violin? :huh:", "Solution_4": "I'll just say my teacher is staff at the North Carolina School of the Arts. And I'm not really that great at violin...not \"seriously good\" at least.", "Solution_5": "DUDE I'M GOING TO THE NORTH CAROLINA SCHOOL OF THE ARTS ON AUGUST 25TH. :o \r\n\r\nIf his name is Kevin Lawrence (my teacher's friend and I think he's a student if Galamian too), then I've had a lesson with him and I think he's going to be my teacher when I go learn there. \r\n\r\nThis is extra extra extra scary... :o", "Solution_6": "Kevin Lawrence? I think he was a faculty at the Green Mountain Chamber Music Festival", "Solution_7": "I like Fischer-Dieskau best", "Solution_8": "Yay for Heifetz. According to my violin teacher, you need good ears to appreciate Heifetz' music. His sound is unique, and is recognizable from anywhere.", "Solution_9": "[quote=\"Quickster94\"]Kevin Lawrence? I think he was a faculty at the Green Mountain Chamber Music Festival[/quote]\r\n\r\nYeah I think so. (He tried to get me to go to the Green Mountain Music camp thing but I was like, \"nah\" :D ) In the school months though he works at the North Carolina School of the Arts.", "Solution_10": "oh, ok, my piano teacher was is a faculty at the festival, the only one actually from vermont i think :maybe:", "Solution_11": "What happened to Yehudi Menuhin?", "Solution_12": "Whoops. Sorry. :blush: \r\n\r\nI wish to keep my violin teacher anonymous. It would probably do him/her shame to make it announced that I study under that person. :rotfl: :rotfl:", "Solution_13": "[quote=\"qwertythecucumber\"]Whoops. Sorry. :blush: \n\nI wish to keep my violin teacher anonymous. It would probably do him/her shame to make it announced that I study under that person. :rotfl: :rotfl:[/quote]\r\n\r\nHaha wow you're really self criticizing of yourself.\r\n\r\nHa for a second I almost forgot who Yehudi Menuhin was.\r\n\r\nAs I said thought, there's probably too many legendary violinists to list them all. :lol:", "Solution_14": "Heifetz + Oistrakh for me", "Solution_15": "Jascha Heifetz is my fave. \r\n\r\nIsaac Stern is good too... Carnegie Hall is here only because of his generous support. :lol:", "Solution_16": "dude, what? Leaving Oistrakh off a list of \"favorite old violinists\" is like leaving God off of a list of \"most influential beings.\"", "Solution_17": "Unless you don't believe in God :)", "Solution_18": "No. Not \"unless you don't believe in God.\"", "Solution_19": "Omigosh! I can't believe I forgot Oistrakh! :wallbash_red: :wallbash_red: :wallbash_red:", "Solution_20": "lol I don't actually know anything about violin, just sticking my atheist comments in where i can.", "Solution_21": "I don't believe that one person's atheistic views can change the influence of God on the rest of the world. (but that's just me)", "Solution_22": "I don't either, but I just like to stick those kind of comments in to distinguish myself as an atheist :O", "Solution_23": "What about Milstein? But I agree, there's so many brilliant violinists it would be impossible to name all of them.", "Solution_24": "I like Milstein too!!! I definately think he should have been on this, but since he wasn't I voted for Heifetz.", "Solution_25": "If you're going to mention oistrakh then I think that you should probably include kogan. Obviously, both of these men were exceptional at what they did so its really hard to tell who's best. It's interesting to note how the soviet background affected both of these musicians. Overall, I believe that Kogan was overshadowed by Oistrakh due to the politics of the time.\r\n\r\nListen to this:\r\n\r\n[youtube]xsFJS4I_05E[/youtube]", "Solution_26": "What about Rabin?", "Solution_27": "^ i never really liked his playing... he was very talented, but never one of my favorites." } { "Tag": [], "Problem": "Solve for B given V = (1/3) bh", "Solution_1": "[quote=\"Mark9182\"]Solve for B given V = (1/3) bh[/quote]\r\n\r\n[hide] \n$V = \\frac{1}{3}bh$\n\n$\\frac{V}{h}= \\frac{1}{3}b$\n\n$\\frac{3V}{h}= b$\n[/hide]\r\n\r\nis that what you mean by solving for B?", "Solution_2": "Thanks for your help", "Solution_3": "Yes that is exactly what I thought the answer would be" } { "Tag": [ "percent" ], "Problem": "During a company's first full year in business, the price of oil increased $ 50\\%$. In the second year of business, the price of oil went down $ 50\\%$ from its price at the beginning of that year. Its price at the end of the second year is what percent of its price at the beginning of the first year?", "Solution_1": "$ 1 \\times 1.5 \\times 0.5\\equal{}0.75 \\implies \\boxed{75 \\%}$." } { "Tag": [], "Problem": "I stumbled on this nice problem in AoPS Volume 1, and solved it. I think it would be good on countdown. Here it is:\r\n\r\n$ \\text{Find }x\\text{ if }2^{16^x} \\equal{} 16^{2^x}$\r\n\r\nCan you solve it under 45 seconds?", "Solution_1": "It's really just a matter of setting $ 16\\equal{}2^{4}$.\r\n\r\nThen both bases are 2's so they are unimportant.", "Solution_2": "$ 16^{2^x}\\equal{}2^{4\\cdot2^x}\\equal{}2^{2^{x\\plus{}2}}$\r\n\r\n$ 2^{16^x}\\equal{}2^{2^{x\\plus{}2}}$\r\n\r\n$ 16^x\\equal{}2^{x\\plus{}2}$\r\n\r\n$ 2^{4x}\\equal{}2^{x\\plus{}2}$\r\n\r\n4x=x+2\r\n3x=2\r\nx=2/3" } { "Tag": [ "probability" ], "Problem": "Three students play a card game. Assuming replacements are allowed, what is the probability the three students pick exactly the same card? What is the probability that all three students pick different cards? (Show work and necessary steps).", "Solution_1": "Well it depends on the number of cards, which you do not contain in the problem.\r\nHowever, if you let n=# of cards, you would do it in the following way.\r\n\r\n[hide=\"1 Solution\"](1/n)^3[/hide]\n[hide=\"2 Solution\"]The first student could pick any card. The next student could pick any card but that card. The last student could pick any card but those two.\n\nSo Probability = (n-1)(n-2)/(n 2 )[/hide]", "Solution_2": "[hide]If there are n unique cards, then it doesn't matter what the first student picks. There's a 1/n shot the second picks the same and 1/n shot the 3rd picks the same. So 1/n 2 \n\nAgain, n unique cards. Doesn't matter what the first person picks. The second has a (n-1)/n shot. The third has a (n-2)/n shot. (n-1)(n-2)/n 2 [/hide]" } { "Tag": [ "algebra", "polynomial", "rotation", "geometry", "geometric transformation", "function", "complex analysis" ], "Problem": "Hello\r\nLet $ A$ and $ B$ be polynomials with real coefficients such that all roots of $ A\\plus{}iB$ have negative imaginary parts.\r\nShow that for all $ u,v \\in \\mathbb R$ , all the roots of $ uA\\plus{}vB$ are real.", "Solution_1": "This is better suited for the complex analysis subforum, so I'm moving it there. \r\n[hide=\"Hint\"]\nUse argument principle in one of the half-planes to conclude that the curve $ t\\mapsto (A(t), B(t))$, $ t\\in\\mathbb R$ rotates around the origin many times[/hide]", "Solution_2": "I don't know about that argument principle could you try to explain without it ? =/\r\n\r\nActualy i don't know anything about complex analysis that's why i posted it in real analysis\r\nbut how would the fact that the curve (A(t),B(t)) rotates a lot of time around the origin help proving the result ?", "Solution_3": "If a curve makes $ n$ full rotations around the origin, it has to cross each line through the origin $ 2n$ times. The number $ n$ here can be either an integer or a half-integer.\r\n\r\nThe argument principle is a very simple thing: it tells you that if the function doesn't vanish inside a closed curve $ \\Gamma$, then the increment of the argument over $ \\Gamma$ is $ 0$. In our case, we need to consider a contour $ \\Gamma$ like this:\r\n[asy]D((-3,0)--(3,0));\nD((0,-1)--(0,3));\nD(MP(\"-R\",(-2,0))--MP(\"R\",(2,0))--CR((0,0),2,0,180),red+linewidth(1));[/asy]\r\nIf the maximal degree of $ A$, $ B$, is $ n$, then the argument of $ A(z)\\plus{}iB(z)$ increases by $ \\pi n$ when $ z$ travels over half-circle (only the highest degree term matters there) whence it has to decrease by the same $ \\pi n$ when you travel over $ [\\minus{}R,R]$, which means that the curve rotates about the origin $ n/2$ times. You can also see it without the argument principle by factoring and looking at the behavior of each linear factor (it rotates by $ \\pi$ clockwise) but I still wanted to bring up the argument principle anyway.\r\n\r\nOf course, you need to be a bit careful since the last statement is not exact (you have a small error near endpoints), but it still proves the result for almost all $ (u,v)$ and then you can just use the fact that complex roots are stable.\r\n\r\nI wonder if there exists some purely real-analytic (or, better, algebraic) solution.", "Solution_4": "Why does the argument of A(z) + i*B(z) increase by Pi*n when z travels over the half-circle ? :maybe:", "Solution_5": "Because, up to a small error, it is just the argument of $ cz^n$ there.", "Solution_6": "But then where do you use the fact that all roots have negative imaginary parts ?\r\nEDIT :Okay i got it sorry, thanks :D" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for $s,n\\in N$ we have \r\n$\\sum_{r=1}^{s}\\left(\\frac{a_r}{a_r+a_{r+1}}\\right)^n\\ge\\frac{s}{2^n}$.\r\nWhere $a_{s+1}=a_1$ and $a_r>0$ \r\n[hide]For $n=s=3$ we have Vietnamese TSE 05 [/hide]", "Solution_1": "[quote=\"Beat\"]Prove that for $s,n\\in N$ we have \n$\\sum_{r=1}^{s}\\left(\\frac{a_r}{a_r+a_{r+1}}\\right)^n\\ge\\frac{s}{2^n}$.\nWhere $a_{s+1}=a_1$ and $a_r>0$ \n[hide]For $n=s=3$ we have Vietnamese TSE 05 [/hide][/quote]\r\n\r\nOf course, It's not true.\r\n\r\nThe right problem has occured in my book.", "Solution_2": "Excuse, me.\r\nThe problem: Find all natural n and s for which this is true" } { "Tag": [ "inequalities", "parameterization", "inequalities proposed" ], "Problem": "Let $ {x,y,z}\\in {R^{ \\plus{} }},k\\in{N},k\\ge{2},$prove that\r\n\\[ (x^k \\plus{} 1)(y^k \\plus{} 1)(z^k \\plus{} 1)\\ge{\\frac {3^{3\\minus{} k}}{4}}(xy \\plus{} yz \\plus{} zx)^k,\\]\r\nand determine the condition of equation.\r\n\r\n\r\nIf you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.", "Solution_1": "[quote=\"fjwxcsl\"]Let $ {x,y,z}\\in {R^{ \\plus{} }},k\\in{N},k\\ge{2},$prove that\n\\[ (x^k \\plus{} 1)(y^k \\plus{} 1)(z^k \\plus{} 1)\\ge{\\frac {3^{k\\minus{}1}}{4}}(xy \\plus{} yz \\plus{} zx)^k,\\]\nand determine the condition of equation.\n\n\nIf you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.[/quote]\r\nI think you meant to prove\r\n$ (x^k\\plus{}1)(y^k\\plus{}1)(z^k\\plus{}1) \\ge \\frac{3^{3\\minus{}k}}{4}(xy\\plus{}yz\\plus{}zx)^k,$ (*)\r\nnot\r\n$ (x^k\\plus{}1)(y^k\\plus{}1)(z^k\\plus{}1) \\ge \\frac{3^{k\\minus{}1}}{4}(xy\\plus{}yz\\plus{}zx)^k.$ (**)\r\nBecause (*) holds, but (**) does not. For (*), we can prove it as follows\r\n\r\nSince $ \\frac{k}{2} \\ge 1,$ for any positive real numbers $ a,b,c$, we have\r\n$ \\left( \\frac{a^{\\frac{k}{2}}\\plus{}b^{\\frac{k}{2}}\\plus{}c^{\\frac{k}{2}}}{3}\\right)^{\\frac{2}{k}} \\ge \\frac{a\\plus{}b\\plus{}c}{3},$\r\nor\r\n$ \\left( {a^{\\frac{k}{2}}\\plus{}b^{\\frac{k}{2}}\\plus{}c^{\\frac{k}{2}}}\\right)^{2} \\ge 3^{2\\minus{}k}(a\\plus{}b\\plus{}c)^k.$\r\nSetting $ a\\equal{}xy,b\\equal{}yz$ and $ c\\equal{}zx,$ we get\r\n$ \\left( \\sqrt{x^ky^k}\\plus{}\\sqrt{y^kz^k}\\plus{}\\sqrt{z^kx^k}\\right)^2 \\ge 3^{2\\minus{}k}(xy\\plus{}yz\\plus{}zx)^k.$\r\nAccording to this inequality, we see that it suffices to prove that\r\n$ (u^2\\plus{}1)(v^2\\plus{}1)(w^2\\plus{}1) \\ge \\frac{3}{4} (uv\\plus{}vw\\plus{}wu)^2,$\r\nwhere $ u\\equal{}\\sqrt{x^k},v\\equal{}\\sqrt{y^k},w\\equal{}\\sqrt{z^k}.$\r\n\r\nNow, using the Cauchy-Schwarz Inequality, we have\r\n$ (uv\\plus{}vw\\plus{}wu)^2\\equal{}[u(v\\plus{}w)\\plus{}vw]^2 \\le (u^2\\plus{}1)[(v\\plus{}w)^2\\plus{}v^2w^2].$\r\nTherefore, it is enough to prove that\r\n$ 4(v^2\\plus{}1)(w^2\\plus{}1) \\ge 3[(v\\plus{}w)^2\\plus{}v^2w^2],$\r\nor\r\n$ (v\\minus{}w)^2\\plus{}(vw\\minus{}2)^2 \\ge 0,$\r\nwhich is obviously true.", "Solution_2": "[quote=\"can_hang2007\"][quote=\"fjwxcsl\"]Let $ {x,y,z}\\in {R^{ \\plus{} }},k\\in{N},k\\ge{2},$prove that\n\\[ (x^k \\plus{} 1)(y^k \\plus{} 1)(z^k \\plus{} 1)\\ge{\\frac {3^{k \\minus{} 1}}{4}}(xy \\plus{} yz \\plus{} zx)^k,\\]\nand determine the condition of equation.\n\n\nIf you feel difficult, please just to prove the inequality for some specific values of parameter,which is holds.[/quote]\nI think you meant to prove\n$ (x^k \\plus{} 1)(y^k \\plus{} 1)(z^k \\plus{} 1) \\ge \\frac {3^{3 \\minus{} k}}{4}(xy \\plus{} yz \\plus{} zx)^k,$ (*)\nnot\n$ (x^k \\plus{} 1)(y^k \\plus{} 1)(z^k \\plus{} 1) \\ge \\frac {3^{k \\minus{} 1}}{4}(xy \\plus{} yz \\plus{} zx)^k.$ (**)\nBecause (*) holds, but (**) does not. For (*), we can prove it as follows\n\nSince $ \\frac {k}{2} \\ge 1,$ for any positive real numbers $ a,b,c$, we have\n$ \\left( \\frac {a^{\\frac {k}{2}} \\plus{} b^{\\frac {k}{2}} \\plus{} c^{\\frac {k}{2}}}{3}\\right)^{\\frac {2}{k}} \\ge \\frac {a \\plus{} b \\plus{} c}{3},$\nor\n$ \\left( {a^{\\frac {k}{2}} \\plus{} b^{\\frac {k}{2}} \\plus{} c^{\\frac {k}{2}}}\\right)^{2} \\ge 3^{2 \\minus{} k}(a \\plus{} b \\plus{} c)^k.$\nSetting $ a \\equal{} xy,b \\equal{} yz$ and $ c \\equal{} zx,$ we get\n$ \\left( \\sqrt {x^ky^k} \\plus{} \\sqrt {y^kz^k} \\plus{} \\sqrt {z^kx^k}\\right)^2 \\ge 3^{2 \\minus{} k}(xy \\plus{} yz \\plus{} zx)^k.$\nAccording to this inequality, we see that it suffices to prove that\n$ (u^2 \\plus{} 1)(v^2 \\plus{} 1)(w^2 \\plus{} 1) \\ge \\frac {3}{4} (uv \\plus{} vw \\plus{} wu)^2,$\nwhere $ u \\equal{} \\sqrt {x^k},v \\equal{} \\sqrt {y^k},w \\equal{} \\sqrt {z^k}.$\n\nNow, using the Cauchy-Schwarz Inequality, we have\n$ (uv \\plus{} vw \\plus{} wu)^2 \\equal{} [u(v \\plus{} w) \\plus{} vw]^2 \\le (u^2 \\plus{} 1)[(v \\plus{} w)^2 \\plus{} v^2w^2].$\nTherefore, it is enough to prove that\n$ 4(v^2 \\plus{} 1)(w^2 \\plus{} 1) \\ge 3[(v \\plus{} w)^2 \\plus{} v^2w^2],$\nor\n$ (v \\minus{} w)^2 \\plus{} (vw \\minus{} 2)^2 \\ge 0,$\nwhich is obviously true.[/quote]\r\n\r\n\r\n\r\n\r\n\r\n\r\nEdited,thank you, you are right! Very nice proof,can_hang2007!" } { "Tag": [ "IMO" ], "Problem": "just wondering, how's the olympiads 2002 project going? I heard it was finished about half a year ago, still I haven't really heard of it since then. Or has the project stopped since the merge with AoPS?", "Solution_1": "Actually, as far as I know, the deadline was postponed to October 2004. BTW, for Valentin, Orlando, Pierre et al.: I am preparing another file with two solutions by Peter Scholze and one by me...\r\n\r\n darij", "Solution_2": "If you still need people to look for more solutions, I've got nothing better to do right now :)", "Solution_3": "What's that ?", "Solution_4": "I guess mathlinks is gonna publish a book with all the olympiads from 2002. :)", "Solution_5": "To dig up an old cow: is this project still going? :)", "Solution_6": "I'm afraid not, (and I don't think this is the right place to discuss OP 02)." } { "Tag": [ "AMC", "AIME", "\\/closed" ], "Problem": "I still haven't seen the message board for AIME seminar...\r\n\r\nis it not up yet??", "Solution_1": "[quote=\"pkerichang\"]I still haven't seen the message board for AIME seminar...\n\nis it not up yet??[/quote]\r\nI don't see it either, you're not the only one.", "Solution_2": "Yeah, I don't see it either. Where would it show up when it does?", "Solution_3": "I don't think there is one since class starts tommorow.\r\nMaybe I'm mistaken though.", "Solution_4": "Mmm, I doubt it:\r\n[quote=\"Email I got today from AoPS\"]Also, there is a message board set up especially for this class. Click on \"Community\" on the left side of the website, then scroll down to the \"Online School\" section of the Forum, in which you should see the \"AIME Problem Seminar\" message board.[/quote]", "Solution_5": "My bad - you should see it now." } { "Tag": [], "Problem": "HI everybody!!! how did you all do your board exams??? :D", "Solution_1": "See the other thread abt CBSE Board exams or sth :) \r\nAnd why would Einstein be bothered abt the boards? :P :rotfl:", "Solution_2": "u must have read al's biography(excerpt) previous year!", "Solution_3": "[quote=\"rituraj007\"]u must have read al's biography(excerpt) previous year![/quote]\r\n\r\nIt was in 12th i believe :D And ofc their history is too fascinating to avoid a peek :D", "Solution_4": ":oops: i was telling to albert einstein!!!!" } { "Tag": [ "HMMT", "Stanford", "college" ], "Problem": "Anyone doing it from the West Coast??? I am interested.", "Solution_1": "Rice and Stanford also hold very similar competitions, I believe. You could always try doing one of them instead of making that huge commute." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Hi all,\r\n\r\nHaving some trouble solving this problem.\r\n\r\nA group of 20 students, including 3 particular girls and 4 particular boys, are to be lined up in two rows with 10 students each. In how many ways can this be done if the 3 particular girls must be in the front row while the 4 particular boys be in the back?\r\n\r\nThe answer is (14C7)x(10!)^2. The (10!)^2 I have some understanding of, but I don't understand how the 14 came about in the binomial coefficient. The 7 I have more understanding of as well.\r\n\r\nThanks folks.", "Solution_1": "14 should be $ 20 \\minus{} 4 \\minus{} 3 \\equal{} 13$ -- that term counts the number of ways to choose seven people to join the three selected girls in front.", "Solution_2": "Can't believe it was a typo! Thanks!" } { "Tag": [ "algebra", "polynomial", "algorithm", "abstract algebra", "vector", "Ring Theory", "superior algebra" ], "Problem": "Consider the ideal generated by $$ over $Z[x]$, where $p$ is a prime number. prove it is not generated by less than $n+1$ polynomials.\r\n\r\nCan we characterize other ideals in $Z[x]$ whit the same property?", "Solution_1": "Let $I=$. $I$ is a maximal ideal of $\\mathbb{Z}[X]$. $I^n=$.\r\nThe elements of $I^n$ have the form $\\sum_{i=0}^np^{n-i}X^if_i(X)$ with $f_i\\in\\mathbb{Z}[X]\\ \\forall i=\\overline{1,n}$.\r\nBecause $(X\\mathbb{Z})\\cdot I^n\\subset I\\cdot I^n=I^{n+1}$, it is easy to see that $\\sum_{i=0}^np^{n-i}X^if_i(X)=\\sum_{i=0}^np^{n-i}X^if_i(0) (mod\\ I^{n+1})$. Therefore $I^n/I^{n+1}$ has a structure of $\\mathbb{Z}$-module.\r\nBecause $(p\\mathbb{Z})\\cdot I^n\\subset I\\cdot I^n=I^{n+1}$, it follows that $(p\\mathbb{Z})\\cdot I^n/I^{n+1}=0$. Therefore $I^n/I^{n+1}$ is an $\\mathbb{Z}_p$ vectorspace. It is obviously generated by $n+1$ elements, therefore $\\dim_{\\mathbb{Z}_p}I^n/I^{n+1}\\leq n+1$.\r\n\r\nDefine $\\varphi: \\mathbb{Z}_p^{n+1}\\to I^n/I^{n+1}$ by $\\varphi(a_0,\\ldots,a_n)=\\sum_{i=0}^na_ip^{n-i}x^i$. ($x$ denotes the class of $X$ mod $I^{n+1}$).\r\nWe prove that $\\varphi$ is injective. Let $\\varphi(a_0,\\ldots,a_n)=\\sum_{i=0}^na_ip^{n-i}x^i=0$. Then $\\sum_{i=0}^na_ip^{n-i}X^i\\in I^{n+1}$ (this time we see the elements $(a_i)_{i=\\overline{0,n}}$ as elements of $\\mathbb{Z}$, not residues mod p).\r\nSo we have $\\sum_{i=0}^na_ip^{n-i}X^i=\\sum_{j=0}^{n+1}p^{n+1-j}X^jf_j(X)$ for some $f_0,\\ldots,f_{n+1}\\in\\mathbb{Z}[X]$.\r\nReduce this mod p and obtain $a_nX^n=X^{n+1}f_{n+1}(X)$ in $\\mathbb{Z}_p[X]$ which implies $a_n=f_{n+1}=0 (mod p)$. \r\nThis means that we have $\\sum_{i=0}^{n-1}a_ip^{n-i}X^i\\in I^{n+1}\\Rightarrow \\sum_{i=0}^{n-1}a_ip^{n-i-1}X^i\\in I^n$ and we repeat this algorithm to prove $p|a_i\\ \\forall i=\\overline{0,n}$. This proves that $\\varphi$ is injective, so $dim_{\\mathbb{Z}_p}I^n/I^{n+1}\\geq n+1$.\r\nTherefore $\\dim_{\\mathbb{Z}_p}I^n/I^{n+1}=n+1$.\r\n\r\nAssume for a contradiction that $I^n$ is generated by $n$ elements. The images of these elements give a set of generatros for $I^n/I^{n+1}$ as $\\mathbb{Z}$-module and $\\mathbb{Z}_p$-module. This conflicts with the dimension of $I^n/I^{n+1}$ over $\\mathbb{Z}_p$.", "Solution_2": "Yes, we find a similar Isomorphism, but it took me a lot to realise it worked, I mean, that the images work like a basis as a Vector space over $Z_p$. But it seems you did it to easy!!! We really had to work hard to prove the rest. Anyway there is a lot of theory I dont understand from your solution, do you ahve a good reference?", "Solution_3": "I don't think that what I did is easy. I don't quite yet understand my idea to consider $I^n/I^{n+1}$.\r\nGood reference? I don't know. It is quite basic stuff in commutative ring theory like the fundamental isomorphism theorem, and you need some modules theory to uderstand the restriction/extension of scalars. I think you know most of this anyway if you got to where you said you did. Maybe some practice.\r\nAfter all this I sure hope my solution is correct.", "Solution_4": "I think the idea of considering $I_n/I_{n+1}$ comes from the fact that we already ahd the clue of maping $I_n$ somehow to $Z_p^{n+1}$ So I first thought about some kind of projection, but i also was thinking about some inductive argument so I tried to relate $I_n$ and $I_{n+1}$ and I saw how only independent terms survived to my projection and later I found it was the action of seing the first ideal mod $I_{n+1}$, but later that multiples of $p$ where ruled out was a surprise, so I found a set i thought i could easily map to $Z_p^{n+1}$ but i didnt filled the details so it wasnt working, but yesterday in class we finished it with some nice ideas but somehow more elementary than yours, we dont know so much theory (I mean , I)." } { "Tag": [ "AMC", "AMC 12", "AMC 12 B", "function" ], "Problem": "What is an inverse function? I couldn't do #13 since i didn't know what inverse function is. I am an 8th-grader so my knowledge in math is not great.", "Solution_1": "what was the question? to do inverse funcitons algebraically i usually just switch the y and the x and then solve for y...I don't know if the question you are thinking of is as simple as this. but that's usually what i aim to do.\r\nbeans", "Solution_2": "wow i had no clue how to do this problem...spent too much time on it...\r\n\r\nf(x) turned out to be -x-1, making the answer -2\r\n\r\nmaybe someone has a solution manual?\r\n\r\ninverse functions are like f(f^-1(x)) = x, so like x/2 and 2x would be inverses of each other", "Solution_3": "the question was:\r\n\r\nf(x) = ax+b, f^-1(x) = bx + a, solve for a+b", "Solution_4": "You can think of an inverse function, f-1(x), of a function f(x) as a function that \"undoes\" f(x). In other words, it's a function such that f-1(f(x)) = x.\r\n\r\nFor example, the inverse of f(x) = x2 is f-1(x) = :sqrt:x since f-1(f(x)) = :sqrt:(x2) = x.", "Solution_5": "The Problem:\n\n\n\nf(x) = ax + b \n\nf-1(x) = bx + a. \n\n\n\nFind a+b.\n\n\n\nSolution (in spoiler):\n\n[hide]\n\n
\n\nx = f(f-1(x)) = a(bx+a) + b \n\n              = abx + a2 + b\n\n
\n\nso the RHS is identically equal to the LHS, therefore \n\nab = 1, and a^2 + b = 0.\n\n\n\n=> b = -a2 = > -a3 = 1 = > a = -1.\n\n=> -b = 1 = > b = -1.\n\n\n\ntherefore a + b = -1 + -1 = -2.\n\n[/hide]", "Solution_6": "Geometrically, I can up with 1. I don't know where I messed up but I know it's wrong. :(", "Solution_7": "i plugged in 0 for x, to find f(0)=b. Then f(b)^-1 = b^2 + a = 0, so b^2=-a.\r\n\r\nThen i plugged in 1 for x, to get f(1)=a+b. then f(a+b)^-1 = b(a+b) + a = 1. Then substituted -b^2=a in that equation and solved.", "Solution_8": "\"f^-1(f(x)) = x. \" is the best way to do it\r\n\r\nI did something weird that I think(hope) is correct.\r\nThe inverse function of f(x)=ax+b is equal to (x-b)/a\r\nAnd we know (x-b)/a = bx+a\r\n\r\nSo 1/a=b and -b/a=a\r\nSo we can solve that a^3=-1 and then finding b=-1\r\n\r\nUm...I found a=-1 and then I read the question as only asking for a so I just circled B and moved on! Gah!" } { "Tag": [ "integration", "geometry", "3D geometry", "sphere" ], "Problem": "Here is another conductor problem which I have difficulty with. It is from Introduction to Electrodynamics by Griffith.\r\n\r\nI am not cheating, because this is not homework for me. It is just sth I'd like to understand.\r\n\r\nPart b and d and e, I can basically guess. But I really need a quantitative and qualitative understanding of why my answer might be wrong or correct.\r\n\r\nThanks so much!", "Solution_1": "a) is wrong. You drew the right Gaussian surface, but it's not true that $ Q_{enc}$ is just $ q_a$: you are forgetting about the induced charge. The reasoning is as follows:\r\n\r\nAt the surface of your Gaussian surface, the E-field is zero everywhere, since that is \"in the meat of a conductor\". Remember, within a conductor E is always zero! So we have $ \\int \\vec{E}\\cdot d\\vec{A} \\equal{} \\int 0\\cdot d\\vec{A} \\equal{} 0$, which means $ Q_{enc} \\equal{} 0$. But $ Q_{enc} \\equal{} q_a \\plus{} q_{induced}$, therefore the induced charge (on the surface of the sphere of radius a) equals $ \\minus{} q_a$. Hence $ \\sigma_a \\equal{} \\frac { \\minus{} q_a}{4\\pi\\epsilon_0 r_a^2}$, and with the same arguments $ \\sigma_b \\equal{} \\frac { \\minus{} q_b}{4\\pi\\epsilon_0 r_b^2}$.\r\n\r\n(the answers differ only in a minus sign, but the arguments are pretty different).\r\n\r\n[b]Btw, you didn't compute $ \\sigma_R$.[/b]\r\nIf you do that, I'll help you with the next questions (the depend in some way on your understanding of question a).", "Solution_2": "Very good, BlueVelvet. Ops: before \"some_other_one\" notice the radius in the $ \\sigma_b$ denominator should be $ r_b$ or better, b for this and a for the other to follow strictly Griffith (an unimportant glitch, but just in case...) \r\n\r\nYou're almost there, integration.\r\n\r\nDr. No should read this \" 2.5.3 The Surface Charge on a Conductor; ...\" In Griffith.", "Solution_3": "Thanks, I corrected the typo.", "Solution_4": "Thanks guys for bearing my stupidity :D It's always nice to get a check. BTW, for part a, I basically forgot to put the minus, but I knew it, because Qe, as you said, had to be 0.", "Solution_5": "[quote=\"integration\"]but I knew it, because Qe, as you said, had to be 0.[/quote]Ok, but why did you write $ \\int \\vec{E}\\cdot d\\vec{A}\\equal{}\\frac{q_a}{\\epsilon_0}$? Because that's not correct ;)\r\n\r\nb) and c) are correct.\r\nd) is not: There can only be an electric force if there's an electric field. But in both cavities there is no E-field (except the one produced by the charges themselves of course). So the answer is just zero force on q_a and q_b.\r\n\r\ne): you are right that $ \\sigma_R$ will change, but what about the E-field outside?", "Solution_6": "As far as I recall , in such cases the charges can be assumed to be at the center and the field will just be like that of a uniformly charged sphere.But I cant understand the logic and dont know the proof either. :(", "Solution_7": "Ok, but why did you write ? Because that's not correct ...\r\n\r\nThat's just a bad habit I have, b/c whenever I see a charge inside a conducting cavity, I know it will cause a charge of opposite sign to be polarized on the inner surface.\r\n\r\nd. So basically the conductor is analogous to a faraday cage? I once heard that you can turn on a crazy Van der graaff, charged with bunches of coulomb's next to you, but as long as you stay inside a conducting cage, you cannot \"feel\" the outside electric field, because charge polarization on the cage causes it to cancel the outer electric field, on the inside.\r\n\r\nSo you're correct about no force, but do you have a kind of intuitive explanation?\r\n\r\ne. Well, if sigma r changes, the field outside will be the superposition of the answer in (b), with a different sigma r, and the field of the outer charge. Is that correct?\r\n\r\nThx.", "Solution_8": "[quote=\"integration\"]d. So basically the conductor is analogous to a faraday cage? I once heard that you can turn on a crazy Van der graaff, charged with bunches of coulomb's next to you, but as long as you stay inside a conducting cage, you cannot \"feel\" the outside electric field, because charge polarization on the cage causes it to cancel the outer electric field, on the inside.\n\nSo you're correct about no force, but do you have a kind of intuitive explanation?[/quote]\nWell, I think yours is pretty intuitive: as long as you are inside a conducting cage (a [i]perfect [/i]conductor, and of course you are not touching it), the conductor makes sure there will be no E-field inside it. It certainly is comparable with a Faraday cage, in fact this [i]is[/i] a Faraday cage.\n[quote]e. Well, if sigma r changes, the field outside will be the superposition of the answer in (b), with a different sigma r, and the field of the outer charge. Is that correct?[/quote]Yep, that's what I was thinking too." } { "Tag": [ "LaTeX" ], "Problem": "how would you type a direct sum symbol in mathmode?", "Solution_1": "$ \\sum\\frac{1}{a\\plus{}b}$\r\n\r\n\\sum\\frac1{a+b}\r\n\r\n$ \\sum_{\\text{sym}}abc$\r\n\r\n\\sum_{\\text{sym}}abc\r\n\r\n$ \\sum_{k\\equal{}0}^{\\infty}\\frac{x^{k}}{k!}$\r\n\r\n\\sum_{k=0}^\\infty\\frac{x^k}{k!}\r\n\r\nDid you mean something like that?", "Solution_2": "the direct summand, it's an encircled plus sign. look at the link:\r\n\r\nhttp://www.sosmath.com/CBB/latexrender/pictures/85f87586990c2b3c10d850c862f148f2.gif", "Solution_3": "\\oplus\r\n\r\n$ \\oplus$", "Solution_4": "thanks, i appreciate it." } { "Tag": [], "Problem": "What is everyone's state cutoff score? MA is 37.", "Solution_1": "KY: 32,33, or 34 (i'm sure its one of those scores)", "Solution_2": "We've already got one of these, I'll lock this because the other has more posts ATM." } { "Tag": [ "USAMTS", "email" ], "Problem": "I am submitting my solutions online, do I have to submit my entry form by snail mail?\r\nIs it okay to submit my solutions, then my entry form, as long as both come in on time?", "Solution_1": "not a GOOD SIGNATURE", "Solution_2": "[quote=\"rt_08\"]not a GOOD SIGNATURE[/quote]\r\n\r\n... that was random.\r\n\r\nThe USAMTS entry form specifies that you should send the form in separately through mail or fax (no email!) to the USAMTS people by the round 1 due date (10/14/08), so I don't think you have to send the entry form first as long as it gets there on time.\r\n\r\nWhy would you want to send in the solutions first? Have you already submitted them before you sent in the entry form?", "Solution_3": "Thanks.\r\n\r\nYeah I finished my solutions and I want to send them in, I was just wondering if I should wait until after my entry form goes in.", "Solution_4": "[quote=\"buzzer11\"]I am submitting my solutions online, do I have to submit my entry form by snail mail?\nIs it okay to submit my solutions, then my entry form, as long as both come in on time?[/quote]\r\nI think it's more appropriate to submit your entry form first, then your solutions. Entry forms can be submitted in any way, mail or fax. As long as USAMTS receives it with the proper signatures, then you're fine. :)", "Solution_5": "buzzer's sig is sick i dont know what rt_08 is talking about.....", "Solution_6": "Haha, thanks mathemonster and thanks everyone for the replies...\r\n\r\nI faxed in my entry form today and then my solutions tonight so I think i should be okay!" } { "Tag": [ "inequalities", "geometry", "inradius", "inequalities unsolved" ], "Problem": "Prove that in a triangel ABC :\r\n (tan(A/2))^(2 \\sqrt 2) + (tan(B/2))^(2 \\sqrt 2) + (tan(C/2))^(2 \\sqrt 2) > = 3^(1- \\sqrt 2)", "Solution_1": "I have just found a good solution using Jensen's inequality. But I don't like it because I must prove Jensen again when using it. Please help me find another way.", "Solution_2": "[quote=\"darkmaster\"]Prove that in a triangel ABC :\n (tan(A/2))^(2 \\sqrt 2) + (tan(B/2))^(2 \\sqrt 2) + (tan(C/2))^(2 \\sqrt 2) > = 3^(1- \\sqrt 2)[/quote]\r\n\r\nWell, let t = 2 \\sqrt 2. Then, your inequality can be rewritten as\r\n\r\n (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t > = 3^(1 - t/2).\r\n\r\nI will prove this inequality for all t >= 1.\r\n\r\nAt first, by the power mean inequality,\r\n\r\n ( ( (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t ) / 3 ) ^ (1/t) >= ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3.\r\n\r\nIn other words,\r\n\r\n (tan(A/2))^t + (tan(B/2))^t + (tan(C/2))^t >= 3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t.\r\n\r\nIf we are now able to show that\r\n\r\n 3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3^(1 - t/2),\r\n\r\nthen we are done. Perform some equivalence transformations with this inequality:\r\n\r\n 3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3^(1 - t/2)\r\n<==> 3 * ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 3 / (sqrt(3))^t\r\n<==> ( ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 ) ^ t >= 1 / (sqrt(3))^t\r\n<==> ( tan(A/2) + tan(B/2) + tan(C/2) ) / 3 >= 1 / sqrt(3)\r\n<==> tan(A/2) + tan(B/2) + tan(C/2) >= sqrt(3).\r\n\r\nNow, this inequality is well-known and can be proven without Jensen. For instance, call s = (a + b + c) / 2 the semiperimeter and r the inradius of triangle ABC, then the inequality becomes\r\n\r\n r / (s-a) + r / (s-b) + r / (s-c) >= sqrt(3),\r\n\r\nor\r\n\r\n 1 / (s-a) + 1 / (s-b) + 1 / (s-c) >= sqrt(3) / r.\r\n\r\nThis was proven on page 5 of [url=http://www.mathlinks.ro/download.php?id=115]my MathLinks Contest 2 PDF file[/url].\r\n\r\n Darij", "Solution_3": "It's wonderful. Thanks a lot." } { "Tag": [ "geometry", "congruent triangles" ], "Problem": "ACE angle is 180 degrees\r\nLet there be $T$ the middle of $BC$.\r\nThe point where $AB$ intersects $ET$ is ${S}$.\r\nProve that $SB=\\frac{CD}{3}$", "Solution_1": "It's possible to make a diagram where segment $AB$ doesn't intersect segment $ET$. :wink:", "Solution_2": "[quote=\"alex__ib\"]$\\angle ACE = 180^\\circ$\nLet $T$ be the midpoint of $BC$.\n$S$ is the intersection of $AB$ and $ET$.\nProve that $SB=\\frac{CD}{3}$[/quote]\r\n\r\nLet's assume $B$ is not on $AE$ and $AB$ is not parallel to $ET$.\r\n\r\nWhat is $CD$?? :maybe:", "Solution_3": "let there be ABC and DCE two equilateral congruent triangles\r\nlook at the description thingy or the title :wink: \r\n\r\n[quote=\"vishalarul\"]It's possible to make a diagram where segment $AB$ doesn't intersect segment $ET$. :wink:[/quote]\r\nHe probably means line $AB$ and line $ET$\r\nThose always intersect :wink:" } { "Tag": [ "Duke", "college" ], "Problem": "just wondering what you guys are going to be doing...whether practicing more maths/other subjects, cooling off, preparing for IMO/other olympiads, just to know what you guys are planning to do... :D \r\n\r\nlet me start off, i'll be going to the beach this weekend and then with my cousins we are going to Six Flags!!! this is my first time so i should enjoy it :play_ball: :icecream: :trampoline:", "Solution_1": "i'll probably be staying home doing the usual. probably get some fireworks as well. nothin' special.", "Solution_2": "[quote=\"furious\"]i'll probably be staying home doing the usual. probably get some fireworks as well. nothin' special.[/quote]\r\n\r\nSo will I :D No Rhode Island for me this year. :(", "Solution_3": "i'm gonna stay at home :(", "Solution_4": "my first 4th of july in NC will be my first firework setting 4th of july ever. (Cali doesn't let you buy fireworks; what a bummer :( )", "Solution_5": "i'm gonna go get ready to go to camp at duke so i have to buy stuff to bring there and study", "Solution_6": "I wonder what will happen for 4th july when I go to HCSSiM....hmmmm...", "Solution_7": "I'm staying home or watching War of the Worlds", "Solution_8": "I am probably coming back from a wedding to which I am going to tomorrow!", "Solution_9": "[quote=\"math92\"]I'm staying home or watching War of the Worlds[/quote]\r\nim sorry but i think that was an overrated movie, i didnt like the ending! too cheesy\r\nthe movie i recommend? [color=blue]Batman Begins[/color]", "Solution_10": "[quote=\"Palytoxin\"]just wondering what you guys are going to be doing...whether practicing more maths/other subjects, cooling off, preparing for IMO/other olympiads, just to know what you guys are planning to do... :D [/quote]I'll be flying across the big water to check out the Fireworks in SD :)", "Solution_11": "I'm doing math!!! :D :lol: :lol: :D \r\nI'm also going on vacation, but I don't know where yet :(", "Solution_12": "War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the \"cheasy\" ending read the book. If anything is overrated its Batman.\r\n\r\nEdit: Ok for the weekend pick War of the Worlds its a much more enjoyable movie and I reccommend it more than Batman \r\n\r\nP.S. Read the book first", "Solution_13": "[quote=\"sirkuku\"]War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the \"cheasy\" ending read the book. If anything is overrated its Batman.[/quote]I think you got the topic wrong ... :?", "Solution_14": "[quote=\"sirkuku\"]War...of...the worlds OVERRATED it is one of the better movies out there, and if you really want to know about the \"cheasy\" ending read the book. If anything is overrated its Batman.[/quote]\r\nwe are talking about what we are doing over the weekend not what movie is overated.\r\nMy sister, colts18, went to some resort in pensylvania called the Nemacolin. She rode in a hummer, played paintball guns, rode in a fourth of july float, and did rock climbing." } { "Tag": [ "calculus", "derivative", "function", "calculus computations" ], "Problem": "The rate of change of a population depends on the current population $ P$ and is given by $ \\frac{d}{dt}\\equal{}kP(L\\minus{}P)$ with $ K$ and $ L$ being positive constants. \r\n\r\nHow do you find for the nonnegative values of $ P$ is where population is increasing, decreasing, and constant?\r\n\r\nThe double derivative of the function?", "Solution_1": "The current population is given by $ P$, so the rate of change is the derivative of $ P$ with respect to time (as mentioned). When $ P$ is increasing, decreasing, or constant, the first derivative will be positive, negative, or zero (respectively), because it represents instantaneous rate of change.\r\n\r\nIntroducing second derivatives is asking for rate of rate of change, which isn't what you want.", "Solution_2": "So I guess the issue I have is how do you find the derivative and second derivative without $ t$ being a part of the equation you are taking the derivative of." } { "Tag": [ "geometry", "trapezoid", "rectangle" ], "Problem": "Figure $ABCD$ is a trapezoid with $AB || DC, AB = 5, BC = 3 \\sqrt 2, \\measuredangle BCD = 45^\\circ$, and $\\measuredangle CDA = 60^\\circ$. The length of $DC$ is\n\n$\\textbf{(A) }7 + \\frac{2}{3} \\sqrt{3}\\qquad\n\\textbf{(B) }8\\qquad\n\\textbf{(C) }9 \\frac{1}{2}\\qquad\n\\textbf{(D) }8 + \\sqrt 3\\qquad\n\\textbf{(E) }8 + 3 \\sqrt 3$", "Solution_1": "[hide=\"Answer\"]Because we see the angle measures 45 and 60, we look for special right triangles. By splitting the trapezoid into a 30,60,90 triangle, 445,45,90 triangle, and a rectangle, we can find CD piece by piece.\n\n$CD=\\sqrt{3}+5+3=8+\\sqrt{3}$\nThe answer is D.\n[/hide]", "Solution_2": "[hide=\"Answer\"]We have $BC=3\\sqrt{2}$, and $\\angle BCD=45^\\circ$, so, labeling the altitude to $B$ from $DC$ as $BF$, we have $BF=3$ and $CF=3$. Labeling the altitude from $A$ to $DC$ as $AE$, we have $AE=3$, and since $\\angle CDA=60^\\circ$, then $DE=\\sqrt{3}$, so we have $5+3+\\sqrt{3}=8+\\sqrt{3}\\Rightarrow \\boxed{D}$.[/hide]", "Solution_3": "[hide]$DC=AB+x+y$, well looking at the triangles you are given, which are $45,45,90$ and $30,60,90$, you can see that the height of the trapezoid is $3$, therefore $x=3$ and $y=\\sqrt{3}$, therefore $\\boxed{DC=8+\\sqrt{3}}$[/hide]" } { "Tag": [], "Problem": "Wooooooo!!!!!!", "Solution_1": "Um... so what?" } { "Tag": [ "geometry", "LaTeX" ], "Problem": "Just wondering, do our answer to geometry questions need to include the diagram, or can we just use not given labels and not worry about including it.\r\n\r\nThanks for your help.", "Solution_1": "same questions daniel had. it seems impossible for me to include any diagram in my answers.", "Solution_2": "If you are going to add your own points or labels, then you should definitely include a diagram. You can read the $\\LaTeX$ tutorial on the AoPS website for information on how to include diagrams in your $\\LaTeX$ files.", "Solution_3": "[quote=\"bookworm271828\"]Just wondering, do our answer to geometry questions need to include the diagram, or can we just use not given labels and not worry about including it.\n\nThanks for your help.[/quote]\r\n\r\nYou don't need to have a diagram, but it sure does help if it's at all different from the problem.", "Solution_4": "Regarding diagrams, if you're too lazy to use the xy-pic environment and whatnot to generate diagrams (like me), then you can simply make them in paint, save them as JPG files in the same folder as your LaTeX document, and use the command \\includegraphics{filename} in your LaTeX document to embed them. It's not the most compact or the prettiest form, but it's by far the simplest.\r\n\r\nStill, I recommend reading the picture section in the LaTeX tutorials. That's what it's there for!" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Suppose a function $f:$ $N \\rightarrow N$, $ff(n) = kn , k \\in R$\r\nProve that\r\n$\\frac{2k}{k+1}n \\leq f(n) \\leq \\frac{k+1}{2}n$\r\nAnd find an example of $f(n)$ when $k = 3$", "Solution_1": "[quote=\"Kalimdor\"]Suppose a function $ f:$ $ N \\rightarrow N$, $ ff(n) \\equal{} kn , k \\in R$\nProve that\n$ \\frac {2k}{k \\plus{} 1}n \\leq f(n) \\leq \\frac {k \\plus{} 1}{2}n$\nAnd find an example of $ f(n)$ when $ k \\equal{} 3$[/quote]\r\n\r\nI think this is wrong.\r\n\r\nFor $ k\\equal{}3$, for example, define $ f(n)$ as :\r\n\r\nIf $ n\\equal{}3^p5^{2q}r$ (with $ \\gcd(r,15)\\equal{}1$), $ f(n)\\equal{}5n\\equal{}3^p5^{2q\\plus{}1}r$\r\n\r\nIf $ n\\equal{}3^p5^{2q\\plus{}1}r$ (with $ \\gcd(r,15)\\equal{}1$), $ f(n)\\equal{}\\frac{3n}{5}\\equal{}3^{p\\plus{}1}5^{2q}r$\r\n\r\nClearly $ f(f(n))\\equal{}3n$ and we never have $ \\frac {3n}{2} \\leq f(n) \\leq 2n$" } { "Tag": [], "Problem": "If x represents a two-digit positvie integer, compute the number of values of x such taht x^2+3x+2 is divisible by 6.", "Solution_1": "[hide]Factor: $(x+1)(x+2)$. One of these is always even, so you just have to find all two digit numbers congruent to 1 or 2 mod 3.[/hide]", "Solution_2": "[quote=\"Phelpedo\"][hide]Factor: $(x+1)(x+2)$. One of these is always even, so you just have to find all two digit numbers congruent to 1 or 2 mod 3.[/hide][/quote]\r\n\r\n(x+1)/6 = n, where n is an integer\r\nx+1=6n\r\nx= 6n -1\r\n\r\nOr,\r\n\r\n(x+2)/6 = n, where n is an integer\r\nx+2 = 6n\r\nx=6n-2", "Solution_3": "I think you forgot the case where (x+1)|3 and (X+2)|2 or vice versa", "Solution_4": "A small error, neelnal. Note that the product of 2 numbers can be divisible by 6 without either of those two numbers being divisible by 6. The most basic example of this: 2*3 = 6.", "Solution_5": "[hide=\"I think its time we got the answer up here\"]\nall positive integers congruent to 1 or 2 mod 3 satisfy the eqn\nso 66 of the first 99 positive integers work\n6 of those are single digit\nso our answer should be [b]60[/b][/hide]", "Solution_6": "Ooh. Ok yes. I'm sorry I made that mistake!\r\n\r\nStupid me!! :rotfl:", "Solution_7": "[hide=\"hint #1\"]Don't forget to factor![/hide]\n\n[hide=\"hint #2\"]What in general are we looking for?[/hide]\n\n[hide=\"hint #3\"]Use modulo arithmatic.[/hide]\n\n\n\n\n[hide=\"solution\"]We can factor the equation to get (x+1)(x+2). We are assured that the product of those two is even. We need that product to also be a multiple of three. The only time that can't be reached is when x is a multiple of three. Otherwise, the first equation is a multiple of 6. There are 33 of the first 99 natural numbers that are multiples of three. 3 of those are single digit multiples. Since there are 90 two digit numbers, there are 90-30=[b]60[/b] values of x that make the first equation a multiple of 6.[/hide]" } { "Tag": [], "Problem": "http://www.cbseresults.nic.in", "Solution_1": "CBSE is probably trying to take the load off its database with 10 lakh people trying to find their results..\r\n\r\nHence realasing the results a week earlier.. :lol:", "Solution_2": "Haha. I am sure 90% of delhi doesnt know about this since they are gone to give CEE today. :p", "Solution_3": "c dis http://www.hindu.com/2009/05/30/stories/2009053058900300.htm\r\n\r\nis it true results will cum at 2 pm", "Solution_4": "I got AIR 215.\r\n\r\nState rank : 26", "Solution_5": "air 1124\r\nwow man \r\ni am pretty damn consistent :)", "Solution_6": "hey i was getting lesser marks than u..how can i get AIR 778??\r\n\r\nState rank 88..nice :P", "Solution_7": "People in Delhi really perform well.\r\n\r\nI got AIR 1151\r\n State rank 41 ( :P Not much competition in chennai compared to Delhi)\r\n\r\nTamil Nadu state 1 is V Prashant( AIR 8 :w00t: )", "Solution_8": "I think except chennai there is not much knowledge among students about engineering entrances. all they aim is state boards and get into local colleges.", "Solution_9": "Congrats to everyone!", "Solution_10": "air 74\r\nstate 3 (TN )\r\nmarks 377\r\n\r\n\r\nhow much did prashanth get \r\n\r\nand whos air 1", "Solution_11": "It's Nitin Jain once again :omighty:", "Solution_12": "Nitin jain got 417.\r\n\r\nAIR 3(Shyam Upadhyay of Chandigarh..nice friend of mine :oops: ) got 409", "Solution_13": "are these marks or rank?????? :o", "Solution_14": "marks. :lol:" } { "Tag": [], "Problem": "Debra began playing a computer game at 6:52 p.m. and she finished\nat 7:37 p.m. the same day. How many minutes over her 30-minute\nlimit did she play?", "Solution_1": "We can see that her limit would give her a stop at 7:22 pm. But 7:37 pm is $ \\boxed{15}$ over this limit.", "Solution_2": "[b]another way to do this is by\n\nbeginning - 6:52 +30=6:82 82-60=22 \n\nso she should have stopped by 7 : 22 but she stopped at 7 : 3 7 so subtracting we get the answer 15 or $\\boxed{15}$ [/b]" } { "Tag": [ "inequalities" ], "Problem": "Prove that (a+b)(a+c) is bigger than or equals to 2 * the root of abc(a+b+c) when a, b and c are real, positive numbers.\r\n\r\nI don't know how to type that in one line with square roots and other things, how do I do that?\r\n\r\nAnd is there a place where I can find some inequalities to practice on? I'm not so good at them.", "Solution_1": "Regarding your inequality the case when $a=0$ is trivial so if $a\\neq 0$ then the inequality is equivalent to:\r\n\\[ 2\\sqrt{\\alpha\\beta(1+\\alpha+\\beta)}\\leq (1+\\alpha)(1+\\beta) \\] \r\n$\\alpha,\\beta\\geq 0$ and in this case you just have to apply the lemma $2\\sqrt{ab}\\leq (a+b)/2$ by considering $a=\\alpha\\beta$ and $b=1+\\alpha+\\beta$ you'll have the result directly ;).\r\nNow for ressources related to inequalities ,there are several ones on the web if you have ghostview then you can check the packet on inequalities on this link :) :\r\nhttp://www.unl.edu/amc/a-activities/a4-for-students/s-index.html", "Solution_2": "How do you know that they are equivalent? Sorry, I'm new when it comes to proving inequalities. And I can't open the inequality packets, do I need a program to open them?", "Solution_3": "[quote=\"MoHaMeD\"]Regarding your inequality the case when $a=0$ is trivial so if $a\\neq 0$ then the inequality is equivalent to:\n\\[ 2\\sqrt{\\alpha\\beta(1+\\alpha+\\beta)}\\leq (1+\\alpha)(1+\\beta) \\] \n$\\alpha,\\beta\\geq 0$ and in this case you just have to apply the lemma $2\\sqrt{ab}\\leq (a+b)$ by considering $a=\\alpha\\beta$ and $b=1+\\alpha+\\beta$ you'll have the result directly.[/quote]\r\nif $ a \\neq 0$ then divide the two terms of the inequality by $a$ take $\\alpha=b/a$ and $\\beta=c/a$ this is what i meant by \"equivalent\".\r\nfor the packet yes you need $Ghostview$ to view it try to ask a teacher of yours who's interested in surfing the web i guess he should have it.", "Solution_4": "[quote=\"Masamune\"]$(a+b)(a+c) \\ge 2\\sqrt{abc(a+b+c)} \\ \\ a, b, c \\in \\mathbb{R^+}$[/quote]$\\frac{(a+b)(a+c)}{2} \\ge \\sqrt{abc(a+b+c)}$ expanding, $\\frac{a^2+ab+ac+bc}{2} \\ge \\sqrt{abc(a+b+c)}$, factoring, $\\frac{a(a+b+c)+bc}{2} \\ge \\sqrt{abc(a+b+c)}$ which is true by AM-GM." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that:\r\n\r\n\r\n\r\n $f(x+y)-f(x-y) = 4xy$ for all $x$ and $y$ non zero in $\\mathbb{R}$", "Solution_1": "Too easy or am I wrong?\r\n[hide=\"Hidden text\"]$x=y=\\frac{t}{2}$ $\\to$ $f(t)=t^2+f(0)$ for every $t$ non zero in $\\mathbb{R}$.\n\nSo the only solutions are $f(t)=t^2+C$ for any $C$ in $\\mathbb{R}$.[/hide]", "Solution_2": "[quote=\"lordWings\"]Too easy or am I wrong?\n[hide=\"Hidden text\"]$x=y=\\frac{t}{2}$ $\\to$ $f(t)=t^2+f(0)$ for every $t$ non zero in $\\mathbb{R}$.\n\nSo the only solutions are $f(t)=t^2+C$ for any $C$ in $\\mathbb{R}$.[/hide][/quote]\r\n\r\n\r\nGood Response if $f$ is defined in zero. But it is not!\r\n Try again :oops:", "Solution_3": "[quote=\"afaouzi\"]$f : \\mathbb{R} \\to \\mathbb{R}$[/quote]\n[quote=\"afaouzi\"]Good Response if $f$ is defined in zero. But it is not![/quote]\r\nSo it's $f : \\mathbb{R^*} \\to \\mathbb{R}$. OK, new problem. :P I'll try later.", "Solution_4": "$x=\\frac{z+t}{2}$, $y=\\frac{z-t}{2}$ $\\to$ $f(z)-f(t)=z^2-t^2$ for every $t,z\\in\\mathbb{R^*}$.\r\n$z$ and $t$ are independent, so we must have $f(z)=z^2+C$ for every $z\\in\\mathbb{R^*}$." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Prove that for $\\{i,j,m,n\\} \\subset \\mathbb{N} $\r\n\r\n\r\n$\\sum_{j=1}^n \\prod_{k=1}^m (j+k-1)=\\frac{1}{m+1}\\prod_{k=1}^{m+1}(n+k-1)$", "Solution_1": "as such the equality follows almost trivially by induction, but here's a combinatorial argument as well:(we prove \r\n$(m+1)\\sum_{j=1}^n\\prod_{k=1}^m(j+k-1)=(n+m)(n+m-1)\\cdots(n+1)n$)\r\n the RHS counts precisely the number of $m+1$-tuples from the set $\\{1,2,\\ldots,m+n\\}$ with all entries pairwise distinct.as for the LHS, we count the number of tuples with $j+m$ as its largest entry in the tuple, where $j$ runs from $1$ to $n$. since there are $m+1$ positions where $j+m$ can occur and $ (\\prod_{k=1}^m (j+k-1)$ such tuples, this is precisely the LHS." } { "Tag": [ "function", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Is the set of functions $ K \\equal{} \\{\\cos nt: n\\in N\\}$ compact or closed, or both? (in $ C^0([0,1],\\mathbf{R})$, under norm $ |f|\\equal{}\\sup_{t\\in[0,1]} f(t)$)", "Solution_1": "Does the sequence $ \\cos nt$ have a convergent subsequence?" } { "Tag": [ "calculus", "integration", "geometry", "rectangle", "complex analysis", "complex analysis unsolved" ], "Problem": "$ \\int_0^\\infty \\frac{e^{i \\beta x}}{e^x \\minus{} e^{\\minus{}x}} dx \\equal{} ?$", "Solution_1": "The integral may be written as\r\n\r\n$ \\frac{1}{2}\\int_{0}^{\\infty} e^{i\\beta x}\\sinh x dx$\r\n\r\nwhich may be integrated by parts (twice).", "Solution_2": "No it can't because $ \\sinh x \\neq \\frac{2}{e^x \\minus{} e^{\\minus{}x}}$.", "Solution_3": "But it is equal to $ \\frac{e^{x} \\minus{} e^{\\minus{}x}}{2}$.", "Solution_4": "Yeah, $ \\frac{e^{x}\\minus{}e^{\\minus{}x}}{2} \\equal{} \\sinh x$ but in integral there is: $ \\int \\frac{e^{i \\beta x}}{2 \\sinh x} dx$", "Solution_5": "Oops! Back to the drawing board.\r\n\r\nI presume you've considered the rectangle with vertices at $ \\minus{}R$, $ R$, $ R\\plus{} \\pi i$ and $ \\minus{}R \\plus{} \\pi i$ and the residue theorem.", "Solution_6": "[quote=\"@petko\"]$ \\int_0^\\infty \\frac {e^{i \\beta x}}{e^x \\minus{} e^{ \\minus{} x}} dx \\equal{} ?$[/quote]\r\n\r\nI think the integrand is asymptotic to $ \\frac {1}{x}$ near zero and so diverge." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given n possitive real numbers $a_1;a_2;...;a_n$\r\nProve the ineq :\r\n$(\\sqrt{2})^n(a_1+a_2)(a_2+a_3)...(a_n+a_1)<(a_1+a_2+a_3)(a_2+a_3+a_4)...(a_n+a_1+a_2)$\r\nCan someone replace $\\sqrt{2}$ by another number for the best ineq ?\r\nI think it is 1.5 but I can't prove it :(", "Solution_1": "1. Case even [i]n[/i]. For $a_1=a_3=...=a_{n-1}=0$ and $a_2=a_4=...=a_n=1$, we get $A\\le 2^{n/2}$.\r\n\r\n2. Case odd [i]n[/i]. For $a_1=a_3=...=a_{n-2}=0$ and $a_2=a_4=...=a_{n-1}=a_n=1$, we get $A \\le 2^{(n+1)/2}$.", "Solution_2": "Just note that $(a_1+a_2+a_3)^2\\geq(2a_1+a_2)(2a_3+a_2)$ and $(2a_1+a_2)(2a_2+a_1)\\geq 2(a_1+a_2)^2$\r\n\r\nBy the way, I believe I just saw this problem 2~3 weeks ago in a competition in Taiwan. Did you get this problem from a friend in Taiwan?\r\n\r\nP.S. This solution is not mine. One of my friends gave me it that time.", "Solution_3": "Dear Darkseer! I get this problem in a old Russian MO. \r\n$(2a_1+a_2)(a_1+2a_2)=2a_1^2+2a_2^2+4a_1a_2+a_1a_2>2(a_1+a_2)^2$\r\nThe equal didn't exist in that ineq. So we can only prove the base problem but I want to get a stronger ineq. For $a_1=a_2=...=a_n$ we'll get $A\\le (1,5)^n$", "Solution_4": "I conjecture that the best $A$ for odd $n$ is $A=\\frac{3^n}{2^n}$ for $n=3$ and $n=5$, and $A=2^{(n+1)/2}$ for odd $n\\ge7$.\r\nCan anybody to prove or disprove this one for $n=5$?", "Solution_5": "A nice inequality is also the following:\r\n\r\n$(1+\\frac{2a}{b+c})(1+\\frac{2b}{c+d})(1+\\frac{2c}{d+a})(1+\\frac{2d}{a+b}) \\ge 9.$\r\n\r\nI wait a nice solution. :)", "Solution_6": "[quote=\"Vasc\"]A nice inequality is also the following:\n\n$(1+\\frac{2a}{b+c})(1+\\frac{2b}{c+d})(1+\\frac{2c}{d+a})(1+\\frac{2d}{a+b}) \\ge 9.$\n\nI wait a nice solution. :)[/quote]\r\nDo we have any conditions on $a, b, c, d$?", "Solution_7": "I think, they are to be positive...", "Solution_8": "In the given context, $a,b,c,d$ are clearly non-negative.", "Solution_9": "Very nice problem.\r\nI'm very glad to find out a short solution only use AM-GM!", "Solution_10": "Post your solution, Hungkhtn.\r\nAnd think to the stronger inequality\r\n$(1+\\frac{3a}{b+c})(1+\\frac{3b}{c+d})(1+\\frac{3c}{d+a})(1+\\frac{3d}{a+b}) \\ge 16.$", "Solution_11": "$\\displaystyle (1+\\frac{2a}{b+c})(1+\\frac{2b}{c+d})(1+\\frac{2c}{d+a})(1+\\frac{2d}{a+b}) \\ge 9$\r\n\r\nclearly,it's equal with: \r\n\r\n$\\displaystyle (1+\\frac{a+c}{a+b})(1+\\frac{a+c}{c+d})(1+\\frac{b+d}{b+c})(1+\\frac{b+d}{a+d}) \\ge 9$\r\n\r\nAnd easy to prove: \r\n\r\n$\\displaystyle (1+\\frac{a+c}{a+b})(1+\\frac{a+c}{c+d}) \\ge (1+\\frac{2a+2c}{a+b+c+d})^2$\r\n\r\n$\\displaystyle (1+\\frac{b+d}{b+c})(1+\\frac{b+d}{a+d}) \\ge (1+\\frac{2b+2d}{a+b+c+d})^2$\r\n\r\nBy AM-GM\r\nLet\r\n$\\displaystyle x=\\frac{a+c}{a+b+c+d}$\r\n$\\displaystyle y=\\frac{b+d}{a+b+c+d}$\r\n\r\nThen $x+y=1$ and $0 \\le x,y \\le 1$\r\n\r\n$\\displaystyle (1+2x)(1+2y) \\ge 3 \\leftrightarrow 1+2(x+y)+xy \\ge 3$\r\n\r\nTrue,and the equal hold with $a=c=0,b=d$ or $b=d=0,a=c$\r\n\r\nVasc,can your post your solution?", "Solution_12": "Yes, but later, because my solution is also valid for the inequality \r\n\r\n$(1+\\frac{3a}{b+c})(1+\\frac{3b}{c+d})(1+\\frac{3c}{d+a})(1+\\frac{3d}{a+b}) \\ge 16.$\r\n\r\nVery nice your solution for the previous inequality!", "Solution_13": "What do you mean \"later\" for you, Vasc ?\r\n\r\nA day, a week, a month, a year, .........\r\n\r\n\r\nThank you. :)", "Solution_14": "A month, Manlio.\r\nBut I hope somebody give a solution in the meantime.", "Solution_15": "[quote=\"Vasc\"]Yes, but later, because my solution is also valid for the inequality \n\n$(1+\\frac{3a}{b+c})(1+\\frac{3b}{c+d})(1+\\frac{3c}{d+a})(1+\\frac{3d}{a+b}) \\ge 16.$\n\nVery nice your solution for the previous inequality![/quote]\r\n\r\nCan we replace 3 to other number,4 or 5?\r\nIs your solution enough to complete this problem,Vasc?", "Solution_16": "Yes, we can replace 3 with 4 or 5.\r\nMy solution works for all these cases.", "Solution_17": "Ok,I've just found the general ineq:\r\n\r\n$(1+\\frac{ka}{b+c})(1+\\frac{kb}{c+d})(1+\\frac{kc}{d+a})(1+\\frac{kd}{a+b}) \\ge (1+k)^2$", "Solution_18": "Of course, for $k$ positive.\r\nIt follows from well known inequality\r\n$\\sum\\frac{a}{b+c} \\ge 1$.\r\n\r\nand from\r\n\r\n$\\sum\\frac{ab}{(b+c)(c+d)} \\ge 1$.\r\n\r\nCongratulations!", "Solution_19": "VASE :can you show the solution to \r\n$\\sum\\frac{a}{b+c} \\ge 1$.\r\n\r\nand \r\n\r\n$\\sum\\frac{ab}{(b+c)(c+d)} \\ge 1$.\r\n\r\nI have some difficult in solving these.thank you in advance", "Solution_20": "I think \\[\r\n\\frac{a}{{b + c}} + \\frac{b}{{c + d}} + \\frac{c}{{d + a}} + \\frac{d}{{a + b}} \\ge 2\r\n\\]\r\n :?", "Solution_21": "[quote=\"Vasc\"]Of course, for $k$ positive.\nIt follows from well known inequality\n$\\sum\\frac{a}{b+c} \\ge 1$.\n\nand from\n\n$\\sum\\frac{ab}{(b+c)(c+d)} \\ge 1$.\n\nCongratulations![/quote]\r\n\r\nOk,My solution is that.But it should be:\r\n$\\sum\\frac{a}{b+c} \\ge 2$.\r\n\r\neasy by Cauchy-Swaczer.\r\n\r\nAnd the rest inequality:\r\n \r\n$\\sum\\frac{ab}{(b+c)(c+d)} \\ge 1$.\r\n\r\nI proof it by expanding.I can't sovle it by AM-Gm or Cauchy!\r\n\r\nVasc,I want to see your solution for it,if it doesn't use expanding?", "Solution_22": "I didn't use expanding.\r\nI used the inequalities\r\n$x_1+x_3 \\ge 2x_1x_3$, \r\n$x_2+x_4 \\ge 2x_2x_4$,\r\nwhere $x_1=a/(b+c)$ etc.", "Solution_23": "Can you please explain better your idea, Vasc? :?\r\n\r\n\r\nFor the first ineq you have\r\n\r\n$\\frac{a}{b+c}+\\frac{b}{c+d}+\\frac{c}{d+a}+\\frac{d}{a+b} \\geq 2\\frac{ac}{(b+c)(d+a)}+2\\frac{bd}{(c+d)(a+b)}$\r\n\r\nwhich is not greater than 2 ?????????????????????????????\r\n\r\nAnd for the second I really don't see how to apply your idea.!!!!!!!!!!!!!!!!!!", "Solution_24": "My solution for the second inequality is the following:\r\n\r\n$(x_1+x_3)(x_2+x_4)+x_1x_3+x_2x_4=(x_2+x_4)(x_1+x_3)/2+x_1x_3+(x_1+x_3)(x_2+x_4)/2+x_2x_4 \\ge (x_2+x_4+1)x_1x_3+(x_1+x_3+1)x_2x_4=1$\r\n\r\n ;)", "Solution_25": "Thank you very much, Vasc :)", "Solution_26": "[quote=\"Vasc\"]My solution for the second inequality is the following:\n\n$(x_1+x_3)(x_2+x_4)+x_1x_3+x_2x_4=(x_2+x_4)(x_1+x_3)/2+x_1x_3+(x_1+x_3)(x_2+x_4)/2+x_2x_4 \\ge (x_2+x_4+1)x_1x_3+(x_1+x_3+1)x_2x_4=1$\n\n ;)[/quote]\r\nvery very nice.thank you", "Solution_27": "Inequality \r\n\\[ \\left( \\frac{a_1 + a_2 + a_3}{a_1 + a_2} \\right) \r\n \\left( \\frac{a_2 + a_3 + a_4}{a_2 + a_3} \\right) \\dots\r\n \\left( \\frac{a_n + a_1 + a_2}{a_n + a_1} \\right) \\geq \r\n \\left( \\! \\sqrt{2} \\, \\right)^n \\]\r\nis not from old Russian MO, it's my problems from 239 school MO, 2000. For \r\nan even n the constant $\\sqrt{2}$ in the best. For odd it can be improved. \r\nWe can put constant 3/2 for n=3 and n=5 only. Case n=5 is my problems from \r\n239 school MO, 2000. The best constnt for odd numbers greater than 5 is \r\nunknown. Details you can see at my articles \"About Mongolian Inequality\",\r\nthey are published at journal \"Matematichaskoe Prosveschenie\", 2003 \r\n(condensed version) and at Problems of SPb MO, 2002 (full version). You also \r\ncan find articles on my www. Unfortunately they are in Russian only.", "Solution_28": "I think the following conjecture is valid for $n=5$ and $p \\le 5/4$:\r\n\r\n$\\prod_{cyclic}(1+\\frac{pa_1}{a_2+a_3}) \\ge (1+p/2)^5$.", "Solution_29": "[quote=\"Vasc\"]I think the following conjecture is valid for $n=5$ and $p \\le 5/4$:\n\n$\\prod_{cyclic}(1+\\frac{pa_1}{a_2+a_3}) \\ge (1+p/2)^5$.[/quote]\r\n\r\nVery nice!And you have found a nice solution,Vasc?", "Solution_30": "No solution ! :(", "Solution_31": "[quote=\"hungkhtn\"]\n...\nAnd easy to prove: \n\n$\\displaystyle (1+\\frac{a+c}{a+b})(1+\\frac{a+c}{c+d}) \\ge (1+\\frac{2a+2c}{a+b+c+d})^2$\n\n$\\displaystyle (1+\\frac{b+d}{b+c})(1+\\frac{b+d}{a+d}) \\ge (1+\\frac{2b+2d}{a+b+c+d})^2$\n...\n[/quote]\r\n\r\nCan You show proof?", "Solution_32": "[quote=\"Vasc\"]I think the following conjecture is valid for $ n \\equal{} 5$ and $ p \\le 5/4$:\n\n$ \\prod_{cyclic}(1 \\plus{} \\frac {pa_1}{a_2 \\plus{} a_3}) \\ge (1 \\plus{} p/2)^5$.[/quote]\r\n\r\n\r\nHave you proved it\uff1f" } { "Tag": [ "real analysis", "inequalities" ], "Problem": "Fie $A$ o multime inclusa in $[0,1].$ Atunci multimea este Lebesgue masurabila daca si numai daca\r\n \\[ \\lambda^{*} (A) +\\lambda^{*} ([0,1]-A) = 1. \\]", "Solution_1": "The direct implication is obvious. \r\nFor the indirect one, we have to show that: $\\lambda^{*}(E) \\geq \\lambda^{*}(E \\cap A) + \\lambda^{*}(E-A)$ for every set $E$ of reals (the subadditivity of $\\lambda^{*}$ giving the reverse inequality). We can assume $E \\subseteq [0,1]$ since the $\\sigma$-algebra on $[0,1]$ given by the trace of the Lebesgue $\\sigma$-algebra on $\\mathbb{R}$ is exactly the Lebesgue $\\sigma$-algebra on $[0,1]$.\r\nSuppose the result to be false. Then $\\exists E \\subseteq [0,1], \\lambda^{*}(E) < \\lambda^{*}(E \\cap A) + \\lambda^{*}(E-A)$. Hence by definition of $\\lambda^{*}$ there exists an open (in $[0,1]$) set $O$ (containing $E$) satisfying $\\lambda^{*}(O) < \\lambda^{*}(O \\cap A) + \\lambda^{*}(O-A)$. But then:\r\n$\\lambda^{*}(O^{c} \\cap [0,1])+\\lambda^{*}(O) < \\lambda^{*}(O^{c} \\cap A) + \\lambda^{*}(O^{c} \\cap [0,1]-A) + \\lambda^{*}(O \\cap A) + \\lambda^{*}(O-A)$,\r\n$1 < [\\lambda^{*}(O^{c} \\cap A)+\\lambda^{*}(O \\cap A)]+[\\lambda^{*}(O^{c} \\cap [0,1]-A)+\\lambda^{*}(O-A)]$,\r\n$1 < \\lambda^{*}(A)+\\lambda^{*}(A^{c}\\cap [0,1]) = 1$, which is absurd.\r\nFor the last inequality we used the measurability of $O$ and $O^{c}\\cap [0,1]$.", "Solution_2": "A similar one: define the inner measure $m_{*}(E)=sup(\\{m(U); \\ U \\subseteq E, \\ U \\ measurable\\})$ where $m$ is the Lebesgue measure on $\\mathbb{R}$. Show that $m_{*}(E)=1-m^{*}([0,1]-E)$ for all $E \\subseteq [0,1]$, where $m^{*}$ denotes the Lebesgue outer measure." } { "Tag": [ "vector", "probability", "linear algebra", "matrix", "function", "integration", "probability and stats" ], "Problem": "how would you do to generate a random vector $ (X_1, X_2, \\ldots, X_n)$ such that\r\n1/ $ P(X_i\\equal{}1) \\equal{} 1\\minus{}P(X_i\\equal{}0)\\equal{}p_i$ is a given \r\n2/ $ Cov(X_i X_j)\\equal{}\\sigma_{i,j}$ is given\r\nas efficiently as possible ?", "Solution_1": "please someone answer this..", "Solution_2": "I will assume for simplicity that the probability of success for all $ X_i$ is $ p$. One possible approach to simulate vector $ X$ is this: let $ \\xi_1,\\ldots,\\xi_n$ be iid Bernoulli with probablity of success $ p$. Let $ X_1\\equal{}\\xi_1$. Let $ \\eta_2$ be independent of $ \\xi_i$ s.t. $ P(\\eta_2\\equal{}1)\\equal{}q_1, \\ P(\\eta_2\\equal{}2)\\equal{}1\\minus{}q_1$, and define $ X_2\\equal{}\\xi_{\\eta}$. It is easy to check that if $ q_1\\equal{}\\frac{\\sigma_{12}}{p(1\\minus{}p)}$, then we get the desired covariation. Now do the same for $ X_3$: this time $ \\eta_3$ can take 3 values, and the probabilities assigned to them are the solution of the system of linear equations of order 3.\r\nAnyway, this is pretty straightforward and probably not the most effective procedure.", "Solution_3": "Yustas, I must admit that I don't understand what have you written :(. Can you ( or someone else ) please explain in detail how to solve alekk's problem?\r\n\r\nI would also like to add another question; suppose that we want to generate a random vector $ (X_1,\\dots,X_n)$, but let's assume that $ X_i$ (*) are continuous variables and we are given the covariance matrix $ \\Sigma$. How would you do it?\r\nWe could express the distribution function $ \\mathbb{F}_{(X_1,\\dots,X_n)}(x_1,\\dots,x_n) \\equal{} \\int_{t_i\\leq x_i,\\i \\equal{} 1,\\dots n}f(t_1,\\dots,t_n,\\Sigma)dt_1\\dots dt_n$, then simulate random variable $ U(\\omega)$ from uniform distribution and invert $ \\mathbb{F}_{(X_1,\\dots,X_n)}^{ \\minus{} 1}(U(\\omega))$.\r\nBut that seems to complicated, and it would be challenging to write a program that would do that in general situations...\u00b8\r\n\r\nEdit: correction: (*) we assume that $ (X_1,\\dots,X_n)$ is continuous, that is, it has a density function.", "Solution_4": "Basically, the idea is to generate our sequence by transforming the array of independent random variables $ \\xi_i$ by assigning the random numbers $ \\eta_i$, that is to look at $ \\xi_{\\eta_i}$. I gave some details above.", "Solution_5": "1) The problem is NP-hard. Moreover, even determining whether there exists a distribution of $ (X_1,\\dots, X_n)$ with the given covariance matrix is NP hard (each $ X_i$ is a 0-1 random variable).\r\n\r\n2) Also the covariance matrix doesn't uniquely define the distribution of $ (X_1,\\dots, X_n)$.\r\n\r\n3) The problem becomes easy if we don't require that each $ X_i$ is a 0-1 random variable. Then we can sample vector $ X$ from a multivariate Gaussian distribution ( http://en.wikipedia.org/wiki/Multivariate_normal_distribution ).", "Solution_6": "do you have any reference for the NP-hard thing ?[/youtube]", "Solution_7": "I will sketch the proof. Let us consider a special case when all $ p_i \\equal{} 1/2$. For simplicity (of notation), we assume that each $ X_i$ takes two values -1 and 1 --- each w.p. 1/2.\r\n\r\nConsider the set $ C$ of covariance matrices corresponding to different joint distributions of $ (X_1,\\dots,X_n)$. Note that the set $ C$ is convex: if $ A\\in C$ is the covariance matrix of vector $ X$ and $ B\\in C$ is the covariance matrix of vector $ Y$, then $ (A \\plus{} B)/2$ is the covariance of the random vector\r\n\\[ Z \\equal{} \\begin{cases} X, & \\text{w.p. } 1/2; \\\\\r\nY, & \\text{w.p. } 1/2. \\end{cases}\r\n\\]\r\nAssume that we are given a graph $ G \\equal{} (V,E)$ on the set $ V \\equal{} \\{1,\\dots,n\\}$. Note that each vector $ X$ defines a cut on $ G$ (i.e. partitioning of its vertices into two parts): $ S \\equal{} \\{i: X_i \\equal{} 1\\}$ and $ T \\equal{} \\{i: X_i \\equal{} \\minus{} 1\\}$. \r\n\r\nMoreover, for every cut $ V \\equal{} S\\cup T$ in $ G$, there is a corresponding random vector $ X_{S,T}$:\r\n\\[ X_{S,T} \\equal{} \\begin{cases} \\chi, & \\text{w.p. } 1/2; \\\\\r\n\\minus{} \\chi, & \\text{w.p. } 1/2. \\end{cases}\r\n\\]\r\nwhere\r\n\\[ \\chi_i \\equal{} \\begin{cases} 1, & \\text{if } i\\in S; \\\\\r\n\\minus{} 1, & \\text{if } i\\in T. \\end{cases}\r\n\\]\r\nWe compute the covariance matrix $ A^{S,T}$ of $ X_{S,T}$:\r\n\\[ a_{ij} \\equal{} \\begin{cases} 1, & \\text{if } i\\in S, j\\in S \\text{ or } i\\in T, j\\in T \\\\\r\n\\minus{} 1, & \\text{if } i\\in S, j\\in T \\text { or } i\\in T, j\\in S. \\end{cases}\r\n\\]\r\nIt is easy to see that matrices $ A^{S,T}$ form the set of vertices of $ C$. Thus the set $ C$ is called [b]the cut polytope[/b] in the literature.\r\n\r\nFinally, we compute the number of cut edges (i.e. edges going from $ S$ to $ T$):\r\n\\[ f(A) \\equal{} \\sum_{(i,j)\\in E} \\frac {1 \\minus{} a_{ij}}{2}\r\n\\]\r\nNote that the number of cut edges is given by a linear function.\r\n\r\nIf we could check whether a point belongs to $ C$ we would be able find the maximum value of a linear function over $ C$ (see e.g. http://www-math.mit.edu/~vempala/papers/adamanneal.pdf ). In particular, we would be able to find the vertex of $ C$ that maximizes $ f$ --- or in other words, we would be able to find the maximum cut in $ G$. But the Maximum Cut Problem is NP-hard [Richard M. Karp (1972). \"Reducibility Among Combinatorial Problems\"].\r\n\r\nBy the way, there is a lot of research on the cut polytope and its \"relaxations\" in computer science.", "Solution_8": "thank you very much! that was very interesting !" } { "Tag": [ "probability", "probability and stats" ], "Problem": "There are $ n$ persons and $ n$ seats available for them.They are all assigned seat numbers written on a ticket given to them.Now there is one person(say $ Mr.X$) on whose ticket the number is not mentioned.\r\n\r\nThe people are told to go and sit one by one,$ Mr.X$ being the first person.\r\n\r\nThe rule is that if $ Mr.X$ doesn't know the number on the ticket then he can sit on any seat chosen at random by him.\r\n\r\nAnd if the person who comes after $ Mr.X$(say $ Mr.Y$) sees that $ Mr.X$ is sitting on the seat meant for him then $ Mr.Y$ can sit on any seat he likes chosen at random, otherwise $ Mr. Y$ will sit on the seat assigned to him on the ticket.Then $ Mr.Z$ comes and if he sees that his seat is occupied then he can sit on any seat he likes otherwise he will sit on the seat assigned to him.This process continues till the last person has been seated. \r\n\r\nFind the probability that the last person is sitting on the seat assigned for him.", "Solution_1": "I think the above problem is akin to the one posted in the thread below: \r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278908[/url]\r\n\r\nP.S.: A solution had been posted in the 'link' that I had provided." } { "Tag": [ "geometry", "incenter", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "Given quadrilateral $ABCD$ inscribed $(O)$ $AC\\cap BD=I$ the circle $(C)$ tangent the rays (not segment )$ID,IC$ at $M,N$ and tangent external $(O)$, Let $I_{a},I_{b}$ be excenters of triangle $ADC$ and $BDC$ corespond angle $A,B$ resp,Prove that $M,N,I_{a},I_{b}$ are colinear. :D", "Solution_1": "Applying Thebault theorem for triangle $ADC$ and $BDC$we have the above result. This problem is still right when the circle $(C)$ tangent internal $(O)$.\r\nThen $M,N$ and the incenters of triangle $ADC$ and $BDC$ are collinear.", "Solution_2": "Do you understand my problem,Huy\u1ec1n V\u0169 ?\r\n[quote=\"Wolfram MathWorld\"]Let $ABC$ be a triangle and $D$ a point on the side $BC$. Let $I$ be the incenter, $P$ the center of the circle tangent to the circumcircle and segments $AD$ and $BD$, $Q$ the center of the circle tangent to the circumcircle and segments $AD$ and $CD$. Then Th\u00e9bault's theorem states that $P, Q$ and $I$ are collinear.[/quote]\r\nActually \"Th\u00e9bault's theorem\" is only corollary of mine, It is three colinear centers while my problem need to prove tangents points and centers, more the relations theorem is for internal tangent circles while mine is for external!", "Solution_3": "Oh sorry .The problem that I wrote is the lemma to prove Theault theorem.\r\nYou can see the lemma here. This lemma is right when (C') tangent external or internal the circle (O). http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf\r\nWhen (C') tangent external (O) the tangents point and excenter are collinear (*)\r\nWhen (C') tangent internal (O) the tangents point and incenter are collinear (**)\r\nYou can solve (*) similarly the solution for lemma in the file or use (**) to solve (*)\r\nThis is my solution for (*)\r\n[b]Lemma:[/b]\r\nLet $I$ be the point on $AC$ of triangle $ADC$. The circle $(C')$ tangents the ray $ID,IC$ at $M,N$ and tangents external the circumcircle $(O)$ of triangle $ADC$ at $K$.Let $Ia$ be excenter of triangle $ADC$ corespond angle $A$. Prove that $M,N,Ia$ are collinear\r\n\r\n[b]Solution[/b]\r\n$KN,KM$ cuts $(O)$ at $P,Q.$\r\n$K$ is the similar point of $(O)$ and $(C')$ $(1)$ so $C'N // OP$. but $C'N \\bot AC$\r\n so $OP \\bot AC$. So $P$ is the midpoint of arc $AC(2)$\r\n$MN$ cuts $DP$ at $Ia$\r\n$\\angle PCN=180-\\angle PCA=180-\\angle PAC$ (because of (2)) $=\\angle PKC$\r\nSo Triangle $PKC$ and $PCN$ are similar\r\n$PC^{2}=PK*PN (4)$\r\nwe have $MN//PQ$ (because of (1))\r\nSo $\\angle IaMK=\\angle PQK=\\angle PDK=\\angle IaDK$\r\nThus $M,D,K,Ia$ are cyclic. So $\\angle DIaK=\\angle DMK=\\angle KNM$\r\nSo that triangle $PIaK$ and $PNIa$ are similar\r\nso $PIa^{2}=PK*PN (5)$\r\nFrom (4) and (5) $PIa=PC=PA.$\r\nSo $A,C,Ia$ are on the circle center $P$ and the radius $PA.$\r\nWe have $\\angle CAIa=1/2 \\angle CPIa=1/2 \\angle CAD$ \r\nSo $AIa$ is the angle bisector $CDA$ or $Ia$ is the excenter of triangle $ACD$ $q.e.d$\r\nApplying this lemma for the main problem we have solution", "Solution_4": "See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41667]Concyclic points with triangle incenter[/url] (external tangency and excenter) and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=118385]Circles and circles[/url] (internal tangency and incenter)." } { "Tag": [], "Problem": "In how many ways can 36 be written as the product\r\na \u00d7 b \u00d7 c \u00d7 d, where a, b, c and d are positive integers such\r\nthat a \u2264 b \u2264 c \u2264 d ?", "Solution_1": "We are asked to find in how many\r\nways 36 can be written as the\r\nproduct a \u00d7 b \u00d7 c \u00d7 d, where a, b, c\r\nand d are positive integers such that\r\na \u2264 b \u2264 c \u2264 d.\r\nFirst factor 36:\r\n36 = 2 \u00d7 2 \u00d7 3 \u00d7 3\r\nThe factors of 36 are:\r\n1, 2, 3, 4, 6, 9, 12, 18 and 36\r\nSuppose the highest value we use is\r\n36.\r\n1 \u00d7 1 \u00d7 1 \u00d7 36\r\nCan\u2019t do anything more with 36.\r\nNow 18.\r\n1 \u00d7 1 \u00d7 2 \u00d7 18\r\nNow 12\r\n1 \u00d7 1 \u00d7 3 \u00d7 12\r\nNow 9:\r\n1 \u00d7 1 \u00d7 4 \u00d7 9\r\n1 \u00d7 2 \u00d7 2 \u00d7 9\r\nNow 6:\r\n1 \u00d7 1 \u00d7 6 \u00d7 6\r\n1 \u00d7 2 \u00d7 3 \u00d7 6\r\nNow 4:\r\n1 \u00d7 3 \u00d7 3 \u00d7 4\r\nNow 3:\r\n2 \u00d7 2 \u00d7 3 \u00d7 3\r\nWe\u2019re done. Let\u2019s count them up: There are 9. so the ans is 9, i think.[/i]", "Solution_2": "[quote=\"HiDN428\"]We are asked to find in how many\nways 36 can be written as the\nproduct a \u00d7 b \u00d7 c \u00d7 d, where a, b, c\nand d are positive integers such that\na \u2264 b \u2264 c \u2264 d.\nFirst factor 36:\n36 = 2 \u00d7 2 \u00d7 3 \u00d7 3\nThe factors of 36 are:\n1, 2, 3, 4, 6, 9, 12, 18 and 36\nSuppose the highest value we use is\n36.\n1 \u00d7 1 \u00d7 1 \u00d7 36\nCan\u2019t do anything more with 36.\nNow 18.\n1 \u00d7 1 \u00d7 2 \u00d7 18\nNow 12\n1 \u00d7 1 \u00d7 3 \u00d7 12\nNow 9:\n1 \u00d7 1 \u00d7 4 \u00d7 9\n1 \u00d7 2 \u00d7 2 \u00d7 9\nNow 6:\n1 \u00d7 1 \u00d7 6 \u00d7 6\n1 \u00d7 2 \u00d7 3 \u00d7 6\nNow 4:\n1 \u00d7 3 \u00d7 3 \u00d7 4\nNow 3:\n2 \u00d7 2 \u00d7 3 \u00d7 3\nWe\u2019re done. Let\u2019s count them up: There are 9. so the ans is 9, i think.[/i][/quote]\r\n\r\n\r\n\r\nMeh..I kinda feel stupid replying to my own post but.. This is how I did it.\r\n\r\nFirst factor 36: \r\n36 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \r\nThe factors of 36 are: \r\n1, 2, 3, 4, 6, 9, 12, 18 and 36 \r\nSuppose the highest value we use is \r\n36. \r\n1 \u00d7 1 \u00d7 1 \u00d7 36 \r\nCan\u2019t do anything more with 36. \r\nNow 18. \r\n1 \u00d7 1 \u00d7 2 \u00d7 18 \r\nNow 12 \r\n1 \u00d7 1 \u00d7 3 \u00d7 12 \r\nNow 9: \r\n1 \u00d7 1 \u00d7 4 \u00d7 9 \r\n1 \u00d7 2 \u00d7 2 \u00d7 9 \r\nNow 6: \r\n1 \u00d7 1 \u00d7 6 \u00d7 6 \r\n1 \u00d7 2 \u00d7 3 \u00d7 6 \r\nNow 4: \r\n1 \u00d7 3 \u00d7 3 \u00d7 4 \r\nNow 3: \r\n2 \u00d7 2 \u00d7 3 \u00d7 3 \r\nSo there are a total of 9 ways. \r\nOk, is there an easier way..lol?", "Solution_3": "haha, idk, i think the best way is just to list them out because there really isn't that much..." } { "Tag": [ "probability", "videos" ], "Problem": "A COMPUTER SOFTWARE PACKAGE GENERATES RANDOM NUMBERS THAT ARE NORMALLY DISTRIBUTED WITH A MEAN AND STANDARD DEVIATION THAT ARE ENTERED BY THE USER. SUPPOSE THAT THE USER ENTERS A MEAN OF 400 AND A STANDARD DEVIATION OF 100. FIND THE PROBABILITY THAT A VALUE WILL BE\r\n\r\n1.) BETWEEN 525 AND 650\r\n2.) ABOVE 490\r\n3.) BELOW 535\r\n4.) BELOW 357\r\n5.) ABOVE 619\r\n6.) BETWEEN 340 AND 714\r\n\r\nPlease help me how can I answer with this one. Even steps will help. Thanks in advance!", "Solution_1": "Ew, statistics. Brings back memories of statistics videos of MathPath. If only I had payed attention once in a while during those videos, I would most likely remember the formula for standard deviations.\r\n\r\nI guess my post is spam then.", "Solution_2": "Normalize the scores by $\\frac{X-\\mu}{\\sigma}$ and then look up the p-value in a z table.", "Solution_3": "Hokkage, can you show me one example please? I have a z table here..", "Solution_4": "Hokkage means to find the z-scores, by using $z=/frac{x-\\mu}{\\sigma}$, where $x$ is the given values. For example, for $x=525$, $z=\\frac{525-400}{100}=1.25$. So part 1) becomes: Find $P(1.25y>0$ and $2(x+y)\\geq 5\\sqrt{xy}$. Prove that $x\\geq 4y$.", "Solution_1": "yes,it is small.\r\njust $\\sqrt{\\frac x y}=a$,$a>1$\r\nwe have $a+\\frac 1 a \\ge \\frac 5 2$\r\nso $a \\ge 2$\r\n...", "Solution_2": "You're right zhaobin\r\n$\\ 2(x+y) \\geq 5 \\sqrt{xy} \\Rightarrow x-\\frac{5}{2} \\sqrt{xy}+y \\geq 0$\r\n$\\ \\rightarrow (\\sqrt{x}-2\\sqrt{y})(\\sqrt{y}-2\\sqrt{x}) \\geq 0$\r\nwhich shows your idea: $\\ \\sqrt{\\frac{x}{y}} \\geq 2$" } { "Tag": [ "algorithm", "search" ], "Problem": "I have been making suggestion for this Section for a while :? \r\nNobody seems to notice :)\r\nAnyway, who would like to see something like this, please vote.", "Solution_1": "Yeah, could be a good idea. At least, considering it would be mainly about algorythms and not really about coding language, as the algorythm search comes pretty close to the thinker-math questions :D", "Solution_2": "Especially graph theory and other counting stuff :)", "Solution_3": "Excellent idea!", "Solution_4": "You guys must convince more than 9 people to vote :) (and especially to vote yes :P) ...", "Solution_5": "I don't get people that vote NO. Does it trouble them in any way? I know that I am wrong , but when I put the option No, I didn't think there would be such votes. Otherwise, the only answer I would have allowed would be YES :) :D", "Solution_6": "well I have no idea, but not even half the people said yes, so :( sorry ..", "Solution_7": "Though programming is a part of my profession (I lecture some programming cources) I think MathLinks is not right place for subj.", "Solution_8": "Why not Myth?\r\nShouldnt Mathlinks expand?\r\n\r\nI dont understand the people who voted no...i'm sorry...", "Solution_9": "[quote=\"PiDeltaPhi\"]Why not Myth?\nShouldnt Mathlinks expand?\n\nI dont understand the people who voted no...i'm sorry...[/quote] Well for the moment it's hard maitaining the site with only math people :) expanding to CS would mean a major increase in web traffic, and I am not (yet) prepared for this :) \r\n\r\nHowever, people can discuss CS subjects in the Bable Tower :) :D", "Solution_10": "[quote=\"Valentin Vornicu\"][quote=\"PiDeltaPhi\"]Why not Myth?\nShouldnt Mathlinks expand?\n\nI dont understand the people who voted no...i'm sorry...[/quote] Well for the moment it's hard maitaining the site with only math people :) expanding to CS would mean a major increase in web traffic, and I am not (yet) prepared for this :) \n\nHowever, people can discuss CS subjects in the Bable Tower :) :D[/quote]\r\n\r\nI understand that :)", "Solution_11": "programming rules! :D :D :D", "Solution_12": "A forum for computer sciences and informatics already exists. And once again, please do not revive old threads.\r\n\r\nLocked." } { "Tag": [ "calculus", "function", "limit", "logarithms", "floor function", "calculus computations" ], "Problem": "This is a small arithmetic trifle, not a serious problem.\r\n\r\nI was writing up a classroom handout for a calculus class on the \"Hierarchy of Size,\" which is something I think quite important to the understanding of calculus. As an example, I talked about the function $ f(x) \\equal{} x^{1000}e^{ \\minus{} .001x}$ on $ [0,\\infty).$ I wrote that although the function increases at first and reaches an impressively high maximum value, it still eventually decreases and $ \\lim_{x\\to\\infty}x^{1000}e^{ \\minus{} .001x} \\equal{} 0.$ \r\n\r\nBut then, to get the effect I intended, I needed to name that maximum value.\r\n\r\nSo: Let $ M \\equal{} \\max_{x\\in[0,\\infty)}x^{1000}e^{ \\minus{} .001x}.$\r\n\r\n1. Using only pencil-and-paper arithmetic - no calculator or computer - estimate the order of magnitude of $ M.$\r\n\r\n2. Using only an ordinary floating-point device (calculator or spreadsheet), give a reasonable floating-point approximation of $ M.$\r\n\r\nDo not use anything (Mathematica, Maple, TI-89, etc.) that uses arbitrary-precision arithmetic; the restriction is to ordinary floating-point arithmetic.", "Solution_1": "[hide=\"Hint for 1.\"]Two of the many approximations I have memorized over the years: $ \\ln 10\\approxeq2.303$ and $ \\log_{10}e\\approxeq.4343.$[/hide]", "Solution_2": "$ f_n(x)\\equal{}x^{10^n}e^{\\minus{}x10^{\\minus{}n}}$\r\n\r\n$ f_n'(x) \\equal{} 10^n x^{10^n\\minus{}1}e^{\\minus{}x10^{\\minus{}n}}\\minus{}10^{\\minus{}n}x^{10^n}e^{\\minus{}x10^{\\minus{}n}}\\equal{}(\\frac{10^n }{x}\\minus{}\\frac{1}{10^n})f_n(x)$\r\n\r\n$ f_n'(x)\\equal{}0 \\to x\\equal{}10^{2n}$\r\n\r\n$ f_n(10^{2n})\\equal{}(10^{2n})^{10^n}e^{\\minus{}10^{2n}10^{\\minus{}n}}\\equal{}\\left( \\frac{10^{2n}}{e}\\right)^{10^n}$\r\n\r\nfor your n=3,\r\n\r\n$ f_3(10^6)\\equal{}\\left( \\frac{100,000}{e}\\right)^{1000}$", "Solution_3": "Ah, but the point was a base-10 approximate representation of that number.", "Solution_4": "uhhh then how about\r\n\r\n$ \\frac{10}{e}\\approx 4, 2^{10} \\approx 10^3$\r\n\r\n$ \\sup_{x\\in [0,\\infty )} f_n(x) \\approx (4\\cdot 10^{2n\\minus{}1})^{10^n} \\equal{} 2^{2\\cdot 10^{n}}10^{(2n\\minus{}1)10^n} \\approx (10^3)^{2\\cdot 10^{n\\minus{}1}}10^{(2n\\minus{}1)10^n}$\r\n\r\n$ \\equal{}10^{(2n\\minus{}1)10^n\\plus{}6\\cdot 10^{n\\minus{}1}}\\equal{}10^{4(5n\\minus{}1)10^{n\\minus{}1}}$\r\n\r\n$ \\to \\sup_{x\\in [0,\\infty )} f_3(x) \\approx 10^{5600}$", "Solution_5": "You are sligtly better off with $ e^7\\approx 1.1\\cdot 10^3$, $ 1.1^{10} \\approx e$, so\r\n\\[ e^{1000}\\approx 1.1^{143}\\cdot 10^{429} /e\\approx 1.1^3 e^{13}\\cdot 10^{429}\\approx 1.1^5\\cdot 10^{435}/e\\]\r\nso\r\n\\[ \\frac{10^{6000}}{e^{1000}}\\approx 10^{5565}\\sqrt e\\approx 1.6\\cdot 10^{5565}.\\]\r\nThe (relative) error is about $ \\frac{1}{250}\\cdot 143$ coming mainly from the estimate of $ e^7$, which is really $ 1096.\\text{something}$ (the thing I memorized long ago when trying to find a power of $ e$ close to a power of $ 10$), so one may be pretty sure that $ 10^{5565}$ is a fairly close guess (unless I made some stupid mistake in my arithmetic).", "Solution_6": "Both of you have been going after question 1. Here's my version of that, depending on my long-ago-memorized $ \\log e\\approxeq.4343.$ (I'm using $ \\log$ here to mean $ \\log_{10}.$)\r\n\r\n$ M\\equal{}(10^6)^{1000}e^{\\minus{}1000}\\equal{}10^{6000}e^{\\minus{}1000}.$\r\n\r\n$ \\log M\\equal{}6000\\minus{}1000\\log e\\approx 6000\\minus{}434.3\\equal{}5565.7.$\r\n\r\nSince $ \\log 2\\equal{}.3$ (which is the same fact as $ 2^{10}\\equal{}10^3$) implies that $ \\log 5\\equal{}.7,$ that gives us an estimate for $ M$ of $ 5\\times 10^{5565}.$\r\n\r\nSo fedja definitely has the correct order of magnitude. To settle the issue of whether $ 1.6$ or $ 5$ is the better number up front requires us to move on to question 2, in which we do get to use an ordinary calculator or spreadsheet. There, the computation I want to do this: let $ a\\equal{}(6000\\minus{}1000\\log e)\\minus{}\\lfloor 6000\\minus{}1000\\log e\\rfloor,$ and then we let $ b\\equal{}10^a.$ We report the answer as $ b\\times 10^{5565}.$ This gives us $ 5.07596\\times 10^{5565}.$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be nonnegative real numbers.\r\n\r\n(a) If $ a \\plus{} b \\plus{} c \\equal{} 3$, then\r\n\r\n$ \\frac {7a^2 \\plus{} 9}{b^2 \\plus{} c^2} \\plus{} \\frac {7b^2 \\plus{} 9}{c^2 \\plus{} a^2} \\plus{} \\frac {7c^2 \\plus{} 9}{a^2 \\plus{} b^2}\\ge 24$;\r\n\r\n(b) If $ ab \\plus{} bc \\plus{} ca \\equal{} 3$, then\r\n\r\n$ \\frac {4a^2 \\plus{} 3}{b^2 \\plus{} c^2} \\plus{} \\frac {4b^2 \\plus{} 3}{c^2 \\plus{} a^2} \\plus{} \\frac {4c^2 \\plus{} 3}{a^2 \\plus{} b^2}\\ge \\frac {21}{3}$;\r\n\r\n(c) If $ abc \\equal{} 1$, then\r\n\r\n$ \\frac {3a^2 \\plus{} 1}{b^2 \\plus{} c^2} \\plus{} \\frac {3b^2 \\plus{} 1}{c^2 \\plus{} a^2} \\plus{} \\frac {3c^2 \\plus{} 1}{a^2 \\plus{} b^2}\\ge 6$.", "Solution_1": "[quote=\"Vasc\"]Let $ a,b,c$ be nonnegative real numbers.\n\n(a) If $ a \\plus{} b \\plus{} c \\equal{} 3$, then\n\n$ \\frac {7a^2 \\plus{} 9}{b^2 \\plus{} c^2} \\plus{} \\frac {7b^2 \\plus{} 9}{c^2 \\plus{} a^2} \\plus{} \\frac {7c^2 \\plus{} 9}{a^2 \\plus{} b^2}\\ge 24$(1)\n\n[/quote]\r\n\r\n$ (1) \\Leftrightarrow \\sum S_{c}(a\\minus{}b)^2 \\geq 0$\r\n\r\nWith : \r\n\r\n$ S_{a}\\equal{}4b^4\\plus{}7b^2c^2\\plus{}10b^3c\\plus{}10bc^3\\plus{}4c^4\\minus{}a^2(a^2\\plus{}3b^2\\plus{}3c^2)$\r\n\r\n$ S_{b}\\equal{}4c^4\\plus{}7c^2a^2\\plus{}10c^3a\\plus{}10ca^3\\plus{}4a^4\\minus{}b^2(b^2\\plus{}3c^2\\plus{}3a^2)$\r\n\r\n$ S_{c}\\equal{}4a^4\\plus{}7a^2b^2\\plus{}10a^3b\\plus{}10ab^3\\plus{}4b^4\\minus{}c^2(c^2\\plus{}3a^2\\plus{}3b^2)$\r\n\r\nAssume : $ a\\leq b \\leq c$ .\r\n\r\nWe have :\r\n\r\n$ S_{a}\\equal{}3b^4\\plus{}4b^2c^2\\plus{}c^4\\plus{}10bc^3\\plus{}(b\\minus{}a)(b\\plus{}a)(b^2\\plus{}a^2) \\geq 0$ \r\n\r\n$ S_{b}\\equal{}(c\\minus{}b)(c\\plus{}b)(c^2\\plus{}b^2)\\plus{}(c^2\\plus{}a^2)(3a^2\\plus{}10ca\\plus{}3(c\\minus{}b)(c\\plus{}b)) \\geq 0$\r\n\r\nAnd :\r\n\r\n$ b^2S_{c}\\plus{}c^2S_{b}\\equal{}a^2(7b^4\\plus{}7c^4\\minus{}6b^2c^2)\\plus{}4(b^2\\plus{}c^2)(a^4\\plus{}(b^2\\minus{}c^2)^{2})\\plus{}10a(a^2b^3\\plus{}b^5\\plus{}c^3a^2\\plus{}c^5) \\geq 0$\r\n\r\nWe have done . :)", "Solution_2": "These are very nice (see point (a) and (b) above)\r\n\r\nLet $ a,b,c$ be nonnegative real numbers.\r\n\r\n(a) If $ a\\plus{}b\\plus{}c\\equal{}3$, then\r\n\r\n$ \\frac {4a^2 \\minus{} 27}{b^2 \\plus{} c^2} \\plus{} \\frac {4b^2 \\minus{} 27}{c^2 \\plus{} a^2} \\plus{} \\frac {4c^2 \\minus{} 27}{a^2 \\plus{} b^2} \\plus{} \\frac {69}{2}\\ge 0$;\r\n\r\n(b) If $ ab \\plus{} bc \\plus{} ca \\equal{} 3$, then\r\n\r\n$ \\frac {a^2 \\minus{} 18}{b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} 18}{c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} 18}{a^2 \\plus{} b^2} \\plus{} \\frac {51}{2}\\ge 0$.\r\n\r\nEquality in each inequality occurs for $ a \\equal{} b \\equal{} c$, and also for $ \\frac a{5} \\equal{} b \\equal{} c$ or any cyclic permutation.", "Solution_3": "[quote=\"Vasc\"]These are very nice (see point (a) and (b) above)\n\nLet $ a,b,c$ be nonnegative real numbers.\n\n(a) If $ a \\plus{} b \\plus{} c \\equal{} 3$, then\n\n$ \\frac {4a^2 \\minus{} 27}{b^2 \\plus{} c^2} \\plus{} \\frac {4b^2 \\minus{} 27}{c^2 \\plus{} a^2} \\plus{} \\frac {4c^2 \\minus{} 27}{a^2 \\plus{} b^2} \\plus{} \\frac {69}{2}\\ge 0$;\n\n(b) If $ ab \\plus{} bc \\plus{} ca \\equal{} 3$, then\n\n$ \\frac {a^2 \\minus{} 18}{b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} 18}{c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} 18}{a^2 \\plus{} b^2} \\plus{} \\frac {51}{2}\\ge 0$.\n\nEquality in each inequality occurs for $ a \\equal{} b \\equal{} c$, and also for $ \\frac a{5} \\equal{} b \\equal{} c$ or any cyclic permutation.[/quote]\r\n(a) and (b) in this problem are equivalence. Similar, (a) and (b) in the first problem are equivalence. :) I proved them by mixing variable methods. :)", "Solution_4": "[quote=\"Vasc\"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Then\n\n$ \\frac {a^2 \\minus{} 18}{b^2 \\plus{} c^2} \\plus{} \\frac {b^2 \\minus{} 18}{c^2 \\plus{} a^2} \\plus{} \\frac {c^2 \\minus{} 18}{a^2 \\plus{} b^2} \\plus{} \\frac {51}{2}\\ge 0$,\n\nwith equality for $ a \\equal{} b \\equal{} c$, and also for $ \\frac a{5} \\equal{} b \\equal{} c$ or any cyclic permutation.[/quote]\r\nThe general problem is to find $ m$ for given nonnegative $ k$ such that\r\n\r\n$ \\sum \\frac {ma^2 \\plus{} ab \\plus{} bc \\plus{} ca}{ka^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac {3(m \\plus{} 3)}{k \\plus{} 2}$.\r\n\r\nOther two interesting particular cases (for $ m\\equal{}1$) are the following:\r\n\r\n$ (1) \\ \\ \\ \\ \\ \\sum\\frac {(a \\plus{} b)(a \\plus{} c)}{7a^2 \\plus{} b^2 \\plus{} c^2} \\le \\frac 4{3}$,\r\n\r\nwith equality for $ a \\equal{} b \\equal{} c$, and also for $ \\frac {5a}{4} \\equal{} b \\equal{} c$ or any cyclic permutation;\r\n\r\n$ (2) \\ \\ \\ \\ \\ \\sum\\frac {(a \\plus{} b)(a \\plus{} c)}{a^2 \\plus{} 4b^2 \\plus{} 4c^2} \\le \\frac 4{3}$,\r\n\r\nwith equality for $ a \\equal{} b \\equal{} c$, and also for $ \\frac {2a}{7} \\equal{} b \\equal{} c$ or any cyclic permutation.\r\n\r\nThe case $ m \\equal{} 0$ is also interesting.", "Solution_5": "[img]http://s7.sinaimg.cn/middle/0018zOAxgy71G08DAmW66&690[/img]" } { "Tag": [ "vector", "absolute value", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose that $u_1,...,u_k$ are some unitary vectors in an euclidian space of dimension $n$ such that the absolute value of $$ is the same for all $i<>j$ and it is different from 0 and 1. Prove that $k\\leq n(n+1)/2$. Can you find an optimal bound ( I don't have answer for the last question and I believe it is very difficult).", "Solution_1": "Agregation Composition Math\u00e9matiques G\u00e9n\u00e9rales 1979 \r\n\r\nHarazi there were severals questions before asked to prove this inequality.", "Solution_2": "I know, I know and I give you the same answer as in the post with hyperbolic matrices. You know, solving problems is not just giving the reference. ;)" } { "Tag": [ "algorithm" ], "Problem": "(1)\r\nshazde koochooloo rooye sayyareye koochaki zendegi mikonad ke shoaaash $60$ metr ast.roozi ba shoroo az rooye khatte esteva,$30\\pi$ metr be shargh,$20\\pi$ metr be shomal,$30\\pi$ metr be gharb va saranjam $20\\pi$ metr be jonoob miravad.shazde koochoolo ta makaane avvalash chand metr faasele darad?\r\n1)sefr \r\n2)$30\\pi$\r\n3)$20\\pi$\r\n4)$40\\pi$\r\n5)$15\\pi$\r\n\r\n(2)\r\nzaraayebe chand jomleie $P$ sahih ast.tahte kodaam sharaayete zir $P$ nemitavaanad risheye sahih dashte bashad?\r\n1)$P(5)=5,P(6)=6$\r\n2)$P(6)=5,P(5)=5$\r\n3)$P(5)=6,P(6)=6$\r\n4)$P(6)=5,P(5)=6$\r\n5)tahte harkodam az in sharayet ,$P$ mitavanad risheye sahih dashte bashad.", "Solution_1": "salam narahat nabash inshalah ghabooli man va doostam sare olampiad basij intori shodim\r\nalbate soalate emrooz benazaram sakhttar as ghabl bood\r\n7)mostatili dar safhe ba roose (0,0),(0,100),(150,0),(150,100) hast. chand khate movazi ba ghotr gozara az (0,0)azlae motatil ra dar do noghtey ba moghtasate sahih ghat mikonad? khode ghotr ham hast?\r\n1)99 2)100 3)199 4)200 5)300\r\n man khodam migam 1)99 chera ke : x/100 =y/150 x= 2,4,6,.....100 dar natige tedadeghtrha mishe\r\n 50 +50 -1(ghotr dobar) =99", "Solution_2": "Ber nazare manam kheili fekre khoobie.inke javaabaaro injaa benevisim.Faghat yeki zahmat bekeshe hamashoono injaa benevise..Dar zemn be nazare manam soaalaaye emsaal ye meghdaar gheire moteaaref bood.fek mikonam nomreye ghabooli 7-8taa soaale doros baashe.", "Solution_3": "2 ta az soalha ra ke midoonam too emtehan oomade bood male man bood yeki jame adade jadvale $10\\times10$ va digari ham tamame $n$ hayi ra biyabid ke $\\sqrt{n-1}+\\sqrt{n+15}$ sahih bashad. Har do soalhaye asooni boodand shayad soalate digari az manham too emtehan oomade bashe. :) :)", "Solution_4": "dar shekle roo be roo mikhahim az noghte A be B beravim.masir ha dar jahate felesh yek tarafe hastand.be chand tarigh mitavanim in kar ra anjam dahim?\r\nman khodam 58 be dast avordam.idamam ine ke az noghte B be aghab bargardim yani hesab konim az har kodam az noghat be chand tarigh mitavan be B raft. :)", "Solution_5": "kodam bozorgtar ast?\r\n1)2 be tavane 431\r\n2)3 be tavane 421\r\n3)4 be tavane 321\r\n4)21 be tavane 43\r\n5)31 be tavane 42\r\nagar eshtebah nakarde basham gozine 2 dorost hast. :)", "Solution_6": "The answer is $\\left(\\begin{array}{c}6\\\\2,2,2\\end{array}\\right)=90$\r\n\r\nFor the number of paths from $A$ to $B$.", "Solution_7": "Etefghan jove gozineha hast gozineha ebaratand az :\r\n\r\n$a)60\\\\ b)90\\\\ c)36\\\\ d)54\\\\ e)45$", "Solution_8": "(3)algorithme zir ra rooye chand jomleie $P(x)$ ejraa kardeim.\r\n1.$d$ ra baraabare darajeye $P$ gharaar bede va agar $d<1$ be satre 4 boro.\r\n2.$a$ ra baraabare zaribe $x^d$ dar $P$ gharaar bede va $P(x)-a{x^{d-1}}(x+2)$ ra be jaye $P(x)$ bogzaar.\r\n3.be satre 1 boro.\r\n4.$P$ ra chap kon.\r\npas az ejraaye algorithm adade $1384$ chap shode ast.$P$ dar aghaaz kodaam boode ast?\r\n1)$x^{10}-45x^3$\r\n2)$13{x^7}-2$\r\n3)${x^11}-83{x^3}$\r\n4)$16{x^8}+x$\r\n5)$-x^{11}+84$\r\n\r\nbebakhshid ke man halle khodamo nemizaram chon man aslan khodamo ghabool nadaram.man dochare yek jange darooni hastam....", "Solution_9": "kermi shekamoo mikhahadhameye mivehaye derakhte rooberoo ra bekhorad( :blush: ) va dar eyne hal kamtarin masaafat ra tey konad kodam makan behtar ast ?\r\n1)a 3)b 3)c 4)d 5)e\r\n man migam b aya doroste?!", "Solution_10": "masaleye aakhar:\r\nshekle rooberoo majmooe javabhaye kodam yek az moaadelaate zir ast :\r\n man zadam 4 :\r\n1) y=0 , x yek javaab dare\r\n2)ba taghire x be -x moaadele sabet mimoone vali dar shekl ekhtelafe noghate barkhorde nemoodar ba mehvar moghtasaat ta markaz mashhoode\r\n3)y sefr nemishavad valiiiiiiii\r\n4)+\r\n5)x va y motagharenand vali be nemoodar nemikhore !!", "Solution_11": "bacheha in soal chi mishe?\r\ntasaode hesabi az addade avvale ba ghadre nesbate n^2+1 ke n addadi tabii ast hadeaksar chand ozv darad???\r\n1)3 2)4 3)5 4)6 5)tasaode ba har teddade ozv vojood darad", "Solution_12": "az shahre a jadeii mostaghim kharej shode ke 2 shahre b va c dotarafash gharar darand majmooe faseleye b va c az jadee hadeaksar chand kilometr ast midanim ke faseleye shahre a az 2 shahre digar 60 va 50 kilometr ast va faseleye do shahre b va c 40 kilometr ast?\r\n1)30 2)40 3)50 4)60 5)70", "Solution_13": "dorbini zire havapeyma nasb shode havapeyma rooye masir khatti dar hal oj giri ast zamin ra mosatah farz konid masahate filmbardari shode tabee daraje chandi az jabejayi makani havapeyma ast?\r\n1)1 2)2 3)3 4)4 5)chand jomleii nist", "Solution_14": "nimsaz haye dakhelli mosalas ABC dayere mohiti an ra dar noghat A' ,B',C' ghat mikonad agar I' markaze dayere mohati dakhelli mosalase A'B'C' bashad andaze zaviye B'I'C' barabare kodam gozine hast?\r\n1)90+ B+C/4 2)180-B+C/4 3)2A+B-C 4)180-A 5)2A", "Solution_15": "javaabhaaro injaa neveshtam :) \r\n\r\n[url=http://math-olympiad.persianblog.com/]math-olympiad.persianblog.com/[/url]" } { "Tag": [], "Problem": "hey\r\n I found a website where you could buy the fundamentals of physics full solution manual for just 20 dollars. It is really cool. All problems are solved and no one is missing. I was able to do all my assignments and prepare for my exams. I recommend this to everyone. Just click the link [u]http://windsor.kijiji.ca/c-buy-and-sell-books-Fundamentals-of-physics-full-solution-manual-8TH-EDITION-W0QQAdIdZ76404065[/u]", "Solution_1": "That's pretty cool. Can you by any chance find the solutions manual for the Halliday/Resnick/Krane book?", "Solution_2": "I downloaded the full solution manual (maybe not) from eSnips for free as well as the main text. I'm not sure if it's complete though.", "Solution_3": "ofc \r\nsoln ,book, and other good books\r\nall r available at http://www.4shared.com" } { "Tag": [ "Divisor Functions" ], "Problem": "Show that if the equation $\\phi(x)=n$ has one solution, it always has a second solution, $n$ being given and $x$ being the unknown.", "Solution_1": "Up to my knowledge, this problem is open (see http://mathworld.wolfram.com/CarmichaelsTotientFunctionConjecture.html ).", "Solution_2": "That is strange... Hojoo, do you have the book in which this appears, or could you check it with the person that contributed this?", "Solution_3": "[quote=\"Peter\"]Show that if the equation $\\phi(x)=n$ has one solution, it always has a second solution, $n$ being given and $x$ being the unknown.[/quote]\r\n\r\nI'm not sure... but it seems that I also included some open problems :)", "Solution_4": "I have asked this problem on ML.Actually,I have tried to solve this one for a day.so disppointed.I think I have no ability to solve an open problem..." } { "Tag": [ "email", "group theory", "abstract algebra", "superior algebra" ], "Problem": "Find all groups that have exactly 6 subgroups. Our algebra teacher from Bucharest said last year that this is a monstruous problem. And I think he's right.", "Solution_1": "You can send your answer in french or english\r\n\r\nAlain Tissier (Professeur MP*)\r\n17, avenue des Ormes, 93370 Montfermeil \r\nFrance\r\n\r\nor by email \r\n\r\nal.tissier@laposte.net\r\n\r\nYou must mention this is the Question 488 in \r\nRMS n2; Mars 2004; page 173 \r\n\r\n\r\nIf your answer is nice RMS will publish it.", "Solution_2": "Do the $6$ subgroups include the improper ones? (i.e. the group itself and the group formed only by the unit)", "Solution_3": "Assuming the $6$ subgroups do include the trivial subgroup and the group itself, I believe the only solutions are $\\mathbb Z_3\\times\\mathbb Z_3,S_3$, and all groups of the form $\\mathbb Z_{p^5}$, where $p$ is prime. I was on my way to post what looked like a correct solution, when my lights went out and I lost my message, so I won't be posting it again any time soon :). There's nothing fancy about it anyway: we just need to check a few cases.", "Solution_4": "One more set of obvious solutions: $\\mathbb{Z}_{pq^2}$", "Solution_5": "[quote=\"grobber\"]Do the $6$ subgroups include the improper ones? (i.e. the group itself and the group formed only by the unit)[/quote]\r\nWrong terminology! Only the group itself is called an improper subgroup. The subgroup formed only by the unit is the trivial subgroup and is a proper subgroup, unless $G=1$.", "Solution_6": "[quote=\"jmerry\"]One more set of obvious solutions: $\\mathbb{Z}_{pq^2}$[/quote]\r\n\r\nOh, yes, I forgot about those. I hope there are no more. Actually, I kind of stormed through the cases that needed to be checked, and there might have been some slip-ups such as this one.", "Solution_7": "You did miss one: the quaternion group.\r\n\r\nHere's an argument:\r\nFirst, we consider cyclic groups. It is clear that a cyclic group has six subgroups when it has order $p^5$ or $p^2q$ for distinct primes $p,q$.\r\n\r\nNow, suppose $|G|=n$ and the largest order of an element is $m$, with $m3$, the subgroup of order $m$ is normal and we have $3+3\\frac{m-1}2$ subgroups, which is too many.\r\nCase 2: $n=2m$. Here, the elements outside the subgroup generate at least two cyclic subgroups, so either $m$ is prime or $m$ is the square of a prime. Considering $n=2p$ with $p>2$, we have the dihedral group, with $p+3$ subgroups. This works for $p=3$, giving us $S_3$.\r\nIn the $n=2p^2$ case, with $p>2$, the group is again dihedral, with more than $p^2+3$ subgroups; this doesn't work.\r\nThis leaves $n=4$ or $n=8$ to check separately. This check gives us the quaternion group, and no others.\r\n\r\nRecapping the full list: \r\n$\\mathbb{Z}_{p^5},\\mathbb{Z}_{p^2q},\\mathbb{Z}_3^2,S_3,Q$." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Hi, \r\nI have a question which is about the positive definite. Wonder if anyone of you could give a little help, thx. \r\n\r\nIf $A$ and $X$ are square matrice, $A$ is posotive definite , and $X$ is nonsingular then the matrix $B=X^T A X$ is also positive definite.\r\n\r\nI guess I might need to use the Cholesky Theorem to prove it, right? Any hint to help? \r\nThx", "Solution_1": "No need. For any $y\\neq0$, $y^T(X^TAX)y=(Xy)^TA(Xy)>0$ and $A$ is positive definite.\r\n\r\nNever overlook the power of the definition." } { "Tag": [ "search", "geometry solved", "geometry" ], "Problem": "let (C) be a circle inwich its center is inaccessible .only by using compass find the center of (C).", "Solution_1": "Come one, this is the third time you're posting this today!!\r\n\r\nAnd you keep confusing \"campos\" and \"compass\".", "Solution_2": "A little search will do it :\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=24888 .", "Solution_3": "oh.. sorry Arne :blush: .\r\n\r\nu know.thats because of that i posted all them in one minute.and by that time i didnt know the different betwin\r\n\r\n(campos)and (compass) :rotfl:", "Solution_4": "to see the solution here is alink: http://www.mathlinks.ro/Forum/viewtopic.php?t=57006" } { "Tag": [ "induction", "function", "vector", "calculus", "derivative", "real analysis", "real analysis theorems" ], "Problem": "show:\r\n$ \\partial^{\\alpha}(fg)\\equal{}\\sum_{\\beta\\plus{}\\gamma\\equal{}\\alpha}\\frac{\\alpha !}{\\beta ! \\gamma !} \\partial ^{\\beta}f\\cdot\\partial ^{\\gamma} g$\r\n\r\ninformation on Multi index notation: http://en.wikipedia.org/wiki/Multi-index_notation", "Solution_1": "Information on mathematical induction:\r\nhttp://en.wikipedia.org/wiki/Mathematical_induction", "Solution_2": "[quote=\"mlok\"]Information on mathematical induction:\nhttp://en.wikipedia.org/wiki/Mathematical_induction[/quote]\r\n\r\nI never said that f and g were functions of one variable.\r\nIf you want to use induction, there's two things you have to show, first it holds for all n, where n is the number variables. but you also have to show that this holds for all vectors $ \\alpha$ no matter its components.\r\n\r\nplease correct me if im wrong, but this doesnt seem trivial to me.", "Solution_3": "It's still induction- we're just using the partial ordering on the index set rather than the strict ordering on $ \\mathbb{N}$. Assume it holds for all smaller indices, differentiate, and get the recursion using the ordinary product rule. The base case here is no derivatives at all.\r\n\r\nAlternately, you could reduce to several copies the single-variable version; partial derivative operators commute, so we can decompose each $ \\frac{\\partial^{n_i}}{\\partial x_i^{n_i}}$ and combine the results.", "Solution_4": "This is Exercise 1.5.3 from Evans, PDE book.\r\n\r\nLet me elaborate on the inductive step. Assume, this formula\r\n\r\n$ \\partial^{\\alpha}(fg)\\equal{}\\sum_{\\beta\\plus{}\\gamma\\equal{}\\alpha}\\frac{\\alpha !}{\\beta !\\gamma !}\\partial^{\\beta}f\\cdot\\partial^{\\gamma}g \\equal{} \r\n\\sum_{\\beta \\le \\alpha}{\\alpha \\choose \\beta}\\partial^{\\beta}f\\cdot\\partial^{\\alpha \\minus{} \\beta}g$\r\n\r\nholds. Remark: there are $ (\\alpha_1 \\plus{} 1) \\cdot ... \\cdot (\\alpha_n \\plus{} 1)$ summands in this sum. Now we prove that \r\n\r\n$ \\partial^{\\alpha \\plus{} e_i}(fg) \\equal{} \r\n\\partial_{x_i}\\sum_{\\beta \\le \\alpha}{\\alpha \\choose \\beta}\\partial^{\\beta}f\\cdot\\partial^{\\alpha \\minus{} \\beta}g \\equal{}\r\n\\sum_{\\beta \\le \\alpha} \\left[ {\\alpha \\choose \\beta} \\partial^{\\beta\\plus{}e_i}f\\cdot\\partial^{\\alpha \\minus{} \\beta}g \\plus{} {\\alpha \\choose \\beta}\\partial^{\\beta}f\\cdot\\partial^{\\alpha \\minus{} \\beta \\plus{} e_i}g \\right] \\equal{}\r\n\\left. \\sum_{\\beta \\le \\alpha \\plus{} e_i}{\\alpha \\plus{} e_i \\choose \\beta}\\partial^{\\beta}f\\cdot\\partial^{\\alpha \\plus{} e_i \\minus{} \\beta}g \\right.$.\r\n\r\nIndeed, the equality is true because each term of the form $ \\partial^{\\beta}f\\cdot\\partial^{\\alpha \\plus{} e_i \\minus{} \\beta}g$ comes from either a term $ \\partial^{\\beta \\minus{} e_i}f\\cdot\\partial^{\\alpha \\minus{} (\\beta \\minus{} e_i)}g$ or a term $ \\partial^{\\beta}f\\cdot\\partial^{\\alpha \\minus{} \\beta}g$ in the large sum expression (except two terms of a view $ f\\partial^{\\alpha \\plus{} e_i}g$ and $ g\\partial^{\\alpha \\plus{} e_i}f$). And multipliers near those terms are combined as\r\n\r\n$ {\\alpha \\choose \\beta \\minus{} e_i} \\plus{} {\\alpha \\choose \\beta} \\equal{} {\\alpha \\plus{} e_i \\choose \\beta}$.\r\n\r\nThe last is verified in a straightforward fashion, remembering the definition\r\n\r\n$ {\\alpha \\choose \\beta} \\equal{} \\frac {\\alpha!} {\\beta! (\\alpha \\minus{} \\beta)!}$, where $ \\alpha! \\equal{} \\alpha_1! \\cdot ... \\cdot \\alpha_n!$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let triangle ABC.Prove that $sin^{3}A+sin^{3}B+sin^{3}C\\leq\\frac{9\\sqrt{3}}{8}$", "Solution_1": "Wrong: take $A\\to 90^{\\circ}$, $B\\to 90^{\\circ}$, $C\\to 0^{\\circ}$.\r\n\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=33518 or http://www.mathlinks.ro/Forum/viewtopic.php?t=48222 for the best bound.\r\n\r\n darij" } { "Tag": [], "Problem": "If $x>y$, then how many values of $(x,y)$ can there be if\r\n\r\n$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{24}$", "Solution_1": "[quote=\"da ban man\"]If $x>y$, then how many values of $(x,y)$ can there be if\n\n$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{24}$[/quote]\r\n\r\n[hide]Getting rid of the denominators,\n$24x+24y = xy$\n$xy-24x-24y+484 = 576$\n$(x-24)(y-24) = 24^{2}$,\nfrom SFFT.\n\nNotice that some pair of factors produce a solution; if $f_{1}f_{2}= 24^{2}$, $(x,y) = (24+f_{1}, 24+f_{2})$, where $f_{1}>f_{2}$.\n\nNotice that the right has $21$ positive factors, and $21$ negative factors. So the number of factors where $f_{1}>f_{2}$ is $20$. Thus, there are $\\boxed{20\\textrm{ solutions}}$[/hide]\r\n\r\nEDIT: thanks, 13375P34K43V312.", "Solution_2": "[hide]\n$xy-24x-24y=0$\n$(x-24)(y-24)=576$\n\n$y-24\\in\\{1,2,3,4,6,8,9,12,16,18,-32,-36,-48,-64,-72,-96,-144,-192,-288,-576\\}$\n\nso 20\n[/hide]\r\n\r\nvishalarul, y doesn't have to be positive.", "Solution_3": "Alright, yea thats right,\r\n\r\nNow what if (x,y) are positive integers also satisfying x>y.", "Solution_4": "[quote=\"da ban man\"]Alright, yea thats right,\n\nNow what if (x,y) are positive integers also satisfying x>y.[/quote]\r\n\r\nIt would by $10$ solutions, as I mentioned earlier.", "Solution_5": "[quote=\"vishalarul\"][quote=\"da ban man\"]Alright, yea thats right,\n\nNow what if (x,y) are positive integers also satisfying x>y.[/quote]\n\nIt would by $10$ solutions, as I mentioned earlier.[/quote]\r\npositive and [b]integers[/b]. (and x>y)" } { "Tag": [ "geometry", "rectangle", "calculus", "integration", "derivative", "quadratics", "modular arithmetic" ], "Problem": "A rectangular piece of of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called [i]basic [/i]if \r\n\r\n(i) all four sides of the rectangle are segments of drawn line segments, and \r\n(ii) no segments of drawn lines lie inside the rectangle.\r\n\r\nGiven that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.", "Solution_1": "[hide]So we realize that drawing x vertical lines and y horizontal lines, the number of basic rectangles we have is (x-1)(y-1). The easiest possible case to see is 223 vertical and 223 horizontal lines as (4+5)223 = 2007. Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation $(222-4x)(222+5x)$ which you want to maximize. Expanded, this gives $-20x^{2}+222x+222^{2}$. From $-\\frac{b}{2a}$ you get that the vertex is at $x=\\frac{111}{20}$. This is not an integer though, so you see that when x=5, you have $-20*25+222*5+222^{2}$ and that when x=6, you have $-20*36+222*6+222^{2}$. 222 is greater than -20*11, so the maximum integral value for x occurs when $x=6$. Now you just evaluate $-20*36+222*6+222^{2}\\mod 1000$ which is $\\boxed{896}$.[/hide]\r\n\r\nEdit: Fixed missing backslash", "Solution_2": "counting the edges of the piece of paper as lines that have been drawn seems unecessary but since i counted only the lines inside i lost the point.", "Solution_3": "isn't it (x-2)(y-2) not (x-1)(y-1)", "Solution_4": "actually, since you draw x horizontal lines, it partitions the paper into x+1 partitions. 2 don't count, so you get x-1. same goes for y.", "Solution_5": "Your solutions make sense, except I have one vital question: Where in the problem does it say that the lines you draw have to be from one side of the piece of paper to the other? I read this problem several times and couldn't understand it, because it only gives that the lines are \"parallel\" to the edges of the piece of paper, and as far as I could tell, you could have a bunch of infinitely small lines intersecting in a tiny grid, and there is no upward bound for the number of basic rectangles formed. I'm very mad at what I percieve to be an inaccurately written question, though I'd appreciate it if someone would explain what I missed. Good math after that question, though.", "Solution_6": "They're lines, not line segments.", "Solution_7": "if your'e gonna distinguish between line and line segment, length wouldn't really make sense for an infinite line...", "Solution_8": "I thought this was a horribly written problem.... I just could not understand it, and it's not like I'm *that* much of a noob. Usually I understand enough of the problem to know what part of it makes it too hard for me to do, but with this one it was vaguely worded enough that I couldn't be sure.", "Solution_9": "I got $a$ in terms of $b$ and took a derivative to minimize :ninja: but it worked, so I'm not complaining.\r\n\r\n[hide=\"My solution\"]\n\n$4a+5b=2007\\Rightarrow a=\\frac{2007-5b}{4}$ from the condition\n\nso the # of squares is $\\left(\\frac{2007-5b-4}{4}\\right)(b-1) = \\left(\\frac{2003-5b}{4}\\right)(b-1)$\n\nnow $\\frac{d}{db}$(expression for # of squares)$=\\frac{-5b+1004}{2}\\Rightarrow b=\\frac{1004}{5}, a=\\frac{1003}{4}$\n\nNow we check the two closest things that are 3 mod 4 to 201 (since b must be 3 mod 4 because 2007=3 mod 4) which are 199 and 203, and $b=199$ gives the maximum # of squares which is\n\n\\[252\\times198=\\fbox{49896}\\][/hide]", "Solution_10": "Yeah, I originally thought about the difference between lines and line segments, but then it started talking about the total \"length\" of the lines, so clearly it meant line segments (as krutyteklown said). So I am still baffled.\r\nAnd mrosett and I agree again! This is kind of cool.", "Solution_11": "[quote=\"not_trig\"]I got $a$ in terms of $b$ and took a derivative to minimize :ninja: but it worked, so I'm not complaining.\n\n[hide=\"My solution\"]\n\n$4a+5b=2007\\Rightarrow a=\\frac{2007-5b}{4}$ from the condition\n\nso the # of squares is $\\left(\\frac{2007-5b-4}{4}\\right)(b-1) = \\left(\\frac{2003-5b}{4}\\right)(b-1)$\n\nnow $\\frac{d}{db}$(expression for # of squares)$=\\frac{-5b+1004}{2}\\Rightarrow b=\\frac{1004}{5}, a=\\frac{1003}{4}$\n\nNow we check the two closest things that are 3 mod 4 to 201 (since b must be 3 mod 4 because 2007=3 mod 4) which are 199 and 203, and $b=199$ gives the maximum # of squares which is\n\\[252\\times198=\\fbox{49896}\\]\n[/hide][/quote]\n\nRather than take the derivative, a pre-calc method is to expand it and use $\\frac{-b}{2a}$.\n\n[hide=\"Solution\"]\nFOIL this to get a quadratic, $-\\frac{5}{4}x^{2}+502x-\\frac{2003}4$ (changed bs to xs b/c thats how i typed it up on the wiki). Use $\\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \\frac{-502}{-2 \\cdot \\frac{5}{4}}= \\frac{1004}5 \\approx 201$. However, this gives a non-integral answer for $y$. The closest two values that work are $(199,253)$ and $(203,248)$. \n\nWe see that $252 \\cdot 198 = 49896 > 202 \\cdot 247 = 49894$. The solution is $896$.\n[/hide]", "Solution_12": "[hide=\"another solution\"]Orient the rectangle so its width is 5. Suppose we have $a$ vertical lines and $b$ horizontal lines. We are given that $4a+5b=2007$, and that we want to maximize $(a-1)(b-1)$. Consider this equation modulo 4. This makes $b\\equiv3\\pmod{4}$, so $b=4y+3$ for $y\\in\\mathbb{Z}$. Likewise, modulo 5 gives $a\\equiv3\\pmod5$, so set $a=5x+3$ with $x\\in\\mathbb{Z}$.\n\nSubstituting, our equation becomes $(20x+12)+(20y+15)=2007$, so $x+y=99$. We want to maximize $(5x+2)(4y+2)$. We can rewrite this as $20(x+.4)(y+.5)$. These last two numbers sum to a constant, so their product will be maximized when their difference is minimized. This occurs for $x=50$ and $y=49$, which makes $(5x+2)(4y+2)=(252)(198)=50400-504=49896$, which ends in $\\fbox{896}$.[/hide]", "Solution_13": "wow that's a good one :D \r\n\r\nvery nice... much better than mine, lol", "Solution_14": "Dangit... I missed that the rectangles touching the sides of the rectangle aren't basic rectangles, so I got $800$ as the answer.", "Solution_15": "i still don't get it. is there a list of solutions on the AMC website? or even better, could someone come up with a down to earth, dumbed-down explanation for me? thanks", "Solution_16": "This is completely after the fact, and I didn't even take the AIME II (took I), but I think that they wanted the fact that you draw \"lines\" and not \"line segments\" to mean they extend infinitely in each direction, but you only draw on the paper, so you'd be drawing from one side of the paper to the other.", "Solution_17": "ya, that's what i did. then i got the equation $4x+5y = 2007$ where $x$ and $y$ are the # of lines drawn in each direction. is that even right?", "Solution_18": "Yeah, that's how they got the 223 lines in each direction that they used to derive their quadratic...", "Solution_19": "Sorry to revive this thread but I think my solution does not use any messy quadratic formulas and can be done very quickly.\n[hide=\"Solution\"]\nCall the number of horizontal lines $x$ and the number of vertical lines $y$. Then we have that $5x+4y=2007$ and we are trying to maximize $(x-1)(y-1)$. Now, let $a=x-1$ and $b=y-1$, then we have that $5a+4b=1998$ and we are trying to maximize $ab$. By $AM-GM$, we have that $\\frac{5a+4b}{2}\\ge\\sqrt{(20)(xy)}$ with equality when $4a=5b$. Therefore, we wish to maximize the geometric mean so we want to be as close to equality as possible. Since $a$ and $b$ are integers, $a=198, b=252$. Therefore, the answer is $896$.\n[/hide]" } { "Tag": [ "inequalities", "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "$A,B,C \\in M_n(C)$ such that $\\forall k \\in \\{1;2;3\\}, \\; A^k=B+kC$ \r\n\r\nProve that $rank(A)+rank(A^2-2A+I_n)= n$", "Solution_1": "[quote=\"Moubinool\"]$A,B,C \\in M_n(C)$ such that $\\forall k \\in \\{1;2;3\\}, \\; A^k=B+kC$ \nProve that $rank(A)+rank(A^2-2A+I_n)= n$[/quote]\r\n\r\nHey, beautiful... I liked the problem, very smart (but somehow easy)\r\n\r\nFirst than all, note that the statement is equivalent to proof that $null(A)+null(A^2-2A+I_n)= n$. Now, note that if $F,G\\in M_n$ such that $FG=0$ then $null(F)+null(G)\\geq n$ because, suppose that $null(B)=p$, then, $ker(B)^\\perp\\subseteq ker(A)$ (it follows also $rank(B)\\leq null(A)$), and the inequality follows. \r\n\r\nNow, note that $A(A^2-2A+I_n)=(A^3-A^2)-(A^2-A)=C-C=0$, hence $null(A)+null(A^2-2A+I_n) \\geq n$. \r\n\r\nNow, note that if $x\\neq0$ ($x\\in\\mathbb{C}^n$), then, if $Ax=0$, then $(A^2-2A+I_n)x=x\\neq0$, hence, $null(A) \\leq rank(A^2-2A+I_n)$. Also, if $(A^2-2A+I_n)x=0\\Rightarrow (A-2I_n)Ax=x\\neq0\\Rightarrow Ax\\neq0\\Rightarrow null(A^2-2A+I_n) \\leq rank(A)$, hence \r\n\r\n\\[null(A)+null(A^2-2A+I_n)\\leq rank(A)+rank(A^2-2A+I_n)\\]\r\n\r\n\\[\\Rightarrow null(A)+null(A^2-2A+I_n) \\leq n\\]\r\n\r\nand therefore\r\n\r\n$null(A)+null(A^2-2A+I_n)=rank(A)+rank(A^2-2A+I_n)=n$.\r\n\r\nThere are some nice properties about this matrices, like the fact that $AB=A$ and $AC=C$, and some others.\r\n\r\nBest regards,", "Solution_2": "$A,B,C \\in M_n(C)$ such that $\\forall k \\in \\{1;2;3\\}, \\; A^k=B+kC$ \r\n\r\nProve that \r\n\r\n$(P_{A}(1)-det(A))B+P'_{A}(1)C+det(A)I_n=0$ where $P_{A}$ is characteristic \r\n\r\npolynomial of $A$" } { "Tag": [ "geometry", "calculus", "trigonometry", "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO" ], "Problem": "I'm learning geometry right now.\r\nNext year I get to take Algebra 2, but I have a chance to take a class over the summer or take a test to pass it and move onto Pre-calc(I think..)\r\nI need you guys' help who has faced or know what to do.\r\nIs Algebra 2 important? or Would you say that it's not as necessary or important.", "Solution_1": "Algebra 2 is extremely important, particularly in calculus but as well in pre-calc/trig. I would skip Geometry though :D .", "Solution_2": "If you can (and actually will) take Algebra II over the summer I say go ahead and do it, and move on to pre-Calc next year. Although I agree that Geometry pretty much sucked, if you're in the US and plan on going to college, there are many random geometry parts to the ACT and SAT that you should know, so don't skip it... Besides, it keeps popping up in both my Physics and Calc classes this year. Argh...\r\n\r\nAlgebra II is important for pre-Calc and Calc. You need it. If you think that you can learn it in it's entirety over the summer, go for it. Just testing out of it only knowing the material on the test isn't such a good idea though - there are likely things not on the test that you need to know... If you plan to teach yourself (which works, I've done it before) make sure that you really do it, don't just kinda read the book and answer a couple of the easier questions... I'm sure that not everyone is a slacker like me, but be warned. Without deadlines, sometimes things don't get done for some people (like me!).\r\n\r\nIf there's a class your school is offering over the summer, and you want to do it, by all means go for it! Good luck!", "Solution_3": "Uh. I never actually learned Algebra II, in school or otherwise, and I'm doing quite fine in BC Calc, thank you very much. I picked up what I didn't know from Algebra II after I got into Precalc by doing AoPS II, haha. Though you should probably go for the summer thing anyway, just in case you don't want to/can't BS through the Precalc entrance exam.", "Solution_4": "I wouldn't reccomend spending the time during the summer learning Algebra II, you could be doing much better things with your time.\r\n\r\nI'd say if you really are like doing your geometry homework in like 10 minutes every night, then you should spend maybe an hour a week studying Algebra II material in place of the excess time you have after finishing geometry homework.", "Solution_5": "Don't have much that much experience on this, but I'd say go ahead, if you want to spend the time over the summer preparing. My brother skipped Alg II by taking a summer course.", "Solution_6": "Just saying, Pre-Cal is just a big review of Algebra II and Geometry. So you really shouldn't take it unless you have too.", "Solution_7": "It's not a review. You do trig and a little bit of calc.", "Solution_8": "In my Algebra II class we did trig and Simple Calc like summation. Maybe it was different from the other Algebra II classes.", "Solution_9": "Hmmm... So what you guys all agree upon is that Algebra 2 is essential for students like me to get to the higher level of Math, and shouldn't think that it's really not important to skip it, right?", "Solution_10": "Well, in my opinion, I think Algebra II is important for you to graduate from high school and get into a good college. It's one of the things that determine the foundation of your future college admission. (and for a good score on the SAT/ACT)", "Solution_11": "At my school, Algebra II is pretty ignorant.", "Solution_12": "Algebra 2 is essential material for anyone. For some exceptional students (like many of us), it's possible to learn that material without actually taking the class. That's not the same as skipping it.\r\nI learned Alg. 2 and trig while I was taking geometry in 9th grade. The teacher let me borrow the textbook for algebra, and I derived most of the trigonometry myself.\r\n\r\nAfter that, I studied some precalculus informally in the summer, and took calculus in 10th grade.", "Solution_13": "[quote=\"jmerry\"]Algebra 2 is essential material for anyone. For some exceptional students (like many of us), it's possible to learn that material without actually taking the class. That's not the same as skipping it.\nI learned Alg. 2 and trig while I was taking geometry in 9th grade. The teacher let me borrow the textbook for algebra, and I derived most of the trigonometry myself.\n\nAfter that, I studied some precalculus informally in the summer, and took calculus in 10th grade.[/quote]\r\n\r\nWow... You must have spent bunch of your time doing those 3 subjects at once..", "Solution_14": "I would have thought jmerry had mostly learned the material of Algebra II while in Algebra I in the 8th grade. Also: he was at National Mathcounts as an 8th grader and a USAMO qualifier as a 9th grader, so this isn't necessarily something that just anyone could do.", "Solution_15": "You're in 8th grade, right?\r\n\r\nYou shouldn't skip Algebra II because it makes the colleges look like your wayy too serious :P Like other people said, you can always spend time doing fun, cool things during summer. Or, if you take Algebra II you can study Geometry (if you really wanted to), Pre-Calc, Trig, anything higher than your level over the summer.\r\n\r\nIt's your choice. I'm just recommending you don't skip it. You'll regret it. Geometry wise :D", "Solution_16": "[quote=\"D3m0n Shad0w\"]You're in 8th grade, right?\n\nYou shouldn't skip Algebra II because it makes the colleges look like your wayy too serious :P Like other people said, you can always spend time doing fun, cool things during summer. Or, if you take Algebra II you can study Geometry (if you really wanted to), Pre-Calc, Trig, anything higher than your level over the summer.\n\nIt's your choice. I'm just recommending you don't skip it. You'll regret it. Geometry wise :D[/quote]\r\n\r\nYes, I am an 8th grader..and I'll take your word along with others.\r\nThanks!", "Solution_17": "[quote=\"D3m0n Shad0w\"]You're in 8th grade, right?\n\nYou shouldn't skip Algebra II because it makes the colleges look like your wayy too serious :P Like other people said, you can always spend time doing fun, cool things during summer. Or, if you take Algebra II you can study Geometry (if you really wanted to), Pre-Calc, Trig, anything higher than your level over the summer.\n\nIt's your choice. I'm just recommending you don't skip it. You'll regret it. Geometry wise :D[/quote]\r\n\r\n*Cough* says the fifth grader. \r\n\r\nAs a general rule of thumb, don't skip a math subject unless your schedule demands it. You'll have an easy year for math so you can concentrate on the subjects you need work on. \r\n\r\nI plan to take AP physics in junior year but to do that I need to be finished calc AB. So in a sense, I [i]have[/i] to skip a math course. I prefer skipping calc than pre-calc because personally, I suck at trig.", "Solution_18": "Algebra 2 was a really fun year. (everything you learn is fun) I don't suggest you skip it.\r\nMaybe skipping Precalculus would be a better idea, if you need to skip a year so that you could put credits in certain electives.", "Solution_19": "[quote=\"ProtestanT\"]Algebra 2 was a really fun year. (everything you learn is fun) I don't suggest you skip it.\nMaybe skipping Precalculus would be a better idea, if you need to skip a year so that you could put credits in certain electives.[/quote]\r\nCan you briefly tell me what you learn in there? \r\nThat would be really helpful.", "Solution_20": "[quote=\"DonkeyKong\"][quote=\"ProtestanT\"]Algebra 2 was a really fun year. (everything you learn is fun) I don't suggest you skip it.\nMaybe skipping Precalculus would be a better idea, if you need to skip a year so that you could put credits in certain electives.[/quote]\nCan you briefly tell me what you learn in there? \nThat would be really helpful.[/quote]\r\nbriefly, we did alot of graphing, polynomial functions, finding roots, minimums and maximums, factoring, synthetic division, rational functions, conic sections (starting with parabolas), transcendamo or hwoever u spell it (exponential + logarithmic) functions, systems of linear equations, simple matrices (finding determinants, operations on matrices, gaussian elimination, inverses, etc), probability, and enumerative combinatorics (counting).\r\nOur teacher did not teach us trignometry.", "Solution_21": "[quote=\"ProtestanT\"][quote=\"DonkeyKong\"][quote=\"ProtestanT\"]Algebra 2 was a really fun year. (everything you learn is fun) I don't suggest you skip it.\nMaybe skipping Precalculus would be a better idea, if you need to skip a year so that you could put credits in certain electives.[/quote]\nCan you briefly tell me what you learn in there? \nThat would be really helpful.[/quote]\nbriefly, we did alot of graphing, polynomial functions, finding roots, minimums and maximums, factoring, synthetic division, rational functions, conic sections (starting with parabolas), transcendamo or hwoever u spell it (exponential + logarithmic) functions, systems of linear equations, simple matrices (finding determinants, operations on matrices, gaussian elimination, inverses, etc), probability, and enumerative combinatorics (counting).\nOur teacher did not teach us trignometry.[/quote]\r\n\r\nThose sound important..\r\n\r\nI think you take a peek at Trig. in the Geometry...", "Solution_22": "From my experience, the first semester of Alg. 2 is review of Alg. 1, and the first semester of Pre-Calc (or Analysis, whatever they call it where you live) is review of Alg. 2\r\n\r\n\r\nIf you really learn the material for Algebra 2 and feel like you could be spending more time on math HW, go ahead. I tested out of Geometry and Analysis, and i've been getting along fine (but i did quite a bit of extra-curricular Geometry)\r\n\r\nIn fact, i got a B- on the Analysis test-out, but still got an A in CalcBC", "Solution_23": "I say- Dont skip it, unless you have taken algebra 1; 2 times. Because its good practice(I bet) and you learn some new stuff. :ninja:", "Solution_24": "In general, I think you should never skip algebra 2. Its important for your college future.", "Solution_25": "my precalc class was basically my alg 2 class+trig, just a bit faster and less explicated. explain exactly what ur schools alg 2 class and precalc class cover 4 accurate suggestions, b/c w/o that info theres no way 2 know if you'd be hurt by skipping it." } { "Tag": [ "floor function" ], "Problem": "Prove or disprove that there exists a positive real number $u$ such that $\\lfloor u^n \\rfloor -n$ is an even integer for all positive integer $n$.", "Solution_1": "Let $ u\\equal{}1\\minus{}\\sqrt{2}$\r\n$ [u^{2n}]\\equal{}0$\r\n$ [u^{2n\\plus{}1}]\\equal{}\\minus{}1$\r\nImply that $ [u^n]\\minus{}n\\equiv 0(\\mod 2)$", "Solution_2": "Why $ [u^{2n}]\\equal{}0$ and $ [u^{2n\\plus{}1}]\\equal{}\\minus{}1$? :maybe:", "Solution_3": "It's since $ \\minus{}12 +1)(x 2 +2 2 ).....(x 2 +y 2 )=z 2", "Solution_1": "x = 1, y = 3, z = 10. Any others?" } { "Tag": [ "projective geometry", "conics", "ellipse", "geometry", "angle bisector", "geometry solved" ], "Problem": "In quadrilateral ABCD BCxAD at P, ABxCD at Q.\r\n\r\n Find locus of points X so that 0$, we have\r\n$ |f(x) \\minus{} f(3)|$\r\n$ \\equal{}\\frac{|x\\minus{}3|}{|3x \\plus{}9|}$ \r\n$ \\leq \\frac{|x\\minus{}3|}{3|x| \\plus{}9}$ \r\n$ <\\frac{|x\\minus{}3|}{15}$ if $ |x\\minus{}3|<1$\r\n$ <\\frac{15\\epsilon}{15}$ if $ |x\\minus{}3|<15\\epsilon$\r\n$ <\\epsilon$.", "Solution_2": "Since we wish to consider values of x close to 3, putting $ \\delta< 1$ is reasonable since that puts x within 1 of 3. Also, $ 0<|x\\minus{}3|<\\delta< 1$ implies that $ \\frac{1}{|x\\plus{}3|}<\\frac{1}{5}$.", "Solution_3": "Okay, thanks I see the whole $ \\delta < 1$ thing now. And I see how to get that $ \\frac {1}{7} < \\frac {1}{|x - 3|} < \\frac {1}{5}$ but I'm [i]really[/i] confused as to how you can just add that to the previous inequality to get:\r\n\\[ \\frac {1}{3}|x - 3|\\frac {1}{|x + 3|} < \\frac {1}{3}|x - 3|\\frac {1}{5} < \\varepsilon\r\n\\]\r\nIt seems to me like in the \"middle expression\" ($ \\frac {1}{3}|x - 3|\\frac {1}{5}$), $ \\frac {1}{|x + 3|}$ was simply replaced by $ \\frac {1}{5}$. I simply don't get the logic behind that...\r\n\r\nEdit: I think I've got the hang of it now and a few practice problems are helping alot. I know $ \\delta=1$ and $ \\delta=15\\varepsilon$ but I don't get why $ \\delta=\\min{\\left\\{1,15\\varepsilon\\right\\}}$, mainly why we want to take the minimum value of $ \\delta$ in these types of cases.", "Solution_4": "[quote=\"akech\"]\n$ \\frac {|x \\minus{} 3|}{|3x \\plus{} 9|} \\leq \\frac {|x \\minus{} 3|}{3|x| \\plus{} 9}$ \n[/quote]\r\n\r\nSo you are claiming $ 3|x| \\plus{} 9 \\leq |3x \\plus{} 9|$? Doesnt that go against triangle inequality?", "Solution_5": "[quote=\"Blkmage\"]Okay, thanks I see the whole $ \\delta < 1$ thing now. And I see how to get that $ \\frac {1}{7} < \\frac {1}{|x - 3|} < \\frac {1}{5}$ but I'm [i]really[/i] confused as to how you can just add that to the previous inequality to get:\n\\[ \\frac {1}{3}|x - 3|\\frac {1}{|x + 3|} < \\frac {1}{3}|x - 3|\\frac {1}{5} < \\varepsilon\n\\]\nIt seems to me like in the \"middle expression\" ($ \\frac {1}{3}|x - 3|\\frac {1}{5}$), $ \\frac {1}{|x + 3|}$ was simply replaced by $ \\frac {1}{5}$. I simply don't get the logic behind that...\n\nEdit: I think I've got the hang of it now and a few practice problems are helping alot. I know $ \\delta = 1$ and $ \\delta = 15\\varepsilon$ but I don't get why $ \\delta = \\min{\\left\\{1,15\\varepsilon\\right\\}}$, mainly why we want to take the minimum value of $ \\delta$ in these types of cases.[/quote]From $ \\frac{1}{|x+3|}<\\frac15$, multiply both sides by $ \\frac13|x-3|$.", "Solution_6": "[quote=\"isomorphism\"]So you are claiming $ 3|x| \\plus{} 9 \\leq |3x \\plus{} 9|$? Doesnt that go against triangle inequality?[/quote]\r\n\r\nSince everything is positive, $ 3|x| \\plus{} 9 \\geq |3x \\plus{} 9|$ implies $ \\frac{1}{3|x| \\plus{} 9} \\leq \\frac{1}{|3x \\plus{} 9|}$. When you have a ratio of two positive quantities, making the denominator larger makes the fraction smaller.", "Solution_7": "[quote=\"JBL\"][quote=\"isomorphism\"]So you are claiming $ 3|x| \\plus{} 9 \\leq |3x \\plus{} 9|$? Doesnt that go against triangle inequality?[/quote]\n\nSince everything is positive, $ 3|x| \\plus{} 9 \\geq |3x \\plus{} 9|$ implies $ \\frac {1}{3|x| \\plus{} 9} \\leq \\frac {1}{|3x \\plus{} 9|}$. When you have a ratio of two positive quantities, making the denominator larger makes the fraction smaller.[/quote]\r\n\r\nYes but see #2, he claims otherwise :(", "Solution_8": "You are right, isomorphism. The mistake there is acknowledged and that step was not needed." } { "Tag": [], "Problem": "If a three-digit number of the form $ 1N1$ is divided by $ N$, the quotient is of the form $ 2N$ remainder 5. What is the value of $ N$?", "Solution_1": "We see that $ 1N1$=$ N(2N)\\plus{}5$, so $ 100\\plus{}10N\\plus{}1\\equal{}2N^2\\plus{}5$, and our answer is $ \\boxed{N\\equal{}6}$", "Solution_2": "$100+10N+1=N(20+N)+5 \\\\ N^{2}+10-96=0 \\\\ (N+16)(N-6)=0 \\\\ N= -16,6$ \nWe cannot have a negative N and thus we choose $ \\boxed {6}$." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "we have circumscribed quadrilateral $ ABCD$ that the center of its circle is $ I$.$ M$ is on $ DI$ such that $ AM$ is perpendicular to $ AB$ and $ N$ is on $ BI$ such that$ AN$ is perpendicular to $ AD$.prove that $ AC$ is perpendicular to $ MN$", "Solution_1": "i find out too solution for my problem.does any body have nice solution", "Solution_2": "Hi,\r\n\r\nCan you post your solution? Thanks.", "Solution_3": "Maybe i missed something , the result seems false, changing positon of C on the circle doesn't affect the rest of the configuration so it can't be $ MN$ perpendicular on $ AC$", "Solution_4": "solution 1. lemma. if $ ABCD$ be a circumscribed quadrilateral and the circle tangent to $ AB,BC,CD,AD$ at $ X,Y,Z,W$ then $ AC,BD,XZ,WY$ are concur at one point and if the intersection named $ P$ then $ (AP)(CW)\\equal{}(CP)(AX)$ and $ (BP)(DX)\\equal{}(DP)(BY)$ \r\nfrom this lemma we can conclude that if $ L,Q$ be the projections of $ B,D$ on $ AC$ then (with Thales) $ (BL)(DX)\\equal{}(DQ)(BY)$.so if in our problem the projection of $ M$ on $ AC$ named $ R$ then $ (AR)(BY)\\equal{}r(BL)$ (by Thales) and if $ R_1$ be similiary like $ R$ for $ N$ then $ (AR_1)(DX)\\equal{}r(DQ)$.awe proved that $ AR\\equal{}AR_1$ so the problem is proved\r\nsol 2.we must prove that $ (AM)^2\\minus{}(CM)^2\\equal{}(AN)^2\\minus{}(CN)^2$.but its not easy to prove it without anything. we must draw perpendicular from $ A$ to $ CD$ and $ CB$" } { "Tag": [ "inequalities" ], "Problem": "Prove that $ 3(a^{2}\\plus{}b^{2}\\plus{}c^{2})>\\equal{}(a\\plus{}b\\plus{}c)^{2}$ :)", "Solution_1": "It is reqd to prove on simplification that $ a^2 \\plus{} b^2 \\plus{} c^2 \\geq a \\cdot b \\plus{} b \\cdot c \\plus{} c \\cdot a$\r\nThis is proved by the AM GM inequality :)", "Solution_2": "It's obviously true by Cauchy-Schwarz :wink:", "Solution_3": "hmm................... I didn't know that .\r\n\r\nWhat are this AM and GM inequality? Can you tell me please?", "Solution_4": "The AM-GM (arithmetic mean-geometric mean) inequality states that for non-negative reals $ r_1, r_2, ... r_n$ we have\r\n\r\n$ \\frac{r_1 \\plus{} r_2 \\plus{} ... \\plus{} r_n}{n} \\ge \\sqrt[n]{r_1 r_2 ... r_n}$.\r\n\r\nmadness is talking about the sum of the inequalities\r\n\r\n$ a^2 \\plus{} b^2 \\ge 2ab$\r\n$ b^2 \\plus{} c^2 \\ge 2bc$\r\n$ c^2 \\plus{} a^2 \\ge 2ca$\r\n\r\nwhich proves the desired inequality. The inequality he gives is well-known and can also be proven by Rearrangement. apollo gives the direct solution\r\n\r\n$ (1^2 \\plus{} 1^2 \\plus{} 1^2)(a^2 \\plus{} b^2 \\plus{} c^2) \\ge (1 \\cdot a \\plus{} 1 \\cdot b \\plus{} 1 \\cdot c)^2$\r\n\r\nby [url=http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality]Cauchy-Schwarz[/url]. These are the first techniques introduced in dealing with inequalities.", "Solution_5": "See this for the inequality http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means\r\n\r\nFor this case, it will be like\r\n\r\n$ \\sum (a^2 \\plus{} b^2) \\geq 2 \\sum \\sqrt {a^2 \\cdot b^2}$\r\n\r\n$ \\implies \\sum (a^2 \\plus{} b^2) \\geq 2 \\sum a \\cdot b$\r\n\r\nHence $ \\sum a^2 \\geq \\sum a \\cdot b$\r\n\r\nEDIT: Sorry t0rajir0u didnt see that you had posted it :)", "Solution_6": "By Cauchy Schwarz Inequality\r\n\r\n$ (a^2 \\plus{} b^2 \\plus{} c^2) (1^2 \\plus{} 1^2 \\plus{} 1^2) \\geq (a \\plus{} b \\plus{} c)^2$\r\n\r\nBy Chebyshev's Inequality, assuming WLOG a>b>c,\r\n\r\n$ 3(a*a \\plus{} b*b \\plus{} c*c) \\geq (a \\plus{} b \\plus{} c) (a \\plus{} b \\plus{} c) \\equal{} (a \\plus{} b \\plus{} c)^2$", "Solution_7": "And here: http://www.artofproblemsolving.com/Wiki/index.php/AMGM", "Solution_8": "Or by simple transformations:\r\n$ 3(a^{2} \\plus{} b^{2} \\plus{} c^{2})\\geq (a \\plus{} b \\plus{} c)^{2}$\r\n$ 2a^{2} \\plus{} 2b^{2} \\plus{} 2c^{2}\\geq 2ab \\plus{} 2bc \\plus{} 2ca$\r\n$ a^{2} \\minus{} 2ab \\plus{} b^{2} \\plus{} b^{2} \\minus{} 2bc \\plus{} c^{2} \\plus{} c^{2} \\minus{} 2ca \\plus{} a^{2}\\geq 0$\r\n$ (a \\minus{} b)^{2} \\plus{} (b \\minus{} c)^{2} \\plus{} (c \\minus{} a)^{2}\\geq 0$ :wink: with equality if and only if $ a \\equal{} b \\equal{} c$", "Solution_9": "Yup, that's what 2-var AM-GM is about :) \r\n\r\nBy the way, the last $ c^2$ in your 3rd line of math should be an $ a^2$", "Solution_10": "Similar one: \r\nProve that for all reals $ a,$ $ b$ and $ c$ the following inequality holds:\r\n\\[ 9(a^4\\plus{}b^4\\plus{}c^4)^2\\geq(a^5\\plus{}b^5\\plus{}c^5)(a\\plus{}b\\plus{}c)^3\\]", "Solution_11": "Cauchy Schwarz? :maybe: :)", "Solution_12": "Nice one arqady :)\r\n\r\nLet $ s_i=a^i+b^i+c^i$\r\n\r\nIt is equivalent to\r\n\r\n$ 9^4s_4^8 \\geq s_5^4s_1^12$\r\n\r\nRepeated (alot) use of CS\r\n\r\n\\begin{align*}\r\n(s_5s_1^3)^4 &\\leq s_1^4s_2^{10}s_3^29^2 \\\\\r\n& \\leq s_1^5s_2^{11}9^2 \\\\\r\n& = (s_1)^4s_1(s_2)^5s_29^2 \\\\\r\n& \\leq 9^6 s_2^4s_4^5s_1 \\\\\r\n& \\leq 9^6s_4^7 s_1 \\\\\r\n& \\leq 9^4s_4^8 \\end{align*}\r\n\r\n:)", "Solution_13": "I think only Chebyshev and power mean is required:\r\nWe have to prove:\r\n$ 9(a^4 \\plus{} b^4 \\plus{} c^4)^2\\geq(a^5 \\plus{} b^5 \\plus{} c^5)(a \\plus{} b \\plus{} c)^3$\r\nBy Chebyshevs inequality we have:\r\n$ \\frac {a^{4} \\plus{} b^{4} \\plus{} c^{4}}{3} \\equal{} \\frac {a^{5}*\\frac {1}{a} \\plus{} b^{5}*\\frac {1}{b} \\plus{} c^{5}*\\frac {1}{c}}{3}\\geq \\frac {a^{5} \\plus{} b^{5} \\plus{} c^{5}}{3}*\\frac {\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}}{3}$\r\nthus it remains to prove:\r\n$ (a^{5} \\plus{} b^{5} \\plus{} c^{5})(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})^{2}\\geq (a \\plus{} b \\plus{} c)^{3}$\r\nby AM-HM we get:\r\n$ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c}\\geq \\frac {9}{a \\plus{} b \\plus{} c}$ thus it suffices to prove that:\r\n$ (a^{5} \\plus{} b^{5} \\plus{} c^{5})*9^{2}\\geq (a \\plus{} b \\plus{} c)^{5}$ which is true, because the last inequality is equivalent to:\r\n$ \\sqrt [5]{\\frac {a^{5} \\plus{} b^{5} \\plus{} c^{5}}{3}}\\geq \\frac {a \\plus{} b \\plus{} c}{3}$ which is a direct conclusion from the generalizec power mean Q.E.D. :wink:\r\n\r\nEDIT: Nice solution from Cachy-Schwarz :)", "Solution_14": "[quote=\"SimonM\"]\n\n$ (s_5s_1^3)^4 \\& \\leq s_1^4s_2^{10}s_3^29^2$\n[/quote]\nI think, it's wrong. :wink: \n[quote=\"polskimisiek\"]\nBy Chebyshevs inequality we have:\n$ \\frac {a^{5}*\\frac {1}{a} + b^{5}*\\frac {1}{b} + c^{5}*\\frac {1}{c}}{3}\\geq \\frac {a^{5} + b^{5} + c^{5}}{3}*\\frac {\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c}}{3}$\n[/quote]\r\nI think, it's wrong. :wink:", "Solution_15": "Aaah, I see. I totally forgot aout the necessary condition $ a\\geq b\\geq c$ and $ \\frac {1}{a}\\geq \\frac {1}{b}\\geq \\frac {1}{c}$ which unfortunately cannot be true :oops: :ninja:" } { "Tag": [ "topology", "advanced fields", "advanced fields theorems" ], "Problem": "In R[sup]n[/sup] and S[sup]n[/sup], when we have a homeomorphism on a space X, we can create a homotopy whose endpoints match the homeomorphism (image of Xx{0} is the identity; image of Xx{1} matches image of homeomorphism).\r\n\r\nIn T^2 (the 2-dimensional torus), many homeomorphisms cannot be \"traced\" by a homotopy (e.g., any combination of meridional and longitudinal twists).\r\n\r\n1. In which spaces can any homeomorphism be \"traced\" by a homotopy, and why?\r\n\r\n2. Can we identify a certain group of homeomorphisms for each space outside of which any homeomorphism can be \"traced\" by a homotopy?\r\n\r\nI have a suspicion that the first question is something relatively basic and that I've seen the reason before, and a suspicion that the second question is not well-defined. Anyone who can point me in a helpful direction gets a cookie. : D\r\n\r\n\r\nEdit:[i]I suppose this is technically inherently topological, and solved, so doesn't belong in the unsolved algebra questions folder. Oops.[/i] >.>\r\n\r\nFurther note: [i]This question came up when thinking about knots on a torus, while working through Rolfsen's [u]Knots and Links[/u][/i].", "Solution_1": "I think you are interested in the [url=http://en.wikipedia.org/wiki/Mapping_class_group]mapping class group[/url].", "Solution_2": "Ah, thank you, exactly what I was looking for." } { "Tag": [ "AMC", "AIME", "floor function", "logarithms" ], "Problem": "Find n such that\r\n$\\lfloor\\log_{2}1\\rfloor+\\lfloor\\log_{2}2+\\rfloor+\\lfloor\\log_{2}3\\rfloor+...+\\lfloor\\log_{2}n\\rfloor=1994$", "Solution_1": "way too easy to be here...", "Solution_2": "posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=73948#p73948]here[/url]" } { "Tag": [ "geometry", "3D geometry", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Assume that M is compact, non-empty, perfect, and homeomorphic to its Cartesian square, M =~ M x M. Must M be homeomorphic to the Cantor set, the Hilbert cube, or some combination of them?", "Solution_1": "I'm just wondering. Do you consider these three problems in topology boring or should I post some more?", "Solution_2": "Please post as many topology problems as you want. Topology is not boring.\r\nWhat is the Hilbert cube?\r\nWhat do you understand by \"some combination of them\"?", "Solution_3": "I found out what Hilbert cube is. It is [0,1] |N|.", "Solution_4": "http://mathworld.wolfram.com/HilbertCube.html", "Solution_5": "Hmm, I read your post before, in school, I was going to write a reply at home, sorry that I didn't! I still don't have any answer to this question, seems hard.", "Solution_6": "Which topological set do you call to be perfect?" } { "Tag": [ "combinatorics solved", "combinatorics" ], "Problem": "4n points are arranged around a circle. The points are colored alternately yellow and blue. The yellow points are divided into pairs and each pair is joined by a yellow line segment. Similarly for the blue points. At most two segments meet at any point inside the circle. Show that there are at least n points of intersection between a yellow segment and a blue segment.\r\n\r\n[i]This is from the Swedish Math Olympiad in 1989.[/i]", "Solution_1": "Consider two intersecting yellow segments AC and BD. Then the number of points of intersection of any line with AB abd CD does not exceed the number of intersection points of the line with AC and BD. \r\nSo if we substitute of yellow segments AC and BD with AB \u0438 CD the number of intersection points of yellow and blue will no increase and the number of intersection points of yellow lines will decrease. \r\nIf we make certain number of such operations all yellow segments will not meet each other. It is sufficient to prove the statement only in this case.\r\nLook at any yellow segment L. Other yellow segments do not meet it, so both semiplanes created by L contain an even number of yellow points and as consequence an odd number of blue points. Therefore there is a blue segment that intersects L. So the number of yellow-blue intersections is at least the number of yellow segments that is n.", "Solution_2": "Good work, zortech!" } { "Tag": [ "inequalities", "geometry", "geometric transformation", "reflection", "inequalities proposed" ], "Problem": "Let $a,b,c$ be positive real numbers. Prove that\r\n \r\n$\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}-3 \\ge 18(\\frac a{b}+\\frac b{c}+\\frac c{a}-\\frac b{a}-\\frac c{b}-\\frac a{c})$,\r\n\r\nwith equality for $a=b=c$, and also for $a=\\frac b{2}=\\frac c{4}$ or any cyclic permutation thereof.", "Solution_1": "I think this one is just another way of formulating the well known.\r\n\r\nLet $a,b,c$ be positve real numbers such that $abc=1$. Prove that\r\n\\[a^{2}+b^{2}+c^{2}\\geq 3+18(a+b+c-ab-bc-ca).\\]", "Solution_2": "yes but why is it well known?\r\nhas it been posted before? :maybe:", "Solution_3": "See the article \"Four Applications of RCF and LCF Theorems\" in Mathematical Reflections, 1, 2007:\r\nhttp://reflections.awesomemath.org/2007_1.html\r\nI am still waiting a nice solution.", "Solution_4": "OK, VASC. It is a typical problem of mixing variable method. I solved it here", "Solution_5": "Dear hungkhtn: How can i buy your book? :?:", "Solution_6": "Only the Vietnamese are available now. It was published in 2006, but the English version is totally different. I think that Gil Publishing house will announce this book before IMO 2007 in Vietnam, so please waiting. Thank you for concern. :lol:", "Solution_7": "So how can Vietnamese orders your book? It's very hard to buy book from Gil in our country, hungkhtn!", "Solution_8": "[quote=\"The soul of rock\"]So how can Vietnamese orders your book? It's very hard to buy book from Gil in our country, hungkhtn![/quote]\r\n\r\nI think in this case, we need help of Gil. Anyway, it is not a right place to discuss about my book, so let me say that we should stop now. Please waiting when Mircea Lascu announces it, or we may discuss through PM Message." } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Find a possible formula for a function $ m(x)$ such that $ m'(x) \\equal{} x^5 \\cdot e^{x^6}$.\r\n\r\n[hide]I came up with m'(x) = e^{x^5}^{6} but I don't think it works[/hide].", "Solution_1": "$ m(x)\\equal{}\\frac{1}{6}\\cdot e^{x^6}$.", "Solution_2": "[quote=\"jeez123\"]Find a possible formula for a function $ m(x)$ such that $ m'(x) \\equal{} x^5 \\cdot e^{x^6}$.\n[/quote]\r\n\r\n$ m'(x) \\equal{} x^5 \\cdot e^{x^6}$ becomes: $ m'(u) \\equal{} \\frac {1}{6}e^{u}du$, with $ u \\equal{} x^{6}$\r\nYou now integrate $ m'(u)$ to get: $ m(u) \\equal{} \\frac {1}{6}e^{u} \\plus{} c$, where $ c$ is a constant. Now replace $ u$ with $ x^{6}$ to get the desired result." } { "Tag": [], "Problem": "Rationalize the denominator: $ \\frac1{2\\sqrt7}$.", "Solution_1": "$ \\frac{1}{2\\sqrt{7}} \\equal{} \\frac{1}{2\\sqrt{7}} \\cdot \\frac{\\sqrt{7}}{\\sqrt{7}} \\equal{} \\frac{\\sqrt{7}}{2 \\sqrt{7^2}} \\equal{} \\boxed{\\frac{\\sqrt{7}}{14}}$" } { "Tag": [ "calculus", "integration", "function", "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "1a. Let S be the solid 0 <= x <= 2, \u2212x <= y <= x, 0 <= z <= x^2 \u2212 y^2. Find the volume of S\r\n b. A scoop is in the shape of the parabolic surface \u22121 <= x <= 1, y = x2, 0 <= z <= 2, with\r\none end covered by the region \u22121 <= x <= 1, x^2 <= y <= 1, z = 0. How much water can the\r\nscoop hold?\r\n\r\n2. Let f be differentiable on [a,b] with |f '(x)| <= M for all x in [a,b]. Let I = int (a to b)f and let L[size=75]n[/size] be the Left Endpoint Approximation for I on n equal-sized subintervals. Prove that \r\n|I - L[size=75]n[/size]| <= (M(b-a)^2)/2n", "Solution_1": "The general method for proving accuracy of numerical integration methods:\r\nStep 1: Do it for one interval, not subdivided. Use Taylor's theorem or something similar to estimate the function with a polynomial that the numerical method gets right.\r\nStep 2: Use the result on each subinterval in your new division. Add up the estimated errors for the total." } { "Tag": [ "calculus", "derivative", "function", "trigonometry", "calculus computations" ], "Problem": "Obtain the derivative of the following function:\r\n\r\nf(x) = exp(sin2x)", "Solution_1": "This thread is in the wrong location. The [b]High School Basics[/b] forum is for algebra/geometry/precalculus topics mainly. This should probably be placed in the [b]Calculus Computations and Tutorials[/b] subforum. :)", "Solution_2": "all right I didn't know...I will post it there instead....thanks for telling me...", "Solution_3": "$(e^{\\sin 2x})'=2 \\cos 2x \\cdot e^{\\sin 2x}$" } { "Tag": [ "geometry", "3D geometry", "function", "trigonometry" ], "Problem": "I don't know how to solve this one \\[x^{6}=20000\\]\r\n\r\nAny help or hint?", "Solution_1": "In wich set you want to solve that equation ? $R$ , or $C$, or $Z$ ? :roll:", "Solution_2": "are you just trying to solve for x? Cause that's all I see, just use logs.", "Solution_3": "For $R$, I would just take $\\pm\\sqrt[6]{20000}$, for $C$ I would use De Moive's theorem.", "Solution_4": "I suppose it is impossible for $Z$ so i would like a $R$ solution and then a $C$ one.\r\n\r\n@besttate I'm afraid $\\pm \\sqrt[6]{20000}$ wouldn't help. And i am not prepared enough to use such theorems like De Moivre's.\r\n\r\nCan you help me?", "Solution_5": "In reals, note that the cube root function is one-to-one, so we want to solve\r\n\r\n$x^{2}= \\sqrt[3]{20000}= 10 \\sqrt[3]{20}= 27.144...$\r\n\r\nWhich has roots\r\n\r\n$x = \\pm \\sqrt[6]{20000}= \\pm 5.2100...$\r\n\r\nThe other four complex cube roots are complex multiples of this one, but their form is not easy to derive without the use of de Moivre's theorem.", "Solution_6": "I do not know how to do the problem in any other way, below is using the methods described above. Hopefully these can help you in some way.\r\n\r\nIn $R$, simplifying $\\pm\\sqrt[6]{20000}$, we have $\\pm2\\sqrt[6]{3125}$\r\n\r\nIn $C$, using De Moivre's theorem, we write $20000$ in polar notation as $20000\\cos0+20000i\\sin0$, or \"$20000$cis$0$\" ($A$cis$\\theta$ simply means $A\\cos\\theta+Ai\\sin\\theta$).\r\n\r\nFirst, we calculate $20000^{1/6}$cis$0\\pi/6$, which becomes $2\\sqrt[6]{3125}$.\r\n\r\nNext, we calculate $20000^{1/6}$cis$2\\pi/6$, which becomes $\\sqrt[6]{3125}+i\\sqrt{3}\\sqrt[6]{3125}$\r\n\r\nNext, we calculate $20000^{1/6}$cis$4\\pi/6$, which becomes $-\\sqrt[6]{3125}+i\\sqrt{3}\\sqrt[6]{3125}$\r\n\r\nNext, we calculate $20000^{1/6}$cis$6\\pi/6$, which becomes $-2\\sqrt[6]{3125}$\r\n\r\nNext, we calculate $20000^{1/6}$cis$8\\pi/6$, which becomes $-\\sqrt[6]{3125}-i\\sqrt{3}\\sqrt[6]{3125}$\r\n\r\nFinally, we calculate $20000^{1/6}$cis$10\\pi/6$, which becomes $\\sqrt[6]{3125}-i\\sqrt{3}\\sqrt[6]{3125}$\r\n\r\nIn summary, the 2 real solutions are $\\pm2\\sqrt[6]{3125}$ and the 4 complex solutions are the 4 combinations of $\\pm\\sqrt[6]{3125}\\pm i\\sqrt{3}\\sqrt[6]{3125}$.\r\n\r\nHope this helps a little :D ." } { "Tag": [ "integration", "LaTeX" ], "Problem": "I need to display a union over a set; however the \\bigcup{}{} command won't let me put i \\in I in the lower argument. Can someone tell me how to get around this, please?", "Solution_1": "Use the same syntax as for \\int and \\sum with _ for lower limit and ^ for upper limit.\r\n\\bigcup_{i \\in I} A_i gives $ \\bigcup_{i \\in I} A_i$ and \\bigcup_{i =1}^n A_i gives $ \\bigcup_{i \\equal{}1}^n A_i$", "Solution_2": "Thanks, stevem.\r\n\r\nI was putting the second argument on, but left it empty; and I didn't use the subscript!---I got something ugly." } { "Tag": [ "vector", "linear algebra", "matrix", "Support", "algebra", "polynomial" ], "Problem": "Hey, I got some questions, please help\r\n\r\n1. I remember in first year LA we preformed ERO's (elementary row operations) on matrices without linear transforms in mind. if we have $ Tx\\equal{}s$ with $ x$ our unknown vector ERO's change our $ T$ and $ s$ such that $ x$ stays the same, if I remember correctly. \r\nWhat is the relevance of this to the study linear transforms? \r\nHow would you prove an ERO leaves the solution to the equation $ Tx\\equal{}s$ unchanged in the language of linear transforms?\r\n\r\n2. Given a matrix T: say $ \\begin{pmatrix}\r\n1 & 1 \\\\\r\n\\minus{}1 & 2\r\n\\end{pmatrix}$ how do we find its upper-triangular form, if it exists.\r\n\r\n3. How would we use the exponential of a matrix to solve the DE:\r\n\r\n$ u'\\equal{}5u\\minus{}2v$\r\n$ v'\\equal{}7u\\minus{}3v$", "Solution_1": "1: A row operation is equivalent to multiplication on the left by some specific matrix $ R$. If $ Tx \\equal{} s$, then $ RTx \\equal{} Rs$. Conversely, if $ R$ is invertible and $ RTx \\equal{} Rs$, $ Tx \\equal{} s$. Row interchanges, shears in which we add a multiple of one row to another, and multiplying a row by a nonzero constant are all invertible.\r\n\r\n2: What do you mean, \"its upper-triangular form\"? There are many ways for matrices to be related to each other, which will give wildly different answers.\r\nYou're probably talking about row/echelon form, which is intrinsically linked to row operations. Even then- that's not really unique, unless you have very strict rules about when to stop. Flesh that out, and it leads to LU-factorization; how to write a matrix as a product $ LU$ with $ L$ lower triangular and $ U$ upper triangular, with one of them having ones on the diagonal. That form isn't quite universal, either; if you needed row interchanges, there's also a multiplication by a permutation matrix $ P$.\r\n\r\n3. The solution to $ y'(t) \\equal{} Ay$ ($ A$ constant) is $ y \\equal{} e^{At}y_0$. Now, how do you actually calculate that matrix exponential? Diagonalize, or at least put it in Jordan form. Similarity works well with everything.", "Solution_2": "1. A row operation (elementary or not) corresponds to left-multiplying by an invertible transformation. So row operations turn $ Tx \\equal{} s$ into $ PTx \\equal{} Ps.$ To support the claim that this leaves the solutions set unchanged, we should explicitly invoke the fact that $ P$ is invertible.\r\n\r\n2. When you say \"its upper triangular form\" are you talking about similarity? If you are, then we should find the characteristic polynomial and hence the eigenvalues. For a $ 2\\times 2$ matrix $ A$ the characteristic polynomial is $ \\lambda^2 \\minus{} (\\text{trace}\\,A)\\lambda \\plus{} \\det A.$ In this case, that's $ \\lambda^2 \\minus{} 3\\lambda \\plus{} 3.$ This has only complex roots, so this matrix cannot be diagonalized or triangularized over the reals, but it can be diagonalized over the complex numbers.\r\n\r\n3. If the differential equation is $ y' \\equal{} Ay$ with the initial condition $ y_0$ (where $ y$ and $ y_0$ are vectors and $ A$ is a square matrix), then the solution is $ y(t) \\equal{} e^{tA}y_0.$\r\n\r\nFor a formula for finding $ e^{tA}$ for any $ 2\\times 2$ matrix $ A$, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310425]here.[/url]\r\n\r\n\r\nEdit: partly redundant with jmerry's post; composed simultaneously. Except that we came to different interpretations of \"its upper triangular form\", which just serves to emphasize jmerry's point about that being ambiguous." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "show that the co-efficient of $ z^{2}$ in the series expansion about $ z\\equal{}0$ of\r\n\r\n$ \\frac{1}{z\\cdot (z\\plus{}1)\\cdot\\Gamma{(z)}}$ \r\n\r\nis $ \\boxed{1\\minus{}\\frac{1}{2}\\,\\gamma\\,(2\\minus{}\\gamma)\\minus{}\\frac{1}{2}\\,\\zeta{(2)}}$ :D", "Solution_1": "read Ga/7 .( first relation)\r\nhttp://dlmf.nist.gov/Contents/GA/7/\r\nI think after you can do your problem yourself" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $ n\\in N,n\\geq2,a_{i},b_{i}\\in R,1\\leq i\\leq n$,such that $ \\sum_{i\\equal{}1}^{n}a_{i}^{2}\\equal{}\\sum_{i\\equal{}1}^{n}b_{i}^{2}\\equal{}1,\\sum_{i\\equal{}1}^{n}a_{i}b_{i}\\equal{}0$.\r\nProve that $ (\\sum_{i\\equal{}1}^{n}a_{i})^{2}\\plus{}(\\sum_{i\\equal{}1}^{n}b_{i})^{2}\\leq n$", "Solution_1": "See here: [url]http://www.mathlinks.ro/viewtopic.php?p=860141#p860141[/url]" } { "Tag": [ "linear algebra", "matrix", "geometry", "rectangle", "vector", "logarithms" ], "Problem": "Can someone explain exactly what a matrice is and how to do magic with it? :D", "Solution_1": "[quote=\"now a ranger\"]Can someone explain exactly what a matrice is and how to do magic with it? :D[/quote]\r\nA [b]matrix[/b] is a rectangle array of numbers. (e.g. 1 3 7\r\n 6 8 3)", "Solution_2": "dude\r\nby magic, do you mean magic squares or something?\r\ni haven't encountered real matrix magic.\r\nfuncia", "Solution_3": "can someone elaborate further? Not getting anything!", "Solution_4": ":lol: \r\nhow about this:\r\nhttp://en.wikipedia.org/wiki/Matrix_%28mathematics%29\r\ni would google matrix and magic if i were you", "Solution_5": "I hope i can learn this today. =()", "Solution_6": "yeah well i have no idea if you have access \r\nto a library but if one nearby had\r\nthis book i liked :\r\nIntroduction to matrices and vectors\r\n by Jacob Schwartz", "Solution_7": "A matrix is like a data table, except you can preform operations with it:\r\n\r\n$\\begin{bmatrix}a & b \\\\ c & d \\end{bmatrix}\\times 2$\r\n\r\n$\\begin{bmatrix}a & b \\\\ c & d \\end{bmatrix}\\times \\begin{bmatrix}e & f \\\\ g & h \\end{bmatrix}$", "Solution_8": "Read AoPS II.", "Solution_9": "We were told that matrices and logarithms can greatly simplify calculations. I know that logs can come in handy but even after learning matrices, I'm not sure of how they can help u in big calculations. How do they??", "Solution_10": "Matricies help with solving systems of equations. Like with 4 variables :wink: .", "Solution_11": "But how do they do that?? :huh: \r\nIf you don't mind, can you show me an example?\r\nThanx!!", "Solution_12": "Say for example:\r\n\r\n$2x+3y=7$\r\n\r\n$x-9y=13$\r\n\r\nI'm only going to use 2 variables because i don't have a lot of time =)\r\n\r\nset up the matrix as follows\r\n\r\n$\\begin{bmatrix}2 & 3 \\\\ 1 &-9 \\end{bmatrix}$\r\n\r\nnow the equation is Ax=B, where A is the above matrix. and B is the solutions:\r\n\r\n$A = \\begin{bmatrix}2 & 3 \\\\ 1 &-9 \\end{bmatrix}$\r\n\r\nand \r\n\r\n$B= \\begin{bmatrix}7 \\\\ 13 \\end{bmatrix}$ \r\n\r\nTo find x, which is your variables-\r\n\r\n$X=\\begin{bmatrix}x \\\\ y \\end{bmatrix}$.\r\n\r\nSolve for x using $A^{-1}X=B \\end{bmatrix}$.\r\n\r\nsorry for not explaining this very well. i have to go now I will make it better when i come back" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "how do you integrate y=$ \\sqrt{4\\minus{}x^2}$?\r\n\r\nthanks :)", "Solution_1": "hello, substitute $ x\\equal{}2\\sin(t)$\r\nSonnhard.", "Solution_2": "[hide=\"Solution?\"]\\begin{align*} \\int\\sqrt {4 - x^2}\\;dx & = \\int\\sqrt {4 - 4\\sin^2t}\\;dt \\\\\n& = 2\\int\\sqrt {\\cos^2t}\\;dt \\\\\n& = 2\\int\\cos{t}\\;dt \\\\\n& = 2\\sin{t} + C \\\\\n& = \\boxed{x + C} \\end{align*}[/hide]", "Solution_3": "$ dx \\neq dt$. You forgot to change that part of the integrand.", "Solution_4": "EDIT: oops...\r\n\r\n$ x \\equal{} 2\\sin t \\implies dx \\equal{} 2\\cos t\\; dt$\r\n\r\n$ \\int \\sqrt {4 \\minus{} x^2} \\; dx \\equal{} \\int 2\\sqrt {1 \\minus{} \\sin^2 t} \\; 2\\cos t \\; dt \\equal{} \\int 2 \\cos t \\; 2\\cos t\\; dt$\r\n\r\nwhich is $ \\int 4\\cos^2 t \\; dt$\r\n\r\nhow do you go from there?", "Solution_5": "You substituted $ \\cos$ instead of $ \\sin$ under the square root sign. You'll need to use the double angle formulae to evaluate the $ \\cos^2$", "Solution_6": "ugh im lost...\r\n\r\ncould somebody please post a solution??", "Solution_7": "Wait! Is $ \\sqrt{A^2}\\equal{}A$ for real number $ A$ always hold?" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $ A$ be a $ nxn$ matrix. Show that $ \\det\\left[\\begin{array}{cc} A & A^2\\\\\r\nA^2 & A^3\r\n\\end{array} \\right]$ is zero.", "Solution_1": "This is a simple application of the following lemma.\r\n\r\n$ \\textbf{Lemma}$ Let $ C$, $ D$, $ X$, $ Y$, be $ n\\times n$ complex matrices. If $ C$ commutes with $ X$ then \r\n\r\n$ \\det\\left[\\begin{array}{cc} C & Y\\\\\r\nX & D\r\n\\end{array} \\right]\\equal{}\\det(CD\\minus{}XY).$", "Solution_2": "Let $ B$ be the $ 2n\\times 2n$ matrix $ \\begin{bmatrix}A&A^2\\\\A^2&A^3\\end{bmatrix}$.\r\n\r\nThen not only is $ B$ singular, but $ \\text{rank}(B)\\equal{}\\text{rank}(A)\\le n.$\r\n\r\nProof by row and column reduction:\r\n\r\n$ \\begin{bmatrix}I&0\\\\\\minus{}A&I\\end{bmatrix} \\begin{bmatrix}A&A^2\\\\A^2&A^3\\end{bmatrix} \\begin{bmatrix}I&\\minus{}A\\\\0&I\\end{bmatrix}\\equal{} \\begin{bmatrix}A&0\\\\0&0\\end{bmatrix}$" } { "Tag": [ "linear algebra", "matrix", "integration", "function", "search", "limit", "calculus" ], "Problem": "I've been working on this for long, however, till now I still cannot verify it except for the confidence of it being true.\r\n\r\nSuppose $ A_n$ is n * n sized real matrix, and the (i, j) element $ a_{ij} \\equal{} \\int_0^1f ^ {(i)}f ^ {(j)}dx$, where i, j = 0, 1, ... , n - 1, f is an arbitrary real function in $ C^{\\infty}[0, 1]$, $ f^{(i)} \\equal{} \\frac{d^if}{dx^i}$.\r\n\r\nMy problem is [b]how to verify the serie {$ \\lambda_n$} tends to 0, where $ \\lambda_n$ is the smallest eigenvalue of $ A_n$[/b], and how fast is the serie tending to 0? An estimation is most welcome.\r\n\r\nBy the way, by some numerical tests, I am almost sure it is true (tending to 0), however, if otherwise is correct, please also give me an example. Thanks a lot.", "Solution_1": "A first step: use the search function on this forum to search for the words \"Hilbert Matrix.\" Do any of the links you find that way help you?", "Solution_2": "I checked the searching results, yet it seems not working. And it seems the matrix $ A_n$ is not a [b]Hilbert Matrix[/b] but a [b]Hanker Matrix[/b].\r\n\r\nLet me suggest something, hope it could help.\r\n\r\nSuppose $ \\lambda_1 > \\lambda_2 > \\cdots > \\lambda_n$ is the n eigenvalues of $ A_n$. Since $ \\lambda_1$ might be much larger than $ \\lambda_n$,\r\n\r\n$ \\lambda_n$ could be \"covered\" by the trace of $ A_n$, i.e. tr(A).\r\n\r\nSo let's consider tr($ A^{\\minus{}1}_n$) which equals to $ \\frac{1}{\\lambda_1} \\plus{} \\cdots \\plus{} \\frac{1}{\\lambda_n}$. Thus we have\r\n\r\n$ tr(A^{\\minus{}1}_n) \\equal{} \\frac{1}{\\lambda_1} \\plus{} \\cdots \\plus{} \\frac{1}{\\lambda_n} < \\frac{n}{\\lambda_n}$.\r\n\r\nSo $ \\lambda_n < \\frac{n}{tr(A^{\\minus{}1}_n)}$.\r\n\r\nAnd then comes our problem, who can give an estimate of tr($ A^{\\minus{}1}_n$)?\r\n\r\nWhy $ \\lim_{n\\rightarrow\\infty} \\frac{n}{tr(A^{\\minus{}1}_n)} \\equal{} 0$?\r\n\r\nHope someone can help me. Thanks a lot.", "Solution_3": "Basically, the question boils down to whether we can have a smooth function $ f$ such that $ \\int_0^1 \\Bigl|\\sum_k c_kf^{(k)}(x)\\Bigr|^2\\,dx\\ge \\sum|c_k|^2$ for every set of coefficients $ c_k$. The answer is \"Yes, of course\" because we can easily add fast oscillating bumps to the derivatives inductively to ensure sufficient amount of linear independence. On the other hand, the question whether one can have an [b]analytic[/b] function like that is a little bit more interesting ;)", "Solution_4": "Sorry, I couldn't understand you clearly.\r\n\r\nBut it is obvious that a lot of smooth function satisfy my assertion, for example $ \\sin{x}$, and the $ A_n$ constructed by $ \\sin{x}$ has the minimal eigenvalue which will tend to be 0 with the increase of n.\r\n\r\nSo my problem is [b]for any function[/b] $ f \\in C^\\infty[0, 1]$, the $ A_n$ constructed by the $ f$ will behave as I've said in my first post.\r\n\r\nAnd also could you give some details on how to reduce my question to the inequality you suggested?\r\n\r\nThank you for your reply. I really appreciate it.\r\n\r\n[quote=\"fedja\"]Basically, the question boils down to whether we can have a smooth function $ f$ such that $ \\int_0^1 \\Bigl|\\sum_k c_kf^{(k)}(x)\\Bigr|^2\\,dx\\ge \\sum|c_k|^2$ for every set of coefficients $ c_k$. The answer is \"Yes, of course\" because we can easily add fast oscillating bumps to the derivatives inductively to ensure sufficient amount of linear independence. On the other hand, the question whether one can have an [b]analytic[/b] function like that is a little bit more interesting ;)[/quote]", "Solution_5": "Sorry for the brevity of my previous post. The idea is to build a function whose derivatives are sufficiently linearly independent. What you can do for that is to add fast oscillating terms to high order derivatives that disappear after integration, so they do not show up in the low order derivatives. The simplest fast oscillating function is $ \\cos \\lambda x$ with big $ \\lambda$. So, it is natural to build a series like\r\n\\[ f(x)\\equal{}\\sum_{k\\ge 1}\\lambda_k^{\\minus{}k\\plus{}\\frac 12}\\cos \\lambda_kx\\]\r\nwith very fast increasing $ \\lambda_k$ ($ \\lambda_k\\equal{}2^{k!}$ should be enough). When you differentiate this series $ n$ times, you'll get $ \\lambda_{n}^{1/2}$ times $ \\sin$ or $ \\cos$ of $ \\lambda_n x$ as the dominating term in the series for the $ n$-th derivative. So, $ f^{(n)}$ will be large and essentially orthogonal on $ [0,1]$ and you can easily derive from here that you cannot find their linear combination that is small in the $ L^2$ norm. This function is, of course, horrible: it is $ C^\\infty$ but very far from being analytic. That's why I asked my question about analytic functions. On the other hand, I suspect that even for as simple function as $ \\frac 1{1\\plus{}x}$, the eigenvalues do not tend to $ 0$: the factorials in the numerator far outweigh the powers of numbers less than $ 2$ in the denominator. Try to verify if it is true yourself. The functions like $ \\sin x$ and polynomials are exceptional in the sense that their derivatives are linearly dependent, so quite soon the least eigenvalue becomes exactly $ 0$.\r\n\r\nAs to your last question, it is an immediate consequence of the standard theorem that the minimal eigenvalue of a non-negative definite real symmetric matrix is just the minimum of the corresponding quadratic form over the unit ball. In the direction you need it here, it is completely trivial: just consider $ \\langle Ax,x\\rangle$ where $ x$ is the eigenvector corresponding to the minimal eigenvalue.", "Solution_6": "Oops, don't know what happened to me :oops: : my remark about $ \\frac 1{1 \\plus{} x}$ is an ultimate nonsense (by the way, who can explain why?). As a matter of fact, I now believe that all analytic functions are OK. At least, if we consider the $ 1$-periodic analytic functions, we do have a positive result. The idea is to show that $ f'$ is in the linear span of the higher order derivatives. Suppose not. Then there is an $ L^2$ $ 1$-periodic function $ \\psi$ such that $ \\int_0^1\\psi( \\minus{} t) f^{(n)}(t)\\,dt \\equal{} 0$ for all $ n > 1$ but is not $ 0$ for $ n \\equal{} 1$. But what is written is just the $ n$-th derivative of the convolution $ f*\\psi$ at $ 0$. Since that convolution is analytic and has all but the first 2 derivatives $ 0$ at $ 0$, it must be a non-constant polynomial of degree $ 1$ but a polynomial of degree $ 1$ cannot be $ 1$-periodic! The question about non-periodic functions is more complicated. Let me think of it for a while :). \r\n\r\nAt least, it is clear now that the problem has hardly anything to do with linear algebra. So I'm moving it :P\r\n\r\nEdit 1: We can also show that the statement is true for the functions analytic in a sufficiently large domain containing $ [0,1]$. Assume for instance that $ f$ is analytic in the disk of radius 3 around the origin. There are two possibilities: \r\na) $ f',f'',\\dots$ span $ L^2[0,1]$. Then $ f$ can be approximated by their linear combinations and we are done.\r\nb) There is a function $ \\psi\\in L^2[0,1]$ orthogonal to all derivatives. Let $ F\\equal{}f'$. Consider $ \\int_0^1 F(x\\plus{}t)\\psi(t)\\,dt$. On one hand, we can write $ F(x\\plus{}t)\\equal{}\\sum_{k\\ge 0}\\frac {x^k}{k!}F^{(k)}(t)$, which immediately implies that the integral must be $ 0$. On the other hand, we can also write $ F(x\\plus{}t)\\equal{}\\sum_{k\\ge 0}\\frac {t^k}{k!}F^{(k)}(x)$, resulting in the identity $ \\sum_k c_k F^{(k)}(x)\\equal{}0$ with $ c_k\\equal{}\\frac 1{k!}\\int_0^1 t^k\\psi(t)\\,dt$. Since $ t^k$ span $ L^2[0,1]$, not all $ c_k$ are zero and we are done again. \r\n\r\nThe more I think about this problem, the more I like it :) The right answer should, probably, be that the statement is true in every quasianalytic class and for every non-quasianalytic class there is a counterexample. But this is just a wild speculation and, honestly speaking, I don't have any more time to think of this problem today :(", "Solution_7": "Yes! You're right. In fact now I believe my statement is correct among the analytic functions. But there is still a little problem, that is about the $ C ^ \\infty$ functions. To tell the truth, your counterexample seems not so persuading, or it just has something I haven't understood yet. Anyway thanks a lot.\r\n\r\nAnd by the way, could you also help answer my second question in my initial post? The question is can we get an estimate about the speed of the minimal eigenvalue tending to zero? For example, it will be very nice if we can draw the conclusion that the minimal eigenvalue $ \\lambda_n$ of $ A_n$ holds $ \\lambda_n < \\frac{1}{n}$. Of course, this is obviously not possible, and I guess the correct estimate should involve $ f ^ {(n)}$, and so this is just an example.\r\n\r\nAnd at last I'd like to say your proof of the analytic functions is very useful for my current work. Thank you very much!\r\n\r\n[quote=\"fedja\"]Oops, don't know what happened to me :oops: : my remark about $ \\frac 1{1 \\plus{} x}$ is an ultimate nonsense (by the way, who can explain why?). As a matter of fact, I now believe that all analytic functions are OK. At least, if we consider the $ 1$-periodic analytic functions, we do have a positive result. The idea is to show that $ f'$ is in the linear span of the higher order derivatives. Suppose not. Then there is an $ L^2$ $ 1$-periodic function $ \\psi$ such that $ \\int_0^1\\psi( \\minus{} t) f^{(n)}(t)\\,dt \\equal{} 0$ for all $ n > 1$ but is not $ 0$ for $ n \\equal{} 1$. But what is written is just the $ n$-th derivative of the convolution $ f*\\psi$ at $ 0$. Since that convolution is analytic and has all but the first 2 derivatives $ 0$ at $ 0$, it must be a non-constant polynomial of degree $ 1$ but a polynomial of degree $ 1$ cannot be $ 1$-periodic! The question about non-periodic functions is more complicated. Let me think of it for a while :). \n\nAt least, it is clear now that the problem has hardly anything to do with linear algebra. So I'm moving it :P\n\nEdit 1: We can also show that the statement is true for the functions analytic in a sufficiently large domain containing $ [0,1]$. Assume for instance that $ f$ is analytic in the disk of radius 3 around the origin. There are two possibilities: \na) $ f',f'',\\dots$ span $ L^2[0,1]$. Then $ f$ can be approximated by their linear combinations and we are done.\nb) There is a function $ \\psi\\in L^2[0,1]$ orthogonal to all derivatives. Let $ F \\equal{} f'$. Consider $ \\int_0^1 F(x \\plus{} t)\\psi(t)\\,dt$. On one hand, we can write $ F(x \\plus{} t) \\equal{} \\sum_{k\\ge 0}\\frac {x^k}{k!}F^{(k)}(t)$, which immediately implies that the integral must be $ 0$. On the other hand, we can also write $ F(x \\plus{} t) \\equal{} \\sum_{k\\ge 0}\\frac {t^k}{k!}F^{(k)}(x)$, resulting in the identity $ \\sum_k c_k F^{(k)}(x) \\equal{} 0$ with $ c_k \\equal{} \\frac 1{k!}\\int_0^1 t^k\\psi(t)\\,dt$. Since $ t^k$ span $ L^2[0,1]$, not all $ c_k$ are zero and we are done again. \n\nThe more I think about this problem, the more I like it :) The right answer should, probably, be that the statement is true in every quasianalytic class and for every non-quasianalytic class there is a counterexample. But this is just a wild speculation and, honestly speaking, I don't have any more time to think of this problem today :([/quote]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For abritary $ \\alpha, \\beta, x, y \\in \\mathbb{R}$ prove that:\r\n\r\n$ (2 \\alpha ^2 \\plus{} 2 \\alpha \\beta \\plus{} \\beta ^2)(2 x ^2 \\plus{} 2 xy \\plus{} y ^2) \\ge (2 \\alpha x \\plus{} \\alpha y \\plus{} \\beta x \\plus{} \\beta y)^2$", "Solution_1": "yes easy but sometimes it's usful, just see $ (2 \\alpha ^2 \\plus{} 2 \\alpha \\beta \\plus{} \\beta ^2)(2 x ^2 \\plus{} 2 xy \\plus{} y ^2) \\equal{} ( \\alpha ^2 \\plus{} ( \\alpha \\plus{} \\beta )^2)( x ^2 \\plus{} (x \\plus{} y)^2)$ \r\n$ \\ge (\\alpha x \\plus{} (\\alpha \\plus{} \\beta)(x \\plus{} y))^2$.", "Solution_2": "hello Mathias, your inequality is equivalent to $ (x\\beta\\minus{}y\\alpha)^2\\geq0$ and this is allways true for all\r\nreal $ x,y, \\alpha$ and $ \\beta$.\r\nSonnhard.", "Solution_3": "Nice solutions :)\r\n\r\nThe way I created the problem:\r\nI chose an inner product space $ \\left \\langle \\left ( \\begin{array}{c} a_1 \\\\ a_2 \\end{array} \\right ) , \\left ( \\begin{array}{c} b_1 \\\\ b_2 \\end{array} \\right ) \\right \\rangle \\equal{} 2a_1b_1 \\plus{} a_1b_2 \\plus{} a_2b_1 \\plus{} a_2b_2$. Then choosing $ u \\equal{} (\\alpha, \\beta)$ and $ v \\equal{} (x,y)$ it just becomes $ \\langle u, u \\rangle ^{1/2} \\langle v, v \\rangle ^{1/2} \\ge \\langle u, v \\rangle$ which is Cauchy Schwartz :)", "Solution_4": "Mathias, this simple inequality is really useful in killing tougher inequalties. I have killed one form a book using this.\r\nThanks anyway." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let sequences ${x_{n}}$ and ${y_{n}}$ be converge. Prove that\r\n\\[\\lim_{n\\to\\infty}\\frac1n\\sum_{i=1}^{n}x_{i}y_{n+1-i}= \\lim_{n\\to\\infty}x_{n}y_{n}\\]", "Solution_1": "Here is an idea:\r\n\r\nLet $z_{n}=\\frac{x_{n}y_{1}+\\cdots x_{1}y_{n}}{n}=\\frac{(x_{1}-x)y_{n}+\\cdots+(x_{n}-x)y_{1}}{n}+x\\frac{(y_{1}-y)+\\cdots (y_{n}-y)}{n}$\r\n\r\n$+xy$, where $x=\\lim x_{n}$ and $y=\\lim y_{n}$.\r\nThus it is sufficient to prove that\r\n\r\nthe two sequences converge to 0.\r\n\r\nNow, $|\\frac{(x_{1}-x)y_{n}+\\cdots+(x_{n}-x)y_{1}}{n}|\\leq M\\frac{|x_{1}-x|+\\cdots+|x_{n}-x|}{n}\\rightarrow 0$, where $M=\\sup y_{n}$\r\nThe second sequence also converges to 0.\r\nThe result follows." } { "Tag": [], "Problem": "I am not sure where the appropriate place to post this is, if it's in the wrong spot then feel free to move it.\r\n\r\nI was at a local math competition and one of the problems was something like \"Find all values of X such that 9^X+3(1-2X)-4=0\" \r\n\r\nI found that X=0 and that 3^2X+3(1-2X)=4\r\n\r\nThe other answer is X=1/2, but I would like to know if there is a way in which I could've found them systematically and been sure that there were no other possibilities.", "Solution_1": "Do you mean $ 9^x \\plus{} 3^{1 \\minus{} 2x} \\minus{} 4 \\equal{} 0$? \r\n\r\nWrite as $ 9^x \\plus{} \\frac{3}{9^x} \\minus{} 4 \\equal{} 0$ and substitute $ y \\equal{} 9^x$.", "Solution_2": "That is what I meant. Thanks for the answer, it works quite well" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ A$ be a set of real numbers has property for every distinct numbers $ a,b,c \\in A$ then $ a^2\\plus{}bc$ is rational numbers.Prove that there exist a positive number $ M$ such that $ a*M^{1/2}$ is rational for any $ a \\in A$", "Solution_1": "This is INMO 2008 problem 3 (Though not original in that contest also).\r\n[url=http://www.isid.ac.in/~rbb/inmosol_08.pdf]Click here[/url] for the solution." } { "Tag": [ "geometry", "3D geometry", "sphere", "search", "number theory unsolved", "number theory" ], "Problem": "The Unforgettable 6th June 1975\r\n\r\n[b][color=darkred][Please don't remove these post and link. They are very important for our humanity!][/color][/b]\r\n\r\nIt was my ninth attempt to make contact. The sky was perfectly clear. Once again I observed the silence in the clearing.\r\n\r\nThere we have it. They were there. Everything was fine.\r\n\r\nThere was an agreed upon signal between us. I knew that I could concentrate on sending my thoughts by telepathy because they had let me know that they were receiving them. Then there was another shooting star which crossed the sky horizontally, once again a third one in the same direction.\r\n\r\nMy attention was drawn to lines of fire which stood out clearly against the night sky. It was about half past ten at night. I could feel a certain presence. As observed, even studied. I didn't take offence to their actions, even the contrary. It was then that a small ball of light, which burst forth from the vastness of the sky, filled my heart with joy.\r\n\r\nWhat a firework!\r\n\r\nHalf an hour went by without me seeing other closer signs. The last one having still been very high in space. Suddenly, as if to contradict me, I saw very clearly above me the shape of a space ship. It was circular, equipped at the back with three small wings. This block of light had just disappeared as if by magic. In its view, I my gesturing signals but I quickly understood that it was useless. The craft was far too far away to understand what I wanted. However, it was responding well to my telepathic messages. Everything was happening so high up that I too wanted to be there. Why couldn't they land here where there was no one to disturb us? I came to the tragic conclusion that they had no intention of contacting me but that they had shown themselves to encourage my actions. I was angered by their attitude and I decided to break purely and simply any relationship with them. I directed my terms to those who were high up there.\r\n\r\n\"I thank you for all that you have done and for showing yourselves to me at this time. However, you must understand that I cannot consciously conceive your presence by such imprecise and distant apparitions. I know that you are very good to us and that without such kindness we would have ceased to exist or would have become slaves. But what is the use of my presence in these places if you don't do anything concrete to explain it to me. I see my presence here as being useless. If there is anything to be done later, do not count on me. Therefore adieu (to God), or rather aux dieux (to the gods), to you\"\r\n\r\nI raised both my arms to the sky in a sign of peace. Then, I prepared myself to leave the place. That's when an extraordinary thing happened. I had only just turned my head away when I noticed a strange orangey-red star above the hill in front of me. However, I still hadn't totally understood this mysterious phenomenon. My ability to understand was inhibited by its focusing effect. With great speed the star came straight at me and became a large sphere with a diameter of 3 metres. \r\n\r\nI kept my feet voluntarily rooted to the ground because under no circumstances should I have run away from what was to happen. The ball stopped implacably. The object pivoted around itself. It was now a small dome. I hardly had enough time to notice its detail because an idiotic but very natural reflex made me protect my face with my left arm for just a moment during the approach, out of belief that it would crash into the ground and get me. However, I had enough time to notice that there was at no point a change in colour or noise, even as little as there was. The object was still in front of me, suspended in the air.\r\n\r\nQuickly my understanding signalled this apparition as being caused by an extra-terrestrial and therefore conscious of what was within, I no longer remembered anything. Then I saw again the object which was reversing several metres at a very slow pace. Suddenly, it raised itself to an altitude of 300 metres, changed colour several times, and then left on a right angle, going horizontally before disappearing in a flash a few hundred metres away in the direction of the town. When the object was no longer visible, there were in the sky four white trails. Everything happened in such a fairytale way that I closed my eyes, pinched myself and slapped myself to know whether this was all real. Therefore, only if when having opened my eyes I would have seen the trails left by the engine would I have believed in my observation. I lowered my head, controlled the rhythm of my breathing and clenched my fists. Then I opened my eyelids, my head straight and tilted towards the sky, I saw the impossible. The four parallel and horizontal lines were really there. My mental state following this wonderful encounter is impossible to describe, it was like being drunk. For the first time in my life, I had seen while in possession of all of my faculties, a time and space belonging to the intersection of two worlds. Mine and one from elsewhere. I had seen indisputable proof of its reality. Who knows what happened during this encounter, I don't remember. Was there a being inside? All that I could know was that the ball could reappeared whenever it thought it to be a good time because while going down the path, I saw it cross over once again high in the sky.\r\n\r\n[img]http://www.antollma.info/astral/Colombier.jpg[/img] \r\n\r\nGreetings,\r\nAntoll MA\r\nALL THE PICTURES ON: [url=http://www.antollma.info]www.antollma.info[/url]", "Solution_1": ":rotfl: \r\n\r\n@all moderators, admins: please don't delete it, it's cool :D :D :D\r\n\r\nWe need some fun between math ;) [and the G&FF is not funny]", "Solution_2": "I agree.Cool and funny!BTW, Lots of numbers and math terms in the story :lol:\r\nI have also a very funny one:\r\nGoogle up for 'Polly Nomial\".It's a first thing found by a search engine. :)", "Solution_3": "Even though he(or she) made an account just for this story, it is really funny! :rotfl: \r\n\r\nCan you send me some pictures of the inside of the spaceship if you can?" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "LetABC with $ A \\leq B \\leq 60^0 \\leq C \\leq90^0$\r\nFIND MIN:P=cosA+cosB+cosC", "Solution_1": "hello, the searched minimum ist $ P_{Min}\\equal{}1$ for $ A\\equal{}B\\equal{}\\frac{\\pi}{3}$ and $ C\\equal{}\\frac{\\pi}{2}$.\r\nSonnhard." } { "Tag": [ "algebra", "polynomial", "function" ], "Problem": "Let from $ f : R\\to R$ such that $ f(x)\\equal{}x^{3}\\minus{}3x^{2}\\plus{}3x$ .Prove the existance of $ f^{\\minus{}1}$ and find $ f^{\\minus{}1}(x)$", "Solution_1": "[url=http://imageshack.us][img]http://img456.imageshack.us/img456/3226/funcxy0.gif[/img][/url]", "Solution_2": "Well nice solution [b]gokhankececi[/b], a long way for proving the existence of $ f^{\\minus{}1}$ we must prove that for all $ x_{1},x_{2}\\in\\mathbb{R}$,\r\n\r\n$ f\\left(x_{1}\\right)\\equal{}f\\left(x_{2}\\right)\\Longrightarrow x_{1}\\equal{}x_{2}$\r\n\r\n$ x^{3}_{1}\\plus{}3x^{2}_{1}\\plus{}3x_{1}\\equal{}x^{3}_{2}\\plus{}3x^{2}_{2}\\plus{}3x_{2}\\Longrightarrow$\r\n\r\n$ \\left(x_{1}\\minus{}x_{2}\\right)\\left(x^{2}_{1}\\plus{}x_{1}x_{2}\\plus{}x^{2}_{2}\\right)\\plus{}3\\left(x_{1}\\minus{}x_{2}\\right)\\left(x_{1}\\plus{}x_{2}\\right)\\plus{}3\\left(x_{1}\\minus{}x_{2}\\right)\\equal{}0$\r\n\r\n$ \\Longrightarrow\\left(x_{1}\\minus{}x_{2}\\right)\\left(x^{2}_{1}\\plus{}x_{1}x_{2}\\plus{}x^{2}_{2}\\plus{}3x_{1}\\plus{}3x_{2}\\plus{}3\\right)\\equal{}0$\r\n\r\nNow we have $ x_{1}\\minus{}x_{2}\\equal{}0$ which gives $ x_{1}\\equal{}x_{2}$ or we must have $ x^{2}_{1}\\plus{}x_{1}x_{2}\\plus{}x^{2}_{2}\\plus{}3x_{1}\\plus{}3x_{2}\\plus{}3\\equal{}0$\r\n\r\nwe assume this second equation as a degree$ 2$ polynomial with $ x_{2}$ its variable and $ x_{1}$ a constant number.\r\n\r\nNow $ \\Delta\\equal{}\\minus{}3\\left(x_{1}\\plus{}1\\right)\\leq0$ here for $ \\Delta<0$ we have no solution and we have $ \\Delta\\equal{}0$ only if $ x_{1}\\equal{}\\minus{}1$ then we have $ x^{2}_{1}\\plus{}x_{1}x_{2}\\plus{}x^{2}_{2}\\plus{}3x_{1}\\plus{}3x_{2}\\plus{}3\\equal{}x^{2}_{2}\\plus{}2x_{2}\\plus{}1\\equal{}0$ then $ x_{2}\\equal{}\\minus{}1$\r\n\r\nwhich is the same reasult as $ x_{1}\\equal{}x_{2}$ and Done! :) \r\n\r\nthis was the general way for roving the existence of $ f^{\\minus{}1}$, actually it is not a nice solution for this function! solution given by [b]gokhankececi[/b] is really nice here! :)" } { "Tag": [ "geometry", "rectangle", "AMC" ], "Problem": "In rectangle $ABCD$, angle $C$ is trisected by $\\overline{CF}$ and $\\overline{CE}$, where $E$ is on $\\overline{AB}$, $F$ is on $\\overline{AD}$, $BE = 6,$ and $AF = 2$. Which of the following is closest to the area of the rectangle $ABCD$?\n[asy]\nsize(140);\npair A, B, C, D, E, F, X, Y;\nreal length = 12.5;\nreal width = 10; \nA = origin;\nB = (length, 0);\nC = (length, width);\nD = (0, width);\nX = rotate(330, C)*B;\nE = extension(C, X, A, B);\nY = rotate(30, C)*D;\nF = extension(C, Y, A, D);\ndraw(E--C--F);\nlabel(\"$2$\", A--F, dir(180));\nlabel(\"$6$\", E--B, dir(270));\n\ndraw(A--B--C--D--cycle);\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(F);\nlabel(\"$A$\", A, dir(225));\nlabel(\"$B$\", B, dir(315));\nlabel(\"$C$\", C, dir(45));\nlabel(\"$D$\", D, dir(135));\nlabel(\"$E$\", E, dir(270));\nlabel(\"$F$\", F, dir(180));\n[/asy]\n$\\textbf{(A)} \\ 110 \\qquad \\textbf{(B)} \\ 120 \\qquad \\textbf{(C)} \\ 130 \\qquad \\textbf{(D)} \\ 140 \\qquad \\textbf{(E)} \\ 150$", "Solution_1": "[hide]\n\nTrisecting $\\angle BCE$ gives two $30-60-90$ triangles and one quadrilateral. Looking at $\\triangle ABC$, we see that $BC$ is $6\\sqrt3$ and thus, $FD$ is $6\\sqrt3 -2$. \n\nBy $\\triangle FDC$, $DC$ is $\\sqrt 3(6\\sqrt3 - 2)$. \n\nThe dimensions of rectangle $ABCD$ is $6\\sqrt3$ by $\\sqrt 3(6\\sqrt3 - 2)$ and the area is closest to $E) 150$[/hide]", "Solution_2": "Here's the diagram that was included with this problem, not to scale: \n[asy]\nsize(150);\npair A, B, C, D, E, F, X, Y;\nreal length = 12.5;\nreal width = 10; \nA = origin;\nB = (length, 0);\nC = (length, width);\nD = (0, width);\nX = rotate(330, C)*B;\nE = extension(C, X, A, B);\nY = rotate(30, C)*D;\nF = extension(C, Y, A, D);\ndraw(E--C--F);\nlabel(\"$2$\", A--F, dir(180));\nlabel(\"$6$\", E--B, dir(270));\n\ndraw(A--B--C--D--cycle);\ndot(A);dot(B);dot(C);dot(D);dot(E);dot(F);\nlabel(\"$A$\", A, dir(225));\nlabel(\"$B$\", B, dir(315));\nlabel(\"$C$\", C, dir(45));\nlabel(\"$D$\", D, dir(135));\nlabel(\"$E$\", E, dir(270));\nlabel(\"$F$\", F, dir(180));\n[/asy]\nThis is for the purpose of keeping the Contest Collections up-to-date with the diagrams that came with the original contest." } { "Tag": [ "geometry", "angle bisector", "algebra solved", "algebra" ], "Problem": "Prove sin 3 = (1+ \\sqrt 5)* \\sqrt (2+ \\sqrt 3) + (1- \\sqrt 3)* \\sqrt (5+ \\sqrt 5).", "Solution_1": "This is really strange. I wonder how on earth did someone find this value.", "Solution_2": "Well, I think it's just an application of the fact that 3 = 18 - 15,\r\nand of course the values of sin(15), cos(15), sin(18) and cos(18) are well-known.", "Solution_3": "Boy, I'm really idiot. You're right, Arne, it's not so strange after all. :blush:", "Solution_4": "Hi Arne,\r\n\r\nI did not know that you know all these values. Otherwise it would not have been so easy... :D", "Solution_5": "We learnt about the values of sin(15) and cos(15) at school.\r\n\r\nThe values of sin(18) and cos(18) were derived in our IMO training last year,\r\nas a simple application of complex numbers.\r\nHowever, there is also a clever geometric approach to find these values.", "Solution_6": "So please enlighten us regarding the geometric approach. :)", "Solution_7": "Let ABC be a triangle such that 2(18) it's easy from here." } { "Tag": [ "geometry", "3D geometry", "sphere", "induction", "geometry solved" ], "Problem": "There are given mn + 1 points such that among any m + 1 of them there are two within distance 1 from each other. Prove that there exists a sphere of radius 1 containing at least n + 1 points.", "Solution_1": "It is an easy one :\r\nSuppose, for a contradiction, that no sphere with radius 1 does contain more than n points.\r\nLet M_1,...,M_(mn+1) be the points, and for each i, let S_i be the sphere with center M_i and radius 1.\r\nThus, S_1 contains non more than n points. It follows that there is a point which does not belongs to S_1. Wlog, say M_2. Then M_1M_2 > 1.\r\nSince S_1 \\cup S_2 does not contains more than 2n of the points, it follows that, if m > 1, there exists a point, say M_3 which is not in S_1 \\cup S_2.\r\nNote that, it follows that M_1M_3 > 1 and M_2M_3 > 1.\r\nAnd so on, we construct by induction a sequence M_1,...M_m,M_(m+1) such that M_iM_j > 1 for each i < j. A contradiction.\r\n\r\nPierre." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "A sequence is defined as follows:\r\n\r\n$ t_1\\equal{}3^8$\r\n$ t_2\\equal{}1000$\r\n\r\n$ t_n \\equal{} \\frac{1\\plus{}t_{n\\minus{}1}}{t_{n\\minus{}2}}$\r\n\r\n\r\nWhat's the sum of the numerator and the denominator of $ t_{2008}$?", "Solution_1": "[quote=\"Gabuntu94\"]A sequence is defined as follows:\n\n$ t_1 \\equal{} 3^8$\n$ t_2 \\equal{} 1000$\n\n$ t_n \\equal{} \\frac {1 \\plus{} t_{n \\minus{} 1}}{t_{n \\minus{} 2}}$\n\n\nWhat's the sum of the numerator and the denominator of $ t_{2008}$?[/quote]\r\n\r\n$ t_3\\equal{}\\frac{1\\plus{}t_2}{t_1}$\r\n\r\n$ t_4\\equal{}\\frac{1\\plus{}t_3}{t_2}$ $ \\equal{}\\frac{1\\plus{}\\frac{1\\plus{}t_2}{t_1}}{t_2}$ $ \\equal{}\\frac{1\\plus{}t_1\\plus{}t_2}{t_1t_2}$\r\n\r\n$ t_5\\equal{}\\frac{1\\plus{}t_4}{t_3}$ $ \\equal{}\\frac{1\\plus{}\\frac{1\\plus{}t_1\\plus{}t_2}{t_1t_2}}{\\frac{1\\plus{}t_2}{t_1}}$ $ \\equal{}\\frac{1\\plus{}t_1\\plus{}t_2\\plus{}t_1t_2}{t_2(1\\plus{}t_2)}$ $ \\equal{}\\frac{1\\plus{}t_1}{t_2}$\r\n\r\n$ t_6\\equal{}\\frac{1\\plus{}t_5}{t_4}$ $ \\equal{}\\frac{1\\plus{}\\frac{1\\plus{}t_1}{t_2}}{\\frac{1\\plus{}t_1\\plus{}t_2}{t_1t_2}}$ $ \\equal{}t_1$\r\n\r\n$ t_7\\equal{}\\frac{1\\plus{}t_6}{t_5}$ $ \\equal{}\\frac{1\\plus{}t_1}{\\frac{1\\plus{}t_1}{t_2}}$ $ \\equal{}t_2$\r\n\r\nAnd so $ t_{n\\plus{}5}\\equal{}t_n$ $ \\forall n$ and so $ t_{2008}\\equal{}t_3\\equal{}\\frac{1\\plus{}t_2}{t_1}$ $ \\equal{}\\frac{1001}{6561}$\r\n\r\nHence the result : $ \\boxed{7562}$", "Solution_2": "Wow! :10: \r\n\r\nThank you very much pco! I probably got something wrong with the calculations :ewpu:" } { "Tag": [], "Problem": "$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = 3 \r\n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}$ = 9 \r\n$x+3 = 9$ \r\n$x=6$ \r\n\r\n-------------------------------------------\r\n\r\n$3= 9-x$\r\n$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\r\n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\r\n$x+3 = 81-18x+x^{2}$\r\n$x^{2}-19x-78=0$\r\n$(x-13)(x-6) = 0$\r\n$x = 13 or 6$\r\n\r\nWhy doesn't 13 work if you sub into original expression ?? :S", "Solution_1": "because $(x-a)(x-b)=0\\Longrightarrow x=a\\ \\ \\text{or}\\ \\ x=b$ :wink:", "Solution_2": "[quote=\"bos1234\"]$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = 3 \n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}$ = 9 \n$x+3 = 9$ \n$x=6$ \n\n-------------------------------------------\n\n$3= 9-x$\n$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\n$x+3 = 81-18x+x^{2}$\n$x^{2}-19x-78=0$\n$(x-13)(x-6) = 0$\n$x = 13 or 6$\n\nWhy doesn't 13 work if you sub into original expression ?? :S[/quote]\r\n\r\nIt's called an [b]extraneous[/b] root. \r\n\r\nTry to solve this equation: $x-\\sqrt{x}= 6$. You should get $9$ and $4$ as solutions, but $4$ doesn't work. This sometimes happens when you square both sides of an equation.", "Solution_3": "[quote=\"bos1234\"]$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = 3 \n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}$ = 9 \n$x+3 = 9$ \n$x=6$ \n\n-------------------------------------------\n\n$3= 9-x$\n$\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\n$x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+\\sqrt{x+....}}}}}$ = $(9-x)^{2}$\n$x+3 = 81-18x+x^{2}$\n$x^{2}-19x-78=0$\n$(x-13)(x-6) = 0$\n$x = 13 or 6$\n\nWhy doesn't 13 work if you sub into original expression ?? :S[/quote]\r\nBecause $\\sqrt{13}>3$\r\nSo obviously it doesn't work.\r\nWhen you raise both sides of an equation to a power, often times extra roots are created.\r\nFor example, $x=-2 \\rightarrow x^{2}=4$, but x=2 is false.", "Solution_4": "formally, extraneous roots exist because $f(x) = x^{2}$ is a non-bijective mapping from $\\mathbb{R}$ to $\\mathbb{R}^{+}$. also, $f(x)x^{2k}$ for $k$ a positive integer also has this property. But if $f(x) = x^{2k+1}$ for $k$ a positive integer, then this is a bijective mapping from $\\mathbb{R}$ to $\\mathbb{R}$." } { "Tag": [ "function", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "The continuous function and twice differentiable function $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ satisfies $2007^{2}\\cdot f(x)+f''(x)=0$. Prove that there exist two such real numbers $k$ and $l$ such that $f(x)=l\\cdot\\sin(2007x)+k\\cdot\\cos(2007x)$.", "Solution_1": "Just a second order differential equation...\r\n :)", "Solution_2": "refer this\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136624[/url]", "Solution_3": "That's a terrible contest problem. It's trivial to anyone with appropriate advanced material (uniqueness theorems for differential equations) and impenetrable to anyone else.", "Solution_4": "Terrible question indeed. ODE's of this exact form can be classified neatly without any DE theory, though. Let $u$ and $v$ be solutions to $y''+\\alpha y=0$. Then there exists a real number number $k$ such that $u'v-uv'=k$. (Compute the derivative of $u'v-uv'$). Now given two solutions $u, v$ such that the Wronskian (let's be honest) above is not zero, it is easily shown by elementary means (use the above) that any solution of the equation is a linear combination of $u$ and $v$.", "Solution_5": "Outrageous problem indeed! :lol: \r\nHowever, I was thinking about some particular functions, maybe : \\[g(x)=\\frac{f^{\\prime}(x)\\cos(2007x)}{2007}+f(x)\\sin(2007x).\\]\r\nThen, taking some derivatives :maybe: stuff like that...\r\nMaybe it's not that bad after all; what do you think guys? :)" } { "Tag": [], "Problem": "An integer n is such that the tens digit of a^2 is odd. What is the units digit of a^2?\r\n\r\n[hide=\" answer\"]\n\n6\n\n[/hide]", "Solution_1": "[quote=\"AndrewTom\"]An integer n is such that the tens digit of a^2 is odd. What is the units digit of a^2?[/quote]\r\n\r\nConsider $ \\ a \\equal{} 10m \\plus{} n,m\\in\\mathbb N_0,\\ n\\in\\{\\ 0,\\ 1,\\ 2,\\ 3,\\ 4,\\ 5,\\ 6,\\ 7,\\ 8,\\ 9,\\}$\r\n\r\nNow $ \\ a^2 \\equal{} 100n^2 \\plus{} 20mn \\plus{} n^2$\r\n\r\nNote tens of $ \\ 20mn$ will be even$ \\ \\ \\implies n^2$ must give odd carryover to make tens of $ \\ a^2$ odd.\r\n\r\n$ \\implies n^2 \\equal{} 16,36$\r\n\r\n$ \\implies$ Units digit of $ \\ a^2 \\equal{} 6$", "Solution_2": "Thanks, makar. :)", "Solution_3": "The units digit can be 0,4, or 6\r\n\r\nThe tens digit can be 1,3,5,7 or 9. Notice that if the units digit is 6, the number is of the form 4k in all cases whereas the same isnt true if the last digit is 0 or 4 (eg. take the last two digits as 34. then the number is of the form 4k+2 which cannot be a perfect square).\r\n\r\nHence the unit's digit is 6" } { "Tag": [ "national olympiad" ], "Problem": "Baltic Way 2005\r\n\r\nProblems", "Solution_1": "The website of it :\r\n\r\nhttp://www.math.su.se/bw2005/problems.shtml\r\n\r\nSolutions" } { "Tag": [ "inequalities", "triangle inequality", "inequalities unsolved" ], "Problem": "how do u prove this\r\nusing the triangle inequality, \r\nthat 1) |A| -|B| <= |A-B|\r\n2) max (a,b) = (|a +b| + |a-b|) /2\r\n3) min (a,b) = (|a +b| - |a-b|) /2\r\n\r\n\r\nasap plz", "Solution_1": "Don't post the same problem in multiple forums. We're not here to do your homework for you.\r\n\r\n1) Is $\\vert a\\vert\\leq \\vert a-b\\vert +\\vert b\\vert$ true?\r\n\r\n2,3) Consider the cases $a>b$, $b>a$ separately.", "Solution_2": "pussy invader?\r\n\r\npardon?", "Solution_3": "help plz", "Solution_4": "This, good pussy invader, is not a site for homework help.", "Solution_5": "yes, I decided to look the other way with respect to the name.", "Solution_6": "[quote=\"blahblahblah\"]yes, I decided to look the other way with respect to the name.[/quote]\r\n\r\nWhy? It's so....brilliant? Classy? Charming? I can't really think of enough nice words to describe it.\r\n\r\nSomeone should really ip ban this guy.\r\n\r\nEdit: The man changed his name." } { "Tag": [ "function", "trigonometry", "topology", "real analysis", "real analysis solved" ], "Problem": "Show that $f:R->R$ is continous if and only if $f(I)$ is an interval for every interval $I$ in $R$ and $f$-1$(x)$ is a closed set$($Every limit of a sequence belongs to the set$)$.", "Solution_1": "It seems to me like $f$ can take open intervals to closed intervals as well :?.", "Solution_2": ":D Really Killer ! Very interesting these new mathematics !! :D", "Solution_3": "Actually, I think this might have something to do with: $f$ is coutinuous iff $f^{-1}(U)$ is open for any open set $U$ of $\\mathbb R$. This is actually used as the definition of a continuous function from $X$ to $Y$ (topological spaces in general).", "Solution_4": "I did not find a mistake till now, so here it goes:\r\n First, I will show that f takes intervals into intervals. It suffices to show that f(a) and f(b) are in the closure of $ f((a,b))$ for all a and b. Suppose we could find a and b such that this does not happen and assume for exampple (all other possibilities are similarly discussed) that $ f((a,b))=(m,M)$ and $ f(b)0$. Take any number $ c>b$ and write $ f((a,c))=f((a,b))|{f(b)}| f((b,c))$, where | means union. Since this union is an open interval, it is clear that $ m-r^$ must be in $ f((b,c))$. Thus making c closer and closer arround b we obtain a sequence $ c_n->b$ such that $m-r$ is in $ f((b,c_n))$. Write $ m-r=f(x_n)$ and deduce that $x_n$ is in the preimage of $ m-r$ for all r. Since this is closed, b is also in the preimage and thus $ f(b)=m-r$ false. Thus we have proved that f takes intervals to intervals and so it has Darboux property.\r\n Assume now that f is not continuous in $x_0$ thus there exists a sequence $ c_n$ convergent to $x_0$ and a $eps>0$ such that $|f(c_n)-f(x_0)|>eps$ for all n. Working with a subsequenceallows us to assume that $ f(c_n)>f(x_0)+eps$ for all n and thus by Darboux property there exists $y_n$ between $x_0$ and $c_n$ such that $ f(x_0)+eps=f(y_n)$. It follows easily that the limit of $y_n$ which is $x_0$ is also in the preimage of $f(x_0)+eps$ and thus $eps=0$, false. Hope it's correct.", "Solution_5": "Here's why I think there's a mistake (I didn't read your solution, Harazi, sorry :)).\r\n\r\nOne of the implications looks like this:\r\n\r\n$f$ is continuous $\\Rightarrow$ $f(I)$ is an open interval for every open interval $I\\subset R$ [b]and[/b] blabla... Define $f(x)=\\sin x$. We have $f((0,2\\pi))=[-1,1]$, so already, one of the sentences in the conjunction is false, meaning that the conjunction is false.\r\n\r\nAm I missing something?", "Solution_6": "Indeed, I did not discuss that implication simply because it is false!" } { "Tag": [ "modular arithmetic", "number theory", "relatively prime" ], "Problem": "m, n are positive integer where (m,n)=1, and {$ r_1, r_2, \\cdots r_\\phi n$},{$ s_1, s_2, \\cdots s_\\phi m$} are reduced residue systems.\r\n\r\nShow that for i, j where $ 1 \\le {i} \\le {\\phi {n}}$ and $ 1\\le {j} \\le {\\phi {m}}$, show that there is only one x in mod mn $ x \\equiv r_i \\pmod{m}$ $ x \\equiv s_j \\pmod{n}$ gcd(x,mn)=1.", "Solution_1": "Yes, we can use the Chinese Remainder Theorem. By the Chinese Remainder Theorem, the system\r\n\r\n$ x \\equiv r_i \\mod{m}$\r\n\r\n$ x \\equiv s_i \\mod{n}$\r\n\r\nhas exactly one solution $ \\pmod{mn}$. Because $ x$ is relatively prime to $ m$ and $ n$, it is relatively prime to $ mn$." } { "Tag": [], "Problem": "hi guys\r\n\r\nany of you taking grade 12 data management?\r\n\r\nhow do you study for it? everyone in my teacher's class did bad in final exam last year, bascially everyone's mark dropped by 5% after final....\r\n\r\n\r\nso how do you guys study for it? our tests are normal but exam is dead heard=(\r\n\r\nany suggestions please?\r\n\r\nthanks a lot!", "Solution_1": "as far as i know, data is one of the easiest gr.12 maths. and to review for the exam, just read your textbook, and you should be fine." } { "Tag": [ "function", "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "Find all the entire functions $f(z)$ such that $(Re f)^3+(Im f)^3=1$. Of course it would be nice to do it non-using Little Picard Theorem", "Solution_1": "It follows just from a judicious application of the Cauchy-Riemann equations, does it not? Let $f(z) = f(x, y) = u(x, y) + iv(x, y)$. Then we have $u^3 + v^3 = 1$ so $u^2 \\frac{\\delta u}{\\delta x} + v^2 \\frac{\\delta v}{\\delta x} = 0 = u^2 \\frac{\\delta u}{\\delta y} + v^2 \\frac{\\delta v}{\\delta y}$. But since $\\frac{\\delta u}{\\delta x} = \\frac{\\delta v}{\\delta y}$ and $\\frac{\\delta u}{\\delta y} = - \\frac{\\delta v}{\\delta x}$, this gives us $u^2 \\frac{\\delta v}{\\delta y} + v^2 \\frac{\\delta v}{\\delta x} = 0$ and $-u^2 \\frac{\\delta v}{\\delta x} + v^2 \\frac{\\delta v}{\\delta y} = 0$. However, multiplying each of these equations by the proper partial of $v$ and adding gives $((\\frac{\\delta v}{\\delta x})^2 + (\\frac{\\delta v}{\\delta y})^2)v^2 = 0$ from which we immediately get that $f$ is constant." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A $ 2n\\times 2n$ chessboard is filled by $ 4n^{2}$ real umbers with sum equal to zero.We know that absulute value of each of these numbers is not grater than $ 1$ .Prove that absulute value of all the numbers from one column or a row doesn't exceed $ n$", "Solution_1": "[quote=\"Nbach\"]A $ 2n\\times 2n$ chessboard is filled by $ 4n^{2}$ real umbers with sum equal to zero.We know that absulute value of each of these numbers is not grater than $ 1$ .Prove that absulute value of all the numbers from one column or a row doesn't exceed $ n$[/quote]\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=41119\n\nSee also the solution of C4 at\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=15622\n\n[EDIT: Thanks to Mathmanman for the following link:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=54037 ]\n\n Darij" } { "Tag": [ "conics", "ellipse" ], "Problem": "Through $B: ( 0, -b )$ draw a chord of $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$ ($a>b>0$). Find the MAXIMUM of the length of the chord.\r\n\r\nEditted.", "Solution_1": "[hide]$(0,-b)$ is on the graph. There is no definate answer, since chords cannot have a 0 length.[/hide]", "Solution_2": "[hide]Erm... not sure if that's possible, because the answer approaches 0 (and 0 can't be the length of a chord, as it's a point)...[/hide]", "Solution_3": "perhaps the chord goes through a focus", "Solution_4": "Or perhaps it should be MAXIMUM length...", "Solution_5": "is it $\\sqrt{a^2+b^2}$?", "Solution_6": "[quote=\"Farenhajt\"]Or perhaps it should be MAXIMUM length...[/quote]\r\nI can't believe that I made this mistake......Yes, MAXIMUM.", "Solution_7": "lol good stuff ;)", "Solution_8": "[hide]\nWell...\n\nWe want to minimize $x^2+(y-b)^2$ given that $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$.\n\nSolve for $x^2$ in the second equation and substitute into the first.\n\nWe now have $a^2\\left(1-\\frac{y^2}{b^2}\\right)+(y-b)^2 = \\left(1-\\frac{a^2}{b^2}\\right)y^2-2by+(a^2+b^2)$.\n\nComplete the square to get $\\left(1-\\frac{a^2}{b^2}\\right)\\left(y-\\frac{b}{1-\\frac{a^2}{b^2}}\\right)^2+\\frac{a^4}{a^2-b^2}$.\n\nBut the square term is negative (since $a > b$), so the maximum chord length occurs at $\\frac{a^2}{\\sqrt{a^2-b^2}}$.[/hide]", "Solution_9": "[quote=\"paladin8\"][hide]\nWell...\n\nWe want to minimize $x^2+(y-b)^2$ given that $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$.\n\nSolve for $x^2$ in the second equation and substitute into the first.\n\nWe now have $a^2\\left(1-\\frac{y^2}{b^2}\\right)+(y-b)^2 = \\left(1-\\frac{a^2}{b^2}\\right)y^2-2by+(a^2+b^2)$.\n\nComplete the square to get $\\left(1-\\frac{a^2}{b^2}\\right)\\left(y-\\frac{b}{1-\\frac{a^2}{b^2}}\\right)^2+\\frac{a^4}{a^2-b^2}$.\n\nBut the square term is negative (since $a > b$), so the maximum chord length occurs at $\\frac{a^2}{\\sqrt{a^2-b^2}}$.[/hide][/quote]\r\n\r\nThat seems fine, but when you substitute the obtained value for $y$ in the ellipse equation, you get $x^2=\\frac{a^4(a^2-2b^2)}{(a^2-b^2)^2}$, which obviously must be discussed for sign, which means that the $y$ value might be outside of the ellipse.", "Solution_10": "[quote=\"Farenhajt\"][quote=\"paladin8\"][hide]\nWell...\n\nWe want to minimize $x^2+(y-b)^2$ given that $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$.\n\nSolve for $x^2$ in the second equation and substitute into the first.\n\nWe now have $a^2\\left(1-\\frac{y^2}{b^2}\\right)+(y-b)^2 = \\left(1-\\frac{a^2}{b^2}\\right)y^2-2by+(a^2+b^2)$.\n\nComplete the square to get $\\left(1-\\frac{a^2}{b^2}\\right)\\left(y-\\frac{b}{1-\\frac{a^2}{b^2}}\\right)^2+\\frac{a^4}{a^2-b^2}$.\n\nBut the square term is negative (since $a > b$), so the maximum chord length occurs at $\\frac{a^2}{\\sqrt{a^2-b^2}}$.[/hide][/quote]\n\nThat seems fine, but when you substitute the obtained value for $y$ in the ellipse equation, you get $x^2=\\frac{a^4(a^2-2b^2)}{(a^2-b^2)^2}$, which obviously must be discussed for sign, which means that the $y$ value might be outside of the ellipse.[/quote]\r\n\r\nHmm, you're right. If the condition were changed to, say, $a > b\\sqrt{2}$, we'd be ok :D.", "Solution_11": "Well, your expression is $d^2=\\frac{a^4}{a^2-b^2}-\\frac{a^2-b^2}{b^2}\\left(y+\\frac{b^3}{a^2-b^2}\\right)^2$, and obviously we must minimize $\\left|y+\\frac{b^3}{a^2-b^2}\\right|$ when $y$ runs over $[-b,b].$ It's easily seen that the minimal value need not be zero, and if it's not, $y$ must be $-b$, hence it's either your solution if $a\\geq b\\sqrt{2}$, or $2b$ if $b\\leq a\\leq b\\sqrt{2}.$\r\n\r\n(Another note: Your final expression isn't even defined for a circular ellipse, in which case we know the solution must be the diameter :))" } { "Tag": [ "geometry", "inequalities", "email", "geometry solved" ], "Problem": "line is a circle or not?", "Solution_1": "Yes, it is a circle with $r\\rightarrow \\infty$", "Solution_2": "Let $d$ be a fixed line and two fixed points $A\\in d$, $B\\not\\in d$, $AB\\perp d$. For any mobile point $M$ which belongs to the ray $[AB$, we consider a circle $w(x)=C(M,x)$, where $x=AM$ (the circle $w(x)$ is always tangent to the line $d$ in the point $A$ and its center $M$ belongs always to the same semiplane $(d,B)$). If $x\\rightarrow \\infty$, then the circle $w(x)\\rightarrow$ the line $d$ (the circle with the infinite radius). If $x\\searrow 0$, then the circle $w(x)\\rightarrow$ the point $A$ (the circle with the null radius).", "Solution_3": "yes,but i think point is not a circle actualy and in fact they are different.", "Solution_4": "I don't say you see a line like a circle, I say you feel it !", "Solution_5": "[quote=\"Virgil Nicula\"]I don't say you see a line like a circle, I say you feel it ![/quote]\r\nvery nice feeling.", "Solution_6": "Why this kind of question is posted here? I see that the Geometry Open Questions is now full of that. \r\n\r\nIn my opinion, it belongs to the Other Problem Solving Section.", "Solution_7": "[quote=\"shobber\"]Why this kind of question is posted here? I see that the Geometry Open Questions is now full of that. \n\nIn my opinion, it belongs to the Other Problem Solving Section.[/quote]\r\n\r\nIt belongs into Geometry Theorems & Formulas, in my opinion. Anyway, I gave a rather detailed answer on this question in http://www.mathlinks.ro/Forum/viewtopic.php?t=65375 .\r\n\r\n Darij", "Solution_8": "jensen - please correct your email immediately. You are choosing to be notified when topics are replied to, but your email is bad, so we are getting swamped with bounced emails. I have PM'ed you 2 weeks ago about this - please correct it soon or we'll have to limit your access to the forum." } { "Tag": [ "number theory open", "number theory" ], "Problem": "Find all x such that $ 2^x$ starts with x.\r\n\r\n($ 2^{10}$ is of course a solution\r\nAre there any other solutions??)", "Solution_1": "See http://oeis.org/A100129" } { "Tag": [ "LaTeX" ], "Problem": "I know that this has been asked before, but I can't find the answer. Page numbers are by solution, not round, right?", "Solution_1": "That's correct - for example, if you use LaTeX, the template file always numbers by solution, i.e. \"page one of problem 2\", etc." } { "Tag": [ "probability", "MATHCOUNTS", "symmetry" ], "Problem": "A bill, after being proposed by a member of Congress, must be passed (meaning it gets majority vote) in both the H.o.R. (House of Representatives) and the Senate before being sent to the President, who can either sign or veto the bill (a.k.a. approve/disapprove the bill). Let's assume that for a certain bill, the Senate has already voted and passed it 51-49. There are 435 members in the House of Representatives and only one president, and each member/president has an equal chance of voting for/against the bill. Find the probability that the President signs (approves) the bill, and it is also passed in the House of Representatives.\r\n\r\nNote: There are no overrides, because the Senate only passed 51-49\r\nNote 2: Don't come up with little details that could affect the problem (such as Lobbyists, or whatever.)\r\n\r\nIf it is unclear, just tell me. I originally made this problem for a Mock MathCounts Test, but I decided that I'm not actually gonna use the test. :ninja:\r\n\r\n[hide]Use symmetry[/hide]", "Solution_1": "To be passed, it needs over 1/2 of the vote right?", "Solution_2": "Pretty much :P", "Solution_3": "There's a $ \\frac{1}{2}$ chance that it passes through the House of Representatives (because the probability that it passes and doesn't pass are equal, and it cannot tie). Then there is a $ \\frac{1}{2}$ chance that the president approves it, so the total probability is $ \\frac{1}{2} \\times \\frac{1}{2}\\equal{}\\boxed{\\frac{1}{4}}$.", "Solution_4": "Dang. You got it. The original purpose of the problem was to intimidate people into skipping the problem." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Prove that if a,b,c\u22650,then $ (a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc)^2$\u2265$ 2(ab\\plus{}bc\\plus{}ca)(ab(a^2\\plus{}b^2)\\plus{}bc(b^2\\plus{}c^2)\\plus{}ca(c^2\\plus{}a^2))$", "Solution_1": "[quote=\"SUPERMAN2\"]Prove that if a,b,c\u22650,then $ (a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc)^2$\u2265$ 2(ab\\plus{}bc\\plus{}ca)(ab(a^2\\plus{}b^2)\\plus{}bc(b^2\\plus{}c^2)\\plus{}ca(c^2\\plus{}a^2))$[/quote]\nLet $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.\nHence, $(a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc)^2\\geq2(ab+ac+bc)\\sum_{cyc}(a^3b+a^3c)\\Leftrightarrow$\n$\\Leftrightarrow(27u^3-27uv^2+6w^3)^2\\geq6v^2(27u^2v^2-18v^4-3uv^2)\\Leftrightarrow f(w^3)\\geq0$,\nwhere $f$ is an increasing function.\nHence, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it's enough to check two cases:\n1. $b=1$, $c=0$, which gives $a^6-2a^4+2a^3-2a^2+1\\geq0$, which is obvious.\n2. $b=c=1$, which gives $a^2(a^2-1)^2\\geq0$. Done!" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "[b]Theorem:[/b] Let $ f_n$ be cauchy in measure the there is a sub-sequence, $ f_{n_k}$ and a function f such that $ f_{n_k} \\rightarrow f$ almost everywhere.\r\n\r\n[b]Proof:[/b] Because $ f_n$ is cauchy in measure we can define a subsequnce $ f_{n_k} \\equal{} g_k$ such that $ m\\{ x : |g_{k \\plus{} 1} \\minus{} g_k| \\geq 2^{ \\minus{} k} \\} < 2^{ \\minus{} k}$. (So, if we take $ g_3$ the set of points where $ g_3$ fails to be within 1/8 of the rest of the functions after it has a measure smaller than 1/8.) Next, define some sets $ E_k \\equal{} \\{ x : |g_{k \\plus{} 1} \\minus{} g_k| \\geq 2^{ \\minus{} k} \\}$. Then let $ F_k \\equal{} \\bigcup_{j \\geq k} E_j$. ($ F_k$ is a set of values of x where $ g_k$ fails to be close enough to the function after $ g_k$ in the sequence.) $ m(F_k) \\leq \\sum_{j \\geq k}^{\\infty} m(E_j) < 2^{ \\minus{} k \\plus{} 1}$. (The measure of each of these $ F_k$ sets is smaller than the sum of the geometric series. It depends on k and we can make it as small as we want by making k larger.)\r\n\r\nNext pick i, j with $ i \\geq j \\geq k$. Now [b]if x is not in [/b]$ F_k$ we have:\r\n\r\n$ |g_i(x) \\minus{} g_j(x)| < 2^{ \\minus{} j \\plus{} 1}$ \r\n\r\n(It is important that we are looking at g(x) since we are working on proving point-wise almost everywhere convergence.) \r\n\r\nNext observe $ m(F) \\equal{} m(\\bigcap_{k \\equal{} 1}^{\\infty} F_k) \\equal{} 0$ and when x is not in F $ g_k(x)$ is a Cauchy sequence. Define f = lim $ g_i(x)$ when x is in F and 0 otherwise. Now f is the function that the sub sequences converses to a.e.\r\n\r\nOK. When I started typing this I had a questions about the proof (I've been looking at it for hours... ) but I think I almost get it now. We are proving point-wise a.e. so at some point in this proof we have fixed x... OK. [b]Why is it important that we have fixed x? [/b]\r\n\r\n*sigh* at any rate...\r\n\r\n------------------\r\nHere is the open question:\r\n\r\nGiven $ f_n \\rightarrow f$ and $ f_n \\rightarrow g$ in measure. Prove f=g a.e.", "Solution_1": "It is not that $ x$ is fixed that it is important but that $ x\\notin F_k$. You want to prove that $ g_k$ converges a.e. To this end, it suffices to explicitly produce a set $ F$ of measure $ 0$ such that $ g_k(x)$ converge for every $ x\\notin F$. The idea is to define $ F\\equal{}\\bigcap_{k\\ge 1} F_k$. In order to show that this set works, one needs to take an arbitrary point outside this set and to check convergence there. The first step is to note that if $ x\\notin F$, then $ x\\notin F_k$ for some $ k$. Then we show that for such $ x$ the sequence $ g_j(x)$ is Cauchy, etc.\r\n\r\nNote that we've proved much more than we asked for, namely, we proved that $ g_k\\to g$ uniformly outside each $ F_k$.", "Solution_2": "[quote=\"thefuturebird\"][\nHere is the open question:\n\nGiven $ f_n \\rightarrow f$ and $ f_n \\rightarrow g$ in measure. Prove f=g a.e.[/quote]\r\n\r\nIs it an open question really? :huh: \r\nsee, e.g. [url=http://books.google.pl/books?id=bfgSwxff_NwC&pg=PA121&lpg=PA121&dq=convergence+in+measure+uniqueness&source=bl&ots=5zi6JCgOHr&sig=SbI75pYRi1xha8J5benT123rYD4&hl=pl&ei=5MA_S5K1DoXp-Qb-jvyuCg&sa=X&oi=book_result&ct=result&resnum=8&ved=0CD8Q6AEwBw#v=onepage&q=convergence%20in%20measure%20uniqueness&f=false]th.1 on p. 121[/url]", "Solution_3": "\"open for me\" :oops:", "Solution_4": "The uniqueness of the limit is always proved in the same way. The first two words are \"Suppose not\". The next step is to give the negation of equality $ f\\ne g$ a quantitative form. In the case of equality a.e., the first substep here is to say that the set ${ E=\\{x: f(x)\\ne g(x)}$ has measure $ \\mu>0$. Unfortunately, this is not fully quantitative yet because the condition $ f(x)\\ne g(x)$ is purely qualitative. We would like to replace it with $ |f(x)-g(x)|>\\delta$ with some $ \\delta>0$ independent on $ x$. We cannot do it on the entire set $ E$ but if we [i]define[/i] $ E_\\delta=\\{x: |f(x)-g(x)|>\\delta\\}$, then we'll get a monotone sequence of sets (the smaller $ \\delta$ is, the larger $ E_\\delta$ is) whose union is $ E$. Thus, $ \\lim_{\\delta\\to 0+}|E_\\delta|=|E|=\\mu$, so $ |E_\\delta|>\\frac\\mu 2$ for some $ \\delta$. Now we are almost done. Just write\r\n\\[ \\frac \\mu 2<|E_\\delta|=|\\{|f-g|>\\delta\\}|\\le |\\{|f-f_n|>\\tfrac \\delta 2\\}|+|\\{|g-f_n|>\\tfrac\\delta 2\\}|\\]\r\nand let $ n\\to\\infty$ to get a contradiction.", "Solution_5": "Thanks so much!\r\n\r\n\"the smaller $ \\delta$ is, the larger $ E_\\delta$ is\"\r\n\r\n...but in fact it will turn out to be 0 we are supposing otherwise for the proof." } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "Let $I=(1,2,..n)$and $(A_i)_{1\\le i \\le {2^n-1}}$ the nonempty subsets of $I$ ordered randomly.Let $B\\in M_{2^n-1}(R)$ the matrix defined by $b_{ij}=1$ when $A_i \\cap A_j \\neq \\emptyset $ and $b_{ij}=0$ otherwise.Show that $det(B)$ does not depend on the chosen order and find it.", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31930]Here it is[/url].", "Solution_2": "Strange.I never thought it comes from there." } { "Tag": [ "FTW", "rate problems", "Pythagorean Theorem", "geometry" ], "Problem": "Kelly drove north for 12 miles and then east for 9 miles at an average rate of 42 miles per hour to arrive at the town of Prime. Brenda left from the same location, at the same time, and drove along a straight road to Prime at an average rate of 45 miles per hour. How many minutes earlier than Kelly did Brenda arrive?", "Solution_1": "[quote=\"GameBot\"]Kelly drove north for 12 miles and then east for 9 miles at an average rate of 42 miles per hour to arrive at the town of Prime. Brenda left from the same location, at the same time, and drove along a straight road to Prime at an average rate of 45 miles per hour. How many minutes earlier than Kelly did Brenda arrive?[/quote]\r\n\r\nKelly drove a total of 21 miles, so it took her 30 minutes to get to Prime. \r\n\r\nIf Brenda left from the same location but drove along a straight road, that means she drove the hypotenuse.\r\n\r\n [geogebra]48d1e123a3a3672aae3133e1e2de0dee6a4176b4[/geogebra] \r\n\r\n(The legs have length 12 and 9.)\r\n\r\nBy the Pythagorean Theorem, Brenda drove $ \\sqrt{12^2 \\plus{} 9^2} \\equal{} 15$ miles, so it took her $ 33\\frac{1}{3}$ minutes. \r\n\r\nThus Kelly arrived $ \\boxed{3\\frac{1}{3}}$ minutes earlier.", "Solution_2": "FTW says that the answer is 10, and the problem asks how much earlier Brenda arrived, not Kelly.", "Solution_3": "Solution: [hide]It took Brenda $\\frac{15}{45}$ of an hour, or $ 20 $ minutes. Since it took Kelly $ 30 $ minutes, $ 30 - 20 = 10 $.[/hide]" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Hey, here's a differential equation that I didn't know how to solve. It 'looks' easy, I bet some of you have seen it before:\r\n$\\frac{d^2y}{dx^2}=A+\\frac{B}{y}-Cy$\r\nA,B,C are positive but I'd be interested in if they were negative, too.", "Solution_1": "It's not simple.\r\n\r\nThis belongs to the following family of equations: $y''=f(y).$ A physical interpretation: an object moves under the influence of a force which depends only on position. (In one dimension, that has to be a conservative force, at least for smooth $f.$) We can solve such an equation using a physical insight based on the conservation of energy.\r\n\r\nUse the symbol $v$ for $y'=\\frac{dy}{dx}.$ Find a primitive for $f.$ That is, find $F(y)$ such that $F'(y)=f(y).$\r\n\r\nThen define $u(x)=\\frac12v^2-F(y).$ Then\r\n\r\n$\\frac{du}{dx}=v\\frac{dv}{dx}-f(y)\\frac{dy}{dx} - v\\left(\\frac{d^2y}{dx^2}-f(y)\\right)=0.$\r\n\r\nThus $u$ is constant and we can write $\\frac12v^2=F(y)+K$ or $\\frac{dy}{dx}=\\pm \\sqrt{2F(y)+K}$ for some constant $K.$ \r\n\r\nThat is an autonomous first order equation, hence separable. But it doesn't have to be pretty.\r\n\r\nIn this particular case, we get:\r\n\r\n$\\frac{dy}{dx}=\\pm\\sqrt{2Ay+2B\\ln y-Cy^2+K}.$\r\n\r\nAnd it's not pretty.", "Solution_2": "That was helpful, thanks." } { "Tag": [ "calculus" ], "Problem": "How do you basically find how many real roots are present in a cubic or fourth degree equation if the coefficients are too large enough for succesive division method and descartes sign rule only gives how many min or max real or positive ex are possible how do you find exactly how many are possible??? :?: :maybe: :ninja: is there any method by calculus ie by taking the nth differential and by doing something or so???? :)", "Solution_1": "For cubic [url=http://en.wikipedia.org/wiki/Cardano%27s_method]this[/url] is the method\r\nFor a Quartic [url=http://en.wikipedia.org/wiki/Quartic_equation]this[/url] is the method\r\nFor Quitic and hence Galois proved that the roots cannot be expressed as combination of cofficients and simple Algaebraic symbols", "Solution_2": "Youll be mad to use those eqns\r\nAnd btw how did ya upsurge Rushil? :maybe: \r\nYou didnt thank me for recommendation :mad:", "Solution_3": "who said i didn't thank u\r\nManasalle nandri ye therivechinde :D \r\nA PM to Valentin \r\n Rushil $ \\Rightarrow$ pardesi :wink:", "Solution_4": "And btw delete the post of Chappli abt me insultin me to the core if u r my friend new mod atleast out of gratitude :D" } { "Tag": [ "algebra", "polynomial", "absolute value", "algebra unsolved" ], "Problem": "1/ Let $P(x)$ be a cubic polynomial with rational coefficients.$q_{1},q_{2},q_{3}$... is a sequence of rationals such that $q_{n}=P(q_{n+1})$ for all positive $n$. Show that for some $k$, we have $q_{n+k}=q_{n}$ for all positive $n$.\r\n\r\n2/$P(x)=(a+x)Q(x)$ is a real polynomial of degree $n$. The largest absolute \r\nvalue of the coefficients of $P(x)$ is $h$ and the largest absolute value of $Q(x)$ is $k$. Prove that $k \\leq nh$", "Solution_1": "is it too hard?" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find n in the following equation:\r\n\r\nn^n^n^...^n = N\r\n(k times)", "Solution_1": "http://mathworld.wolfram.com/PowerTower.html\r\n\r\ni think is impossible to find n in function of N\r\n\r\n\r\nyou can try inversion of series\r\n\r\nbut\r\n\r\nthis is impossible, huauh", "Solution_2": "To be clear as to what it is,\r\n\r\nn^n^n^...^n means ((n^n)^n)^... [to a certain point] from \r\n\r\nthe order of operations, which is (n^(n^2))^... [to a certain \r\n\r\npoint]. But \"(k times)\" is not clear to me. It appears the \r\n\r\nnumber of exponentations is k. If that is true, I have \r\n\r\nx^(x^(k-1)) = N, with me not seeing any explicit solution for x \r\n\r\nexisting in terms of N as Thales418 has already stated \r\n\r\n(thought)." } { "Tag": [ "inequalities", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "1) First One question\r\n\r\nthe canonical basis $ (e_{n})_{n\\in \\mathbb{N}}$ is NOT weakly null in $ \\ell_{1}$.\r\n\r\nBecause if we suppose it is, this would imply (Schur property) that $ (e_{n})_{n\\in \\mathbb{N}}$ converges in norm to $ 0$, which is obviously not true. Am I right?\r\n\r\n\r\n2) Another one.\r\nLet $ p_{1} >\r\np_{2}\r\n> \\cdot \\cdot \\cdot > 1$ be a sequence of real numbers.\r\nSo consider the Banach space $ Z\\equal{} (\\bigoplus_{n\\equal{}1}^{\\infty} \\ell_{p_{n}})_{1},$ If for all $ n\\in \\mathbb{N}$, $ (\\overline{e}_{i,n})_{i\\in \\mathbb{N}}$ denote the canonical basis\r\nfor $ \\ell_{p_{n}}$ AND let $ (e_{i,n})_{i\\in \\mathbb{N}}$ denote its natural copy in $ Z$:\r\n\r\n\\[ e_{i,n} \\equal{} (\\underbrace{0,\\cdots,0}_{n\\minus{}1},\\overline{e}_{i,n},0,\r\n\\cdots) \\in W.\\]\r\n\r\nHow can I prove that $ e_{i,n}\\rightarrow 0$ weakly, when\r\n$ i\\rightarrow \\infty$.??\r\n\r\nIm having problems because the norm on $ Z$ is the 1-norm. Although I know $ p_{n}>1$ for all $ n$.\r\nI hope i made myself clear. Thx", "Solution_1": "Ups I just found the answer in kreyZsig book since $ p_{n} > 1$ the canonical basis converges weakly to $ 0$." } { "Tag": [ "calculus", "integration", "real analysis", "real analysis solved" ], "Problem": "f:[0,infinity)->(0, infinity), increasing, derivable, with a continuous derivative. If a_n= \\int 0 to n dx/(f(x)+f'(x)) <=1 for any n in N prove that there exists a>0 s.t. \\int 0 to n dx/(f(x)) <= a for any n in N.", "Solution_1": "I reckon that this is one of Cristinel Mortici's problems! :D", "Solution_2": "How did you know that? Maybe you have his book? (great book, by the way).\r\nBye! :D", "Solution_3": "1/f(t) \\leq 2/(f(t)+f '(t)) + f '(t)/(f(t)) 2\r\n\r\nintegrate on [0;n]\r\n\r\nBn \\leq 2An + \\int_[0;n] f '(t)/(f(t)) 2 dt\r\n\r\nBn \\leq 2 + 1/f(0) - 1/f(n) \\leq 2+1/f(0)" } { "Tag": [ "calculus", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Compute \\[I = \\int_{-\\frac{\\pi}2}^\\frac{\\pi}2 \\frac1{2003^{x}+1}\\cdot \\frac{\\sin^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}\\, dx \\, .\\]", "Solution_1": "[hide]$I={\\frac{\\pi}4}$[/hide] \r\n\r\n\r\n\r\n\r\n\r\nedit: i made a stupid mistake,i forgot to bring the 2 to the denominator", "Solution_2": "We first note that $\\int_{-a}^{a}f(x)\\, dx = \\int_{0}^{a}\\left( f(x)+f(-x) \\right)\\, dx.$ (I will leave this as an elementary exercise to the interested reader.)\r\n\r\nApplying this result to $I,$ we obtain\r\n\r\n$I = \\int_{0}^{\\frac{\\pi}{2}}\\left( \\frac1{2003^{x}+1}\\cdot \\frac{\\sin^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}+\\frac1{2003^{-x}+1}\\cdot \\frac{\\sin^{2002}(-x)}{\\sin^{2002}(-x)+\\cos^{2002}(-x)}\\right) \\, dx$\r\n\r\n$= \\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sin^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}\\left( \\frac1{2003^{x}+1}+\\frac{2003^{x}}{2003^{x}+1}\\right) \\, dx$\r\n\r\n$= \\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sin^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}\\, dx$\r\n\r\n$= \\int_{0}^{\\frac{\\pi}{2}}\\frac{\\cos^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}\\, dx$ ... [Using the identity $\\int_{0}^{a}f(x)\\, dx = \\int_{0}^{a}f(a-x)\\, dx$]\r\n\r\n$\\Rightarrow 2I = \\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sin^{2002}x+\\cos^{2002}x}{\\sin^{2002}x+\\cos^{2002}x}\\, dx = \\int_{0}^{\\frac{\\pi}{2}}\\, dx$\r\n\r\n$\\Rightarrow \\boxed{I = \\frac{\\pi}{4}}.$", "Solution_3": "[quote=\"FieryHydra\"]We first note that $\\int_{-a}^{a}f(x)\\, dx = \\int_{0}^{a}\\left( f(x)+f(-x) \\right)\\, dx.$ (I will leave this as an elementary exercise to the interested reader.)[/quote]\r\nThanks for the nice trick. I couldn't get started on the problem :blush:" } { "Tag": [ "limit", "logarithms", "algebra proposed", "algebra" ], "Problem": "Find the limitation :\\[ \\lim_{n \\rightarrow \\infty}2\\times \\sqrt[3]{4}\\times\\sqrt[9]{16}\\times...\\times\\sqrt[3^n]{2^{2^n}}. \\]\r\n\r\nHappy new year :lol: :D", "Solution_1": "Take logarithm and we want to find:\r\n$\\sum_{n=0}^\\infty ln(\\sqrt[3^n]{2^{2^n}}) = \\sum_{n=0}^\\infty \\frac{2^n}{3^n}ln(2) = 3 ln(2)$\r\nSo the desired value is $2^3=8$.", "Solution_2": "In fact there is no need to take logarithm. ;) \r\n$\\lim_{n \\rightarrow \\infty}2\\times \\sqrt[3]{4}\\times\\sqrt[9]{16}\\times...\\times\\sqrt[3^n]{2^{2^n}}$\r\n$=\\lim_{n \\rightarrow \\infty}2^1\\times 2^{\\frac23}\\times 2^{\\frac49}\\times\\cdots\\times 2^{(\\frac{2}{3})^n}$\r\n$=\\lim_{n \\rightarrow \\infty}2^{1+\\frac23+(\\frac23)^2+\\cdots +(\\frac23)^n}$\r\n$=2^{\\lim_{n \\rightarrow \\infty}1+\\frac23+(\\frac23)^2+\\cdots +(\\frac23)^n}$\r\n$=2^3$\r\n$=8$ :D" } { "Tag": [ "probability" ], "Problem": "I thought I knew what independent and disjoint (mutually exclusive) events were until I ran across this question on an old AP Stat exam:\n\n\n\nWhich of the following statements is true for 2 events, each with probability greater than 0?\n\n\n\n(A) If the events are mutually exclusive, they must be independent.\n\n(B) If the events are independent, they must be mutually exclusive.\n\n(C) If the events are not mutually exclusive, they must be independent.\n\n(D) If the events are not independent, they must be mutually exclusive.\n\n(E) If the events are mutually exclusive, they cannot be independent.\n\n\n\nThe correct answer is [hide](E)[/hide], but I don't really get why.", "Solution_1": "Try to prove the contrapositive of [hide](E)[/hide].", "Solution_2": "[hide]Can events A and B both be true at the same time?[/hide]", "Solution_3": "Here's the full solution, if you don't want to figure it out yourself: [hide]\n\nA: Mutually exclusive means P(A&B) = 0. Independent means P(A&B) = P(A)P(B) > 0. X\n\nB: Independent means P(A&B) = P(A)P(B) > 0. Mutually exclusive means P(A&B) = 0. X\n\nC: Make event A that Bush \"wins\" the next election, and event B that the country falls to ruin within the next four years. They are clearly not mutually exclusive, but they are clearly not independent.\n\nD: Make event A that Bush \"wins\" the next election, and event B that the country falls to ruin within the next four years. They are clearly not independent, but they are clearly not mutually exclusive.\n\nE: Mutually exclusive means P(A&B) = 0. Independent means P(A&B) = P(A)P(B) > 0. P(A&B) cannot be both 0 and >0, so this is true.\n\n[/hide]", "Solution_4": "Or if you want another way of thinking about it:\n\n[hide]\n\nEvents are independent if one happening doesn't affect whether the other one happens. Being mutually exclusive means one happening means the other one cannot happen. So they can't be independent.\n\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "This problem is pretty easy, but it scared me at the beginning :D:\r\n\r\nProve that the following sentences are equivalent:\r\n\r\n(1): There exist n odd numbers x_1, x_2,.., x_n s.t. their sum is equal to their product.\r\n\r\n(2): n has form 4k+1.", "Solution_1": "If n=4k+1 then 1+1+1+...+1+3+(2k+1)=3(2k+1).\r\n\r\nSuppose x1+x2+...+xn = x1x2...xn - it is an odd number ==> n is an odd number. Consider this equality (mod 4) ==> 3s+(n-s) = 3s where s is the number of xi = 3 (mod 4). So 2s+n = 3s (mod 4). If n = 3 (mod 4) then 2s+3 = 3s (mod 4) - it is wrong equality whenever s is odd or even.", "Solution_2": "Yup, you got it, Myth!" } { "Tag": [ "floor function", "algebra proposed", "algebra" ], "Problem": "Prove that: $ \\sum_{i\\equal{}1}^{n^2} \\lfloor \\frac{i}{3} \\rfloor\\equal{} \\frac{n^2(n^2\\minus{}1)}{6}$\r\nFor all $ n \\in N$.", "Solution_1": "Let \r\n\\[ f(m)\\equal{}\\sum_{i\\equal{}1}^{m}[\\frac{i}{3}]\\equal{}\\frac{m(m\\plus{}1)}{2}\\minus{}\\frac 13 ([\\frac{m\\plus{}2}{3}]\\plus{}2[\\frac{m\\plus{}1}{3}])\\equal{}\\frac{m(m\\minus{}1)}{6}\\plus{}\\theta (m),\\] \\[ \\theta (m)\\equal{}\\frac 13 (m\\minus{}[\\frac{m\\plus{}2}{3}]\\minus{}2[\\frac{m\\plus{}1}{3}]).\\]\r\nObviosly $ \\theta (m\\plus{}3)\\equal{}\\theta (m),\\theta (0)\\equal{}0\\equal{}\\theta (1),\\theta (2)\\equal{}\\minus{}1$\r\nBecause $ n^2\\equal{}0,1\\mod 3$ we get $ \\theta (n^2)\\equal{}0.$", "Solution_2": "Clearly we need only prove that \\[ \\sum_{i\\equal{}1}^{n^2}\\left\\{\\frac{i}{3}\\right\\}\\equal{}\\frac{n^2}{3}\\]\r\nThis is equivalent to showing that the number of $ i\\equiv 2\\mod{3},i\\in[n^2]$ is equal to the number of $ i\\equiv 0\\mod{3}$, which is clear.", "Solution_3": "First note that if the claim holds for $n=3k$ then it holds for $n=3k+1,3k+2$. To prove the claim for $n=3k$, double count the number of lattice points in the interior of the triangle defined by $y=0, y=x/3,x=n^2$. Since the triangle has lattice vertices, we can apply Pick's Theorem: $A=0.5*n^2*(n^2/3)=n^4/6=I+B/2-1$. The LHS of the given equation is $I+B-(n^2+1)$ and it is easy to show that $B=(n^2+1)+n^2/3+(n^2-3)/3=5n^2/3$. The rest is simple algebra." } { "Tag": [], "Problem": "A circle with center $O$ passes through point $A$ and $C$ and intersects the sides $AB$ and $BC$ of the triangle $ABC$ at points $K$ and $N$, respectively. The circunscribed circles of the triangles $ABC$ and $KBN$ intersect at two distinct points $B$ and $M$. Prove that $\\measuredangle OMB=90^\\circ$", "Solution_1": "Let the circumcentre of circle(ABC) be X and the circumcentre of circle(AKN) be Y. Therefore $BY \\perp BC$, $OX \\perp BC$, $BX \\perp KN$, $OY \\perp KN$, thus $BY \\parallel OX$ and $BX \\parallel OY$, thus $BYOX$ is a parallelogram. So XY passes through the midpoint of BO. Since it also passes through the midpoint of BM, so $XY \\parallel OM$. Since $XY \\perp BM$, so $OM \\perp BM$. Thus $\\angle{OMB}=90^{o}$." } { "Tag": [ "calculus", "quadratics", "geometry", "geometric transformation", "FTW" ], "Problem": "How good do you think you are at math?", "Solution_1": "Err. I can crush pretty much everyone in my math class save maybe 2 people (that are AoPSers). \r\n\r\nAnd Albert Einstein was involved a genius in physics not math :P", "Solution_2": "Yeah, there definitely needs to be a bit between owning our maths group and Einstein. As in, I can take everyone in my maths class, but compared to people on here I'm a complete noob.", "Solution_3": "i totally agree. \r\ni was so happy with my amc scores, but comparing to ppl here, they suck", "Solution_4": "Pawning Albert Einstein doesn't mean much seeing that he's not really a mathematician. Still.......", "Solution_5": "[quote=\"BanishedTraitor\"]Pawning Albert Einstein doesn't mean much seeing that he's not really a mathematician. Still.......[/quote]\r\n\r\nHmm... Diraic wasn't [i]really[/i] a mathematician, but I expect he could own any one on here. So could most of the people who worked on QT in the early days.", "Solution_6": "i'd say there's a big gap between 4 and 5..\r\n\r\ni mean.. my class sucks at math... (its geometry.. D:)\r\n\r\nand i can't pwn einsteing... i'd say i'm about 4.3 :D", "Solution_7": "4 was true when I was in school, but no longer, with the IMO bronze medallist in class, and another quite brilliant guy who for some reason never sat for olympiads.\r\n\r\nseeing these, and the fact that I couldn't qualify for IMO, I think it's 1.", "Solution_8": "Well if u read abt Einstein youd knw that he knew advanced calculus also and even though he chose to leave his calculations to his assistants, he was very good in math so he could have pwned everyone here easily except maybe a few teachers in math in this forum :) \r\nIf i was to go ranting like bubka, i have a fellow imotc guy in class but apart frm him i hope n think that i can pwn everyone else in olympiad math :D", "Solution_9": "Homeschooled, but at me math circle, I am one of the top students.\r\n\r\nNotice the key word \"one\". I'm not the top student.", "Solution_10": "[quote=\"1=2\"]Homeschooled, but at me math circle, I am one of the top students.\n\nNotice the key word \"one\". I'm not the top student.[/quote]\r\n\r\nThat's a good thing. Being the best in your class/school/area means that you lose sight of the bigger picture\r\n\r\nI've always sat back, until recently, satisfied with beating my classmates but now that isn't really enough", "Solution_11": "I can out-score all of the kids in my class, with some competition.", "Solution_12": "[quote=\"SimonM\"]Being the best in your class/school/area means that you lose sight of the bigger picture[/quote]\r\nNot really. For people who care more than just about being first, there are plenty of opportunities to network outside of the locality. For example, there aren't many high schoolers around me that i can have a nice math talk with. But that's not too big a problem because there's AoPS. Networks like AoPS remind people that true competition doesn't mean Jack across the hallway or Jill down the street, but Johann across the state or Jienzhang from across the country. It really gives a better perspective.", "Solution_13": "hmm, I think I chose the first option :huh:", "Solution_14": "I pwn everyone including Albert Einstein.", "Solution_15": "My math class is stupid bar 2 kids. AoPS however is beast.", "Solution_16": "It does seem as if the new trend is stupidity. (no offense to anyone xD)", "Solution_17": "Stupidity is the Key to Happiness. :rotfl: See my blog.\r\n\r\nThey don't teach you to do math in school, or even think at all. They train you to become robots. They also don't do a very good job.\r\n\r\nThis is all OK up until about right here in Alg. 1:\r\nExample: For the following quadratic equations, complete the square.\r\nTranslation: Apply the steps you hopefully memorized in class. If you forget one step, you get the question wrong.\r\nAnd here usually the breaking point for # of steps to memorize is reached, and A+s go to D-s and no one likes math anymore (at least this is my experience in school math classes).\r\n\r\nAnd so the people who have bad memory end up on the bottom, the people who have good memory end up in the middle, and the people who know the reasoning behind the methodology end up on top.", "Solution_18": "i'm a 4, because my class is full of idiots...no offense to them, but if they did the ftw, they would seriously have a rating of 1000...\r\n\r\nbut i cant pwn einstein, so i think im a 4.1", "Solution_19": "I hope I revived this???", "Solution_20": "I could PWN Albert Einstein.\n\nTwo-year-old albert einstein, that is.", "Solution_21": "[quote=smartestpanda123]I hope I revived this???[/quote]\n\n-_- gg well played troll lel xd", "Solution_22": "4 :) $ $", "Solution_23": "I'm in the 0 level", "Solution_24": "bump $\\phantom{aaaaaaaaaaaa}$", "Solution_25": "This is a dead thread", "Solution_26": "[quote=crazychinesemaniac]How good do you think you are at math?[/quote]\n\nI voted 5!!!", "Solution_27": "I'm the only one who isn't failing his math class so... ", "Solution_28": "Take a look at my alcumus rating and I've been grinding for 1 year. 14.4. :ewpu: \ncan you assume how bad i am at math??", "Solution_29": "why do the idiots think albert einstein is a mathmetician. He was genius at physics nurdz " } { "Tag": [ "floor function", "number theory unsolved", "number theory" ], "Problem": "Find all the real solutions to the equation\r\n\r\n$ 4x^2 \\minus{} 40 \\lfloor x \\rfloor \\plus{} 51 \\equal{} 0$", "Solution_1": "[quote=\"knol\"]Find all the real solutions to the equation\n\n$ 4x^2 \\minus{} 40 \\lfloor x \\rfloor \\plus{} 51 \\equal{} 0$[/quote]\r\n\r\nLet $ [x]\\equal{}n$. We have $ x^2\\equal{}10n\\minus{}\\frac{51}4$ with the constraint $ n^2\\le 10n\\minus{}\\frac{51}4 <(n\\plus{}1)^2$ ($ n\\ge 2$ and so $ x>0$) and so :\r\n\r\n$ \\iff$ $ n^2\\minus{}10n\\plus{}\\frac{51}4\\le 0$ and $ n^2\\minus{}8n\\plus{}\\frac {55}4>0$\r\n\r\n$ \\iff$ $ (n\\minus{}5)^2\\le \\frac {49}4$ and $ (n\\minus{}4)^2>\\frac 94$\r\n\r\n$ \\iff$ $ |n\\minus{}5|\\le \\frac 72$ and $ |n\\minus{}4|>\\frac 32$\r\n\r\n$ \\iff$ $ n\\in\\{2,6,7,8\\}$\r\n\r\nHence the solutions $ x\\in\\{\\sqrt{20\\minus{}\\frac{51}4},\\sqrt{60\\minus{}\\frac{51}4},$ $ \\sqrt{70\\minus{}\\frac{51}4},\\sqrt{80\\minus{}\\frac{51}4}\\}$\r\n\r\nAnd so $ \\boxed{x\\in\\{\\frac{\\sqrt{29}}2,\\frac{\\sqrt{189}}2,\\frac{\\sqrt{229}}2,\\frac{\\sqrt{269}}2\\}}$" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let $ P(x)$ be a polynomial of degree $ n$ whose roots are $ i\\minus{}1,i\\minus{}2,...,i\\minus{}n$, where $ i^2\\equal{}\\minus{}1$. Let $ R(x),S(x)$ be the polynomials with real coefficients such that $ P(x)\\equal{}R(x)\\plus{}iS(x)$.\r\n\r\nShow that the polynomial $ R$ has $ n$ real roots", "Solution_1": "$ P(x)\\equal{}c\\prod_{j\\equal{}1}^n(x\\minus{}i\\plus{}j)$ with $ c\\ne0$\r\nObviously, we have\r\n$ 2R(x)\\equal{}c\\prod_{j\\equal{}1}^n(x\\minus{}i\\plus{}j)\\plus{}\\bar{c}\\prod_{j\\equal{}1}^n(x\\plus{}i\\plus{}j)$.\r\n\r\nSuppose $ R(a\\plus{}bi)\\equal{}0$ for some $ b\\ne0$, we have\r\n$ c\\prod_{j\\equal{}1}^n(a\\plus{}j\\plus{}(b\\minus{}1)i)\\equal{}\\minus{}\\bar{c}\\prod_{j\\equal{}1}^n(a\\plus{}j\\plus{}(b\\plus{}1)i)$.\r\nBoth sides have same modulus, but \r\n$ (a\\plus{}j)^2\\plus{}(b\\minus{}1)^2<(a\\plus{}j)^2\\plus{}(b\\plus{}1)^2$ when $ b>0$;\r\n$ (a\\plus{}j)^2\\plus{}(b\\minus{}1)^2>(a\\plus{}j)^2\\plus{}(b\\plus{}1)^2$ when $ b<0$.\r\nWe also notice the two modulus can not be zero at the same time, because that requires $ b\\equal{}1$ and $ b\\equal{}\\minus{}1$ at the same time.\r\nHence, we get contradiction and $ b$ has to be $ 0$." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "Can you solve this problem?", "Solution_1": "This looks like it involves circumcircles.", "Solution_2": "No I wrote exactly true.\r\nThere is no circumcircles .", "Solution_3": "I don't think the angle is unique.", "Solution_4": "yea.... according to my graphing program, the angle varies, a similar problem was posted in the intermediate geometry marathon that might be helpful to look at problem #58\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70276&start=220[/url]" } { "Tag": [ "function", "logarithms", "analytic geometry", "graphing lines", "slope", "calculus", "calculus computations" ], "Problem": "Find the necessary and sufficient condition of $ p,\\ q$ for which the line $ y\\equal{}px\\plus{}q$ and the function $ y\\equal{}\\ln x$ have no intersection points.", "Solution_1": "The line and the curve (which is concave down) have the same slope when $ x\\equal{}\\frac1p.$ At that point, $ y\\equal{}1\\plus{}q$ on the line and $ y\\equal{}\\ln\\left(\\frac1p\\right)\\equal{}\\minus{}\\ln p$ on the curve. If the line is still above the curve, meaning that $ 1\\plus{}q>\\minus{}\\ln p,$ then they will never intersect.\r\n\r\nThe necessary and sufficient condition for no intersection is $ \\ln p\\plus{}q\\plus{}1>0.$", "Solution_2": "What's the sign of $ p$?", "Solution_3": "Ah, yes. I was tacitly assuming $ p>0.$ If $ p\\le0,$ there will be an intersection. So the condition should be stated as $ \\ln p\\plus{}q\\plus{}1>0$ and $ p>0.$", "Solution_4": "O.K. Let $ f(x)\\equal{}px\\plus{}q\\minus{}\\ln x\\ (x>0)$, can anyone solve the problem by examining the behaviour of $ f(x)$?" } { "Tag": [], "Problem": "I know I may have posted this before, but just in case.\r\n\r\nGiven that there is a regular n-gon inscribed within a circle of radius 1, how do you find the product of all the sides and diagonals of the triangle? What about sum??", "Solution_1": "[hide=\"Giveaway\"] Roots of unity. [/hide]" } { "Tag": [], "Problem": "Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters. There are 12 players on the team. Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup. How many different starting lineups are possible? (The order of the players in a basketball lineup doesn't matter.)", "Solution_1": "We already know Ace and Zeppo will be on the team, so we only need to choose 3 players out of 10. Thus, our answer is $ \\binom{10}{3} \\equal{} 120$.", "Solution_2": "This would be a more realistic question if they said that all of the other people are the same skill level :dry: ", "Solution_3": "[quote=Possible]This would be a more realistic question if they said that all of the other people are the same skill level :dry:[/quote]\n\nno cuz in sports some team members are better than others", "Solution_4": "[quote=V_V][quote=Possible]This would be a more realistic question if they said that all of the other people are the same skill level :dry:[/quote]\n\nno cuz in sports some team members are better than others[/quote]\n\nExactly but the starting lineup could be the worst 3 players and the 2 best players", "Solution_5": "[hide=Solution]Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is$$ \\binom{10}{3} = \\frac{10 \\times 9 \\times 8}{3 \\times 2 \\times 1} = \\boxed{120}. $$[/hide]", "Solution_6": "Why, oh, why, are problems in the textbook not level 1??????????" } { "Tag": [ "calculus" ], "Problem": "this is gonna be dumb for you physics geniuses :blush: \r\n\r\nA particle initially at rest is subject to 2 forces, one is constant, the other is a retarding force proportional to the particle velocity. Draw the graph of its motion.", "Solution_1": "m*dv/dt = a - bv\r\n\r\nln (a - bv) = -b/m * t + C\r\n\r\nbv = a - Ce^(-b/m * t)\r\n\r\nif v = 0 at time t=0, then C = a\r\n\r\nintegrate again to get\r\n\r\nx(t) = a/bt + am/b^2 * e^(-b/m * t) + C\r\n\r\nFrom there, you can graph it on any graphing calc", "Solution_2": "aw come on, calculus :( ! how about trying simple forces stuff , i mean insstead of this heavy math , how about some staraight forward physics ?? :) \r\n\r\nok,if not the graph ,can someone just tell how would it's motion be?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "[u][b]Problem (zaizai-hoang):[/b][/u]\r\nLet a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\r\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$", "Solution_1": "Here's my proof. \r\n\r\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\r\n\r\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \r\n\r\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\r\n\r\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nso the lemma is proved. \r\n\r\n[b][color=blue]Let's start to prove the problem ! [/color][/b]\r\n\r\nIf $abc>1$ then $ab+bc+ca\\geq 3(abc)^{\\frac{2}{3}}> 3$\r\n\r\nso the condition $ab+bc+ca+abc=4$ can't be satisfied. Therefore, $abc \\leq 1$\r\n\r\n$3(a^{2}+b^{2}+c^{2})+abc \\geq 10$\r\n\r\n$\\Leftrightarrow$ $3(a^{2}+b^{2}+c^{2}+2abc+1 ) \\geq 13+5abc$\r\n\r\nUsing lemma, ETS $6(ab+bc+ca) \\geq 13+5abc$\r\n\r\n$\\Leftrightarrow$ $24-6abc \\geq 13+5abc$\r\n\r\n$\\Leftrightarrow$ $1 \\geq abc$\r\n\r\n$Q.E.D.$\r\n\r\n\r\n\r\nI was fool... :blush: \r\n\r\n[b]Second proof.[/b]\r\n\r\n$abc \\leq 1$ as I showed above. \r\n\r\ncuz $a^{2}+b^{2}+c^{2}\\geq ab+bc+ca$\r\n\r\nETS $3(ab+bc+ca)+abc \\geq 10$.\r\n\r\n$\\Leftrightarrow$ $12-2abc \\geq 10$\r\n\r\n$\\Leftrightarrow$ $1 \\geq abc$\r\n\r\n$Q.E.D.$ \r\n\r\n :blush: :blush:", "Solution_2": "A simpler way ;) (it helps sometimes to know nothing complicated)\r\n\r\nChange a 4 on the RHS to $ab+bc+ac+abc$. We get\r\n$2(a^{2}+b^{2}+c^{2}) \\ge 6$\r\n\r\nBut we need $ab+bc+ac+abc = 4$. If $ab+bc+ac < 3$, then we have $3 > ab+ac+bc \\ge 3(abc)^{2/3}$, whence $abc < 1$ and our condition is contradicted. $ab+bc+ac \\ge 3$.\r\n\r\nWe conclude by $2(a^{2}+b^{2}+c^{2}) \\ge 2(ab+ac+bc) \\ge 6$.\r\n\r\nedit: I see [b]Chang Woo-JIn[/b] has a better solution also, faster than mine! :P", "Solution_3": "[quote=\"zaizai-hoang\"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$[/quote]\r\n\r\nFrom $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\\frac{2p}{q+r}, b=\\frac{2q}{r+p}, c=\\frac{2r}{p+q}$. Applying this, the original one becomes \\[\\sum \\frac{6p^{2}}{(q+r)^{2}}+\\frac{4pqr}{(p+q)(q+r)(r+p)}\\geq 5\\] or \\[6\\sum_{cyc}p^{6}+12\\sum_{sym}p^{5}q+\\sum_{sym}p^{4}q^{2}+14\\sum_{cyc}p^{4}qr \\geq 10 \\sum_{cyc}p^{3}q^{3}+14 \\sum_{sym}p^{3}q^{2}r+24p^{2}q^{2}r^{2}\\] which can be obtained by summation of the following: \\[6\\sum_{cyc}p^{6}\\geq 6\\sum_{cyc}p^{3}q^{3}\\] \\[2\\sum_{sym}p^{5}q \\geq 4\\sum_{cyc}p^{3}q^{3}\\] \\[10\\sum_{sym}p^{5}q \\geq 10 \\sum_{sym}p^{3}q^{2}r\\] \\[\\sum_{sym}p^{4}q^{2}\\geq \\sum_{sym}p^{3}q^{2}r\\] \\[6\\sum_{cyc}p^{4}qr \\geq 3\\sum_{sym}p^{3}q^{2}r\\] \\[8\\sum_{cyc}p^{4}qr \\geq 24p^{2}q^{2}r^{2}.\\]", "Solution_4": "[quote=\"Sung-yoon Kim\"][quote=\"zaizai-hoang\"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$[/quote]\n\nFrom $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\\frac{2p}{q+r}, b=\\frac{2q}{r+p}, c=\\frac{2r}{p+q}$. Applying this, the original one becomes \\[\\sum \\frac{6p^{2}}{(q+r)^{2}}+\\frac{4pqr}{(p+q)(q+r)(r+p)}\\geq 5\\] [/quote]\nThen just use: $8pqr\\ge (p+q)(q+r)(r+p)$ and $\\sum\\frac{p^{2}}{(q+r)^{2}}\\ge \\frac{3}{4}$[/quote]", "Solution_5": "[quote]\nThen just use: $8pqr\\ge (p+q)(q+r)(r+p)$ and $\\sum\\frac{p^{2}}{(q+r)^{2}}\\ge \\frac{3}{4}$[/quote]\r\n\r\nI think you have a mistake. $(p+q)(q+r)(r+p)\\ge 8pqr$ by AM-GM", "Solution_6": "[quote=\"Chang Woo-JIn\"]Here's my proof. \n\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\n\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \n\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\n\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nso the lemma is proved. \n\n :blush: :blush:[/quote]\r\nNice solution. I have a solution to prove that your [b]Lemma[/b]\r\nWLOG we assume\r\n $(a-1)(b-1)\\ge 0 \\Leftrightarrow ab+1-a-b\\ge 0 \\\\ \\Leftrightarrow abc+c-ac-bc\\ge 0$ (because $c\\ge 0$)\r\nNow the inequality equavalent:\r\n$(a^{2}-2ab+b^{2})+2(abc+c-ac-bc)+(c^{2}-2c+1)\\ge 0 \\\\ \\Leftrightarrow (a-b)^{2}+2c(a-1)(b-1)+(c-1)^{2}\\ge$\r\nIt true. :P I think this lemma as same as VMO 2006, group B.\r\nThis problem my solution use mixing variable :D", "Solution_7": "[quote=\"zaizai-hoang\"][quote=\"Chang Woo-JIn\"]Here's my proof. \n\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\n\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \n\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\n\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nso the lemma is proved. \n\n :blush: :blush:[/quote]\nNice solution. I have a solution to prove that your [b]Lemma[/b]\nWLOG we assume\n $(a-1)(b-1)\\ge 0 \\Leftrightarrow ab+1-a-b\\ge 0 \\\\ \\Leftrightarrow abc+c-ac-bc\\ge 0$ (because $c\\ge 0$)\nNow the inequality equavalent:\n$(a^{2}-2ab+b^{2})+2(abc+c-ac-bc)+(c^{2}-2c+1)\\ge 0 \\\\ \\Leftrightarrow (a-b)^{2}+2c(a-1)(b-1)+(c-1)^{2}\\ge$\nIt true. :P I think this lemma as same as VMO 2006, group B.\nThis problem my solution use mixing variable :D[/quote]\r\n\r\nThe lemma was posted by Darij:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19666", "Solution_8": "If $ a,b,c$ are [color=red]real[/color] numbers such that $ abc \\plus{} bc \\plus{} ca \\plus{} ab \\equal{} 4,$ then\r\n\r\n$ abc \\plus{} 3\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right) \\geq 10.$\r\n\r\n[b]Proof[/b] From $ abc \\plus{} bc \\plus{} ca \\plus{} ab \\equal{} 4,$ there's [color=blue]real[/color] numbers $ p,q,r$ such that \r\n\r\n$ a \\equal{} \\frac {2p}{q \\plus{} r}, b \\equal{} \\frac {2q}{r \\plus{} p}, c \\equal{} \\frac {2r}{p \\plus{} q}.$ \r\n\r\nApplying this, the original one becomes\r\n\r\n$ \\frac {4pqr}{(q \\plus{} r)(r \\plus{} p)(p \\plus{} q)} \\plus{} \\sum{\\frac {6p^2}{(q \\plus{} r)^2}}\\geq 5,$\r\n\r\nwhich can be obtained by the following :\r\n\r\n$ \\frac {4pqr}{(q \\plus{} r)(r \\plus{} p)(p \\plus{} q)} \\plus{} \\sum \\frac {6p^{2}}{(q \\plus{} r)^{2}} \\minus{} 5$\r\n\r\n$ \\equal{} \\frac {(p \\plus{} q \\plus{} r)^2\\{5(q \\minus{} r)^2(r \\minus{} p)^2(p \\minus{} q)^2 \\plus{} 3[(q \\plus{} r)(q \\minus{} r)^2 \\plus{} (r \\plus{} p)(r \\minus{} p)^2 \\plus{} (p \\plus{} q)(p \\minus{} q)^2]^2\\}}{(q \\plus{} r)^2(r \\plus{} p)^2(p \\plus{} q)^2[(q \\minus{} r)^2 \\plus{} (r \\minus{} p)^2 \\plus{} (p \\minus{} q)^2]}$\r\n\r\n$ \\plus{} \\sum{\\frac {(q \\minus{} r)^2(q \\plus{} r \\minus{} 2p)^2}{(q \\plus{} r)^2[(q \\minus{} r)^2 \\plus{} (r \\minus{} p)^2 \\plus{} (p \\minus{} q)^2]}}\\geq 0.$\r\n\r\nBy the way, if $ a,b,c$ are [color=green]positive[/color] numbers, then\r\n\r\n$ (a\\plus{}b\\plus{}c)\\left(abc \\plus{} 3a^2 \\plus{}3 b^2 \\plus{} 3c^2\\minus{}10\\right)\\geq 7(abc \\plus{} bc \\plus{} ca \\plus{} ab \\minus{}4).$", "Solution_9": "$ 2(a^2\\plus{}b^2\\plus{}c^2) \\ge 10\\minus{}abc\\equal{}6\\plus{}ab\\plus{}bc\\plus{}ca$\r\n$ (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2\\plus{}2(a^2\\plus{}b^2\\plus{}c^2) \\ge 12$\r\nit's enough to prove ineq:\r\n$ a^2\\plus{}b^2\\plus{}c^2 \\ge 6$\r\nwe have $ ab\\plus{}bc\\plus{}ca\\plus{}abc\\equal{}4$ so \r\n$ \\frac{ \\sqrt{ab} }{2}\\equal{}cos \\alpha$ \r\n$ \\frac{ \\sqrt{bc} }{2}\\equal{}cos \\beta$ \r\n$ \\frac{ \\sqrt{ca} }{2}\\equal{}cos \\gamma$ \r\nwith $ \\alpha \\plus{} \\beta \\plus{} \\gamma \\equal{} \\pi$\r\nnow we have inequality:\r\n$ \\sum_{cyc}^{} \\frac{(cos \\alpha)^2 (cos \\beta)^2}{(cos \\gamma^2)} \\ge \\frac{3}{4}$\r\nit is well known that $ cos^2 \\alpha\\plus{}cos^2 \\beta\\plus{}cos^2 \\gamma \\ge \\frac{3}{2}$\r\nso\r\n$ \\sum_{cyc}^{} \\frac{(cos \\alpha)^2 (cos \\beta)^2}{(cos \\gamma^2)} \\ge cos^2 \\alpha\\plus{}cos^2 \\beta\\plus{}cos^2 \\gamma \\ge \\frac{3}{2}$\r\nDone!" } { "Tag": [ "integration", "logarithms", "calculus", "calculus computations" ], "Problem": "The expression $ \\frac{4x}{(x\\minus{}5)(x\\minus{}3)^{2}}$ is denoted by $ f(x)$.\r\n\r\n(i) Express $ f(x)$ in the form $ \\frac{A}{x\\minus{}5} \\plus{} \\frac{B}{x\\minus{}3} \\plus{} \\frac{C}{(x\\minus{}3)^{2}}$, where $ A$, $ B$ and $ C$ are constants.\r\n\r\n(ii) Hence find the exact value of $ \\int_{1}^{2} f(x)\\,dx$.", "Solution_1": "[quote=\"AndrewTom\"]\n(i) Express $ f(x)$ in the form $ \\frac {A}{x \\minus{} 5} \\plus{} \\frac {B}{x \\minus{} 3} \\plus{} \\frac {C}{(x \\minus{} 3)^{2}}$, where $ A$, $ B$ and $ C$ are constants.\n[/quote]\r\n\r\ncomparing the coefficients, you get $ A \\equal{} 5, B \\equal{} \\minus{}5 , C \\equal{} \\minus{}6$\r\n\r\n\r\n$ \\int f(x).dx \\equal{} \\frac {5}{x \\minus{} 5} \\plus{} \\frac {\\minus{}5}{x \\minus{} 3} \\plus{} \\frac {\\minus{}6}{(x \\minus{} 3)^{2}}$\r\n\r\n\r\n$ \\equal{} 5 ln|x \\minus{} 5| \\minus{} 5 ln|x \\minus{} 3| \\plus{} \\frac{6}{(x\\minus{}3)} \\plus{} C$\r\n\r\n\r\n[b](ii)[/b]\r\n\r\n$ \\int_1^2 f(x)dx \\equal{} 5 ln3 \\minus{} 15ln2 \\minus{} 9$", "Solution_2": "Thanks Nora.91.\r\n\r\nWhen I put in the limits, I get $ 5\\ln(\\frac{3}{2}) \\minus{} 3$.", "Solution_3": "Yeah.. you are correct, AndrewTom\r\n\r\n\r\nmy mistake.. sorry :blush:" } { "Tag": [], "Problem": "The product of two whole numbers is 100000. If neither of the numbers are multiples of 10, what is their sum?", "Solution_1": "[hide]100,000 = \n10^5=\n2^5 * 5^5\n\nSince neither 2^5 nor 5^5 are multiples of 10, we add\n32\nand 3125\n\nto get 3157, our answer.[/hide]" } { "Tag": [ "integration", "calculus", "derivative", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Find the sum of the following series and prove it:\r\n\r\n\r\n$\\sum_{n=1}^{\\infty}\\frac{1}{\\left(\\begin{array}{cc}2n\\\\\\ n\\end{array}\\right)}$", "Solution_1": "You can compare it with the power series for arcsin and find that that the sum is $\\frac13 + \\frac{4}{3\\sqrt{3}}\\arcsin\\left(\\frac12 \\right)$, but I am too lazy to write down the details (it takes me a long time to write down things in latex).", "Solution_2": "If i'm not doing wrong in my computation, it is $\\frac{\\sqrt{3}\\pi}{9}+\\frac{5}{3}$\r\n\r\ncorrect me if i'm wrong.", "Solution_3": "[quote=\"Kalle\"]You can compare it with the power series for arcsin and find that that the sum is $\\frac13 + \\frac{4}{3\\sqrt{3}}\\arcsin\\left(\\frac12 \\right)$, but I am too lazy to write down the details (it takes me a long time to write down things in latex).[/quote]\r\n\r\n\r\nYou are right!But i'd like to see your proof... ;)", "Solution_4": "I've check my computation and found an error on the last line.\r\nI 'll post it here.\r\n\r\nLet $S=\\sum_{n=1}^{\\infty} \\frac{1}{\\binom{2n}{n}}$. lets compute \r\n$\\sum_{n=0}^{\\infty} \\frac{1}{\\binom{2n}{n}}=S+1$\r\n\r\nWe have that \\begin{eqnarray*}\\sum_{n=0}^{\\infty} \\frac{1}{\\binom{2n}{n}} &=& \\sum_{n=0}^{\\infty} \\frac{(n)!(n)!}{(2n)!} \\\\ &=& \\sum_{n=0}^{\\infty} (2n+1) \\frac{\\Gamma(n+1) \\Gamma(n+1) }{\\Gamma(2n+2)} \\\\ &=& (2n+1)B(n+1,n+1) \\end{eqnarray*}\r\n\r\nand we need to compute $\\sum_{n=0}^{\\infty} \\int_{0}^1 (2n+1) \\, t^n (1-t)^n \\, dt$\r\n\r\nNotice that the integran $t^n (1-t)^n$ with $n\\geq 0$ is converges uniformly when $t \\in [0,1)$ by Weierstrass M-Test since we have \r\n\\[ |t^n (1-t)^n | \\leq M^n \\] for $0\\leq t =2 as for a=1 $z_1$=5 is a prime.\n\n", "Solution_15": "Take $a=4k^4$, for some non-zero integer $k$. We have \n$$n^4+4k^4=(n^2+2k^2)-4n^2k^2=(n^2+2nk+2k^2)(n^2-2nk-2k^2)$$\nIf $(n^2-2nk-2k^2)=1$, we have that $(n-k)^2+k^2=1$, so since $k$ is positive, we have that $n=k$. Plugging in, we have that $(n^2+2nk-2k^2)=5n^2$, and which is composite due to how we defined $k$. If $(n^2-2nk-2k^2)$ does not equal one, than $n^2+a^4$ is still prime so we win, as desired. $\\square$", "Solution_16": "@above\nYou made a mistake, because $$n^4+4k^4=(n^2+2k^2)-4n^2k^2=(n^2+2nk+2k^2)(n^2-2nk+2k^2)$$", "Solution_17": "Let $m=p^{2}$ where $p$ is a prime and $p \\equiv -1 \\pmod 4$. If , $n^{4}+p^{2}=q$ where $q$ is a prime then by Fermat's christmas theorem $ q|n$ and $q|p$ which is clearly contradictory $\\square$.", "Solution_18": "[quote=Garou]Let $m=p^{2}$ where $p$ is a prime and $p \\equiv -1 \\pmod 4$. If , $n^{4}+p^{2}=q$ where $q$ is a prime then by Fermat's christmas theorem $ q|n$ and $q|p$ which is clearly contradictory $\\square$.[/quote]\n\nSomething is totaly wrong here:\n\nChoose $p=11$. Then $p$ is a prime and $p \\equiv -1 \\pmod 4$.\nChoose $n=2$.\nThen $q=n^4+p^2=16+121=137$ is prime, \nbut your conclusion $137|2$ and $137|11$ does not hold and there is no contradiction.", "Solution_19": "People seem to be missing the fact that $n^4+4m^4$ is a prime for $n=m=1$.\nSo better take $a=4m^4 \\forall m \\in \\{2, 3 \\dots \\}$" } { "Tag": [ "Euler", "LaTeX" ], "Problem": "\u0391\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 x,y,z isx;yoyn oi \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2:\r\n\\[ {x^2} \\minus{} y \\equal{} {z^2}\\],\\[ {y^2} \\minus{} z \\equal{} {x^2}\\],\\[ {z^2} \\minus{} x \\equal{} {y^2}\\]\r\n\r\n\u039d\u03b1 \u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03bf\u03c4\u03b9 \\[ {x^3} \\plus{} {y^3} \\plus{} {z^3} \\equal{} 0\\]", "Solution_1": "\u0394\u03bf\u03ba\u03af\u03bc\u03b1\u03c3\u03b5 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2. \u0398\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x+y+z=0. \u03a0\u03ac\u03c1\u03b5 \u03ad\u03c0\u03b5\u03b9\u03c4\u03b1 \u03c4\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03cc\u03c4\u03b7\u03c4\u03b1 Euler-Cauchy (\u03c4\u03b7\u03bd \u03ad\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u0386\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2 \u03c4\u03b7\u03c2 \u0391\u0384\u039b\u03c5\u03ba\u03b5\u03af\u03bf\u03c5), \u03ba\u03b1\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x^3+y^3+z^3=0.\r\n\r\n\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7: \u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf 4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c6\u03b5\u03c4\u03b9\u03bd\u03bf\u03cd \u0398\u03b1\u03bb\u03ae \u03c4\u03b7\u03c2 \u0391' \u039b\u03c5\u03ba\u03b5\u03af\u03bf\u03c5.", "Solution_2": "[quote=\"Spribo\"]\u0394\u03bf\u03ba\u03af\u03bc\u03b1\u03c3\u03b5 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2. \u0398\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x+y+z=0. \u03a0\u03ac\u03c1\u03b5 \u03ad\u03c0\u03b5\u03b9\u03c4\u03b1 \u03c4\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03cc\u03c4\u03b7\u03c4\u03b1 Euler-Cauchy (\u03c4\u03b7\u03bd \u03ad\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u0386\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2 \u03c4\u03b7\u03c2 \u0391\u0384\u039b\u03c5\u03ba\u03b5\u03af\u03bf\u03c5), \u03ba\u03b1\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x^3+y^3+z^3=0.\n\n\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7: \u0391\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf 4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03c6\u03b5\u03c4\u03b9\u03bd\u03bf\u03cd \u0398\u03b1\u03bb\u03ae \u03c4\u03b7\u03c2 \u0391' \u039b\u03c5\u03ba\u03b5\u03af\u03bf\u03c5.[/quote]\r\nto kserw aplws me enan filo prospathousame na to lisoume etsi.Alla kai pali de vgainei.", "Solution_3": "\u039c\u03b1, \u03b2\u03b5\u03b2\u03b1\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9. \u0395\u03ac\u03bd \u03b8\u03b5\u03c2 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03cc \u03c4\u03c1\u03cc\u03c0\u03bf, \u03c0\u03ac\u03c1\u03b5 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03bf\u03cd \u03c4\u03b7\u03c2 \u0386\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2 \u03b3\u03b9\u03b1 x^3+y^3+z^3=0, \u03cc\u03c4\u03b1\u03bd x+y+z=0.\r\n\u0391\u03c0' \u03cc\u03c3\u03bf \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03c9, \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5!", "Solution_4": "[quote=\"Spribo\"]\u039c\u03b1, \u03b2\u03b5\u03b2\u03b1\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9. \u0395\u03ac\u03bd \u03b8\u03b5\u03c2 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03cc \u03c4\u03c1\u03cc\u03c0\u03bf, \u03c0\u03ac\u03c1\u03b5 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03bf\u03cd \u03c4\u03b7\u03c2 \u0386\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2 \u03b3\u03b9\u03b1 x^3+y^3+z^3=0, \u03cc\u03c4\u03b1\u03bd x+y+z=0.\n\u0391\u03c0' \u03cc\u03c3\u03bf \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03c9, \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5![/quote]\r\n\u03bc\u03b1 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03ad\u03c2\r\n\u0388\u03c3\u03c4\u03c9 x=5,y=-3,z=-2, \u03c4\u03cc\u03c4\u03b5 x+y+z=0\r\n\u0386\u03c1\u03b1 \\[ {x^3} \\plus{} {y^3} \\plus{} {z^3} \\equal{} 125 \\minus{} 27 \\minus{} 8 \\ne 0\\]", "Solution_5": "\u039c\u03b1 \u03c4\u03b7\u03bd \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1, \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b3\u03ba\u03ac\u03c6\u03b1 \u03c4\u03c1\u03bf\u03bc\u03b5\u03c1\u03ae! \u0395\u03c0\u03b1\u03bd\u03bf\u03c1\u03b8\u03ce\u03bd\u03c9 \u03cc\u03bc\u03c9\u03c2 \u03ac\u03bc\u03b5\u03c3\u03b1:\r\n\u0391\u03c0\u03cc \u03c4\u03bf x+y+z=0, \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 x^3+y^+z^3=3xyz. \u03a4\u03ce\u03c1\u03b1, \u03b8\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 x,y,z \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 0.\r\n\u03a0\u03ac\u03c1\u03b5 \u03b4\u03c5\u03bf-\u03b4\u03c5\u03bf \u03c4\u03b9\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2 (\u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03bb\u03cc\u03b3\u03bf \u03ba\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03cd\u03bc\u03b2\u03bf\u03bb\u03b1, \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 latex) \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 (z+y)(y-z-1)=0, \u03ba\u03b1\u03b9 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 \u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2. \u0395\u03ac\u03bd \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b4\u03c5\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2, \u03b8\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9\u03c2 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03c3\u03c4\u03bf \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03bf\u03cd\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 0.\r\n\u03a4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03af\u03b1\u03c3\u03b1 \u03c3\u03c4\u03bf \u0398\u03b1\u03bb\u03ae! (\u03a4\u03ce\u03c1\u03b1 \u03b4\u03b5\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03b6\u03c9 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2).\r\n\u0395\u03af\u03b4\u03b5\u03c2 \u03c4\u03c9\u03bd \u03c0\u03c1\u03ac\u03be\u03b5\u03c9\u03bd \u03b7 \u03b2\u03b9\u03b1\u03c3\u03cd\u03bd\u03b7, \u03b7 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03b7 \u03bc\u03bd\u03ae\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03b9\u03c3\u03c7\u03c5\u03c1\u03bf\u03b3\u03bd\u03c9\u03bc\u03bf\u03c3\u03cd\u03bd\u03b7...", "Solution_6": "[quote=\"Spribo\"]\u039c\u03b1 \u03c4\u03b7\u03bd \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1, \u03ad\u03ba\u03b1\u03bd\u03b1 \u03b3\u03ba\u03ac\u03c6\u03b1 \u03c4\u03c1\u03bf\u03bc\u03b5\u03c1\u03ae! \u0395\u03c0\u03b1\u03bd\u03bf\u03c1\u03b8\u03ce\u03bd\u03c9 \u03cc\u03bc\u03c9\u03c2 \u03ac\u03bc\u03b5\u03c3\u03b1:\n\u0391\u03c0\u03cc \u03c4\u03bf x+y+z=0, \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 x^3+y^+z^3=3xyz. \u03a4\u03ce\u03c1\u03b1, \u03b8\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03c4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 x,y,z \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf\u03c2 \u03bc\u03b5 0.\n\u03a0\u03ac\u03c1\u03b5 \u03b4\u03c5\u03bf-\u03b4\u03c5\u03bf \u03c4\u03b9\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2 (\u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03bb\u03cc\u03b3\u03bf \u03ba\u03b9 \u03cc\u03c7\u03b9 \u03c3\u03cd\u03bc\u03b2\u03bf\u03bb\u03b1, \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 latex) \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 (z+y)(y-z-1)=0, \u03ba\u03b1\u03b9 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 \u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2. \u0395\u03ac\u03bd \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b4\u03c5\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2, \u03b8\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9\u03c2 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03c3\u03c4\u03bf \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03bf\u03cd\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 0.\n\u03a4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03b1\u03c5\u03c4\u03ae\u03bd \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03af\u03b1\u03c3\u03b1 \u03c3\u03c4\u03bf \u0398\u03b1\u03bb\u03ae! (\u03a4\u03ce\u03c1\u03b1 \u03b4\u03b5\u03bd \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03b6\u03c9 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2).\n\u0395\u03af\u03b4\u03b5\u03c2 \u03c4\u03c9\u03bd \u03c0\u03c1\u03ac\u03be\u03b5\u03c9\u03bd \u03b7 \u03b2\u03b9\u03b1\u03c3\u03cd\u03bd\u03b7, \u03b7 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03b7 \u03bc\u03bd\u03ae\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03b9\u03c3\u03c7\u03c5\u03c1\u03bf\u03b3\u03bd\u03c9\u03bc\u03bf\u03c3\u03cd\u03bd\u03b7...[/quote]\r\n\u03a6\u03b9\u03bb\u03b5 \u03bc\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9. \u0391\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ac\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b2\u03b3\u03b5\u03af \u03cc\u03c4\u03b9 \\[ {x^3} \\plus{} {y^3} \\plus{} {z^3} \\equal{} 0\\]\r\n\u03c3\u03c5\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b5\u03c3 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2", "Solution_7": "\u03a0\u03ac\u03c1\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7. \u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03af\u03b1\u03c3\u03b5 \u03bc\u03b5 y^2. \u0388\u03c1\u03b3\u03b1\u03c3\u03b1\u03b9 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2. \u03a0\u03c1\u03cc\u03c3\u03b8\u03b5\u03c3\u03b5 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7. \u039a\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 x^3+y^3+z^3=0! (\u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x,y,z \u03bc\u03b7 \u03bc\u03b7\u03b4\u03b5\u03bd\u03b9\u03ba\u03bf\u03af, \u03b1\u03bb\u03bb\u03ac, \u03b5\u03ac\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03ae\u03c4\u03b1\u03bd \u03bc\u03b7\u03b4\u03b5\u03bd\u03b9\u03ba\u03cc\u03c2, \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03b8\u03b1 \u03af\u03c3\u03c7\u03c5\u03b5).", "Solution_8": "[quote=\"Spribo\"]\u03a0\u03ac\u03c1\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7. \u03a0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03af\u03b1\u03c3\u03b5 \u03bc\u03b5 y^2. \u0388\u03c1\u03b3\u03b1\u03c3\u03b1\u03b9 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2. \u03a0\u03c1\u03cc\u03c3\u03b8\u03b5\u03c3\u03b5 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7. \u039a\u03b9 \u03ad\u03c7\u03b5\u03b9\u03c2 x^3+y^3+z^3=0! (\u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03b8\u03ad\u03c3\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 x,y,z \u03bc\u03b7 \u03bc\u03b7\u03b4\u03b5\u03bd\u03b9\u03ba\u03bf\u03af, \u03b1\u03bb\u03bb\u03ac, \u03b5\u03ac\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03ae\u03c4\u03b1\u03bd \u03bc\u03b7\u03b4\u03b5\u03bd\u03b9\u03ba\u03cc\u03c2, \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03b8\u03b1 \u03af\u03c3\u03c7\u03c5\u03b5).[/quote]\r\n\u039c\u03b9\u03bb\u03b1\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03c0\u03b5\u03c3\u03b5 \u03c6\u03ad\u03c4\u03bf\u03c2?", "Solution_9": "\u039c\u03b1 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1! \u0391\u03c5\u03c4\u03cc \u03ad\u03c0\u03b5\u03c3\u03b5 \u03c4\u03ad\u03c4\u03b1\u03c1\u03c4\u03bf, \u03bc\u03b5 \u03b4\u03cd\u03bf \u03cc\u03bc\u03c9\u03c2 \u03c5\u03c0\u03bf\u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1. \u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03ad\u03bb\u03b5\u03b3\u03b5 \u03bd.\u03b4.\u03bf. x+y+z=0 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bd.\u03b4.\u03bf x^3+y^3+z^3=0, \u03b5\u03ac\u03bd \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac...", "Solution_10": "[quote=\"Spribo\"]\u039c\u03b1 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1! \u0391\u03c5\u03c4\u03cc \u03ad\u03c0\u03b5\u03c3\u03b5 \u03c4\u03ad\u03c4\u03b1\u03c1\u03c4\u03bf, \u03bc\u03b5 \u03b4\u03cd\u03bf \u03cc\u03bc\u03c9\u03c2 \u03c5\u03c0\u03bf\u03b5\u03c1\u03c9\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1. \u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03ad\u03bb\u03b5\u03b3\u03b5 \u03bd.\u03b4.\u03bf. x+y+z=0 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bd.\u03b4.\u03bf x^3+y^3+z^3=0, \u03b5\u03ac\u03bd \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ac...[/quote]\r\n\u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03ad\u03bb\u03b5\u03b3\u03b5 \u03bd\u03b4\u03bf \u03c7^3+^3+\u03b6^3=3\u03c7\u03c5\u03b6\r\n>\u039a\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bd\u03b4\u03bf \u03b5\u03bd\u03b1\u03c2 \u03b1\u03c0\u03bf \u03c7\u03c5\u03b6 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c2 \u03bc\u03b5 0", "Solution_11": "\u0395\u03bd \u03c0\u03ac\u03c3\u03b5\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9...", "Solution_12": "[quote=\"Spribo\"]\u0395\u03bd \u03c0\u03ac\u03c3\u03b5\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9...[/quote]\r\n\u038c\u03bc\u03c9\u03c2 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03c0\u03ae\u03c1\u03b5\u03c2 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b3\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03bd \u03c0\u03b9\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b1\u03c0\u03bb\u1f04.\r\n\u0391\u03c6\u03bf\u03cd \u03b1\u03c0\u03bf \u03b4\u03b5\u03b9\u03ba\u03bd\u03c5\u03b5\u03b9\u03c2 \u03bf\u03c4\u03b9 \\[ {x^3} \\plus{} {y^3} \\plus{} {z^3} \\equal{} 0\\] \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03c0\u03ad\u03b4\u03b5\u03b9\u03be\u03b5\u03c2 \\[ {x^3} \\plus{} {y^3} \\plus{} {z^3} \\equal{} 3xyz\\]\r\n\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 \\[ 3xyz\\equal{}0\\] \u03ac\u03c1\u03b1 \u03ae \u03c7 \u03ae \u03c5 \u03ae \u03b6 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c2 \u03bc\u03b5 0", "Solution_13": "\u03a4\u03bf \u03be\u03ad\u03c1\u03c9, \u03b1\u03bb\u03bb\u03ac \u03b5\u03bd \u03c4\u03b7 \u03ce\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd \u03b4\u03b5 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03bf\u03c5\u03bd \u03c0\u03bf\u03bb\u03cd \u03c4\u03b7\u03bd \u03ba\u03bf\u03bc\u03c8\u03cc\u03c4\u03b7\u03c4\u03b1...", "Solution_14": "[quote=\"Spribo\"]\u03a4\u03bf \u03be\u03ad\u03c1\u03c9, \u03b1\u03bb\u03bb\u03ac \u03b5\u03bd \u03c4\u03b7 \u03ce\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd \u03b4\u03b5 \u03c3\u03ba\u03b5\u03c6\u03c4\u03cc\u03bc\u03bf\u03c5\u03bd \u03c0\u03bf\u03bb\u03cd \u03c4\u03b7\u03bd \u03ba\u03bf\u03bc\u03c8\u03cc\u03c4\u03b7\u03c4\u03b1...[/quote]\r\n\u039a\u03b1\u03b9 \u03b7 \u03b5\u03bc\u03b5 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b5\u03b9\u03c2 \u03b5\u03b4\u03c9\u03c3\u03b5", "Solution_15": "\u039a\u03b1\u03bb\u03b7\u03bd\u03c5\u03c7\u03c4\u03af\u03b6\u03c9!", "Solution_16": "[quote=\"Spribo\"]\u039a\u03b1\u03bb\u03b7\u03bd\u03c5\u03c7\u03c4\u03af\u03b6\u03c9![/quote]\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2", "Solution_17": "\u0397 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 Spribo \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c3\u03c9\u03c3\u03c4\u03ae. \r\n\r\n\u0398\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n\r\nx^3+y^3+z^3=0\r\n\u0391\u03c6\u03bf\u03cd x+y+z=0 \u03b1\u03c0\u03cc Euler \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 x^3+y^3+z^3=3xyz.\r\n\r\n\u0386\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af 3xyz=0 <=> xyz=0 <=> x=0 \u03ae y=0 \u03ae z=0. \u0391\u03c5\u03c4\u03cc \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b5 \u03bf Spribo.", "Solution_18": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce, \u03b1\u03bb\u03bb\u03ac \u03b1\u03c1\u03c7\u03b9\u03ba\u03ac \u03b5\u03af\u03c7\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2!", "Solution_19": "[quote=\"Spribo\"]\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce, \u03b1\u03bb\u03bb\u03ac \u03b1\u03c1\u03c7\u03b9\u03ba\u03ac \u03b5\u03af\u03c7\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2![/quote]\r\nBasika ti lisi egw tn proteina kat auton tn tropo. Ostoso tn arxiki apodeiksi tn ekane o spirbo kai tn euxaristo" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Let ABC be a triangle with sides lengths are a,b,c. It's well known that if $a^2+b^2=c^2$ then ABC is a right angled triangle. I wonder is there such a triangle exists with a,b,c as its sides satisfying the equation $a^n+b^n=c^n$ and 00$. Then, of course, $c$ is the longest side.\r\n\r\nIf $t>2$, the triangle is acute.\r\nIf $t=2$, the triangle is right.\r\nIf $1 0 \u03ba\u03b1\u03b9 ab(c+d)+cd(a+b) = 2abcd \u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9\r\n\r\n$ (a\\plus{}b)(b\\plus{}c)(c\\plus{}d)(a\\plus{}c)(b\\plus{}d)(a\\plus{}d)\\le a^3b^3c^3d^3$\r\n\r\n \u03a5.\u03a3 \u039c\u03b1\u03bd\u03b9\u03ad \u03c4\u03bf \u03be\u03b5\u03c0\u03ad\u03c4\u03b1\u03be\u03b5\u03c2 \u03c4\u03bf \u03b1\u03c3\u03ba\u03b7\u03c3\u03ac\u03ba\u03b9 vm6 :) \r\n\r\n \u0391\u03c5\u03c4\u03ae \u03b3\u03b9\u03b1 \u03b5\u03c3\u03ad\u03bd\u03b1 ...", "Solution_1": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03ac\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03cc\u03bc\u03bf\u03c1\u03c6\u03b7 \u03b1\u03c6\u03b9\u03ad\u03c1\u03c9\u03c3\u03b7.\r\n\r\n$ ab(c \\plus{} d) \\plus{} cd(a \\plus{} b) \\equal{} 2abcd \\Leftrightarrow \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\plus{} \\frac {1}{d} \\equal{} 2$\r\n\u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bd\u03b1\u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 \u03c4\u03c9\u03bd $ a,b, c, d$ \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03ae \u03c4\u03b7\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 $ a,b, c, d$\r\n\r\n$ 2abcd \\equal{} ab(c \\plus{} d) \\plus{} cd(a \\plus{} b)$ (\u03b1\u03c0\u03cc AM - GM) $ \\ge 2\\sqrt {abcd(a \\plus{} b)(c \\plus{} d)} \\Leftrightarrow$\r\n$ (abcd)^2 \\ge abcd(a \\plus{} b)(c \\plus{} d) \\Leftrightarrow abcd \\ge (a \\plus{} b)(c \\plus{} d)$\r\n\r\n\u039b\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03af\u03b1\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2:\r\n$ abcd \\ge (b \\plus{} c)(a \\plus{} d)$ \u03ba\u03b1\u03b9\r\n$ abcd \\ge (a \\plus{} c)(b \\plus{} d)$\r\n\r\n\u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf\r\n$ (a \\plus{} b)(b \\plus{} c)(c \\plus{} d)(a \\plus{} c)(b \\plus{} d)(a \\plus{} d) \\le a^3b^3c^3d^3$ :lol: \r\n\r\n\r\n\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7:\u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b5\u03c6\u03b1\u03c1\u03bc\u03cc\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd AM - GM \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03b1\u03c3\u03bc\u03bf\u03cd\u03c2\r\n\u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2.", "Solution_2": "\u039c\u03c0\u03c1\u03ac\u03b2\u03bf \u03bc\u03b1\u03bd\u03b9\u03ad ... :) \r\n\r\n\u039c\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03b9\u03b4\u03ad\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 $ (x \\plus{} y)^2\\ge 4xy$.\r\n\u03c3\u03b5 \u03c3\u03c5\u03bd\u03b4\u03c5\u03b1\u03c3\u03bc\u03bf \u03bc\u03b5 \u03c4\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\plus{} \\frac {1}{d} \\equal{} 2$, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ \\left(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\plus{} \\frac {1}{d}\\right)^2\\ge4\\left(\\frac {1}{a} \\plus{} \\frac {1}{b}\\right)\\left(\\frac {1}{c} \\plus{} \\frac {1}{d}\\right)$ \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b4\u03af\u03bd\u03b5\u03b9 $ abcd\\ge (a \\plus{} b)(c \\plus{} d)$ (1)\r\n\r\n\u0395\u03c6\u03b1\u03c1\u03bc\u03cc\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03ac\u03bb\u03bb\u03b5\u03c2 \u03b4\u03cd\u03bf \u03c6\u03bf\u03c1\u03ad\u03c2 \u03bf\u03bc\u03b1\u03b4\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b1 \u03b6\u03b5\u03c5\u03b3\u03ac\u03c1\u03b9\u03b1 \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 : $ abcd\\ge (a \\plus{} c)(b \\plus{} d)$ (2) \u03ba\u03b1\u03b9 $ abcd\\ge (a \\plus{} d)(b \\plus{} c)$ (3)\r\n\r\n\u03a0\u03bf\u03bb/\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b9\u03c2 (1),(2),(3) \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf... :)" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "$ A,B \\in M_{3}(Z)$ , $ det(A)\\equal{}det(B)\\equal{}0$.Prove that $ \\frac{det(A^3\\plus{}B^3)\\plus{}det(A^3\\minus{}B^3)}{2}$ are cub perfect .", "Solution_1": "I think : Let $ A^{3}=X ,B^{3}=Y$\r\n\r\n$ X,Y \\in M_{3} \\mathbb{(R)}$\r\n$ det(X)=0 ,det(Y)=0$\r\nThen : rank(Y) $ \\in$ {1,2}\r\nIf Rank(Y)=1 then : $ det(X-Y)+det(X+Y) = 2.det(X)=0$\r\nIf Rank(Y)=2 then :$ det(X-Y)+det(X+Y) = 2.x_{33}$, $ x_{33}$ is colum 3 ,row 3 of X\r\nYou can prove :roll:", "Solution_2": "If Rank(Y)=1 then : $ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.det(X)$\r\nIf Rank(Y)=2 then :$ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.x_{33}$, $ x_{33}$ for every matrix $ \\in M_{3}(C)$?\r\nCan you prove this pls ?", "Solution_3": "[quote=\"posabogdan\"]If Rank(Y)=1 then : $ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.det(X)$\nIf Rank(Y)=2 then :$ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.x_{33}$, $ x_{33}$ for every matrix $ \\in M_{3}(C)$?\nCan you prove this pls ?[/quote]\r\n\r\nNo, because it's not true. :)\r\n\r\n$ X\\equal{}\\left(\\begin{array}{ccc} 0 & 0 & 0 \\\\\r\n0 & 0 & 1 \\\\\r\n0 & 1 & 0\\end{array}\\right), Y\\equal{}\\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n0 & 0 & 0 \\\\\r\n0 & 0 & 0\\end{array}\\right)$\r\n\r\nThen $ \\det(X\\plus{}Y)\\plus{}\\det(X\\minus{}Y)\\equal{}\\minus{}2$.\r\n\r\nSimilarly if $ Y\\equal{}\\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n0 & 1 & 0 \\\\\r\n0 & 0 & 0\\end{array}\\right)$ for the rank 2 case.", "Solution_4": "Actually, the whole problem is wrong. This is the simplest counterexample I could find:\r\n\r\n$ A\\equal{}\\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n0 & 0 & 0 \\\\\r\n0 & 0 & 0\\end{array}\\right), B\\equal{}\\left(\\begin{array}{ccc} 1 & 1 & 0 \\\\\r\n1 & 1 & 0 \\\\\r\n0 & 0 & 1\\end{array}\\right)$\r\n\r\nThen $ \\frac{\\det(A^3 \\plus{} B^3) \\plus{} \\det(A^3\\minus{}B^3)}{2} \\equal{} 4$.", "Solution_5": "[quote=\"LydianRain\"][quote=\"posabogdan\"]If Rank(Y)=1 then : $ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.det(X)$\nIf Rank(Y)=2 then :$ det(X \\minus{} Y) \\plus{} det(X \\plus{} Y) \\equal{} 2.x_{33}$, $ x_{33}$ for every matrix $ \\in M_{3}(C)$?\nCan you prove this pls ?[/quote]\n\nNo, because it's not true. :)\n\n$ X \\equal{} \\left(\\begin{array}{ccc} 0 & 0 & 0 \\\\\n0 & 0 & 1 \\\\\n0 & 1 & 0\\end{array}\\right), Y \\equal{} \\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\n0 & 0 & 0 \\\\\n0 & 0 & 0\\end{array}\\right)$\n\nThen $ \\det(X \\plus{} Y) \\plus{} \\det(X \\minus{} Y) \\equal{} \\minus{} 2$.\n\nSimilarly if $ Y \\equal{} \\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 0\\end{array}\\right)$ for the rank 2 case.[/quote]\r\n\r\nSomething about me and Mathematica... anyway, this was not actually a counterexample to what posabogdan posted originally, since I mixed up the order of the matrices.\r\n\r\nWhat posabogdan posted about rank 1 matrices was correct, though his statement about rank 2 matrices was wrong. Here is a correct counterexample:\r\n\r\n$ X \\equal{} \\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n0 & 0 & 1 \\\\\r\n0 & 1 & 0\\end{array}\\right), Y \\equal{} \\left(\\begin{array}{ccc} 1 & 0 & 0 \\\\\r\n0 & 1 & 0 \\\\\r\n0 & 0 & 0\\end{array}\\right)$" } { "Tag": [ "function", "algebra", "domain", "calculus", "calculus computations" ], "Problem": "A fixed point of a function f is a number c in its domain such that f(c) = c. (the function doesn't move the input value c; it stays fixed.) Suppose that f is a continuous function with domain [0,1] and range contained in [0,1]. Prove that f must have at least one fixed point.", "Solution_1": "Let $g(x)=f(x)-x$. Since $f(x)$ is continuous, then $g(x)$ is continuous. Note that $g(0)\\ge 0\\ge g(1)$.\r\n\r\nCase 1: $g(0)=0$ or $g(1)=0$.\r\nThis implies one of the endpoints is a fixed point.\r\n\r\nCase 2: $g(0),g(1)\\neq 0$.\r\nBy the intermediate value theorem, there exists $c$ with $0 f(n)$ if $ m > n$, that $ f(n)$ is an integer for an integer n, and that $ f(2) \\equal{} 2$, prove that $ f(k) \\equal{} k$ for integral k.", "Solution_1": "Do you mean $ f :\\mathbb{Z}\\to\\mathbb{Z}$?", "Solution_2": "Well, even if not, we need only prove it for integers.\r\n[hide]We have $ f(1)f(2) \\equal{} f(1*2) \\equal{} f(2)$ so f(1) = 1. Also, $ 2f(0)\\equal{}f(0)f(2)\\equal{}f(0)$, so $ f(0)\\equal{}0$. Furthermore, $ f(\\minus{}1)f(\\minus{}1)\\equal{}f(1)$ and $ f(\\minus{}1)2$, $ p\\plus{}1$ and $ p\\minus{}1$ must be divisible only by primes less than $ p$. \nThis is obvious, as $ (p,p\\plus{}1)$ and $ (p,p\\minus{}1)$ = 1, and all primes bigger than $ p$ are at least $ p\\plus{}2$.\n\nNow, we induct on the primes.\n\nBase case: $ f(2)\\equal{}2$. Because $ f(2)\\equal{}2$ and $ f(4) \\equal{} f(2)f(2) \\equal{} 4$, $ f(3)\\equal{}3$.\nInductive step:Because $ p\\plus{}1$ and $ p\\minus{}1$ are strictly divisible by primes < $ p$, we know that $ f(p\\plus{}1)\\equal{}p\\plus{}1$ and $ f(p\\minus{}1)\\equal{}p\\minus{}1$. Since $ f$ is strictly increasing and takes integers to integers, we have $ f(p)\\equal{}p$ and we are done.[/hide]", "Solution_3": "[quote=\"not_trig\"]Well, even if not, we need only prove it for integers.[/quote]\r\n\r\nI mean as opposed to $ f :\\mathbb{R}\\to\\mathbb{R}$, which would make the proof a lot shorter :)", "Solution_4": "The phrase \"totally multiplicative\" is meaningless on $ \\bf R$." } { "Tag": [ "search", "real analysis", "real analysis unsolved" ], "Problem": "f:R -> R, both f^2, f^3 inf-tely differentiable => f too?", "Solution_1": "There was lot of discussion about this problem on the forum. Please search for it, if I'm not wrong then it's a post by sam-n. I don't remember if the problem was solved or not, but the only (correct) solution I saw spanned over 6 pages, so I'm definitely not going to write it here.", "Solution_2": "Let's put this into an abstract form: we have a smooth map $F\\colon \\mathbb R\\to\\mathbb R^{2}$, namely $t\\mapsto (f^{2}(t),f^{3}(t))$, whose image is contained in the cusped curve $C=\\{(x,y)\\colon x^{3}=y^{2}\\}$. The curve $C$ is covered by $\\pi\\colon \\mathbb R\\to C$, where $\\pi(s)=(s^{2},s^{3})$. The question is: does $F$ lift to a smooth map $\\tilde F\\colon \\mathbb R\\to\\mathbb R$ such that $\\pi\\circ \\tilde F=F$? \r\n\r\nIt surely does in the continuous category. What are the obstructions to lifting maps in the smooth category? :?:" } { "Tag": [ "Putnam", "college contests" ], "Problem": "Define a sequence $ {a_i}$ by $ a_1 \\equal{} 3$ and $ a_{i\\plus{}1} \\equal{} 3^{a_i}$ for $ i \\geq 1$. Which integers between 00 and 99 occur as the last two digits in the decimal expansion of infinitely many $ a_i$?", "Solution_1": "$ \\lambda(100) \\equal{} 20, \\lambda(20) \\equal{} 4, \\lambda(4) \\equal{} 2$. After $ a_4$ the value of $ a_k \\bmod 100$ is fixed; namely,\r\n\r\n$ 3 \\equiv 1 \\bmod 2 \\implies$\r\n$ 3^3 \\equiv 3 \\bmod 4 \\implies$\r\n$ 3^{3^{3}} \\equiv 7 \\bmod 20 \\implies$\r\n$ 3^{3^{3^{3}}} \\equiv \\boxed{87} \\bmod 100$\r\n\r\nby Carmichael's theorem.", "Solution_2": "I've never heard of Carmichael's Theorem. Is there a way to do it with just Euler's Theorem?", "Solution_3": "Of course. It just takes more layers if you use $ \\phi$ instead of the actual exponent of the group.\r\n\r\n$ \\phi(100)\\equal{}40$\r\n$ \\phi(40)\\equal{}16$\r\n$ \\phi(16)\\equal{}8$\r\n$ \\phi(8)\\equal{}4$\r\n$ \\phi(4)\\equal{}2$\r\n$ \\phi(2)\\equal{}1$\r\n\r\n$ a_n\\equal{} 1\\mod 2$\r\n$ a_{n\\plus{}1}\\equal{} 3^1\\equal{}3\\mod 4$\r\n$ a_{n\\plus{}2}\\equal{}3^3\\equal{}3\\mod 8$\r\n$ a_{n\\plus{}3}\\equal{}3^3\\equal{}11\\mod 16$\r\n$ a_{n\\plus{}4}\\equal{}3^{11}\\equal{}27\\mod 40$\r\n$ a_{n\\plus{}5}\\equal{}3^{27}\\equal{}87\\mod 100$\r\nand $ a_m\\equiv 87\\mod 100$ for all $ m\\ge 6$.", "Solution_4": "Unless I am missing something, you actually had to do some work to calculate 3^11 and 3^87 mod 40 and mod 100, right (I'm not complaining--I'm just asking)?\r\n\r\nEDIT: Let me try to answer my own question. So, I guess you use Euler's Theorem in two ways in this problem: first to show that $ a_{n} \\equiv 3^{a_{n \\minus{} 1}\\mod{\\phi(100)}}\\mod 100$ and also to explicitly calculate 3^11 and ^27\r\n\r\nEDIT 2: disregard the first edit. I still do not see how you got 3^11 and 3^27.\r\nAlso, does the value of a_1 matter?", "Solution_5": "$ 3^{8}\\equal{}(3^4)^2\\equal{} 1^2\\equal{}1\\mod 40$, so $ 3^{11}\\equal{} 3^3\\equal{}27\\mod 40$\r\n\r\nI didn't actually calculate $ 3^{27}\\mod 100$ myself; I trusted t0rajir0u for that.\r\nEuler's theorem doesn't help for doing the calculation; we just have to multiply a bunch of numbers. An efficient method:\r\n$ 27\\equal{}16\\plus{}8\\plus{}2\\plus{}1$\r\n$ 3^1\\equal{}3$\r\n$ 3^2\\equal{}9$\r\n$ 3^4\\equal{}9^2\\equal{}81$\r\n$ 3^8\\equal{}81^2\\equal{}61$\r\n$ 3^{16}\\equal{}61^2\\equal{}21$\r\n$ 3^{27}\\equal{}3\\cdot 9\\cdot 61\\cdot 21\\equal{}27\\cdot 81\\equal{}87$\r\n(All calculations mod $ 100$ for this example. This gets you any exponent up to $ 2^m$ with at most $ 2m$ multiplications mod $ n$, plus the cost of getting the binary expansion of $ m$)\r\n\r\nThe value of $ a_1$ doesn't matter. I could have gone back one more step and started at $ a_{n\\minus{}1}\\equal{}0\\mod 1$- which is always true for any starting value, and works in all possible versions of the problem.", "Solution_6": "[quote=\"FibonacciFan\"]I still do not see how you got 3^11 and 3^27.[/quote]\n\nUsing only the Totient Theorem, we must proceed as jmerry has described. Fortunately, the Carmichael Theorem allows us to use $ 3^7$ rather than $ 3^{27}$, which is much simpler :)\n \n[quote=\"FibonacciFan\"]Also, does the value of a_1 matter?[/quote]\r\n\r\nNope. If $ a_1$ is odd then $ a_1 \\equiv 1 \\bmod 2$; if $ a_1$ is even then $ a_2 \\equiv 1 \\bmod 2$. It's the base of the exponent in the recursion that matters." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "let $( {u_{n})}$ be a sequence of positive integers definied by\r\n$u_{0}=1, u_{n+1}=au_{n}+b$\r\nwhere $a,b$ are natural numbers. Prove that for any choice of $a$ and $b$ , the sequence $( u_{n} )_ {{n} \\geq 0}$ contains infintely many composite numbers", "Solution_1": "If a=1 $u_n=1+nb$. Let prime $p|u_{n_0}$ then $p|u_{n_0+kp}$ for all k.\r\nIf $a\\not=1$ sequence $u_n=au_{n-1}+b=a^2u_{n-2}+ab+b=\\dots=a^nu_0+b(a^{n-1}+\\dots+1)$. It give $u_n=a^n+\\frac{b(1-a^n)}{1-a}$. \r\nIf prime $p|u_{n_0}$ then $p|u_{n_0+k|1-a|(p-1)}$ for all k." } { "Tag": [ "LaTeX" ], "Problem": "I had to do a system restore on my computer recently and I'm trying to get TeXnicCenter working again. When I open it the Configuration Wizard pops up. When it says: \"Enter the full path of the directory where the executables (latex, tex, etc.) of your TeX-distribution are located:\"\r\n\r\nWhat do I do here? I've tried many things but nothing works. Also, I have MiKTeX still so that shouldn't be the problem...\r\n\r\nAny help?", "Solution_1": "The path should be something like C:\\texmf\\MiKTeX\\bin\\ which should contain programs such as latex.exe, yap.exe, pdflatex.exe etc etc. Look for these files on your system. \r\n\r\nGenerally speaking LaTeX programs prefers files in folders that have no spaces in their names, so it's not wise to install MiKTeX in Program Files, even if it appears to work.", "Solution_2": "Thanks, that works. But now I can't figure out the paths for the DVI, PS and PDF compiler...", "Solution_3": "It's the same path for the compilers. Do you mean the viewers?\r\nDVI: Uses yap.exe in C:\\texmf\\MiKTeX\\bin\\ \r\nPDF: uses Adobe Reader, depending on where you installed it\r\nPS: uses Ghostscript or Ghostgum \r\n\r\nUse the Wizard which should automatically find MiKTeX. Alt+F7, Wizard" } { "Tag": [ "function", "algebra", "polynomial", "arithmetic sequence", "4th edition" ], "Problem": "umm ... i thought q2 was nice (but a bit annoying) but the typical infinite descent problem. my sol for q3 was a stupid co-ord bash (which doesn't really involve too much computations though :) ). \r\ni have no idea whatsoever how to go about solving q1. it looks totally crazy to me. i think the answer is all functions of form f(x)=x^k where a is a perfect k-th power, and some linear functions. maybe i'm missing some obvious ones tho. \r\ncould someone plz post a sol for q1? and maybe a synthetic one for q3?", "Solution_1": "When you guys will stop telling me that #2 was trivial? I don't see any point why it should be. Anyway, the weaker problem to show that there doesn't exist an infinite arithmetic progression of squares is in Engel's \"Problem-Solving Strategies\", and it's really a simple infinite descent, but...\r\n\r\n#1 was definitiely unattractive to me... I hate those problems where one first has to think for ten minutes what the formula means.\r\n\r\n#3 was very nice - and yes, I have a quite synthetic solution :) See the attached ZIP file. (To the mods: please [b]don't[/b] delete the TeX, DVI and WMF files; they are rather small and some people could use them.)\r\n\r\n Darij", "Solution_2": "Q3 can be done using mostly Pythagoras' theorem and Apollonius': just get everything in terms of $r_1$, $r_2$ and the length of the common chord (I think). My attempt at Q1 was as follows: first, all the coeffs have to be rational; now convert into normal form, and look at the integers that produce $a^{(kd)^2}$ for each natural $k$ where $d$ is the degree of the polynomial. Now for sufficiently large $k$, these have to be multiples of each other (the polynomial basically becomes its first term). You can show this by showing that the next integer up is too big, and the next one down too small. This forces the polynomial to be a power of a linear expression, and its pretty much done from there.\r\n\r\nHow did your infinite descent argument go? I tried for ages and made little progress for Q2.", "Solution_3": "For Fiachra:\r\nFor Q1 i dont see how you deduce from the beggining that all coefficients must be rationals..... But i think the rest of the aproach is ok.\r\n\r\nFor Darij:\r\nI agree that Q2 wasnt trivial. It was hard, but it is a well known result in the web. I found a solution wich is also a descend, but it was not easy.\r\n\r\nAnd i think Q3 was very easy.\r\n\r\nThanks", "Solution_4": "For Q1, to show that all the coefficients are rational, note that we have infinitely many pairs $m,n$ of integers such that $m=f(n)$, and each of these yields a 1st-degree equation for the coefficients. When we solve these simultaneously, we must get rational solutions.", "Solution_5": "About Q2, it can be found in Titu Andreescu, \"An Introduction to Diophantine Equations\". It is shown that there are no 4 squares in an arithmetic progression.\r\nUse the Lemma : The equation \\[x^4-x^2y^2+y^4=z^2\\] has only trivial solutions.", "Solution_6": "It is very ruefully that problem from ML Contest can be such easily found.", "Solution_7": "[quote=\"DusT\"]About Q2, it can be found in Titu Andreescu, [b][Dorin Andrica,][/b] \"An Introduction to Diophantine Equations\". It is shown that there are no 4 squares in an arithmetic progression.[/quote]\r\n\r\nCurses! That book lies on my bookshelf! Now as you say it, I see that the problem was example 3 in paragraph 2.3.1. But I haven't yet managed to systematically read more than the first chapter...\r\n\r\n Darij", "Solution_8": "[quote=\"darij grinberg\"]\n\nCurses! That book lies on my bookshelf! Now as you say it, I see that the problem was example 3 in paragraph 2.3.1. But I haven't yet managed to systematically read more than the first chapter...\n\n Darij[/quote]\r\n\r\nML contests seems to be designed to have a look into new literature. :D", "Solution_9": "I cannot possibly know every other book that had this problem solved. Good for you readers :P :) \r\n\r\n@Myth: it's hard to create 21 (3x7) problems hard and completely original, that cannot be found in any book :P \r\n\r\nProblem no. 3 was my creation, and Problem no. 1 was Harazi's creation.", "Solution_10": "[quote=\"Valentin Vornicu\"]@Myth: it's hard to create 21 (3x7) problems hard and completely original, that cannot be found in any book[/quote]\nI think threre are many people whose opinion you may ask.\n\n[quote]Problem no. 1 was Harazi's creation.[/quote]\r\nIt was clear. Who can also invent [b]such[/b] problem? ;)", "Solution_11": "I am sorry, Myth if you didn't like it. :blush: :blush: :blush: :blush: :blush:.", "Solution_12": "Well, Q2 is not due to Titu Andreescu but to Fermat himself...\r\n\r\nPierre.", "Solution_13": "Yeah, I agree with Valentin, it is not difficult to come up with a problem of your own, but it's VERY difficult to be perfectly original. I think regardless of the origins of the problems and their difficulties, this contest does serve (at the very least) to bring new problems, and it does help a lot (maybe just for me, but...) Nevertheless, I don't think the problem sets deserve too much criticism. (that's, of course, just my trash... :lol: )", "Solution_14": "[quote=\"al.M.V.\"]Yeah, I agree with Valentin, it is not difficult to come up with a problem of your own, but it's VERY difficult to be perfectly original. I think regardless of the origins of the problems and their difficulties, this contest does serve (at the very least) to bring new problems, and it does help a lot (maybe just for me, but...) Nevertheless, I don't think the problem sets deserve too much criticism. (that's, of course, just my trash... :lol: )[/quote]\r\n\r\nI agree with you. I just think that we all should appreciate that Valentin takes his time to create an interesting contest, check solutions and make up rankings. Recently Valentin did not have the time to compile solutions for the contest. That's why I think some forum members as Pierre, Darij, myth etc. could take the source files and generate solutions for it. To share the work might be best. What do you think ? :)", "Solution_15": "Exactly orl!!! :lol: I could not have said that any better! :P \r\n\r\nThanks for understanding. :cool:", "Solution_16": "I know it's a very stupid question, but anyway can somebody help me to rephrase Q1... So, as I understood, A is a set of integers of given form and it's a subset of B. Now what does that vertical line mean? I only know that a statement a divides b is denoted by $a|b$, but I think it's something else here...", "Solution_17": "Ok, I know the statement of the problem is not the happiest one. I prefer the following one: for a given integer $a>1$ find all polynomials with real coefficients f such that the image of the set 1,2,3,... by f contains all the numbers of the form $a^{n^2}$ with n=1,2,3,...", "Solution_18": "Usually in the definition of a set $A = \\{ x \\in \\mathbb{R} \\mid x^2 \\leq 1 \\}$ for example, means that the set $A$ is formed with all the reals $x$ such that (so the | sign stands for \"such that\") $x^2 \\leq 1$." } { "Tag": [ "IMO" ], "Problem": "Should the IMO add a commemoration slumber party to the official list of events?", "Solution_1": "Absolutely! That was the most fun evening at any IMO ever!! AWESOME! Next time they should add some big speakers and a good DJ too :D", "Solution_2": "Heh, but keep it quiet enough for conversation too.\r\n\r\nI think this would be great fun, anyway. I'm all for it.", "Solution_3": "Emily was my favourite weather event ever! I loved it, although when I went to bed at like 4 o'clock, I had no bed to sleep in since it was already token by other persons. Fortunately I'm used to sleeping on the ground." } { "Tag": [ "inequalities", "function", "induction", "inequalities proposed" ], "Problem": "Let a^2+b^2+c^2 = 3, a,b,c >=0, prove or disprove the followings:\r\n\r\n(1) $ a^3 \\plus{} b^3 \\plus{} c^4 \\leq 9$,\r\n\r\n(2) $ a^n \\plus{} b^n \\plus{} c^{n \\plus{} 1} \\leq (\\sqrt {3})^{n \\plus{} 1}$, for all positive integers n and n>=2.", "Solution_1": "someone help. thx.........", "Solution_2": "Solution of (1) :\r\n\r\n$ a^3 \\plus{} b^3 \\plus{} c^4 \\leq 9 \\Longleftrightarrow a^2 \\plus{} b^2 \\plus{} \\sqrt {9 \\minus{} a^3 \\minus{} b^3} \\geq 3$\r\n\r\nWe can assume that $ min(a,b) \\equal{} a$ without loss of generality.\r\n\r\n$ a^2 \\plus{} b^2 \\leq 3$ and $ a \\leq \\frac {\\sqrt {6}}{2}$\r\n\r\n$ f(a) \\equal{} a^2 \\plus{} b^2 \\plus{} \\sqrt {9 \\minus{} a^3 \\minus{} b^3}$\r\n\r\n$ f'(a) \\equal{} 2a \\minus{} \\frac {3a^2}{2\\sqrt {9 \\minus{} a^3 \\minus{} b^3}}$\r\n\r\n$ f'(a) > 0 \\Longleftrightarrow 9a^2 \\plus{} 16a^3 \\plus{} 16b^3 < 144$\r\n\r\nSince $ a^2 \\plus{} b^2 \\leq 3$ \r\n\r\n$ 16b^3 \\leq 16(3 \\minus{} a^2)^{\\frac {3}{2}}$\r\n\r\n$ 9a^2 \\plus{} 16a^3 \\plus{} 16b^3 \\leq 9a^2 \\plus{} 16a^3 \\plus{} 16(3 \\minus{} a^2)^{\\frac {3}{2}} \\leq$\r\n\r\n$ \\frac {27}{2} \\plus{} 12\\sqrt {6} \\plus{} 48\\sqrt {3} \\approx 126.0323 < 144$\r\n\r\nHence $ f(a)$ is an increasing function and $ f(a) \\geq f(0) \\equal{} b^2 \\plus{} \\sqrt {9 \\minus{} b^3}$\r\n\r\nNow if we prove that $ b^2 \\plus{} \\sqrt {9 \\minus{} b^3} \\geq 3$ then the solution is done.\r\n\r\n$ b^2 \\plus{} \\sqrt {9 \\minus{} b^3} \\geq 3$ $ \\Longleftrightarrow$\r\n\r\n$ b^2(b^2 \\plus{} b \\minus{} 6) \\equal{} b^2(b \\plus{} 3)(b \\minus{} 2) \\leq 0$ which is obviously true since $ 0 \\leq b \\leq \\sqrt {3} < 2$ :wink:\r\n\r\nEquality holds if and only if $ a\\equal{}b\\equal{}0$ , $ c\\equal{} \\sqrt{3}$", "Solution_3": "[quote=\"crazyfehmy\"]Solution of (1) :\n\n$ a^3 \\plus{} b^3 \\plus{} c^4 \\leq 9 \\Longleftrightarrow a^2 \\plus{} b^2 \\plus{} \\sqrt {9 \\minus{} a^3 \\minus{} b^3} \\geq 3$.....\n\n[/quote]\r\n\r\nthanks. very clear answer.", "Solution_4": "[quote=\"4865550150\"]Let a^2+b^2+c^2 = 3, a,b,c >=0, prove or disprove the followings:\n\n(1) $ a^3 \\plus{} b^3 \\plus{} c^4 \\leq 9$,\n\n(2) $ a^n \\plus{} b^n \\plus{} c^{n \\plus{} 1} \\leq (\\sqrt {3})^{n \\plus{} 1}$, for all positive integers n and n>=2.[/quote]\r\n\r\n(1), easy to prove that\r\n$ a^3\\plus{}b^3 \\le \\sqrt{(a^2\\plus{}b^2)^3}\\equal{}\\sqrt{(3\\minus{}c^2)^3}$\r\nso we just need to prove that\r\n$ (9\\minus{}c^4)^2 \\ge (3\\minus{}c^2)^3$\r\n$ \\Leftrightarrow \\left(c^2\\minus{}3\\right)^2 \\left(c^2\\plus{}1\\right) \\left(c^2\\plus{}6\\right) \\ge 0$\r\nclear true.\r\n\r\n\r\n\r\nfor (2), I guess, it will holds for any [b]reals[/b] n and $ n \\ge 2$.\r\n\r\nin the same way, just need to prove that\r\n\r\n$ \\left(3^{\\frac{n\\plus{}1}{2}}\\minus{}c^{n\\plus{}1}\\right)^2 \\ge (3\\minus{}c^2)^n, \\quad c \\in [0,\\sqrt{3}] \\qquad (*)$\r\n\r\nbut I always can't prove it... :oops:\r\n\r\n[b]Remark[/b]:\r\n$ n\\equal{}2$, (*) will be $ ({c}^{4}\\plus{}2{c}^{3}\\sqrt {3}\\plus{}8{c}^{2}\\plus{}4c\\sqrt {3}\\plus{}6)(c\\minus{}\\sqrt{3})^2 \\ge 0$;\r\n$ n\\equal{}4$, (*) will be $ (c^8 \\plus{} 2\\sqrt 3 c^7 \\plus{} 8c^6 \\plus{} 10\\sqrt 3 c^5 \\plus{} 48c^4 \\plus{} 48\\sqrt 3 c^3 \\plus{} 90c^2 \\plus{} 36\\sqrt 3 c \\plus{} 54)(c\\minus{}\\sqrt{3})^2 \\ge 0$;\r\n$ n\\equal{}5$, (*) will be $ (c^2\\plus{}6)(c^6\\plus{}c^4\\plus{}12c^2\\plus{}9)(c^2\\minus{}3)^2 \\ge 0$;\r\n...", "Solution_5": "Solution of Kuing's inequality(but while $ n$ is an integer) which is\r\n\r\n$ \\left(3^{\\frac {n \\plus{} 1}{2}} \\minus{} c^{n \\plus{} 1}\\right)^2 \\geq (3 \\minus{} c^2)^{n}$ where $ n \\geq 2$ is an integer and $ 0 \\leq c \\leq \\sqrt {3}$ is a real.\r\n\r\nWe will prove by induction. The inequality is true for $ n \\equal{} 2$ as Kuing said.\r\n\r\nIf we prove that $ \\left(\\frac {3^{\\frac {n \\plus{} 2}{2}} \\minus{} c^{n \\plus{} 2}}{3^{\\frac {n \\plus{} 1}{2}} \\minus{} c^{n \\plus{} 1}}\\right)^2 \\geq 3 \\minus{} c^2$ then the induction is done.\r\n\r\n$ \\left(\\frac {3^{\\frac {n \\plus{} 2}{2}} \\minus{} c^{n \\plus{} 2}}{3^{\\frac {n \\plus{} 1}{2}} \\minus{} c^{n \\plus{} 1}}\\right)^2 \\geq 3 \\minus{} c^2$ $ \\Longleftrightarrow$\r\n\r\n$ 2.c^{2n \\plus{} 2} \\plus{} 2.3^{\\frac {n \\plus{} 3}{2}}.c^{n \\minus{} 1} \\plus{} 3^{n \\plus{} 1} \\geq 3.c^{2n} \\plus{} 2.3^{\\frac {n \\plus{} 2}{2}}.c^n \\plus{} 2.3^{\\frac {n \\plus{} 1}{2}}.c^{n \\plus{} 1}$\r\n\r\nIf we divide both sides by $ c^{2n}$ , inequality is equivalent to \r\n\r\n$ \\left[c \\minus{} \\sqrt {3}.\\left(\\frac {\\sqrt {3}}{c}\\right)^n \\right]^2 \\plus{} 2\\sqrt {3}(\\sqrt {3} \\minus{} c).\\left(\\frac {\\sqrt {3}}{c}\\right)^{n \\plus{} 1} \\geq 3 \\minus{} c^2$\r\n\r\n$ \\Longleftrightarrow$ $ (\\lambda^2 \\minus{} 2\\lambda \\plus{} 2)c^2 \\minus{} 2\\sqrt {3}\\lambda c \\plus{} 6\\lambda \\minus{} 3 \\geq 0$ where $ \\lambda \\equal{} \\left(\\frac {\\sqrt {3}}{c}\\right)^{n \\plus{} 1}$\r\n\r\nThe last inequality is equivalent to\r\n\r\n$ (\\lambda \\minus{} 1)^2c^2 \\plus{} (\\sqrt {3} \\minus{} c)(2\\sqrt {3}\\lambda \\minus{} c \\minus{} \\sqrt {3}) \\geq 0$ \r\n\r\nSince $ c\\leq \\sqrt {3}$ and $ \\lambda \\geq 1$\r\n\r\n$ \\sqrt {3} \\minus{} c \\geq 0$ and $ 2\\sqrt {3}\\lambda \\minus{} c \\minus{} \\sqrt {3} \\equal{} 2\\sqrt {3}(\\lambda \\minus{} 1) \\plus{} (\\sqrt {3} \\minus{} c) \\geq 0$ . Induction is done. :wink:", "Solution_6": "[quote=\"kuing\"]\n\n(1), easy to prove that\n$ a^3 \\plus{} b^3 \\le \\sqrt {(a^2 \\plus{} b^2)^3} \\equal{} \\sqrt {(3 \\minus{} c^2)^3}$.....[/quote]\n\n[quote=\"crazyfehmy\"]Solution of Kuing's inequality(but while $ n$ is an integer) which is\n\n$ \\left(3^{\\frac {n \\plus{} 1}{2}} \\minus{} c^{n \\plus{} 1}\\right)^2 \\geq (3 \\minus{} c^2)^{n}$ where $ n \\geq 2$ is an integer and $ 0 \\leq c \\leq \\sqrt {3}$ is a real.\n\nWe will prove by induction. The inequality is true for $ n \\equal{} 2$ as Kuing said........\n[/quote]\r\n\r\nMy guess is true :lol: . thx again~~~ :)" } { "Tag": [], "Problem": "Visto l'OliSfratto conviene postare anche in questa sezione l'annuncio del Winter Camp. Spargete la voce.\r\n\r\n==============================================\r\n\r\nWinter Camp (Italiano) - Informazioni Generali\r\n\r\nIl Winter Camp 2007 si svolgera' a Pisa. L'arrivo dei partecipanti e' previsto entro la sera di martedi' 23 Gennaio 2007, la partenza nel pomeriggio di domenica 28 Gennaio 2007.\r\n\r\nLo stage e' pensato per studenti che hanno gia' una buona conoscenza del programma olimpico (vedi Schede Olimpiche e lezioni dello stage senior) e sono ad un buon punto della preparazione sugli esercizi. Lo stage consistera' in sessioni di lavoro su problemi di livello internazionale.\r\n\r\nIl Winter Camp si concludera' con una prova finale nelle mattinate di sabato e domenica. [b]Tale prova finale, opportunamente pesata con i risultati di settembre, varra' come prova di selezione per la partecipazione alle \"Balkan Mathematical Olympiads\" nel caso in cui l'Italia venisse invitata a partecipare[/b] (cosa che finora *non* e' accaduta: l'organizzazione sembra piuttosto in ritardo, al punto che non sono ancora nemmeno note le date ufficiali della manifestazione).\r\n\r\nCome gia' accaduto a partire dallo stage Senior 2005, anche al Winter Camp si potra' partecipare come \"spesati\" (spesati dall'organizzazione) o come \"volontari\". Gli aspiranti volontari dovranno fare domanda come indicato nell'apposito paragrafo.\r\n\r\nPer tutti, invitati o volontari, la partecipazione e' da considerare \"sub iudice\", nel senso che tutti devono far pervenire le soluzioni scritte di 8 esercizi, con modalita' indicate nell'ultimo paragrafo. Un'apposita commissione valutera' la qualita' del lavoro svolto e sulla base di questo confermera' o meno la possibilita' di partecipare.\r\n\r\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\r\n\r\nElenco degli \"spesati\"\r\n\r\nCIMETTA Leone\r\nCOLOMBO Maria\r\nCONTI Andrea\r\nDALUISO\tRoberto\r\nDE CAPUA Antonio\r\nFOGARI Andrea\r\nGALEOTTI Mattia\r\nKUZMIN Kirill\r\nLAZZARINO Leslie Lamberto\r\nLO BIANCO Federico\r\nLIN Francesco\r\nLOMBARDO Davide\r\nNARDIN Denis\r\nPATIMO Leonardo\r\nSANTI Elia\r\nTRAGER Matthew\r\nVERTECHI Pietro\r\n\r\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\r\n\r\nInformazioni per i \"volontari\"\r\n\r\nCome fare domanda? Gli interessati dovranno inviare la propria domanda [b]via fax allo 050-2213802 (att. Massimo Gobbino) entro il 6 gennaio 2007[/b].\r\nPer la domanda basta riadattare con i propri dati il fac simile sottostante.\r\n\r\nTutte le domande saranno accolte? Non e' detto.\r\n\r\nCome verranno scelti i partecipanti? Sulla base degli esercizi svolti, dell'eta' (con preferenza ai piu' giovani), e del curriculum olimpico.\r\n\r\nQuanto costa? La partecipazione alle attivita' dello stage (lezioni, sessioni di esercizi, eventuale prova finale) e' gratuita. Tutto il resto, ed in particolare il viaggio, il vitto e l'alloggio, e' a carico del partecipante, sia come spese, sia come organizzazione (ricerca di treni, alberghi, ...). I partecipanti si devono impegnare a seguire tutto lo stage.\r\n\r\nDove posso alloggiare/mangiare? L'organizzazione provvedera' a prenotare un po' di camere in piu' all'Hotel Di Stefano (lo stesso in cui risiederanno gli spesati). I volontari che vorranno potranno usufruirne, ovviamente a pagamento (a settembre 2006 il costo di una sistemazione in camera multipla e' stato di circa 35 euro a testa per notte). I volontari che lo vorranno potranno mangiare insieme agli spesati alla mensa della Scuola Normale Superiore (a settembre 2006 il costo era di 10.00 euro a pasto).\r\n\r\nA cosa ho diritto? A seguire le attivita' dello stage e la prova finale (e' chiaro pero' che i volontari che non hanno partecipato allo stage Senior 2006 saranno svantaggiati nel calcolo del punteggio ai fini dell'eventuale Balkan) alla pari degli altri invitati, ad un attestato di partecipazione.\r\n\r\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\r\n\r\nFAC SIMILE DOMANDA DI PARTECIPAZIONE COME VOLONTARIO\r\n\r\nSono , nato a il . Quest'anno (2006/2007) sto frequentando il ... anno di corso (nota bene: II liceo classico = quarto anno di corso) del di .\r\nIl mio indirizzo e' .\r\n\r\nVorrei seguire le attivita' del Winter Camp Pisa 2007.\r\n\r\nLe mie precedenti esperienze olimpiche sono: .\r\n\r\nSe la mia domanda viene accolta, mi impegno a partecipare per tutta la durata della manifestazione.\r\n\r\nProvvedero' in maniera autonoma ad organizzare il viaggio. Mi piacerebbe alloggiare all'Hotel Di Stefano (a mie spese) e/o mangiare alla mensa SNS (a mie spese) (oppure: provvedero' in maniera autonoma ad organizzare il soggiorno, ed a sostenere le relative spese).\r\n\r\n\r\n\r\nFirma dello studente\r\n\r\nFirma di un genitore (se lo studente e' minorenne)\r\n\r\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\r\n\r\nSpedizione esercizi\r\n\r\nModalita' di spedizione: gli svolgimenti dovranno essere spediti in uno dei seguenti tre modi\r\n\r\n- per via cartacea a Massimo Gobbino, Dipartimento di Matematica Applicata, via F. Buonarroti 1c, 56127 Pisa;\r\n\r\n- per fax allo 050-2213802 all'attenzione di Massimo Gobbino;\r\n\r\n- per posta elettronica ([color=blue]un *unico* file in formato pdf (e dimensioni ragionevoli) per ogni partecipante, con nome WC07 seguito dalle prime 4 lettere del cognome senza spazi[/color]) a m.gobbino@dma.unipi.it.\r\n\r\nScadenze: gli svolgimenti dovranno pervenire entro il [b]6 gennaio 2007[/b]. La conferma dell'invito avverra' a quel punto il prima possibile.\r\n\r\nQuali sono gli esercizi da svolgere: [b]8 esercizi[/b] scelti tra quelli dell'[url=http://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=2006]Iran National Mathematical Olympiad (3rd round 2006)[/url], di cui 2 nel gruppo Algebra, 2 nel gruppo Combinatoria, 2 in Geometria e 2 in Teoria dei Numeri (con esclusione degli esercizi C3, G3 e N1).\r\n\r\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++", "Solution_1": "Qualche piccola aggiunta personale all'annuncio del Winter Camp.\r\n[list][*]Attenzione al fatto che ci sono tre problemi esclusi.\n[*]Attenzione al fatto che alcuni problemi richiedono, anche solo per capire il testo, nozioni al di fuori del programma olimpico solitamente inteso in Italia. In ogni caso in ogni sezione sono almeno 2 i problemi del tutto aderenti al programma standard.\n[*]Attenzione al fatto che i testi di alcuni problemi contengono evidenti errori di stampa (ad esempio l'N5), spesso evidenziati nelle discussioni. [/list]Spero vivamente che i partecipanti al Winter Camp avessero gi\u00e0 familiarit\u00e0 con quei problemi, essendo in vetrina su questo sito da almeno 3 mesi... :wink:", "Solution_2": "Faccio un elenco degli errori trovati fino adesso:\r\n- A1. Chiaramente $m,n$ devono essere positivi.\r\n- G3. (anche se non \u00e8 da fare). Dimostrare che vale quella formula se e soltanto se il triangolo \u00e8 acutangolo o rettangolo.\r\n- N5. Bisogna dimostrare che $\\phi(n) \\mid L(n)T(n)$\r\n- N6. Le frasi dovrebbero iniziare con \"a) $P(x),R(x)$are polynomials...\" e \"b) $P,R$ are polynomials...\"", "Solution_3": "Sgrunt! :bomb: E' chiaro che non verranno al winter Camp[list]\n[*]quelli che hanno spedito gli esercizi per raccomandata (nessuno ha voglia di fare ore di fila all'ufficio postale per ritirarla!);\n[*]quelli che hanno spedito file con gli esercizi senza il nome (prova tu a riconoscerli dopo averli stampati);\n[*]quelli che hanno spedito pi\u00f9 volte o in pi\u00f9 modi (vai a capire quale \u00e8 la versione buona);\n[*]quelli che hanno spedito file con scannerizzazioni illeggibili;\n[*]quelli che hanno chiamato il file in modo diverso da quello raccomandato (il loro file non compare quando cerco con quel criterio);\n[*]quelli che hanno spedito pi\u00f9 esercizi del necessario.[/list]", "Solution_4": "[quote=\"Xamog\"]Sgrunt! :bomb: E' chiaro che non verranno al winter Camp[list]\n[*]quelli che hanno spedito pi\u00f9 esercizi del necessario.[/list][/quote]\n\nGrazie!!\nE se vogliamo essere precisi, sull'e-mail che a me \u00e8 arrivata da Gobbino c'\u00e8 scritto:\n[quote]\nQuali sono gli esercizi da svolgere: 8 esercizi scelti tra quelli dell'Iran National Mathematical Olympiad (3rd round 2006), di cui 2 nel gruppo Algebra, 2 nel gruppo Combinatoria, 2 in Geometria e 2 in Teoria dei Numeri (con esclusione degli esercizi C3, G3 ed N1).\n[/quote]\r\nIo 8 esercizi li ho mandati. Non [b]solo[/b] 8 esercizi, purtroppo.\r\nSe poi ho aggiunto qualche altro esercizio dell'Iran, che mi piaceva / mi era venuto / mi andava di scrivere, non potreste semplicemente ignorarlo?\r\nEro completamente all'oscuro dell'ultima clausola che hai aggiunto!", "Solution_5": "[quote=\"edriv\"]non potreste semplicemente ignorarlo?[/quote]\r\nCercheremo un punizione adeguata. :lol:", "Solution_6": "Una correzione particolarmente larga di maniche :(, e una grossa pazienza di fronte ai disguidi di cui si \u00e8 parlato in un post precedete :wallbash_red: , ha portato alla seguente lista dei convocati per il WC:\r\n\r\n(S) CIMETTA Leone\r\n(S) COLOMBO Maria\r\n(S) CONTI Andrea\r\n(V) COSTA Simone\r\n(S) DALUISO Roberto\r\n(S) DE CAPUA Antonio\r\n(S) FOGARI Andrea\r\n(S) GALEOTTI Mattia\r\n(S) KUZMIN Kirill\r\n(S) LAZZARINO Leslie Lamberto\r\n(V) LEONETTI Paolo\r\n(S) LO BIANCO Federico\r\n(S) LIN Francesco\r\n(S) LOMBARDO Davide\r\n(V) MATTE BON Nicolas\r\n(S) NARDIN Denis\r\n(S) PATIMO Leonardo\r\n(S) SANTI Elia\r\n(S) TRAGER Matthew\r\n(S) VERTECHI Pietro" } { "Tag": [ "function", "college contests" ], "Problem": "Let be given f from C^1[0,1] and f(0) = 0, f(1) = 1, 0 <= f(x) <=1 for all x from [0, 1]. Show that there exist a, b from (0, 1), a <> b such that f'(a).f'(b) = 1.\r\n\r\nNamdung", "Solution_1": "Hmmm... Maybe I'm hasty, but this seems pretty easy: \r\n\r\nFrom Lagrange thm there is a c in (0,1) s.t. f'(c)=1. If f'(x)=1 for all x then we just choose a and b to be 2 different reals from (0,1). \r\n\r\nLet's assume, then, that f'(x) isn't constant. The set M={f'(x) | x in (0,1)} is an interval because the function f'(x) has Darboux' property. If f'(x)>=1 for all x then f(x)-x is an increasing function, and since f(0)-0=f(1)-1 we get f(x)-x=0 for all x, so f(x)=x, so f'(x)=1 for all x, and we've already discussed this. Something similar happens if f'(x)<=1 for all x, so if f'(x) isn't always 1 then there exist 2 numbers u and v in M s.t. u<10[/tex], does the Jensen's inequality for a convex function [tex]f(x)[/tex] become strict too?\r\n\r\nNo. (To see why, try $f(x)=x^2$)", "Solution_3": "What are necessary and sufficient conditions for equality in Jensen's inequality?", "Solution_4": "If the function $f$ is stricly convex then the equality possible $\\iff$ $x_1=...=x_n$.\r\nOtherwise we can find distinct $x_1,..,x_n$ s.t. the equality will hold.\r\n\r\nWhat do you mean when you saying \"the inequality becomes strict\"? We can always attain the eqaulity be putting $x_1=...=x_n$. :?" } { "Tag": [ "geometry", "rectangle", "circumcircle", "analytic geometry", "graphing lines", "slope", "conics" ], "Problem": "$ABCD$ is a rectangle. $A$, $B$ are on $y=2x+m$, $C$, $D$ are on $y^2=4x$. The equation of the circumcircle of $ABCD$ is $x^2+y^2-x-4y-t=0$. \r\n\r\nFind $m$ and $t$.", "Solution_1": "$AB: y=2x+m$\r\n$P: y^2=4x$\r\nThe equation of the circle can be written as\r\n$(x-\\frac{1}{4})^2 + (y-2)^2 = t+\\frac{17}{4}$\r\n\r\nSo, if $(K,R)$ is the circle, then $K=(\\frac{1}{2},2)$ and $R^2 = t+\\frac{17}{4}$\r\n\r\n$C(y_c,y_c)$\r\n$D(y_d,y_d)$\r\n\r\n$AB \\parallel CD$ so the line $CD$ has also slope $2$\r\nso we have $x_c \\neq x_d$ and $y_c \\neq y_d$ (we will use this fact later) \r\n\r\n$(y_c-y_d) = 2(x_c-x_d)$\r\n\r\n$C,D \\in P \\Rightarrow$\r\n\r\n$y^2_c=4x_c, \\ \\ \\ y^2_d=4x_d \\Rightarrow$\r\n\r\n$y^2_c -y^2_d = 4x_c - 4x_d$\r\n$(y_c+y_d)(y_c-y_d) = 4(x_c-x_d)$\r\n$(y_c+y_d)(y_c-y_d) = 2(y_c-y_d)$\r\n$y_c+y_d=2$\r\nbecause $y_c \\neq y_d$\r\n\r\nso, if $M(x_m,y_m)$ is the midpoint of $CD$ then $y_m = \\frac{y_c+y_d}{2} = 1$\r\n\r\n\r\nThe perpendicular bisector of the chord $CD$ is the line $KM$, and has a slope $-\\frac{1}{2}\\Rightarrow$\r\n\r\n$KM: y-2 = -\\frac{1}{2}(x-\\frac{1}{2})$\r\n\r\nApplying $y_m=1$ we get $1-2 = -\\frac{1}{2}(x_m-\\frac{1}{2}) \\Rightarrow$\r\n\r\n$x_m-\\frac{1}{2} = 2 \\Rightarrow x_m=\\frac{5}{2}$\r\n\r\nSo $M(\\frac{5}{2},1)$\r\n\r\nThe line $CD$ has slope $2$ and passes through $M$. So its equation is\r\n\r\n$CD: y-1 = 2(x-\\frac{5}{2}) \\Rightarrow y=1+2x-5 \\Rightarrow$\r\n\r\n$y=2x-4=2(x-2)$\r\n\r\n$C$ is on this line and also on the parabola\r\n\r\n$y_c = 2(x_c-2) \\Rightarrow y^2_c = 4(x_c-2)^2$\r\n\r\n$y^2_c = 4x_c \\Rightarrow$\r\n$4(x_c-2)^2 = 4x_c \\Rightarrow$\r\n$(x_c-2)^2 = x_c \\Rightarrow$\r\n$x^2_c -4x_c+4 = x_c \\Rightarrow$\r\n$x^2_c -5x_c+4 = x_c$ \r\n\r\nThis equation has the roots $4$ and $1$.\r\nThe one is for $C$ and the other for $D$\r\n\r\nLet $x_c=4 \\Rightarrow y_c = 2 \\cdot (4-2) = 4\\Rightarrow$\r\n$C(4,4)$\r\n\r\napplying this values to the initial equation of the circle $x^2+y^2-x-4y=t$ we get:\r\n\r\n$t = 4^2+4^2-4-16 \\Rightarrow t = 12$\r\n\r\nSince $AB \\perp BC$ so $AC$ is a diameter of the circle $\\Rightarrow \\overline{CA} = 2\\overline{CK}$\r\n\r\n$C(4,4)$\r\n${K(\\frac{1}{2},2})$\r\n\r\n$\\Rightarrow A(-3,0)$\r\n\r\n$A \\in y=2x+m \\Rightarrow$\r\n\r\n$0=2(-3)+m \\Rightarrow m=6$" } { "Tag": [ "calculus", "derivative", "factorial", "calculus computations" ], "Problem": "The question asks me to find a formula of $f^{(n)}(x)$\r\n\r\nI'm going to do a few here, then I will write the question of the one I do not understand and show my work of what I've done so far. Numbers 1 through 4, I've solved...it is number 5 I do not know.\r\n\r\n1. $f(x) = x^n$\r\n\r\n\r\nTo solve it, I look at the pattern by figuring out the first couple terms\r\n$f'(x) = nx^{n-1}$\r\n$f''(x) = (n)(n-1)x^{n-2}$\r\n\r\nSo $f^{(n)}(x)=(n!)x^{n-n}$\r\n$=n!$\r\n\r\n\r\n\r\n2. $f(x) = \\dfrac{1}{5x-1}$\r\n\r\n\r\n$f'(x) = -5(x-1)^{-2}$\r\n$f''(x) = 10(x-1)^{-3}$\r\n$f'''(x) = -150(x-1)^{-4}$\r\n\r\n\r\nso $f^{(n)}(x) = (-1)^n (5^{n-1})(n!)(5x-1)^{-(n+1)}$\r\n\r\nThere may be an easier to write this, but I know that this works....\r\n\r\n\r\n\r\n3. $f(x) = e^{2x}$\r\n\r\nThis one it's easy to see that $f^{(n)}(x) = 2^n e^{2x}$\r\n\r\nI wrote the first few out when I solved it, but I don't think it's needed here....\r\n\r\n\r\n\r\n\r\n\r\n4. $f(x) = \\frac1{3x^3}$\r\n\r\n$f'(x) = -x^{-4}$\r\n$f''(x) = 4x^{-5}$\r\n$f'''(x) = -20x^{-6}$\r\n\r\nI found this one to be\r\n$f^{(n)}(x) = (-1)^n \\dfrac{(n+2)!}{6}x^{-(n+3)}$\r\n\r\nThe sign alternates, that's why there is $(-1)^n$, the 6 in the denominator is caused by 3!...\r\nThere may be a cleaner way to write it, but again, I found this to work\r\n\r\n\r\n\r\n\r\n\r\nNow for the one I could not solve...\r\n5. $f(x) = \\sqrt{x}$\r\n\r\nIt looks like it should be so simple, and I am close, but I have not got it....\r\n\r\n$f'(x) = \\frac12x^{-\\frac12}$\r\n$f''(x) = -\\frac14x^{-\\frac32}$\r\n$f'''(x) = \\frac38x^{-\\frac52}$\r\n$f^{(4)}(x) = -\\dfrac{15}{16}x^{-\\frac72}$\r\n$f^{(5)}(x) = \\dfrac{105}{32}x^{-\\dfrac{11}{2}}$\r\n\r\nI have solved for everything, but what really throws me off is the numerator values for the coefficient of x.\r\nThe pattern is 1, 1, 3, 15, 105, 945\r\n\r\nI understand that we are multiplying by consecutive odd numbers, but I cannot figure out how to write the general form of that...\r\n\r\nSo far, I've solved for everything but that:\r\n\r\n$f^{(n)}(x) = (-1)^{n+1} (\\frac12)^n ??? x^{-\\dfrac{2n-1}{2}}$\r\n\r\nThe ??? represents the product of the first n-1 odd numbers....\r\nI feel that it is something really elementary that I am missing....and I feel kinda dumb for not knowing the pattern of the product of odd numbers...I've tried using factorials, but I can't seem to isolate the odd ones...\r\n\r\n\r\nAny help with this part is really appreciated....\r\n\r\n_ankur", "Solution_1": "[quote=\"ankur87\"]$f^{(n)}(x) = (-1)^{n+1} (\\frac12)^n ??? x^{-\\dfrac{2n-1}{2}}$\n[/quote]\r\n$(2n-1)!!$", "Solution_2": "I don't see how that works....\r\nCould you explain that to me....\r\n\r\nIf I stick in $n=2$...I get $(2(2)-1)!!$\r\n$=3!! = 6! = 720$\r\n\r\nWhere does 720 fit into the equation...I am not understanding....", "Solution_3": "should be something like\r\n\r\n$f^{(n)}(x) = \\dfrac{(-1)^{n-1}}{2^{2n-1}}\\dfrac{(2n-2)!}{(n-1)!}x^{-\\dfrac{2n-1}{2}}$", "Solution_4": "[quote=\"ankur87\"]I don't see how that works....\nCould you explain that to me....\n\nIf I stick in $n=2$...I get $(2(2)-1)!!$\n$=3!! = 6! = 720$\n\nWhere does 720 fit into the equation...I am not understanding....[/quote]\r\nYou misunderstand $n!!$. $n!! = n (n-2) (n-4) \\cdots$\r\n$3!! = 3\\cdot 1 = 3$\r\n$4!! = 4\\cdot 2 = 8$\r\n$5!! = 5\\cdot 3 = 15$", "Solution_5": "Thank you both for your help, but I still am not understanding 100% of the concept of $n!!$\r\n\r\nIs the definition $n!! = n (n-2) (n-4) \\cdots$?\r\n\r\nIf so, then I've never seen that before, but it definitely works\r\n\r\nThanks shyong, your answer works also, I am just figuring out the thought process behind it\r\n\r\nAgain, thanks for the help, I really appreciate it", "Solution_6": "Let $\\{x_n \\} = 1, \\ 1 \\cdot 3 \\cdot 5, \\ 1 \\cdot 3 \\cdot 5 \\cdot 7, \\ ... \\ 1 \\cdot 3 \\cdot 5 \\cdot ...(2n+1)$ where $n =1, \\ 2, \\ 3, \\ ...$.\r\n\r\nNote that $1 \\cdot 3 \\cdot 5 = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6}{2 \\cdot 4 \\cdot 6}$ $= \\frac{6!}{2^3 3!}$, and that in general, \\[ 1 \\cdot 3 \\cdot 5 \\cdot ... \\cdot (2n+1) = \\frac{(2n+2)!}{2^n n!} \\]", "Solution_7": "[quote=\"mathisfun1\"]Let $\\{x_n \\} = 1, \\ 1 \\cdot 3 \\cdot 5, \\ 1 \\cdot 3 \\cdot 5 \\cdot 7, \\ ... \\ 1 \\cdot 3 \\cdot 5 \\cdot ...(2n+1)$ where $n =1, \\ 2, \\ 3, \\ ...$.\n\nNote that $1 \\cdot 3 \\cdot 5 = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6}{2 \\cdot 4 \\cdot 6}$ $= \\frac{6!}{2^3 3!}$, and that in general, \\[ 1 \\cdot 3 \\cdot 5 \\cdot ... \\cdot (2n+1) = \\frac{(2n+2)!}{2^n n!} \\][/quote]\r\n\r\nAssuming that I want to do $9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1$\r\n\r\nThat would be $2n+1 = 9 \\Rightarrow n=4$\r\n\r\nI know that $9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1 = 945$\r\n\r\nPlugging $n=4$ into your equation, I get\r\n\r\n$\\frac{(2(4)+2)!}{2^4 4!}$\r\n\r\n$=\\frac{10!}{16 \\cdot 4!}$\r\n\r\n$=\\frac{10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5}{16}$\r\n\r\n$=\\frac{151200}{16}$\r\n\r\n$=9450$\r\n\r\nI think the formula is supposed to be $\\frac{(2n+1)!}{2^n n!}$\r\n\r\nThanks for the explanation, that really cleared things up...It really helped me understand shyong's answer as well\r\n\r\nI appreciate the help, everyone", "Solution_8": "I apologize for the error; $\\frac{(2n+2)!}{2^{n+1} (n+1)!}$ is the correct formula.", "Solution_9": "[quote=\"mathisfun1\"]I apologize for the error; $\\frac{(2n+2)!}{2^{n+1} (n+1)!}$ is the correct formula.[/quote]\r\n\r\nThanks for clearing that up...Soon after that post I realized what I wrote didnt work for other values of n....this one you have is working \r\nSo for all n, this formula will give you the product of odd numbers up to (2n+1)....\r\n\r\nThanks again" } { "Tag": [ "absolute value", "algebra unsolved", "algebra" ], "Problem": "Determine the sum of absolute values for the complex roots of $ 20 x^8 \\plus{} 7i x^7 \\minus{}7ix \\plus{} 20.$", "Solution_1": "$ 20 x^8 \\plus{} 7i x^7\\minus{}7ix \\plus{} 20 \\equal{} 20x^4(x^4\\plus{}1/x^4)\\plus{}7x^4(x^3\\minus{}1/x^3)$\r\nNow first substitute $ y\\equal{}ix$ and one gets\r\n$ 20y^4(y^4\\plus{}1/y^4)\\plus{}7y^4(y^3\\plus{}1/y^3)$\r\nNow a second substitution with $ z\\equal{}y\\plus{}\\frac{1}{y}$ leads to\r\n$ y^4(20z^4\\plus{}7z^3\\minus{}80z^2\\minus{}21z\\plus{}40)$\r\nNow we only have to consider the solutions to the equation\r\n$ (20z^4\\plus{}7z^3\\minus{}80z^2\\minus{}21z\\plus{}40)$\r\n\r\nBy putting z = -2, -1, 0, 1, 2 one sees that this equation has real solutions.\r\nNow looking at the relation between y = a + bi and z\r\n$ z\\equal{}a\\plus{}bi\\plus{}\\frac{1}{a\\plus{}bi} \\equal{} a(1\\plus{}\\frac{1}{a^2\\plus{}b^2})\\plus{}bi(1\\minus{}\\frac{1}{a^2\\plus{}b^2})$\r\n\r\nAs the solutions for z are real we know that\r\n$ 1\\minus{}\\frac{1}{a^2\\plus{}b^2}\\equal{}0 \\implies a^2\\plus{}b^2\\equal{}1$\r\nSo we now know the absolute value of every solution y of the original equation without even knowing the solutions, looks nice and hopefully correct to me.", "Solution_2": "[quote=\"Max(i)\"]$ 20 x^8 \\plus{} 7i x^7 \\minus{} 7ix \\plus{} 20 \\equal{} 20x^4(x^4 \\plus{} 1/x^4) \\plus{} 7x^4(x^3 \\minus{} 1/x^3)$ [/quote]\r\n\r\nI guess you miss an $ i$: $ 20 x^8 \\plus{} 7i x^7 \\minus{} 7ix \\plus{} 20 \\equal{} 20x^4(x^4 \\plus{} 1/x^4) \\plus{} 7i x^4(x^3 \\minus{} 1/x^3)$", "Solution_3": "The answer is 8.", "Solution_4": "Yes I missed an i and I should have written the final answer so thanks to both of you ;-)\r\nDo you have another solution for this problem? Would be nice to see another approach." } { "Tag": [], "Problem": "two friends want alex to help them mow lawns.cindy says that she gets 80 dollars for the lawns she wants alex to cut and will give him 60% of this if he works for her. Banto says that he gets 60 dollars for the lawns he wants alex to cut and will give alex 80% of this. Which friend should alex work for", "Solution_1": "[quote][Unparseable or potentially dangerous latex formula. \nError 6 ][/quote]\r\n\r\ndoes this have something to do with the problem?", "Solution_2": "[quote=\"math92\"]two friends want alex to help them mow lawns.cindy says that she gets 80 dollars for the lawns she wants alex to cut and will give him 60% of this if he works for her. Banto says that he gets 60 dollars for the lawns he wants alex to cut and will give alex 80% of this. Which friend should alex work for?[/quote]\r\n\r\n[hide]Either friend would give alex 48 dollars. [/hide]", "Solution_3": "i am really sorry. The dollar signs triggered latex.", "Solution_4": "I fixed it by quoting your post.", "Solution_5": "[hide]It doesn't matter. They both give the same. If alex wanted he could work for both and get EXTRA MONEY!! :D[/hide]", "Solution_6": "they both give the same, $48" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find $ f: A\\rightarrow A$ such that $ f(x^2\\minus{}y^2)\\plus{}f(2xy)\\equal{}f(x^2\\plus{}y^2)$, for $ A\\equal{}N,Z,Q,R$", "Solution_1": "$ f(x)\\equal{}x^2$ ??", "Solution_2": "For $ A \\equal{} \\mathbb{Z}$ at least, I don't think it's that simple: $ f(x) \\equal{} x^2$ for $ |x| \\neq 1$, $ f(1) \\equal{} f( \\minus{} 1) \\equal{} \\text{anything}$ seems to work.", "Solution_3": "I thought he was asking for an example of such a function :lol:" } { "Tag": [ "probability", "percent", "articles", "geometry", "geometric transformation", "reflection" ], "Problem": "I'm sure this there's 99% probability of this thread running into many controversies. For the sake of Pardesi (pardesi is studying so seriously, so lets not disturb him to come and delete posts and move threads), please dont use strong words. Let us discuss the verdict in general and its implications..\r\n\r\nThe Supreme Court is going to deliver its verdict on the OBC Quota Issue in premier institutions like IIT's, IIM's, etc. \r\nhttp://ibnlive.com/news/judgement-day-sc-to-deliver-obc-verdict-today/63020-3.html\r\n\r\n[hide]On a personal note, I hope they don't implement it this year.[/hide]", "Solution_1": "no da not studying :rotfl: :rotfl:", "Solution_2": "I hope they introduce quota for General category....", "Solution_3": "Breaking News::\r\n\r\nSC Decides to uphold 27% quota in premier institutions http://ibnlive.com/news/sc-upholds-obc-quota-in-educational-institutes/63029-3.html.\r\n\r\nMight even be implemented this year, but excluding the creamy layer. God knows how they will identify the creamy layer though. \r\n\r\n\r\n[hide]Maybe I should convert to OBC[/hide]", "Solution_4": "For all practical purposes,I checked the CNNIBN and it said something about those having annual income of above 2.5 lacs and children of government level 1 and 2 officers as being part of definition of creamy layer.This means even my friends who are bc,st,sc,obc etc wont be able to avail the quota!!(u need atleast an income of more than 2.5 lacs to survive decently in chennai!!!).Well actually this reservation should come as no suprise seeing that the elections are drawing near.... :rotfl:\r\n\r\nIam only unhappy that the judgement has come before april 13 seeing that it is now going to have psycholigical impact on those writing JEE this year:( \r\nHopefully something can be done", "Solution_5": "well actually it hardly matters because now the not so useful IIT's get termed as useless IIT's :P :rotfl: :rotfl:", "Solution_6": "i feel that now it is the general category who really needs reservation in these institutions do you all know one thing in AIIMS only 32 seats are available for forward caste all others go the OBC but still our school PS Senior was able to get one student in Anand venkatraman who got a whooping AIR3 in it!!!! :D :o", "Solution_7": "[quote=\"Ram\"]\n\nIam only unhappy that the judgement has come before april 13 seeing that it is now going to have psycholigical impact on those writing JEE this year:( \nHopefully something can be done[/quote]\r\n\r\nDefinitely there will be a psychological impact. :( I wish the verdict had come after JEE. But I doubt any of us should be worried....Lets just do our best. But the cutoffs will increase for sure.... \r\n\r\nI think 2.5 lacs/annum implies that you can afford good education for your child (school level that is). The quota's purpose was to reach the poor penniless backward people. \r\n\r\nBut the big question is: Will it be implemented fully this year or like a 3tier system of 9+9+9.5 percent per year.. And Mr.A.Singh has promised to increase number of seats which means IIT's std may be diluted. because only what is rare, is precious.", "Solution_8": "obc reservations wont matter to genius ones\r\nwho r assured of a top rank\r\nit will matter to all those who r at the frontiers of qualifying and not qualifying\r\nwho basically aim for qualifying not for a rank :( \r\nso i get to lose my seat if i do some crazy mistakes :o", "Solution_9": "[quote=\"phoenix76\"]obc reservations wont matter to genius ones\nwho r assured of a top rank\nit will matter to all those who r at the frontiers of qualifying and not qualifying\nwho basically aim for qualifying not for a rank :( \nso i get to lose my seat if i do some crazy mistakes :o[/quote]\r\n\r\nYour last two lines I guess, refer to the likes of me. :( Anyway, the reservation will affect your course. Its gonna be really difficult to get into the popular groups like CSC or EEE. If there are 40 seats in IIT-X , then only 20 will be available to the general public.", "Solution_10": "the bottom 2 lines were meant for me\r\n\r\n :(\r\nif u dont qualify i am sure i wouldnt either\r\ni just aim to get into an iit\r\ni m not after a branch :) \r\nits for people like shreyas,chapplis,soumyas,pardesis,madness,rakeshs and varuns :(", "Solution_11": "lets not get into who is on the frontier line, and who is not shall we??\r\ni feel the people who come here, should be able to clear the exam with ease, but thats not the problem. the minimum rank to not compromise on both college and stream, will go up due the reservation (that is u will get a stream not of ur choice, in an IIT not of ur choice, if u r below that rank). and that should be our main concern, but not now, only after the exam!!\r\nso all the best for JEE 2008!!", "Solution_12": "[quote=\"phoenix76\"]the bottom 2 lines were meant for me\n\n :(\nif u dont qualify i am sure i wouldnt either\ni just aim to get into an iit\ni m not after a branch :) \nits for people like shreyas,chapplis,soumyas,pardesis,madness,rakeshs and varuns :([/quote]\r\n\r\nSo you won't even mind Paper Pulp Engineering or Naval Engineering or Cotton Textile Eng?? Nothing wrong in taking those groups but still.....", "Solution_13": "no i wont mind at all\r\ni wont mind going to iit guwahati also", "Solution_14": "[quote=\"phoenix76\"]no i wont mind at all\ni wont mind going to iit guwahati also[/quote]\r\nyou are a true iitian!!!!! :thumbup: :lol:", "Solution_15": "i m pretty realistic\r\ni know where i stand\r\nso i dont expect things which r beyond my reach :)", "Solution_16": "Theres also a proposal for incr no of IITs :mad: \r\nIncreasing no. of IITs will definitely bring down d standards, :mad: \r\nfew weeks bac an old IIT M dean had blasted off d govt for proposing dis in his article in economic times.\r\n[hide]he said theres a craze into politics so y dont u just incr d size of the parliament \nso dt people like soumya get 2 bcom MPS :rotfl: [/hide]", "Solution_17": "[quote=\"phoenix76\"]i m pretty realistic\ni know where i stand\nso i dont expect things which r beyond my reach :)[/quote]\r\nbut i think u underestimate yourself too low u have got great marks in fiitjee compared to people here and why do you need to worry about rank!!!! :)", "Solution_18": "*long gurgling sound coming from the bathroom flush*\r\n\r\n\"What was that?! The toilet??\"\r\n\r\n\"No, not really. It's just the Indian Education system going down the drain\"\r\n\r\n :( :mad:", "Solution_19": "[quote=\"hash_include\"]*long gurgling sound coming from the bathroom flush*\n\n\"What was that?! The toilet??\"\n\n\"No, not really. It's just the Indian Education system going down the drain\"\n\n :( :mad:[/quote]\r\n\r\na perfect begining for the wrong ending!!!! god knows where this gonna doom to!!!! :mad: :o", "Solution_20": "Just Concentrate and do ur best...... If u get a top 100 hundred rank ,, I think nobody can stop u .................", "Solution_21": "quotas doesn't matters 4 AOPS at all :D :roll: :winner_first:", "Solution_22": "[quote=\"rituraj007\"]quotas doesn't matters 4 AOPS at all :D :roll: :winner_first:[/quote]\r\n\r\nquotas matters,even for big guns\r\nlast year closing rank for iit mumbai computers was 47(general)\r\ndon't you think it matters\r\neven in medicals (AIIMS)there are very few seats for general\r\nmost probably ,creamy layer is going to be revised (income level is going to be increased,as can be deduced from reactions of politicians,politics is the only justification of quotas)\r\nreservations should be only on the basis of income, not on caste,it segregates society\r\nafter all, the standard of iit is going to fall due to it\r\nwhat is method of implementation (gradually increasing the quota or single slaughter)?\r\nit is a question for most others present here,till my time full fledged quota would have been be implemented :( :mad:", "Solution_23": "For all you know, the next thing to happen is a new qouta-general category quota-5% reservation", "Solution_24": "This year, we will probably have a new IIT in Andhra...\r\n\r\n8 more IITs coming up and one of them in Orissa....\r\n\r\nPeople will no longer go to private engineering colleges ..Everybody will be an IITian....\r\n\r\nA good method to achieve Communist equality...\r\n\r\nDown with Communism.... :rotfl: :rotfl:", "Solution_25": "[quote=\"SOUMYASHANT NAYAK\"]This year, we will probably have a new IIT in Andhra...\n\n8 more IITs coming up and one of them in Orissa....\n\nPeople will no longer go to private engineering colleges ..Everybody will be an IITian....\n\nA good method to achieve Communist equality...\n\nDown with Communism.... :rotfl: :rotfl:[/quote]\r\n\r\nOfcourse, thats what the government wants. You see IIT'ians contribute a lot to the country's economy. So now more IIT'ians means more money.!! Nice logic eh? :rotfl: :rotfl: \r\n\r\nHail Equality, Hail Socialism, Hail Communism.", "Solution_26": "and ofc, the unsaid words are \" Hail madness \" :D \r\nPersonally, i tried to refrain from commenting from this topic as i didnt want to swear :D", "Solution_27": "There is huge caste-based discrimination in India, there's no denying that. It's just that reservation may not be the best solution to solving the problem that is social inequality. But real solutions like major improvement in the primary education seem very difficult and impractical. So I don't see a better way than reservation to make some progress in the direction of social reform. You cannot deny that there has been tremendous growth over the last fifty years due to the implementation of these policies. \r\nAnyway, those are just my two cents. I guess as a student aspiring for the IIT reservation does make a difference. But - look at it this way - if the student who got through in your place was in your socio-economic position and you were in his/hers would he/she have made the most out of it? More than what you have? That's a hard question to answer. But the fact is that not everybody has the good fortune of being born into a good socio-economic situation. It's those who have done exceptionally well considering their socio-economic context who deserve to get a good education despite having average scores when compared to the competition. \r\nAnyway, that being said - I sincerely hope none of you on AoPS will fall prey to that system, but remember that it's not all that senseless. \r\nAshwath", "Solution_28": "it never is senseless but there remains this fact:\r\nPeople who get formal iit training shouldnt get quota benefits.\r\nIf people who dont have formal education get quota benefits, most of them join the iitss and fail there and in some cases commit suicide, so is the govt sending them their flight ticket to hell? :huh:", "Solution_29": "[quote=\"ashwath.rabindranath\"]There is huge caste-based discrimination in India, there's no denying that. It's just that reservation may not be the best solution to solving the problem that is social inequality. But real solutions like major improvement in the primary education seem very difficult and impractical. So I don't see a better way than reservation to make some progress in the direction of social reform. [b]You cannot deny that there has been tremendous growth over the last fifty years due to the implementation of these policies. [/b]\nAnyway, those are just my two cents. I guess as a student aspiring for the IIT reservation does make a difference. But - look at it this way - if the student who got through in your place was in your socio-economic position and you were in his/hers would he/she have made the most out of it? More than what you have? That's a hard question to answer. But the fact is that not everybody has the good fortune of being born into a good socio-economic situation. It's those who have done exceptionally well considering their socio-economic context who deserve to get a good education despite having average scores when compared to the competition. \nAnyway, that being said - I sincerely hope none of you on AoPS will fall prey to that system, but remember that it's not all that senseless. \nAshwath[/quote]\r\n\r\nwell written, but i must say that i tend to disagree with you.\r\ndo you mean to say ppl belonging to the \"Backward classes\" have not at all done well till now?\r\nOr do you mean to say that all of those who [b]Have[/b] achieved and who do belong to the aforesaid class are all untalented and depend solely on reservations.\r\nI am sure you dont.\r\nFrankly i think in such competitive exams, if a student wants to get through, He will come what may. This kind of compulsory reservation sometimes tends to bring in more people into iit's in the beginning, though a number of them to survive, there are many others who tend to drop out.This is not what i am saying myself but one of my professors in mathematics who has taught in iit bombay for 40 years and has himself been the part of the jee paper setting comitte severally many times.\r\nNow to the lines i have highlighted:\r\n\r\nJust because some development has been made do we become complacent and convince ourselves that yes Reservations are doing some work and we donot have to look for an alternative.\r\nI think you are speaking ignorant of the fact that such decisions regarding quota etc. by whomsoever is indeed to have a good votebank in the following elections\r\nNo more, no less.\r\nSo We must all remember that reservation is certainly not a wise tool to bring the country forward.\r\nNow a little message for others:\r\nPlease do not let such issues perturb you, just do your best in the exams and believe in yourselves. :D", "Solution_30": "Well written rags! I have only one thing to add. I personally feel that our leaders must concentrate more on primary education, build better schools in villages and offer higher salaries to entice teachers to teach in Villagers. Those in the city belonging to middle class can afford decent education, but without primary education, many are suffering. That, according to me is the chief issue which must be addressed by the great leaders if they truly care about social development, but if political gimmicks is what people want, thats what they'll get. :(", "Solution_31": "[quote=\"Euclidean Geometer\"]\n\nwell written, but i must say that i tend to disagree with you.\ndo you mean to say ppl belonging to the \"Backward classes\" have not at all done well till now?\nOr do you mean to say that all of those who [b]Have[/b] achieved and who do belong to the aforesaid class are all untalented and depend solely on reservations.\nI am sure you dont.\n[b]Frankly i think in such competitive exams, if a student wants to get through, He will come what may.[/b] This kind of compulsory reservation sometimes tends to bring in more people into iit's in the beginning, though a number of them to survive, there are many others who tend to drop out.This is not what i am saying myself but one of my professors in mathematics who has taught in iit bombay for 40 years and has himself been the part of the jee paper setting comitte severally many times.\nNow to the lines i have highlighted:\n\nJust because some development has been made do we become complacent and convince ourselves that yes Reservations are doing some work and we donot have to look for an alternative.\nI think you are speaking ignorant of the fact that such decisions regarding quota etc. by whomsoever is indeed to have a good votebank in the following elections\nNo more, no less.\nSo We must all remember that reservation is certainly not a wise tool to bring the country forward.\nNow a little message for others:\nPlease do not let such issues perturb you, just do your best in the exams and believe in yourselves. :D[/quote]\r\n\r\nIf the politicians want to use this to enhance the votebanks - by all means let them do so. As long as the reservations can do some good - I couldn't care less. \r\nAnd yes - i think that without primary education reservation can only serve a very temporary purpose.\r\nYour highlighted comment reflects your lack of knowledge of how bad circumstances can get. We certainly should try to look for an alternative - primary education being a very important one among them, but I think that equal opportunity for all is vital. Anyway, let's not have ideological debates here - that sort of thing can be reserved for a separate topic.", "Solution_32": "[quote=\"ashwath.rabindranath\"]\n\nIf the politicians want to use this to enhance the votebanks - by all means let them do so. As long as the reservations can do some good - I couldn't care less. \nAnd yes - i think that without primary education reservation can only serve a very temporary purpose.\nYour highlighted comment reflects your lack of knowledge of how bad circumstances can get. We certainly should try to look for an alternative - primary education being a very important one among them, but I think that equal opportunity for all is vital. Anyway, let's not have ideological debates here - that sort of thing can be reserved for a separate topic.[/quote]\r\n\r\nIf you had read some of our replies you might have realized that none of us here are against Social Equality. Do you really feel reservation will do good to more than 10 people? Obviously, the creamy layer will fail and again it will reach only those who can afford a good seat in a good college. When have your schemes ever reached the poor?\r\n\r\nWell, I think discrimination today is only based on money: The haves and the havenots. There are so many millions who can't afford 2 square meals a day. Why not provide reservation to them? Irrespective of whether they belong to GE or SC or OBC. And many people misuse OBC Quota. By law you are no longer an OBC if you have converted to other religions but I personally know contradictions.", "Solution_33": "Oh - when I mean reservation I mean socio-economic reservation. There's no question about that.", "Solution_34": "[quote=\"Chronoz\"][quote=\"ashwath.rabindranath\"]\n\nIf the politicians want to use this to enhance the votebanks - by all means let them do so. As long as the reservations can do some good - I couldn't care less. \nAnd yes - i think that without primary education reservation can only serve a very temporary purpose.\nYour highlighted comment reflects your lack of knowledge of how bad circumstances can get. We certainly should try to look for an alternative - primary education being a very important one among them, but I think that equal opportunity for all is vital. Anyway, let's not have ideological debates here - that sort of thing can be reserved for a separate topic.[/quote]\n\nIf you had read some of our replies you might have realized that none of us here are against Social Equality. Do you really feel reservation will do good to more than 10 people? Obviously, the creamy layer will fail and again it will reach only those who can afford a good seat in a good college. When have your schemes ever reached the poor?\n\nWell, I think discrimination today is only based on money: The haves and the havenots. There are so many millions who can't afford 2 square meals a day. Why not provide reservation to them? Irrespective of whether they belong to GE or SC or OBC. And many people misuse OBC Quota. By law you are no longer an OBC if you have converted to other religions but I personally know contradictions.[/quote]\r\n\r\nShre is absolutely right. I ve forgotten to mention that.\r\nThese reservations are ultimately misused by the ones who belong to this class but can afford to get themselves a \"costlier\" seat. it really fails to help the people who really need it.\r\nby providing more and more of reservations, i think we are indirectly saying that they are hopeless and can not do without it.\r\nI have quite a few friends belonging to the \"OBC\" but feel sternly that they are no less than others and needn't reservation beyond a limit.\r\nAnother important thing to be remembered is that quality must not be compromised in lieu of the bulk or quantity of ppl that get educated.\r\nPrimary education must be strong. it must provide every1 the mental courage that they donot need any reservation.\r\n@ ashwath: you said its impractical improving primary education, easier a solution is reservation.\r\nThis i certainly fail to understand.\r\n[hide] a day is surely to come when the so called \"Forward class\" or \"general category\" is gonna start desperately needing reservation, or atleast their fair share [/hide]\r\nNo offence is meant to any1.", "Solution_35": "By the way, the IIT's have decided to implement the reservation policy in phases http://sify.com/news/fullstory.php?id=14650981", "Solution_36": "I matters nothing for me\r\ntill 2010 full reservations would be implemented(i would be among first batch of students giving iit in full 50%reservation)\r\n\r\nbut their is one point i'd ie to make\r\nto give reservations to economically poor is completely right,but is it right to give reservations to well off students(most of reservation beneficials would be among them)\r\ni feel benefit of quota should be given only once to a person and his family not again and again\r\nthen only people who really deserve to get quotas would get them and their condition will improve, but most importantly schooling should be improved as many students who went to iit through quotas failed repeatedly and finally had to leave iit.They have been sufferers and they were not benefitted from quotas,in fact they would have better taken admission in NIT and got their degrees.", "Solution_37": "hey the system is getting down \r\nsee social equality is ok.........\r\nbut then the cream layer gets redefined and u know this creamy layer affects the education of india.it might not matter to ghenuiseslike u all but it does to us who barely hav a chance a get into iits \r\n\r\nthink bout it guys and give me ur further thoughts\r\n :lol:", "Solution_38": "i also think along the same lines as you\r\n\r\nbut it would be better if we don't take this discussion ahead\r\nit may hurt others(now hardly any1 comes here)\r\n\r\nif you have an examination with a question like'Who has done maximum harm to indian education??????'\r\nthe obvious answer would be ARJUN SINGH\r\nin all ways,depreciation in standard of these institutions,creating havoc for students,not giving good students their legitimate share,etc\r\n\r\n :wallbash_red: :diablo: :furious: :starwars:", "Solution_39": "hey skand i ought to be in the same thinking perspective as u \r\nthat s cool bro!\r\n\r\n\"" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "fifty-one small insects are placed inside a square of side 1.Prove that at any moment there are at least three insects which can be covered by a single circular disk of radius 1/7", "Solution_1": "this is very well-known..\r\n\r\n[hide]divide the square into 25 squares, each of size $ \\frac15\\times\\frac15$. by php, at least three insects is contained inside one of the unit squares. but the unit square can be covered by a disk of radius 1/7. q.e.d.[/hide]" } { "Tag": [ "Vieta", "function", "algebra", "polynomial" ], "Problem": "Could anyone please help solving the following system of non-linear equations with 2 variables?..\r\n(I've tried something with [i]Newton's method[/i], but I'm not sure of it. Any other ways of solving it, please?)\r\n[b]\n$ x^{2} \\plus{} y^{2} \\equal{} 41$\n$ x^{3} \\plus{} y^{3} \\equal{} 189$\n[/b]\r\n(now, it may be obvious that a possible solution is $ (x,y) \\equal{} (4,5) \\equal{} (5,4)$, but how do I [i]solve[/i] it?)", "Solution_1": "To prove that is a solution just substitute it back in\r\n\r\n$ 4^{2} \\plus{} 5^{2} \\equal{} 41$\r\n$ 4^{3} \\plus{} 5^{3} \\equal{} 189$\r\n\r\nSo $ (4, 5) , (5, 4)$ satisfy the system.\r\n\r\nAs far as solving I'd say brute force finds the solution the quickest. \r\n\r\nThe most common method I've seen for these is to find the largest and smallest possible $ x$ or $ y$ values.\r\n\r\nAssuming $ x, y \\in \\mathbb{N}$ then, $ x^{2} \\plus{} y^{2} \\equal{} 41 \\rightarrow 0 < x, y < 7$\r\nand from the second $ x^{3} \\plus{} y^{3} \\equal{} 189 \\rightarrow 0 < x, y < 6$. This helps you know what numbers to try and brute force with which may make the process quicker. Sorry if my wording is confusing.", "Solution_2": "[hide=\"I want you to fill in the remainder for practice (method that gives all 3 pairs of solutions)\"]\nLet $ p \\equal{} x \\plus{} y$ and $ q \\equal{} xy$\n\n$ p^2 \\minus{} 2q \\equal{} x^2 \\plus{} y^2 \\equal{} 41$\n\n$ p^3 \\minus{} 3pq \\equal{} x^3 \\plus{} y^3 \\equal{} 189$\n\n$ p^3 \\minus{} 3pq \\equal{} p(p^2 \\minus{} 2q) \\minus{} pq \\equal{} 41p \\minus{} pq \\equal{} p(41 \\minus{} q) \\equal{} 189$\n\nWe know that p is not equal to 0 (since then $ p^3 \\minus{} 3pq \\equal{} 0$ ) so we have\n\n$ q \\equal{} 41 \\minus{} \\frac {189}{p}$\n\nPlugging back into the first eq. we get\n\n$ p^2 \\minus{} 2(41 \\minus{} \\frac {189}{p}) \\equal{} 41$\n\n$ p^2 \\plus{} 2\\frac {189}{p} \\equal{} 123$\n\n$ p^3 \\minus{} 123p \\plus{} 378 \\equal{} 0$\n\nguess the root p=9, (honestly no other way other than cardano cubic formula) and we have:\n\n$ (p \\minus{} 9)(p^2 \\plus{} 9p \\minus{} 42)$\n\nWhich has the other 2 roots $ \\frac { \\minus{} 9\\pm\\sqrt {81 \\plus{} 4*42}}{2} \\equal{} \\frac { \\minus{} 9\\pm\\sqrt {249}}{2}$\n\nNow, once you have p, you find q with one of the first 2 eq. (just plug in).\n\nThen vieta's formula give 3 pairs of solutions (6 if you count them permuted), 2 extraodinarily ugly, one being (4,5)\n\nP.S. when you get p=9, try that first, as it will probably yield easy solutions[/hide]", "Solution_3": "Brute force doesn't work if you remove the restriction that $ x, y$ are integers. rofler has given the general method for solving symmetric systems (where $ x, y$ can be interchanged to give the same system); the change of variables he uses is standard. \r\n\r\nThe most general method is to solve for one variable in terms of the other and plug in, but that's hardly ever nice. Usually nonlinear systems involve a variety of algebraic tricks such as other kinds of changes of variables.", "Solution_4": "thank you all for your time and effort, especially [b]rofler[/b]. I really appreciate your immediate response :) \r\n\r\nI understand the method [i]rofler[/i] has written, but I do have a question though:\r\n[quote=\"rofler\"]\n$ p^3 \\minus{} 123p \\plus{} 378 \\equal{} 0$\n\nguess the root p=9, (honestly no other way other than cardano cubic formula) and we have:\n\n$ (p \\minus{} 9)(p^2 \\plus{} 9p \\minus{} 42)$\n[/quote]\r\nIndeed, [b]p=9[/b] is a root. But [i]how[/i] did you extract it?\r\nAlso, do you apply [i]Cardano cubic formula[/i], that you mentioned?\r\nI searched at Wikipedia [url]http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method[/url] about [i]Cardano's method[/i] of extracting the roots of a cubic function, but unfortunately I can't figure it out whether you apply [i]Cardano's method[/i] or not.", "Solution_5": "If you have a nasty wasty polynomial such as we have, it's good to guess roots. $ 378\\equal{}2\\cdot 3^3\\cdot 7$. We guess 2, 3, 6, and 7, but those don't work. We guess 9 and it works. Yay.", "Solution_6": "We can find that root using the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]rational root theorem[/url].", "Solution_7": "[quote=\"rofler\"][hide=\"I want you to fill in the remainder for practice (method that gives all 3 pairs of solutions)\"]\nLet $ p \\equal{} x \\plus{} y$ and $ q \\equal{} xy$\n\n$ p^2 \\minus{} 2q \\equal{} x^2 \\plus{} y^2 \\equal{} 41$\n\n$ p^3 \\minus{} 3pq \\equal{} x^3 \\plus{} y^3 \\equal{} 189$\n\n$ p^3 \\minus{} 3pq \\equal{} p(p^2 \\minus{} 2q) \\minus{} pq \\equal{} 41p \\minus{} pq \\equal{} p(41 \\minus{} q) \\equal{} 189$\n\nWe know that p is not equal to 0 (since then $ p^3 \\minus{} 3pq \\equal{} 0$ ) so we have\n\n$ q \\equal{} 41 \\minus{} \\frac {189}{p}$\n\nPlugging back into the first eq. we get\n\n$ p^2 \\minus{} 2(41 \\minus{} \\frac {189}{p}) \\equal{} 41$\n\n$ p^2 \\plus{} 2\\frac {189}{p} \\equal{} 123$\n\n$ p^3 \\minus{} 123p \\plus{} 378 \\equal{} 0$\n\nguess the root p=9, (honestly no other way other than cardano cubic formula) and we have:\n\n$ (p \\minus{} 9)(p^2 \\plus{} 9p \\minus{} 42)$\n\nWhich has the other 2 roots $ \\frac { \\minus{} 9\\pm\\sqrt {81 \\plus{} 4*42}}{2} \\equal{} \\frac { \\minus{} 9\\pm\\sqrt {249}}{2}$\n\nNow, once you have p, you find q with one of the first 2 eq. (just plug in).\n\nThen vieta's formula give 3 pairs of solutions (6 if you count them permuted), 2 extraodinarily ugly, one being (4,5)\n\nP.S. when you get p=9, try that first, as it will probably yield easy solutions[/hide][/quote]\r\n\r\nyes , this method works effectively for the problem which is recognized as when you switch x and y by each other , the system remain the same :)\r\n\r\nand by the way , i often use computer or calculation to find root of cubic :P" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "show\r\n\r\n$\\int_{0}^{1}\\int_{0}^{1}\\;\\frac{y}{1-x^{3}y^{3}}\\;dx\\;dy\\;=\\;\\frac{\\pi}{3\\sqrt{3}}$", "Solution_1": "$\\int_{0}^{1}\\int_{0}^{1}\\frac{y}{1-x^{3}y^{3}}=\\left|xy=u,y=v\\right|=\\int_{0}^{1}\\int_{0}^{1}\\frac{v}{1-u^{3}}\\frac{1}{v}\\, dudv = \\int_{0}^{1}\\frac{1}{1-u^{3}}du = \\left|t=1-u^{3}\\right| = \\frac{1}{3}\\int_{0}^{1}\\frac{dt}{(1-t)^{2/3}t}= ?$ but the last integral is not exist !!! :(", "Solution_2": "How about expanding the integrand as a geometric sum?\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112388]Similar problem[/url]", "Solution_3": "$ \\int_{0}^{1}\\int_{0}^{1}\\sum_{k\\equal{}0}^{\\infty}x^{3k}y^{3k\\plus{}1}dx\\, dy\\equal{}\\sum_{k\\equal{}0}^{\\infty}\\frac{1}{3k\\plus{}1}\\frac{1}{3k\\plus{}2}\\equal{}\\sum_{k\\equal{}0}^{\\infty}\\left(\\frac{1}{3k\\plus{}1}\\minus{}\\frac{1}{3k\\plus{}2}\\right)$ $ \\equal{}\\frac{\\pi}{3}\\cot\\frac{\\pi}{3}$", "Solution_4": "[quote=\"misan\"]show\n\n$ \\int_{0}^{1}\\int_{0}^{1}\\;\\frac {y}{1 - x^{3}y^{3}}\\;dx\\;dy\\; = \\;\\frac {\\pi}{3\\sqrt {3}}$[/quote]\n$ \\int_0^1 {\\frac{1}{{1 + u + u^2 }}\\,du} = \\left. {\\frac{2}{{\\sqrt 3 }}\\arctan \\frac{{2u + 1}}{{\\sqrt 3 }}} \\right|_0^1 = \\frac{\\pi }{{3\\sqrt 3 }}.$\n\nBack to the problem:\n\nDefine $ u=xy$ and reverse the integration order\n\n\\begin{eqnarray*}\\int_0^1 {\\int_0^1 {\\frac{y}{{1 - x^3 y^3 }}\\,dx}\\,dy} &=& \\int_0^1 {\\int_0^y {\\frac{1}{{1 - u^3 }}\\,du} \\,dy}\\\\&=& \\int_0^1 {\\int_u^1 {\\frac{1}{{1 - u^3 }}\\,dy} \\,du}\\\\&=& \\int_0^1 {\\frac{1}{{1 + u + u^2 }}\\,du}\\\\&=& \\frac{\\pi }{{3\\sqrt 3 }}\\,\\blacksquare\\end{eqnarray*}\n\n[quote=\"misan\"]toughie...[/quote]\r\n\u00bf?" } { "Tag": [ "function", "complex analysis", "real analysis", "real analysis unsolved" ], "Problem": "This is a very difficult result, but knowing the abilities of some members of the forum, I decided to post it here, hoping someone will come up with a relatively easy solution: \r\n Find all subseries of the series $ \\sum_{n\\geq 1}{\\frac{x^n}{n!}}$ that are bounded for $ x<0$. That is, for which subsets $ A$ of the positive integers is the function $ \\sum_{n\\in A}{\\frac{x^n}{n!}}$ bounded for $ x<0$?", "Solution_1": "Is the answer $ A\\equal{}\\mathbb N$ or $ A\\equal{}\\{4n\\plus{}k,4n\\plus{}k\\plus{}1\\}_{n\\ge 0}$, $ k\\equal{}0,1,2,3$? If so, I'll try to make my half-baked ideas into a rigorous proof; if not, I'll have to think a bit more. My approach is based on Borel transform and Fabry theorem, so it is complex analysis rather than real. Does the known solution follow the same lines?", "Solution_2": "Hi, Fedja, this is the answer! To my shame, I do not know what Fabry theorem is, but what they use is Laplace transform and a theorem of Szego posted by Vess some time ago and which remained unanswered, of course (I have a solution, but it's far too technical to write it on mathlinks): it's about analytic continuation of power series having only a finite number of distinct coefficients." } { "Tag": [ "vector", "Functional Analysis", "linear algebra" ], "Problem": "Let L and K be the liniear subspaces of $R^n$ Euclidean space and dimL \\dim L$ hence there is a basis vector in $K$ that it is not \r\n\r\nin $L$ and this will be orthogonal to any vector in $L$ .", "Solution_4": "Oh,thanx a lot.\r\nIt was so easy and i didn't solve it :blush:", "Solution_5": "Well actually I have a question about the second line of that proof.\r\n\r\n[quote]Then some of the basis vectors are basis of $K$ and some basis of $L$.[/quote] \r\n\r\nIf you start by picking an arbitrary orthogonal basis of $\\mathbb{R}^n$, then it may be true that none of those basis vectors are in $L$. That's a possibility. So I'm not sure what it means to say that \"some of the basis vectors are basis of $K$ and some basis of $L$.\"\r\n\r\nHere's another idea.\r\n\r\nConsider the case where $L$ is a one-dimensional subspace of $\\mathbb{R}^3$ and $K$ is a two-dimensional subspace of $\\mathbb{R}^3$. Consider $L^{\\bot}$, the subspace of $\\mathbb{R}^3$ that is perpendicular to $L$. $L^{\\bot}$ has dimension 2. Since $L^{\\bot}$ and $K$ are both two-dimensional subspaces of $\\mathbb{R}^3$, they must have a non-trivial intersection. This implies that there is a non-zero vector in $K$ which is orthogonal to every vector in $L$.\r\n\r\nI believe this idea can be used to make a good proof for the general case.\r\n\r\nTurns out I'm not using Gram-Schmidt after all.", "Solution_6": "I think that one needs use Gram-Schmidt. But the idea is basically the same.\r\n\r\n\r\n[b]Problem.[/b] [color=blue]Let $L$ and $K$ be subspaces of $\\mathbb R^{n}$ such that $\\textrm{dim}\\, L < \\textrm{dim}\\, K$.\n\nProve that there is a non-zero vector $v$ in $K$ that is orthogonal to any vector from $L$.[/color]\r\n\r\n\r\n[i]Solution.[/i]\r\n\r\nLet $\\ell = \\textrm{dim}\\, L$ and $k = \\textrm{dim}\\, K$. By Gram-Schmidt construct a basis $\\left\\{ e_{1},\\ldots,e_\\ell \\right\\}$ for $L$. Next, extend it to a basis $\\left\\{ e_{1},\\ldots,e_\\ell,e_{\\ell+1},\\ldots,e_{n}\\right\\}$ for $\\mathbb R^{n}$. It's clear that $\\left\\{ e_{\\ell+1},\\ldots,e_{n}\\right\\}$ span the subspace orthogonal to $L$. Let this subspace be $L^\\prime$.\r\n\r\nThen,\r\n\\begin{eqnarray*}\\textrm{dim}\\, \\left( L^\\prime \\cap K \\right) &=& \\textrm{dim}\\, \\left( L^\\prime \\right)+\\textrm{dim}\\, \\left( K \\right)-\\textrm{dim}\\, \\left( L^\\prime+K \\right) \\\\ \\ &=& \\left( n-\\ell \\right)+k-\\textrm{dim}\\, \\left( L^\\prime+K \\right) \\\\ \\ &=& \\left( n-\\textrm{dim}\\, \\left( L^\\prime+K \\right) \\right)+\\left( k-\\ell \\right) \\\\ \\ &\\geqslant& 0+\\left( k-\\ell \\right) > 0. \\end{eqnarray*}\r\n\r\nThus, $L^\\prime \\cap K$ has dimension $\\geqslant 1$, and the conclusion follows. $\\; \\; \\; \\; \\; \\blacksquare$", "Solution_7": "The statement is also true in infinite dimensions: If $L$ and $K$ are closed subspaces of a separable Hilbert space and $\\dim L<\\dim K$ then there exists $u\\in K$ with $u\\in L^\\bot$." } { "Tag": [ "parameterization", "parametric equation" ], "Problem": "For example, how do you convert 2x^2+3y^2=5 to parametric mode", "Solution_1": "x=r cos(theta)\r\ny=r sin (theta)\r\nx squared plus y squared = r squared\r\ntangent(theta) = y/x\r\nthat should be everything you need to know to solve it", "Solution_2": "what you just said was about how Convert Cartesian equations to polar equations, not parametric equations.\r\nParametric equation is like x=at+b y=ct+d...", "Solution_3": "wly3298456, polar equations are a kind of parametric equations. Anything of the form:\r\n$x=f(t)$\r\n$y=g(t)$, counts as a parametric equation.\r\n\r\nPlug in $x=at+b$ in your equation. You cannot find a y of the same form generally.", "Solution_4": "alright...I see..thanks" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all positive integers $x;y$ greater than 1 so that $2xy-1$ divisible by $(x-1)(y-1)$", "Solution_1": "Let $z=x-1,y=t-1$, then $2xy-1=2(z+1)(t+1)-1=2zt+2z+2t+1$. Therefore $zt|2z+2t+1$. It give $z,t$ are odd. If $|z|\\ge |t|\\ge 3$ , then $2t+1=\\pm z$. Therefore $|t|\\leq 2+1=3$.\r\nConsder $t=-3,-1,1,3$ we get all solutions \r\n$t=-1, z=\\pm 1$ \r\n$t=1, z=\\pm 1$ or $t=1,z=\\pm 3$.\r\n$t=-3, z=5$\r\n$t=3, z=7$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "In $ \\triangle ABC$ with no obtuse angle.Prove that ineq:\r\n $ \\frac {(1 \\minus{} cos2A)(1 \\minus{} cos2B)}{1 \\minus{} cos2C}$ + $ \\frac {(1 \\minus{} cos2C)(1 \\minus{} cos2A)}{1 \\minus{} cos2B}$ + $ \\frac {(1 \\minus{} cos2B)(1 \\minus{} cos2C)}{1 \\minus{} cos2A}$ $ \\geq$ $ \\frac{9}{2}$", "Solution_1": "See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=183391", "Solution_2": "thank you very much." } { "Tag": [ "inequalities", "function" ], "Problem": "For any real numbers, x_0, x_1, ..., x_n, find with proof, the minimum value of f[x]=\\sum_{k=1}^{n} |x-x_k|.\r\n\r\nOf course we can monotonize so that x_0<=x_1<=...<=x_n, and it is easily seen that the minimum lies within the interval [x_0,x_n]. After experimenting, I've come up with the conjecture that f[x] attains its minimum if and only if x=x_{n/2}+1 (n even) and on the interval [x_{Floor[n/2]},x_{Ceiling[n/2]}] (n odd), but I am unable to prove this.\r\n\r\nThis question was from Kiran Kedlaya's MOSP 1995 Inequality notes.", "Solution_1": "I am afraid I am not sure I'm reading what you wrote correctly: in the case n even, did you mean x = x_(n/2 + 1) (more likely) or \r\nx = x_(n/2) + 1 (which is what you wrote, I believe, but seems less likely)?", "Solution_2": "Assume that the minimum value is not taken at any of the points x=x_0, x=x_1, .. x=x_n, but is taken at x=a.\n\n\n\nNow all you have to do is see that [hide]if we consider the point a+b for an arbitrarily small b, then the new value of the function is b times (number of x_i less than a - number of x_i more than a) more than the original. Similarly a-b is (number more - number less)*b more. But one of these must give a new minimum, contradiction, or the number more=number less where we can shift over to the nearest x_i and the minimum is taken there, also contradiction.\n\n\n\nSo the minimum value is taken at one of the x_i, and its pretty easy to see it has to be the middle one. And as before, if there are an even number of x_i, it will be in that range.[/hide]" } { "Tag": [ "algebra", "difference of squares", "special factorizations" ], "Problem": "Hi,\r\n\r\nI need some help for solve in N^3 this equation : 1 + 4ab = c^2\r\n\r\nThanks a lot.", "Solution_1": "[hide=\"Solution\"]Taking this mod 4, we see that $ c^2 \\equiv 1 \\mbox{ mod 4}$, so $ c$ must be odd. Letting $ c \\equal{} 2d \\plus{} 1$ for some nonnegative integer $ d$, we see that $ 4d^2 \\plus{} 4d \\plus{} 1 \\equal{} 1 \\plus{} 4ab$, i.e., $ d(d \\plus{} 1) \\equal{} ab$. \n\nIn addition, all numbers $ a, b$ that satisfy $ ab \\equal{} d(d \\plus{} 1)$ with $ c \\equal{} 2d \\plus{} 1$ for some nonnegative integer $ d$ also satisfy the equation, i.e., $ \\{(a, b, c) : \\exists d \\in \\mathbb{Z}^\\plus{} (ab \\equal{} d(d \\plus{} 1) \\mbox{ and } c \\equal{} 2d \\plus{} 1\\}$ is the solution set (I cannot see any way to simplify the condition that $ ab \\equal{} d(d \\plus{} 1)$.) \n[/hide]", "Solution_2": "Another way to achieve the same result is to subtract 1, factor using difference of squares, and then use the fact that (c-1)(c+1) cannot be even if c is even.\r\n\r\nOverkill is much more fun than parity.", "Solution_3": "yet another way is to note that a multiple of four is even and therefore one more than a multiple is odd?", "Solution_4": "[quote=\"Temperal\"]yet another way is to note that a multiple of four is even and therefore one more than a multiple is odd?[/quote]\r\nYeah, on hindsight, I didn't even need mods..." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ n\\in\\mathbb{N}$ and $ p,q$ irrational such that $ a\\plus{}\\sqrt{ab\\plus{}n},b\\plus{}\\sqrt{ab\\plus{}n}\\in\\mathbb{Q}$\r\nProve that $ n$ perfect square and $ a\\plus{}\\sqrt{ab\\plus{}n}\\plus{}b\\plus{}\\sqrt{ab\\plus{}n}\\equal{}0$", "Solution_1": "you mean $ a,b$ irrational..", "Solution_2": "The desired equality is equivalent to $ a \\minus{} b \\equal{} 2 \\sqrt {n}$, and we know already that $ a \\minus{} b$ is rational.\r\n\r\nLet $ a \\minus{} b \\equal{} 2r$ for some rational $ r$ and let $ q \\equal{} a \\plus{} \\sqrt {a(a \\minus{} 2r) \\plus{} n} \\in \\mathbb{Q}$. Then\r\n\r\n$ q \\minus{} a \\equal{} \\sqrt {a(a \\minus{} 2r) \\plus{} n} \\implies$\r\n$ a^2 \\minus{} 2aq \\plus{} q^2 \\equal{} a^2 \\minus{} 2ar \\plus{} n \\implies$\r\n$ 2a(r \\minus{} q) \\plus{} (q^2 \\minus{} n) \\equal{} 0$\r\n\r\nwhich is linear in $ a$. But since $ a$ is irrational, it follows that the coefficients must be zero, which gives the result.\r\n\r\nEdit: This is a cute problem. Source?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let be $ a,b\\in \\mathbb{R}$ . Prove that :\r\n\r\n$ \\left(\\frac {3a \\plus{} b}{2}\\right)^2 \\plus{} \\frac {13b^2 \\plus{} 41}{4}\\ge \\frac {13a \\plus{} 3b}{2}$", "Solution_1": "$ Mx^2\\plus{}Nx\\plus{}P \\ge 0$ and $ N^2\\minus{}4MP \\le 0$" } { "Tag": [ "quadratics", "function" ], "Problem": "The product of three consecutive integers is 8 times their sum. what is the sum of their squares?", "Solution_1": "[hide]$(a-1)a(a+1)=8(a-1+a+a+1)$\n$(a^{2}-1)a=24a$\n$a^{2}=25$\n\n$16+25+36=77$[/hide]", "Solution_2": "[quote=\"davidli\"][hide]$(a-1)a(a+1)=8(a-1+a+a+1)$\n$(a^{2}-1)a=24a$\n$a^{2}=25$\n\n$16+25+36=77$[/hide][/quote]\n[hide] When you divided by $a$, you ruled out the possibility of $a=0$ You have to be careful not to do that. :) \n\nAn alternative answer would be:\n\n$(-1)^{2}+0^{2}+1^{2}=2$[/hide]", "Solution_3": "[quote=\"laughinghead505\"][quote=\"davidli\"][hide]$(a-1)a(a+1)=8(a-1+a+a+1)$\n$(a^{2}-1)a=24a$\n$a^{2}=25$\n\n$16+25+36=77$[/hide][/quote]\n[hide] When you divided by $a$, you ruled out the possibility of $a=0$ You have to be careful not to do that. :) \n\nAn alternative answer would be:\n\n$(-1)^{2}+0^{2}+1^{2}=2$[/hide][/quote]\r\n\r\nOh yeah, that's right. I always forget zero.", "Solution_4": "When I saw this problem I thought that negative numbers would work but then I saw that it was squares so negative doesn't matter.\r\nWhen we learned about quadratic functions in school, our teacher kept telling us \"DON'T FORGET ZERO!!!\" And it worked! :P", "Solution_5": "It kind of surprises me that people forget zero a lot because it is usually the obvious answer.", "Solution_6": "[quote=\"now a ranger\"]The product of three consecutive integers is 8 times their sum. what is the sum of their squares?[/quote]\r\n[hide]\n$24a=(a-1)(a)(a+1)$\n$24a=(a^{2}-a)(a+1)$\n$24a=(a^{3}-a^{2}+a^{2}-a)$\n$24a=a^{3}-a$\n$a^{3}-25a=0$\n$a(a^{2}-25)=0$\n$a(a-5)(a+5)=0$\n$a=0, 5, or-5$\nsum of squares if a = 0: 2\nsum of squares if a = 5 or -5: $16+25+36=77$\nso the answer is either 2 or 77\n\n[/hide]", "Solution_7": "[hide]\n\n$abc=8(a+b+c)$\n\nit is obviously 4,5,6\n\nso, the answer is $16+25+36=77$\n\n[/hide]\r\n\r\ndavidli, please tell me this isn't like urs (didn't look to make sure)\r\n\r\n-jorian", "Solution_8": "[quote=\"jhredsox\"][hide]\n\n$abc=8(a+b+c)$\n\nit is obviously 4,5,6\n\nso, the answer is $16+25+36=77$\n\n[/hide]\n\ndavidli, please tell me this isn't like urs (didn't look to make sure)\n\n-jorian[/quote]\r\nHow is that obvious?", "Solution_9": "[quote=\"bpms\"][quote=\"jhredsox\"][hide]\n\n$abc=8(a+b+c)$\n\nit is obviously 4,5,6\n\nso, the answer is $16+25+36=77$\n\n[/hide]\n\ndavidli, please tell me this isn't like urs (didn't look to make sure)\n\n-jorian[/quote]\nHow is that obvious?[/quote]\r\n\r\nI bet he just looked at the answer and then wrote that.", "Solution_10": "omg, why do you think that? no seriously, i got it myself (not that hard)\r\n\r\noh and it's obvious because idk. i thought it was something else that was obvious, then it took me 10 more secs cause it wasn't, but i was too lazy to remove the wording\r\n\r\n-jorian", "Solution_11": "Okay, jorian, your answers are fine. It was just that last time that your post was annoyingly similar to my post. You don't have to say things like you didn't look at mine.", "Solution_12": "Please post the source next time.\r\n\r\nI think this was AMC 10/12 2002B.", "Solution_13": "i think it's [hide]77[/hide]\r\n\r\nbut i might be miles off", "Solution_14": "[quote=\"Rafaelloaa\"]i think it's [hide]77[/hide]\n\nbut i might be miles off[/quote]\r\nYou should show your work just in case you made a mistake or something. \r\nBTW, there is one other pair...... (other people have made solutions)" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "if a,b,c>0 ,and $ab+bc+ca=1$,\r\nplease find the minimum of $1/(a+b)+1/(a+c)+1/(c+b)$\r\n\r\nthank u everyone :)", "Solution_1": "[quote=\"little boy\"]if a,b,c>0 ,and $ab+bc+ca=1$,\nplease find the minimum of $1/(a+b)+1/(a+c)+1/(c+b)$\n[/quote]\r\n$\\inf\\left(\\frac{1}{a+b}+\\frac{1}{a+c}+\\frac{1}{b+c}\\right)=2.5$ but my solution is very ugly.", "Solution_2": "it doesn't matter\r\ncan u show me some steps :wink:", "Solution_3": "[quote=\"little boy\"]it doesn't matter\ncan u show me some steps :wink:[/quote]\r\nOK. $\\frac{1}{a+b}+\\frac{1}{a+c}+\\frac{1}{b+c}\\geq2.5\\Leftrightarrow\\frac{a^{2}+b^{2}+c^{2}+3}{a+b+c-abc}\\geq2.5\\Leftrightarrow$\r\n$\\Leftrightarrow4(a^{2}+b^{2}+c^{2}+3(ab+ac+bc))^{2}(ab+ac+bc)\\geq$\r\n$\\geq25((a+b+c)(ab+ac+bc)-abc)^{2}\\Leftrightarrow$\r\n$\\Leftrightarrow4\\cdot\\sum_{sym}(a^{5}b-a^{4}b^{2})+3(a-b)^{2}(a-c)^{2}(b-c)^{2}+$\r\n$+8\\cdot\\sum_{cyc}(a^{4}bc+a^{3}b^{2}c+a^{3}c^{2}b+\\frac{7}{3}\\cdot a^{2}b^{2}c^{2})\\geq0.$ :)", "Solution_4": "It was a pleasant surprise to learn of this method\r\nbut i want to know how can u think of it\r\nit's too hard for me :(", "Solution_5": "How did you come up with the number right away? do you use some kind of method? Because you solve nearly every inequality here...can you tell me that method?Please, please... :help:", "Solution_6": "There are many receptions how to receive the sum of squares. In our problem I used that $(a-b)^{2}(a-c)^{2}(b-c)^{2}=$\r\n$=\\sum_{cyc}(a^{4}b^{2}+a^{4}c^{2}-2a^{4}bc-2a^{3}b^{3}+2a^{3}b^{2}c+2a^{3}c^{2}b-2a^{2}b^{2}c^{2}).$\r\n\r\nThis is very useful equality.", "Solution_7": "It's a tough problem which was posted before (ask Soarer). You can find it in Hojoo's notes .", "Solution_8": "upstairs... do u know a nicer solution about this question?\r\nplease write it down ... thank u", "Solution_9": "please search. Besides arqady's miraculous method (really amazing!!), we can use mixing variables to tackle it.", "Solution_10": ":| could u hlep me to find the connection if u don't mind", "Solution_11": "Yes, mixing variables works. But my proof is still ugly:\r\nLet $f(a,b,c)=\\frac{1}{a+b}+\\frac{1}{a+c}+\\frac{1}{b+c}-2.5$ and $g(a,b,c)=ab+ac+bc.$\r\nHence, $g(a,b,c)=g(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a).$\r\nLet $a=\\min\\{a,b,c\\}.$ Then \r\n$f(a,b,c)-f(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a)=$\r\n$=\\frac{1}{a+b}-\\frac{2}{\\sqrt{(a+b)(a+c)}}+\\frac{1}{a+c}-\\left(\\frac{1}{2\\sqrt{(a+b)(a+c)}-2a}-\\frac{1}{b+c}\\right)=$\r\n$=\\left(\\frac{1}{\\sqrt{a+b}}-\\frac{1}{\\sqrt{a+c}}\\right)^{2}-\\frac{(\\sqrt{a+b}-\\sqrt{a+c})^{2}}{2(b+c)(\\sqrt{(a+b)(a+c)-a)}}=$\r\n$=\\frac{(b-c)^{2}(2(b+c)\\sqrt{(a+b)(a+c)}-2a(b+c)-(a+b)(a+c))}{2(a+b)(a+c)(b+c)(\\sqrt{a+b}+\\sqrt{a+c})^{2}(\\sqrt{(a+b)(a+c)}-a)}\\geq0,$\r\nwhich true since, $(b+c)\\sqrt{(a+b)(a+c)}-2a(b+c)\\geq0$ \r\nand $(b+c)\\sqrt{(a+b)(a+c)}-(a+b)(a+c)\\geq0.$\r\nId est, remain to prove that $f(a,b,b)\\geq0$, where $2ab+b^{2}=1.$ I think it is not hard.", "Solution_12": "I remembered Soarer posted another way using also mixing variables (from a Chinese book?) .", "Solution_13": "Yes, I also solved it with mixing variables. But what made you choose a such teribble approach?\r\n[quote=\"arqady\"]Let $f(a,b,c)=\\frac{1}{a+b}+\\frac{1}{a+c}+\\frac{1}{b+c}-2.5$ and $g(a,b,c)=ab+ac+bc.$\nHence, $g(a,b,c)=g(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a).$\n[/quote]", "Solution_14": "[quote=\"Lovasz\"]\nYes, I also solved it with mixing variables. But what made you choose a such teribble approach?\n[quote=\"arqady\"]\nLet $g(a,b,c)=ab+ac+bc.$\nHence, $g(a,b,c)=g(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a).$\n[/quote][/quote]\r\nOnly desire that $g$ did not vary. :)", "Solution_15": "Sorry, I meant this one,\r\n[quote=\"arqady\"] \n$f(a,b,c)-f(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a) \\ge 0.$\n[/quote]", "Solution_16": "[quote=\"Lovasz\"] I meant this one,\n[quote=\"arqady\"] \n$f(a,b,c)-f(a,\\sqrt{(a+b)(a+c)}-a,\\sqrt{(a+b)(a+c)}-a) \\ge 0.$\n[/quote][/quote]\n[quote=\"Lovasz\"] But what made you choose a such teribble approach?\n[/quote]\r\nSuch terrible approach?\r\nProbably I on another am not able. :maybe: :)", "Solution_17": "thank u !!thank u !!\r\ni love u ,arqady :P", "Solution_18": "To Lovasc : The mixing variables method Arqady used is to keep $ab+bc+ca$ unchange . This's the most natural method using mixing variables .", "Solution_19": "[quote=\"little boy\"]:| could u hlep me to find the connection if u don't mind[/quote]\r\nsorry, i can't find it now.", "Solution_20": "here it is:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=31377\r\n\r\n :)", "Solution_21": "[quote=\"romano\"]To Lovasz: The mixing variables method Arqady used is to keep $ab+bc+ca$ unchange . This's the most natural method using mixing variables .[/quote]\nThanks.\n[hide=\" :!: \"]\n\n[quote] Given $a,b,c \\ge 0$ satisfying $ab+bc+ca=1$. Prove that: \\[{1 \\over a+b}+{1 \\over b+c}+{1 \\over c+a}\\ge{5 \\over 2}.\\] [/quote]\n\n[b]Solution:[/b] Let $f(a,b,c)={1 \\over a+b}+{1 \\over b+c}+{1 \\over c+a}$. Then: \\begin{eqnarray*}f(a,b,c)-f\\left(a,\\frac{b+c}{2},\\frac{b+c}{2}\\right)&=&{1 \\over a+b}+{1 \\over a+c}-{2 \\over a+{b+c \\over 2}}\\\\ &=&{1 \\over a+b}+{1 \\over a+c}-{4 \\over 2a+b+c}\\end{eqnarray*} It\u2019s obviously true since ${1 \\over x}+{1 \\over y}\\ge{4 \\over x+y}$.\nSo that: \\[f(a,b,c) \\ge f(a,t,t) \\quad{\\rm where}\\quad t^{2}+2ta=1.\\] It remains to prove that: \\[{2 \\over a+t}+{1 \\over 2t}\\ge{5 \\over 2}\\\\ \\iff (t-1)(5t^{2}-4t+1) \\le 0.\\] Thus... with $t \\le 1$.\nDone.\n[/hide]", "Solution_22": "cheer :roll:", "Solution_23": "I can use Iran 96 to solve this problem but not nice! :blush:", "Solution_24": "For $a,\\ b,\\ c\\geq 0$, \n\n\\[(1)\\ \\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}+\\frac 54(ab+bc+ca)\\geq \\frac{15}4\\]\n\n\\[(2)\\ \\left(\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\right)\\sqrt{ab+bc+ca}\\geq \\frac 52\\]", "Solution_25": "See also here\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=449840&p=3617601#p3617601" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ f: R\\rightarrow Z$ and:\r\n$ i)f(x+1)=f(x)+1;$\r\n$ ii)$ For all $ n \\in N: f(x)=f(\\frac{f(nx)}{n});$\r\nProve that for all $ x \\in Q: f(x)=[x]$ or $ f(x)=[x]+1.$", "Solution_1": "Since f(x+1)=f(x)+1, f(x+2)=f(x+1)+1=f(x)+2, f(x+3)=f(x)+3...\r\nSo, for any natural k, f(x+k)=f(x)+k (1).\r\nIf k and x are both natural, then f(k+x)=f(k)+x (2).\r\nBy (1) and (2), we write: f(x)-f(k)=x-k. So, for any natural pair a and b, f(a)+f(b)=a+b, If a and b are reals, then f(a+b) equal to the most natural, not more then x. So, f(x)=x or f(x)=x+1. By putting it in\r\nthe second given thing, we will make sure, that these functions are what we were looking for.\r\nQ.E.D." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Find a differentiable function $f: \\mathbb{R}\\to\\mathbb{R}$ such that $|f|$ is not differentiable anywhere.", "Solution_1": "$|f|$ is differentiable on the set $\\{x\\colon f(x)\\ne 0\\}$." } { "Tag": [ "trigonometry", "algebra proposed", "algebra" ], "Problem": "Solve the equation:\r\n$ \\Bigg(\\sin^2{x}\\minus{}\\frac{1}{\\sin^2{x}}\\Bigg)^2\\plus{}\\cos^2{x}\\minus{}\\frac{1}{\\cos^2{x}}\\equal{}\\frac{3}{4}$", "Solution_1": "[quote=\"tdl\"]Solve the equation:\n$ \\Bigg(\\sin^2{x} \\minus{} \\frac {1}{\\sin^2{x}}\\Bigg)^2 \\plus{} \\cos^2{x} \\minus{} \\frac {1}{\\cos^2{x}} \\equal{} \\frac {3}{4}$[/quote]\r\nIf you put $ x \\equal{} \\frac{\\pi}{4}$, the equation is satisfied." } { "Tag": [ "logarithms", "calculus", "limit", "algebra unsolved", "algebra" ], "Problem": "Determine the number of solutions of the simultaneous equations $ x^2 \\plus{} y^3 \\equal{} 29$ and $ \\log_3 x \\cdot \\log_2 y \\equal{} 1.$", "Solution_1": "[quote=\"orl\"]Determine the number of solutions of simultaneous equations to $ x^2 \\plus{} y^3 \\equal{} 29$ and $ log_3 x \\cdot log_2 y \\equal{} 1.$[/quote]\r\nLet $ x \\equal{} 3^k$ and $ y \\equal{} 2^j$ we then have $ kj \\equal{} 1 \\iff k \\equal{} \\frac {1}{j}$\r\n\r\nSo $ f(j) \\equal{} \\sqrt [j]{9} \\plus{} 8^j \\equal{} 29$. How do you find $ f'(j)$?\r\n\r\nFor $ j \\in ( \\minus{} \\infty;0]$ there is obviously no solution....\r\nFor $ j \\in [1;\\infty)$ there is only one solution, since $ f(j)$ is increasing.\r\nFor $ j \\in (0;1)$ there is only one solution cause $ f(j)$ is decreasing.,\r\n\r\nI have no idea how to find the solutions.. I'll post, if I get one.\r\n\r\nEdit: $ f'(j) \\equal{} ln(8) \\cdot 8^j \\minus{} \\frac {ln(9) \\cdot \\sqrt [j]{9}}{j^2}$", "Solution_2": "[quote=\"Mathias_DK\"][quote=\"orl\"]Determine the number of solutions of simultaneous equations to $ x^2 \\plus{} y^3 \\equal{} 29$ and $ log_3 x \\cdot log_2 y \\equal{} 1.$[/quote]\nLet $ x \\equal{} 3^k$ and $ y \\equal{} 2^j$ we then have $ kj \\equal{} 1 \\iff k \\equal{} \\frac {1}{j}$\n\nSo $ f(j) \\equal{} \\sqrt [j]{9} \\plus{} 8^j \\equal{} 29$. How do you find $ f'(j)$?\n\nFor $ j \\in ( \\minus{} \\infty;0]$ there is obviously no solution....\nFor $ j \\in [1;\\infty)$ there is only one solution, since $ f(j)$ is increasing.\nFor $ j \\in (0;1)$ there is only one solution cause $ f(j)$ is decreasing.,\n\nI have no idea how to find the solutions.. I'll post, if I get one.\n\nEdit: $ f'(j) \\equal{} ln(8) \\cdot 8^j \\minus{} \\frac {ln(9) \\cdot \\sqrt [j]{9}}{j^2}$[/quote]\r\nOk,good,i think you will solve it,The result is $ 2$pairs $ (x,y)$\r\nBut i cant find clearly $ x\\equal{}...,y\\equal{}...$ who can?", "Solution_3": "[quote=\"Allnames\"][quote=\"Mathias_DK\"][quote=\"orl\"]Determine the number of solutions of simultaneous equations to $ x^2 \\plus{} y^3 \\equal{} 29$ and $ log_3 x \\cdot log_2 y \\equal{} 1.$[/quote]\nLet $ x \\equal{} 3^k$ and $ y \\equal{} 2^j$ we then have $ kj \\equal{} 1 \\iff k \\equal{} \\frac {1}{j}$\n\nSo $ f(j) \\equal{} \\sqrt [j]{9} \\plus{} 8^j \\equal{} 29$. How do you find $ f'(j)$?\n\nFor $ j \\in ( \\minus{} \\infty;0]$ there is obviously no solution....\nFor $ j \\in [1;\\infty)$ there is only one solution, since $ f(j)$ is increasing.\nFor $ j \\in (0;1)$ there is only one solution cause $ f(j)$ is decreasing.,\n\nI have no idea how to find the solutions.. I'll post, if I get one.\n\nEdit: $ f'(j) \\equal{} ln(8) \\cdot 8^j \\minus{} \\frac {ln(9) \\cdot \\sqrt [j]{9}}{j^2}$[/quote]\nOk,good,i think you will solve it,The result is $ 2$pairs $ (x,y)$\nBut i cant find clearly $ x \\equal{} ...,y \\equal{} ...$ who can?[/quote]\r\nI think it's almost impossible to do without the computer(and very easy with). It does'nt even give a \"nice\" expression like $ \\frac{ln(3) \\plus{} ln(2)}{4*4.21^4}$, when i run it through.........", "Solution_4": "hello, from the second equation we get\r\n$ y=e^{\\frac{\\ln(3)\\ln(2)}{\\ln(x)}}$ for $ x>0$,by inserting this in the first equation we have $ x^2+\\left(e^{\\frac{\\ln(3)\\ln(2)}{\\ln(x)}\\right)^3=29}$. I see here no chance to solve this equation explicit with respect to $ x$.\r\nSonnhard.", "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, from the second equation we get\n$ y = e^{\\frac {\\ln(3)\\ln(2)}{\\ln(x)}}$ for $ x > 0$,by inserting this in the first equation we have $ x^2 + \\left(e^{\\frac {\\ln(3)\\ln(2)}{\\ln(x)}\\right)^3 = 29}$. I see here no chance to solve this equation explicit with respect to $ x$.\nSonnhard.[/quote]\r\nBut finding $ x$ is very hard (without computer) Can you?", "Solution_6": "Mathematica says \"The equations appear to involve the variables to be solved for in an essentially non-algebraic way.\"\r\n\r\nWas there some non-calculus solution offered? This one is sort of ugly, and calculus generally isn't required...", "Solution_7": "claim x^3=a,y^3=b,\r\ntherefore loga.logb=6log2.log3=k(say)\r\nand see that a+b=29 and loga.logb=k(some constant)\r\nhence loga.log(29-a)=k\r\nthere are only two values possible for a\r\nand the answer follows two solutions\r\n(is it rigorous enough :wink: :wink: )", "Solution_8": "i have done a calculation mistake but the idea is the same\r\ncorrection:\r\n take x^2=a instead of x^3=a\r\n now the answer follows in the same way", "Solution_9": "$x^2 + y^3 = 29$ and $log_3 x \\cdot log_2 y = 1$ then from $log_3 x \\cdot log_2 y = 1$, we get $x^y=2^3$ , from $x^2 + y^3 = 29$ we get $x=+\\sqrt{29-y^3}$, equating we obtain $(29-y^3)^y=2^6$\nLet $f(y)=(29-y^3)^y$ ,then $\\lim_{y\\rightarrow{0}}f(y)=1$ , $f(2)=21^2>2^6$ , and there exist a $y_0$ for which $f(y_0)=0$.\n$f'(y)=(29-y^3)^y[log(29-y^3)-\\frac{3y^3}{29-y^3}]$ ,then $f'(1)<0$ and $f'(y_0)\\approxeq{0}$, also there is a $y_1$ in $(1,y_0)$ for which $f'(y_1)=0$\nSo $f(y)$ attains two times $2^6$(since $f(2)>2^6$), also after $y>y_0$ the eq. $f(y)<0$\nHence there is two solutions of these equations.", "Solution_10": "[quote=\"orl\"]Determine the number of solutions of simultaneous equations to $ x^2 \\plus{} y^3 \\equal{} 29$ and $ log_3 x \\cdot log_2 y \\equal{} 1.$[/quote]\n[quote]Determine the number of solutions of simultaneous equations to $ x^2 \\plus{} y^3 \\equal{} a (a \\geq 17)$ and $ log_3 x \\cdot log_2 y \\equal{} 1.$\n[/quote]", "Solution_11": "We can find that $y^3+3^{\\log_y{2}}=29$. It is apparent upon graphing the equations (disregard variables, the following graphs are on the $xy$ plane)\n\\[\nx^3=y\n\\] \nand \n\\[\n29-3^{\\log_x{2}} = y\n\\]\nthat there are exactly 2 points of intersection, so the answer is thus $\\boxed{2}$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "We denote $ C(T)$ and $ D(T)$ to be the center set and centroid set respectively of a tree $ T$. For each $ n \\geq 2$, construct a tree $ T$ in which $ \\text{dist}_T(C(T), D(T)) \\geq n$. I don't see how to construct these trees.", "Solution_1": "Start with a left-right path made of $ n\\plus{}1$ vertices, then vertex $ z$, then $ n$ more vertices, last being $ w$. Now add enough leaves, let's say $ 3n$ to the end $ w$. \r\nVertex $ z$ is the center, at distance $ \\leq n\\plus{}1$ from any other vertex, while any other vertex is at distance $ \\geq n\\plus{}2$ from some vertex.\r\nVertex $ w$ is the centroid, with all subtrees of size $ \\leq 2n\\plus{}1$, while any other vertex has some larger subtree (for example $ z$ has a subtree of size $ 4n$).\r\nThe distance between $ z$ and $ w$ is $ n$." } { "Tag": [ "greatest common divisor" ], "Problem": "The lengths in feet of three pieces of timber are 48, 72, and 40. The sawmill operator needs to cut the timber into logs of equal length with no waste. How many feet long is the greatest possible length she can cut?", "Solution_1": "[quote=\"GameBot\"]The lengths in feet of three pieces of timber are 48, 72, and 40. The sawmill operator needs to cut the timber into logs of equal length with no waste. How many feet long is the greatest possible length she can cut?[/quote]\r\n\r\nWe need the GCF, which is 8. Thus, the greatest length she can cut is 64.", "Solution_2": "wait...am i misunderstanding or something? why wouldn't it just be 8? i think i'm failing, but i don't know why...", "Solution_3": "[quote=\"nikeballa96\"]wait...am i misunderstanding or something? why wouldn't it just be 8? i think i'm failing, but i don't know why...[/quote]\r\n\r\nThe largest piece is 72. She can cut 64 off of that to get a piece of length 8." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "found all fonctions N--->N\r\nf(x^3+y^3+z^3)=(f(x^3))+(f(y^3))+(f(z^3))", "Solution_1": "I think it should be $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$, right\u00bf" } { "Tag": [ "IMO", "IMO 2007" ], "Problem": "hello mathlinkers who will go to vietnam this summer.\r\n\r\nwhat do you think of the idea that everyone can bring items like coins or something similar to the IMO to change them? especially those from countries that are a bit smaller ;)\r\n\r\nNaphthalin", "Solution_1": "It's really a very nice idea !", "Solution_2": "Sure, that's a good idea.\r\n\r\nAnd we will bring Canmoo too, but not for exchange! :P", "Solution_3": "Canmoo will be ours!\r\n\r\nBut yes it's a good idea, I still have the coins the slovenian team in mexxico (2005) gave us, together with other stuff, as \"preparation\" for 2006. But doing so with all teams is more fun^^\r\nBut I will not be there :(", "Solution_4": "sorry for the question but what is Canmoo exactly?\r\neven if i come from germany i can bring some nice money. e.g. old money of the german \"democratic\" republic^^\r\n\r\nNaphthalin", "Solution_5": "I suspect Canmoo is the Canadian Moose :moose: that has been traveling around at the IMO for the past years. While previous attempts of getting posesion of the moose have failed, I have designed a master plan this year that will surely lead to the capturing of the moose :w00tb: beware moose guardians!", "Solution_6": "Oh guys very nice topic! During the last four years I have collected so many interesting things from many friends at IMO(particularly a lot of coins :) ,I haven't counted how much it is in total,but I guess I am a millioner :D ). Last year,besides Armenian coins and other stuff, I took with me two botteles of Armenian brandy :blush: ,the people liked it very much though they were drank in the end :oops: .", "Solution_7": "It's really a very nice idea ! :lol: \r\n\r\nI'll bring some with me....... (of course, not only coins... :wink: )", "Solution_8": "We always bring coins (we have a coin that has 1/2 printed on it, and we think it's funny).\r\n\r\n@Valentin Vornicu:\r\nMany teams had the moose for some time, including mine (We had a master plan last year).\r\nBut I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI. \r\nI hope I'm wrong.", "Solution_9": "[quote=\"Fuaran\"]But I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI. \nI hope I'm wrong.[/quote]\r\nThat's quite wrong. :wink:", "Solution_10": "[quote]Fuaran wrote:\nBut I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI.\nI hope I'm wrong.\n\nThat's quite wrong. Wink[/quote]\r\n\r\nYay! \r\nActually, I heard it from a former Canadian participant. Looking forward to see you all at Hanoi!", "Solution_11": "I think I know who it was. :ninja:", "Solution_12": "I don't know what to bring... I will bring the best spaghetti of Italy! One spaghetto for each contestant :D", "Solution_13": "I remember some team bringing food last year as well, Vietnam I think... But they got a HUGE box of spaghetti, everyone got a package :lol: ." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Show that for every prime number $ p>3$ there exist integer numbers $ x, y, k$ such that $ 0<2k p$ so exist $ x,y\\in\\{0,...,\\frac {p \\minus{} 1}{2}\\}$ such that\r\n$ p|x^2 \\plus{} y^2 \\minus{} 3$\r\n$ x^2 \\plus{} y^2 \\minus{} 3 \\equal{} kp$\r\n$ x^2 \\plus{} y^2 \\minus{} 3\\leq 2(\\frac {p \\minus{} 1}{2})^2 \\minus{} 3 \\equal{} \\frac {p^2 \\minus{} 2p \\minus{} 5}{2}$ \r\nSo $ 2k\\leq p \\minus{} 1$ \r\nThis problem claim." } { "Tag": [ "calculus", "derivative", "linear algebra", "matrix", "function", "integration", "trigonometry" ], "Problem": "I don't necessarily need a proof of this, but more of a reason why I should believe it to be true. At this point I don't believe it at all.\r\n\r\nShow that the diagonal elements of the matrix representation in any basis for $ H_1$ for the operator $ D\\equal{}\\frac{\\partial}{\\partial x}$ are purely imaginary.\r\n\r\n$ H_1$ is defined to be the space of all functions such that $ \\int_0^a \\phi^*\\phi \\ dx<\\infty$ (and * is complex conjugate).\r\n\r\nWhy I don't believe it: It seems as if you could use some sort of Fourier sine basis $ \\{\\sqrt{\\frac{a}{2}}\\sin\\left(\\frac{n\\pi x}{a}\\right)\\}$ and get no imaginary numbers in the matrix representation at all.", "Solution_1": "With $ \\sin,\\cos$ basis you get zeros on the diagonal. And $ 0$ counts as a purely imaginary number (it has real part $ 0$).", "Solution_2": "Thanks, that makes sense. This is still a little strange in my mind. I've thought of a couple of approaches, none seem to be working.\r\n\r\nBrute force: Let $ B\\equal{}\\{\\phi_n \\}$ be a basis. Then $ D$ has the matrix form $ D_{qn}\\equal{}\\int \\phi^*_q D \\phi_n dx$ so that the diagonal elements are $ D_{mm}\\equal{}\\int_0^a \\phi_m^*\\frac{\\partial \\phi_m}{\\partial x} dx$. There doesn't seem to be an obvious reason from here that $ \\text{Re}(D_{mm})\\equal{}0$. \r\n\r\nAlternate approach: Diagonalize the matrix via eigenvalues, and use some sort of similarity argument?", "Solution_3": "We have $ D_{mm} \\equal{} \\int_0^a \\overline{\\phi_m(x)}\\phi_m'(x)\\,dx$. Therefore, $ \\overline{D_{mm}} \\equal{} \\int_0^a \\phi_m(x)\\overline{\\phi_m}'(x)\\,dx$. We want to show that $ D_{mm} \\plus{} \\overline{D_{mm}} \\equal{} 0$. By the product rule for derivatives, $ D_{mm} \\plus{} \\overline{D_{mm}} \\equal{} \\int_0^a \\frac {d}{dx}(\\phi_m(x)\\overline{\\phi_m}(x))\\,dx \\equal{} |\\phi_m(a)|^2 \\minus{} |\\phi_m(0)|^2$. \r\n\r\nTherefore, the statement you want to prove is true if all elements of $ H_1$ attain the same value at $ 0$ and at $ a$. Most likely, your definition of $ H_1$ is incomplete.\r\n\r\n(Moral of this story: when you have a differential operator acting on a Hilbert space of functions, be ready to integrate by parts.)", "Solution_4": "Wow, I'm sorry. I actually had gotten to where you ended, and completely forgot about the fact that I knew that $ \\phi(a)\\equal{}\\phi(0)\\equal{}0$ (hence why I only used $ \\sin$ for my basis in the first post). Thanks a lot!" } { "Tag": [ "geometry", "probability", "geometric transformation", "dilation", "function", "trigonometry", "3D geometry" ], "Problem": "Well, it was inevitable that somebody had to start it, so why not me??? After I'm the one who always studies for school.\r\n\r\nSchedule::\r\n\r\nMarch 1 - Chemistry\r\nMarch 7 - Physics\r\nMarch 10 - English\r\nMarch 15 - Biology\r\nMarch 20 - Maths\r\nMarch 31 - Computer Science\r\n\r\nList of PPl who are writing this yr (and in AOPS)\r\n\r\n[hide]\nchapplikka\nchemrock\nCyber-Shreyas\nEucleidan Geometry\nhash_include\nmadness\nnitishexpress\npardesi\nshruthibaskaran\nSOUMYASHANT NAYAK\nsrinitrs\nvaleriummaximum\nvihang\n\nICSE:\nAnitSahu\n\nOther Boards:\nVarunarasu\n[/hide]\r\nSo get the ball rolling", "Solution_1": "Hmm.. Let the hilarity ensue, what else can I say? And I think it was Anit Sahu right?", "Solution_2": "include Raghav in CBSe shre, he is here as Euclidean Geometer", "Solution_3": "You missed Abhimanyu Banerjee.", "Solution_4": "Best of luck to all", "Solution_5": "Have u all finished portions for chem..????\r\nme not even once...", "Solution_6": "Dei...\r\nOnakke idhellam toooo mucha thereela ?\r\nNaan exam ku munnadi oru dharava padichhu mudipena ne pureelai :!: \r\nAny ways,\r\nHave u all been asked to mug egs. of tranquillizers blah blah... diff kinds of drugs :?: \r\nI mean the names \r\nIts bugging and belting", "Solution_7": "naanum orutharava mudikanumnu romba try panituiruken", "Solution_8": "2 Good Questions\r\n\r\n1. What do you mean by the functionality of a monomer?\r\n2. How do the double bonds in natural rubber control its properties?", "Solution_9": "the double bonds in the dont do anything actually.....Once natural rubber is stretched there is no restoring force for it to regain the original sahpe...\r\nsince double bond is functional site...so they help in interlinking the two chains of natural rubber thereby helping to establish the rubber like property", "Solution_10": "[quote=\"valeriummaximum\"]the double bonds in the dont do anything actually.....Once natural rubber is stretched there is no restoring force for it to regain the original sahpe...\nsince double bond is functional site...so they help in interlinking the two chains of natural rubber thereby helping to establish the rubber like property[/quote]\r\n\r\nWrong!! \r\n\r\nClue:\r\n[hide]Think isomerism[/hide]", "Solution_11": "cis and trans?????\r\n\r\nhow is it related to the intermolecular forces shre...", "Solution_12": "Natural rubber is the cis form isnt.... Cis 1, 4 something if I remember what I read for the preboards :D Functionality is the number of functional groups in a monomer?", "Solution_13": "the natural rubber's name is cis-1,4-polyisoprene.....\r\n\r\nCommon Shre, Explain how the configuration affects the intermolecular forces???", "Solution_14": "Ranga, he just said properties you know, never mentioned anything about intermolecular forces.. I think he meant wrong in a way that your answer that the double bonds are useless is wrong.. I maybe wrong, but who cares...\r\n\r\n2 more questions:\r\n\r\n1. During preparation of Nitrogen from ammonium chloride and sodium nitrite, what are the two impurities evolved? How can nitrogen be purified from them?\r\n2. Two ways of preparing S2Cl2", "Solution_15": "elasti got that one it seems! Brutes, his friends had nved sumita arora downright n then they were discussing this morn before him so he got it Cha! :mad: :|", "Solution_16": "[quote] Out due to inheritance[/quote] ur children took away ur marks :rotfl: :rotfl: \r\n\r\nit was not multilevel , not multiple , so all that came into my mind was heirarchial :D \r\n\r\nwhat r we expected to write for that wrong sql question ? i didn't even notice it , so just assumed that it was customer.i_id and item.i_id . Isn't that enough? should we mention that it is wrong?", "Solution_17": "[quote=\"bruteforce\"][quote] Out due to inheritance[/quote] ur children took away ur marks :rotfl: :rotfl: \n\nit was not multilevel , not multiple , so all that came into my mind was heirarchial :D \n\nwhat r we expected to write for that wrong sql question ? i didn't even notice it , so just assumed that it was customer.i_id and item.i_id . Isn't that enough? should we mention that it is wrong?[/quote]\r\n\r\nEven i thought that they had made a mistake so i wrote it by assumption :)", "Solution_18": "I just stated the assumption and found the output.", "Solution_19": "Hierarchical please... Note the spelling :P", "Solution_20": "[quote]it was not multilevel , not multiple , so all that came into my mind was heirarchial [/quote]\r\n\r\nThe spelling is wrong...It should be hierarchical....(anyway,who cares...)\r\n\r\nA centum wouldn't exactly surprise me .... :P :P \r\nbtw,\r\nI had almost got my output wrong assuming 'R' comes after 'P'...\r\nI guess I should revise my alphabets again... :rotfl: \r\nThe paper was a bit too easy...\r\n\r\nEDIT: I see raghav has already mentioned that...", "Solution_21": "Another worst mistake i would have made and almost made was that i was considering the loop for the word MyString :D :o . Then i realised that there was no special chars or whatever in that so i checked again to find that the word was actually sth else and that was the name of the string :rotfl:", "Solution_22": "Great everybody has written hierarchical. :( :((P.S: Spell checker in Firefox is really nice ;))\r\n\r\nI liked that question where you had to swap every odd and even position. There was just 1 trick in that question which is ofcourse not very easy to miss considering that his own example was illustrating the trick. Not like the INOI where he gives pathetic examples. \r\n\r\n@brute: I wrote. Assumption: its I_id and not Item_id and wrote the answer.\r\n\r\nI also liked that 'Write the equivalent POS form' question. A lot of people here just inverted the expression.", "Solution_23": "Yup the POS question was good...I tells how people just mugged up the K mapp and went for exam\r\n\r\nHe used have given a 10 size array ex. for the swapping.. Then most of the students wouyld have not thought of the trick", "Solution_24": "[quote=\"Nitishexpress\"]Yup the POS question was good...I tells how people just mugged up the K mapp and went for exam\n[/quote]\r\n\r\nNow that was both rude and unnecessary :huuh: :mad: Well there are plenty of instances i can show where you 'pros' have made mistakes. That means u are all bloody nvs and bloody bloody hypocrites too. :mad:\r\n\r\nEDIT: Dont get me started, then a whole lot of unnecessary comments will appear without invitation", "Solution_25": "No, Rohit, I interpreted Nitish comments like this. We weren't really taught boolean in a very understanding or conceptual way. I'm sure that nobody in our class can really explain why a K-Map works and why do we base it on gray code.\r\n\r\nThe problem is that none of us really understand SOP and POS, Maxterm and Minterm so well enough to answer twisted questions on that.... The problem is not with you, it was with the teaching.", "Solution_26": "Thalla mmadness, i thought there was only one meaning for that... and that is what shre told... if u make a mistake it doesnt mean that question is a nv question...", "Solution_27": "[quote=\"madness\"][quote=\"Nitishexpress\"]Yup the POS question was good...I tells how people just mugged up the K mapp and went for exam\n[/quote]\n\nNow that was both rude and unnecessary :huuh: :mad: Well there are plenty of instances i can show where you 'pros' have made mistakes. That means u are all bloody nvs and bloody bloody hypocrites too. :mad:\n\nEDIT: Dont get me started, then a whole lot of unnecessary comments will appear without invitation[/quote]\r\n\r\nCOOL DOWN :coolspeak: Yes shre is right. KVKK didn't do us any great favours with his teaching. \"Just NV my text to get 95\" that's all he says.", "Solution_28": "Lalalalala :D", "Solution_29": "[quote=\"Nitishexpress\"]Thalla mmadness, i thought there was only one meaning for that... and that is what shre told... if u make a mistake it doesnt mean that question is a nv question...[/quote]\r\n\r\nhello! :D Sorry da if i wasnt clear, i didnt call that question an nv ques at all, infact that was a good trick question as i told shre.Read my post properly da, even the best of persons make mistakes in logical questions :D thats what i meant. And ok, the msg got over and i cant agree more, esp the networking part :D" } { "Tag": [ "geometry", "perimeter", "trigonometry", "linear algebra", "matrix", "integration", "calculus" ], "Problem": "One a semicircle of radius 1, two points are selected at random and independently, along the perimeter of the semicircle. Together with the midpoint of the diameter, these three points form a triangle. Prove that the expected area of the triangle is $ (2\\plus{}\\pi)^{\\minus{}1}$.", "Solution_1": "\"The midpoint of the diameter\" is a wonderful way of referring to the center of a circle :wink: \r\n\r\n[hide=\"Possible approach\"]\n\nIf I understand you correctly, $ \\alpha$ and $ \\beta$ are chosen uniformly at random in $ [0,\\pi]$, and we are asking for the expected are of the triangle with vertices $ (0,0)$, $ (\\cos\\alpha, \\sin\\alpha)$, $ (\\cos\\beta, \\sin\\beta)$. If this is so, the area can be calculated as a 3x3 determinant\n\n$ \\frac{1}{2} \\begin{vmatrix} 0 & 0 & 1 \\\\ \\cos\\alpha & \\sin\\alpha & 1 \\\\ \\cos\\beta & \\sin\\beta & 1 \\end{vmatrix}$\n\nwe just need to be careful about signs - depending on the order of $ \\alpha$ and $ \\beta$ this could be positive or negative. This can be avoided by forcing, say, $ \\beta \\leq \\alpha$, so calculating the expectation as\n\n$ E \\equal{} \\frac{1}{2 \\pi^2} \\int_{\\alpha\\equal{}0}^\\pi \\int_{\\beta\\equal{}0}^\\alpha (\\cos\\alpha \\sin\\beta \\minus{}\\sin\\alpha \\cos\\beta) \\, d\\beta\\, d\\alpha$\n\nwhich is straightforward. Unfortunately I'm getting a different answer from yours... need to check my calculations.\n\n[/hide]", "Solution_2": "I think you're missing the cases when one of the points is on the diameter.", "Solution_3": "Ah. So I did misread the question. OK, that's easy to fix - just add another integral for the case of one point on the diameter. Also, my use of the determinant formula in the previous solution was just plain silly; the triangle with one vertex at the origin and two vertices at angles $ \\alpha$, $ \\beta$ on the unit circle obviously has area $ \\frac{1}{2} |\\sin (\\alpha\\minus{}\\beta)|$: just rotate it so that $ \\beta\\equal{}0$ and then the base is 1 and the altitude is $ \\sin(\\alpha\\minus{}\\beta)$. \r\n\r\n[hide=\"Correct solution\"]\nThe probability for each point to end up on the circular arc is $ p\\equal{}\\frac{\\pi}{\\pi\\plus{}2}$, and the probability to land on the diameter is $ q\\equal{}\\frac{2}{\\pi\\plus{}2}$. So we have \n\n1. Probability $ p^2$: both points on the arc, expected area is\n\n$ E_1 \\equal{} \\frac{1}{\\pi^2}\\int_0^\\pi \\int_0^\\pi \\frac{1}{2}|\\sin(\\alpha\\minus{}\\beta)|\\, d\\alpha \\, d\\beta$\n\n(the coefficient $ \\frac{1}{\\pi^2}$ just normalizes the integral so that the total probability is 1).\n\n2. Probability $ 2pq$: one point on the arc, one on the diameter, expected area is\n\n$ E_2 \\equal{} \\frac{1}{2\\pi} \\int_{x\\equal{}\\minus{}1}^1 \\int_{\\alpha\\equal{}0}^\\pi \\frac{1}{2} |x \\sin\\alpha| \\, d \\alpha \\, d x$\n\n3. Probability $ q^2$: both points on the diameter, area 0.\n\nAn easy calculation gives $ E_1 \\equal{} \\frac{1}{\\pi}$ and $ E_2 \\equal{} \\frac{1}{2\\pi}$ so\n\n$ E \\equal{} p^2 E_1 \\plus{} 2pq E_2 \\equal{} \\frac{1}{2\\plus{}\\pi}$.\n\n[/hide]", "Solution_4": "Thanks, I was hoping to see a non-calculus solution, but apparently not :P" } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "(From 2002 Belarussian Maths Olympiad)\r\n\r\nProve that there exist infinitely many positive integers which cannot be represented in the form\r\n\r\nx[size=59]1[/size]^3+x[size=59]2[/size]^5+x[size=59]3[/size]^7+x[size=59]4[/size]^9+x[size=59]5[/size]^11\r\n\r\nwhere x1,x2,x3,x4,x5 are positive integers.", "Solution_1": "The number of $k$-th powers $\\leq n$ is $\\leq n^\\frac{1}{k}$.\r\nAssume that two different sums of powers of your form are different, then there are at most\r\n$n^\\frac{1}{3}n^\\frac{1}{5}n^\\frac{1}{7}n^\\frac{1}{9}n^\\frac{1}{11}=x^r,r<1$\r\ndifferent numbers of that type $\\leq n$.\r\nBut $\\lim_{n \\to \\infty} \\frac{x^r}{x} = 0$,\r\nso only a very small part of all integers can be represented in this way, so there are infinitly many that cannot be represented." } { "Tag": [ "geometry" ], "Problem": "Hi All,\r\nI am looking for summer school specially in Europe for summer 2008.\r\nPlease help me on any thing about this kind of programs (websites, name of university,..).", "Solution_1": "a summer school like a mathe camp? or a summer school that caches you up if you fail something? Also it would help to know what area of europe you live in.", "Solution_2": "Hm, you might be able to take an online long-distance [url=http://cty.jhu.edu/]CTY[/url] course. They cover many topics around the high school age, but might be a bit pricy." } { "Tag": [ "geometry", "probability", "Alcumus" ], "Problem": "Point P is selected at random from the interior of the pentagon with vertices A=(0,2) ,B=(4,0) ,C=(2pi+1,0) , D=(2pi=1,4) , and E=(0,4). What is the probability that angle APB is obtuse? Express your answer as a common fraction.", "Solution_1": "You are new, but I don't think this belongs here. I am not sure of the difficulty, so maybe High School Intermediate topics.", "Solution_2": "[quote=\"zheng\"]Point P is selected at random from the interior of the pentagon with vertices A=(0,2) ,B=(4,0) ,C=(2pi+1,0) , D=(2pi=1,4) , and E=(0,4). What is the probability that angle APB is obtuse? Express your answer as a common fraction.[/quote]\r\n\r\nAre you looking for the solution Alcumus provides or a clarification on this solution? I've attached Alcumus's solution for you to take a look out. If you don't understand something in the solution, raise your question in this thread.", "Solution_3": "[quote=\"XK-39\"][quote=\"zheng\"]Point P is selected at random from the interior of the pentagon with vertices A=(0,2) ,B=(4,0) ,C=(2pi+1,0) , D=(2pi=1,4) , and E=(0,4). What is the probability that angle APB is obtuse? Express your answer as a common fraction.[/quote]\n\nAre you looking for the solution Alcumus provides or a clarification on this solution? I've attached Alcumus's solution for you to take a look out. If you don't understand something in the solution, raise your question in this thread.[/quote]\r\n\r\nI was looking for clarification", "Solution_4": "[quote=\"PowerOfPi\"]You are new, but I don't think this belongs here. I am not sure of the difficulty, so maybe High School Intermediate topics.[/quote]\r\n\r\n\r\nHey,\r\nPowerOfPi let me say but it was an alcumus question. Plus i was only looking for clairification.", "Solution_5": "Yeah, it's fine here. Just one of the more difficult Alcumus problems. Plus, clarification on what?", "Solution_6": "[quote=\"pinkmuskrat\"]Plus, clarification on what?[/quote]\r\n\r\nOn the solution(which i got)" } { "Tag": [ "algebra", "polynomial", "number theory open", "number theory" ], "Problem": "The polynomials P(x) = x^2 and Q(x) = x^2-1 differ by one, they also factorise fully over the integers. \r\n\r\nI want to know what are the highest degree polynomials that have this property? Here is a cubic example\r\n\r\nP(x) = (x+1)(4x+1)(4x+1), Q(x) = x(4x+3)(4x+3) => P(x) - Q(x) = 1\r\n\r\nI could go higher but I wont spoil the fun. I dont know a complete solution.", "Solution_1": "$ P(x) \\equal{} x^{2^n}$ and $ Q(x) \\equal{} x^{2^n} \\minus{} 1 \\equal{} (x \\minus{} 1)(x \\plus{} 1)(x^2 \\plus{} 1) \\cdots (x^{2^{n \\minus{} 1}} \\plus{} 1)$ satisfy the conditions for any natural number $ n$\r\n\r\nEDIT: Nevermind I misunderstood the question", "Solution_2": "I should perhaps have been clearer but that's not what I menat by \"fully\" factorise. I meant they must factorise into linear factors\r\ni.e. P(x) = (a1 x + b1)(a2 x + b2) ... (an x + bn) for integers ai, bi, and similarly for Q(x)." } { "Tag": [ "probability" ], "Problem": "If you have 6 pizza toppings to choose from, how many different ways can you choose toppings to make a pizza with at least 3 toppings?\r\nIs the answer 6C3\r\n\r\nUsing the digits 1 through 7 once each, how many numbers can are there where the digits alternate between odd and even?", "Solution_1": "[hide]If you want a pizza with atleast 3 toppings, it can have either 3,4,5, or 6 toppings. Therefore your answer is 6C3+6C4+6C5+6C6\n\nyou must start and end with an odd. 4!3! [/hide]", "Solution_2": "[hide]If you have at least 3 toppings, then you could have 3,4,5, or 6 toppings. This can be done in $\\binom63+\\binom64+\\binom65+\\binom66=\\boxed{42}$ ways.\n\nThe digits must be arranged like:\n\noeoeoeo\n\nThere are $4!$ ways to rearrange the odd numbers and $3!$ ways to rearrange the even numbers. That makes $4!\\cdot3!=144$ total combinations.[/hide]" } { "Tag": [ "articles" ], "Problem": "I understand Google Chat is up and running. Anybody have any experience with it? Should I get it?\r\n\r\nAs I understand it, GC works with a Jabber-enabled client with server talk.google.com, username yourusername@gmail.com, and password yourpassword.", "Solution_1": "Where do I get it?", "Solution_2": "Can you post a link saying something about this, cause googled \"Google Chat,\" but I didn't find anything.", "Solution_3": "http://www.spreadhello.com/\r\n\r\nThat's the closest thing I know of. Uh...but it's not really \"Google Chat\". It's more \"Google's AIM-wannabe\".", "Solution_4": "Yea, I found that, but I didn't think that was what solafidefarms meant.", "Solution_5": "It isn't *officially* released yet, but you can use it with GAIM or iChat. Here's where I heard about it: be warned slashdot has some bad language: http://it.slashdot.org/it/05/08/23/2316223.shtml?tid=217&tid=218\r\n\r\nAlso, for the direct link to the article: http://www.smashsworld.com/2005/08/im-on-google-talk-right-now.php\r\n\r\nThere are some reports that it isn't working any more; ... but it should tomorrow.", "Solution_6": "Ah yes. Hmm...I'm trying to get it to work. Trillian fails me.", "Solution_7": "is it free?", "Solution_8": "trillian and gaim work for most ppl they can talk to msn and yahoo and aim", "Solution_9": "It is free, and is very good iff all your friends have gmail.", "Solution_10": "http://www.google.com/talk/index.html\r\n\r\nIt's alive! :-D And it also works on Trillian and Gaim now.", "Solution_11": "I had problems with Yahoo, I have no clue what gmail is. I like my MSN and AIM..." } { "Tag": [ "ratio", "summer program", "MathPath" ], "Problem": "where comes number of PI :arrow: ", "Solution_1": "It is the ratio of a circle's circumference to its diameter, which I would think is the simplest definition. It can be found in many places, including nature. Some formulas can be found [url=http://mathworld.wolfram.com/PiFormulas.html]here[/url].", "Solution_2": "http://en.wikipedia.org/wiki/Pi", "Solution_3": "$\\pi = 4\\arctan{1}$", "Solution_4": "PRETTY COOL!!!!!!!!", "Solution_5": "There is no need to revive the topic. \nYour comment added nothing.\nPlease refrain from reviving in the future.", "Solution_6": "3.14159265358979323846264338327950288419716939937510582097494459230781640628620\n899862803482534211706798214808651328230664709384460955058223172535940812848111\n745028410270193852110555964462294895493038196442881097566593344612847564823378\n6783165271201909145648566923460348610454326648213393607260249141273724587006606\n3155881748815209209628292540917153643678925903600113305305\n\nMessed up with LaTeX.\nMemorized all this, but if you only knew 3.14159, and memorized a digit a day, you could catch up in only a year. And I'm sure most, if not all of you could memorize this faster.", "Solution_7": "LOL countyguy thAts exactly how far i know.\nAt mathpath 11, someone recited 3321 digits." } { "Tag": [ "USAMTS" ], "Problem": "I am taking USAMTS for the 1st time this year.\r\nI know I shouldn't expect to do too well this year, but what is the average score?", "Solution_1": "I think the median is 40-50, according to the USAMTS webiste. Doing your best is the most important thing though :)" } { "Tag": [ "algebra", "function", "domain", "polynomial", "trigonometry", "calculus", "vector" ], "Problem": "Find f:R+->R+ such that\r\nf(f(x))+f(x)=2x", "Solution_1": "$ f(f(x))\\plus{}f(x)\\equal{}2x$\r\n\r\n$ f(f(f(x)))\\plus{}f(f(x))\\equal{}2f(x)$\r\n\r\n$ f(2x\\minus{}f(x))\\plus{}2x\\minus{}f(x)\\equal{}2f(x)$\r\n\r\n$ 2(2x\\minus{}f(x))\\equal{}2f(x)$\r\n\r\n$ 4x\\minus{}2f(x)\\equal{}2f(x)$\r\n\r\n$ f(x)\\equal{}x$\r\n\r\nHope it's correct!", "Solution_2": "[quote=\"milin\"]$ f(f(x)) \\plus{} f(x) \\equal{} 2x$\n\n$ f(f(f(x))) \\plus{} f(f(x)) \\equal{} 2f(x)$\n\n$ f(2x \\minus{} f(x)) \\plus{} 2x \\minus{} f(x) \\equal{} 2f(x)$\n\n$ 2(2x \\minus{} f(x)) \\equal{} 2f(x)$\n\n$ 4x \\minus{} 2f(x) \\equal{} 2f(x)$\n\n$ f(x) \\equal{} x$\n\nHope it's correct![/quote]\r\n\r\nHow do you go from $ f(2x \\minus{} f(x)) \\plus{} 2x \\minus{} f(x) \\equal{} 2f(x)$ to $ 2(2x \\minus{} f(x)) \\equal{} 2f(x)$ :?: \r\n :cool:", "Solution_3": "[quote=\"milin\"]$ f(f(x)) \\plus{} f(x) \\equal{} 2x$\n\n$ f(f(f(x))) \\plus{} f(f(x)) \\equal{} 2f(x)$\n\n[/quote]\r\n\r\nWhy does $ f$ need to be distributive?", "Solution_4": "I had an explanation in my mind when I did this, but I see now that it is wrong. :(", "Solution_5": "is it wrong?", "Solution_6": "I think this looks better.\r\n\r\n[hide=\"Solution.\"]\n$ f(f(x)) \\plus{} f(x) \\equal{} 2x$\n\nIf $ x$ is in the domain of $ f$, then $ f(x)$ is also in the domain of $ f$. Thus we can substitute $ f(x)$ for $ x$. We are [i]not[/i] assuming that $ f(x)\\equal{}x$:\n\n$ f(f(f(x)))\\plus{}f(f(x))\\equal{}2f(x)$\n\nSince $ f(f(x))\\plus{}f(x)\\equal{}2x\\implies f(x)\\equal{}2x\\minus{}f(f(x))$:\n\n$ f(f(x))\\plus{}2x\\minus{}f(f(x))\\equal{}2(2x\\minus{}f(f(x)))$\n\n$ 2x\\equal{}4x\\minus{}2f(f(x))$\n\n$ f(f(x))\\equal{}x$\n\nSubstitute this back into the original equation:\n\n$ f(f(x))\\plus{}f(x)\\equal{}2x\\implies x\\plus{}f(x)\\equal{}2x \\implies \\boxed{f(x)\\equal{}x}$\n\nThis easily checks as the solution.\n[/hide]", "Solution_7": "[quote=\"jvenezuela716\"]\n\n$ f(f(x)) \\plus{} 2x \\minus{} f(f(x)) \\equal{} 2(2x \\minus{} f(f(x)))$\n\n[/quote]The flaw is in this line.", "Solution_8": "[quote=\"10000th User\"][quote=\"jvenezuela716\"]\n\n$ f(f(x)) \\plus{} 2x \\minus{} f(f(x)) \\equal{} 2(2x \\minus{} f(f(x)))$\n\n[/quote]The flaw is in this line.[/quote]\r\n\r\nThe RHS should be 2x, which makes it an identity. Instead, $ f(x)$ was substituted for $ x$, so it is no surprise we end up with $ x \\equal{} f(x)$.\r\n\r\nI think its obvious the function $ f(x) \\equal{} x$ works. The question is are there others?\r\n\r\n[hide=\"ideas\"]\n$ f(f(x)) \\plus{} f(x) \\equal{} 2x$\n$ f(x) \\plus{} x \\equal{} 2f^{ \\minus{} 1}(x)$\nHence $ f(x) \\equal{} 2f^{ \\minus{} 1}(x) \\plus{} x$ and therefore $ f(f(x)) \\plus{} 2f^{ \\minus{} 1}(x) \\equal{} x$ so $ x \\minus{} f(x) \\plus{} 2f^{ \\minus{} 1}(x) \\equal{} 0$\n\nDon't exactly know what to do from here...\n\nIf we let $ f$ be a polynomial of degree $ n$, $ f(f(x))$ has degree $ n^2$. $ f(f(x)) \\plus{} f(x)$ can only be less than degree $ n^2$ if the leading terms can cancel, which is only remotely possible if $ n \\equal{} n^2$ for which $ n \\equal{} 0$ or $ n \\equal{} 1$, both trivial cases.\n\nIf $ n \\equal{} 0$, $ f(x) \\equal{} c \\implies 2c \\equal{} 2x$ which is not true for all $ x$'s. In the other case, we know that $ n \\equal{} 1 \\implies f(x) \\equal{} ax \\plus{} b \\implies a(ax \\plus{} b) \\plus{} b \\plus{} ax \\plus{} b \\equal{} 2x \\implies (a^2 \\plus{} a)x \\plus{} ab \\plus{} 2b \\equal{} 2x$ and equating coefficients $ a^2 \\plus{} a \\minus{} 2 \\equal{} 0 \\implies (a \\plus{} 2)(a \\minus{} 1) \\equal{} 0$ so $ a \\equal{} \\minus{} 2$ or $ a \\equal{} 1$. Additionally, we have $ ab \\plus{} 2b \\equal{} 0 \\implies b(a \\plus{} 2) \\equal{} 0$ so either $ b \\equal{} 0$ (as in the case of $ a \\equal{} 1$) or $ a \\equal{} \\minus{} 2$, as previously shown.\n\nHence the only polynomials that satisfy the condition are $ y \\equal{} x$ and $ y \\equal{} \\minus{} 2x \\plus{} b$ for all $ b$. Looking back on the conditions, for the second case we must require b>0 and restrict the domain such that $ 0 < \\minus{} 2x \\plus{} b \\implies x < b/2$ so the domain is $ (0,b/2)$ so that the range is $ (0,b)$, while for $ y \\equal{} x$ the domain and the range are $ (0,\\infty)$.\n[/hide]", "Solution_9": "Thank [color=darkblue][b]facis[/b][/color]!", "Solution_10": "thanhnam2902, no problem. However, I think I should remind you that I didn't prove that those were the only solutions but the only polynomial solutions - a pretty important category of functions, but not the only category...\r\n\r\nJust a thought: from Taylor series, we know that most functions can be written as polynomials with an infinite number of terms. Does my work with degrees of polynomials say anything about a Taylor series?\r\n\r\nIf so, then we can rule out exponential, trig, logarithmic functions, functions with x (or functions of x) raised to rational powers, etc - all functions that can be written as Taylor series.\r\n\r\nA less general generalization could deal with extending my work to include all rational functions - certainly doable, though much more complicated.", "Solution_11": "let a,b,c,d >0 Find MIN\r\n$ (a^2\\plus{}b^2\\plus{}10)(c^2\\plus{}d^2\\plus{}10)\\minus{}(a\\plus{}b)(c\\plus{}d)$", "Solution_12": "[quote=\"facis\"]\nThe RHS should be 2x, which makes it an identity. Instead, $ f(x)$ was substituted for $ x$, so it is no surprise we end up with $ x \\equal{} f(x)$.[/quote]\n\nI suppose that is a mistake (still thinking about it), but I don't think your logic is any better:\n\n[quote=\"facis\"]\n$ f(f(x)) \\plus{} f(x) \\equal{} 2x$\n$ f(x) \\plus{} x \\equal{} 2f^{ \\minus{} 1}(x)$[/quote]\r\n\r\nAre you not also assuming that $ x\\equal{}f^{\\minus{}1}(x)$? In fact, you are making an even larger assumption that $ f$ has an inverse in the first place, which need not be the case. (Consider $ f(x)\\equal{}(x\\minus{}2)^2\\implies f(1)\\equal{}f(3)\\equal{}1 \\implies f^{\\minus{}1}(1)$ does not exist.)", "Solution_13": "@jvenezuela716: The point of the reasoning above is more like let $ x\\equal{}f(y)$ but then it really doesn't matter, once we substitute this in everywhere in an equation, if we call the variable $ x$ or $ y$ (but it has caused quite some confusion). I know I did try to make the assumption that $ f$ has an inverse, but i really didn't get anywhere doing it. I was just trying not to get buried in $ f(f(f(...)))$'s but it didn't help at all.\r\n\r\nThe mistake is that somewhere in all the attempted proofs that claim $ f(x)\\equal{}x$ as the only solution is that somewhere, someone accidentally mistook $ x$ for $ f(x)$ or vice -versa. For instance, you took $ f(f(x)) \\plus{} f(x) \\equal{} 2x$ and, thinking of the case where $ f$ had been substituted in for $ x$, proceeded as if $ f(f(x)) \\plus{} f(x) \\equal{} 2f(x)$, thus replacing $ f(x)$ with $ x$ in one part of the equation but not the whole equation, which basically worked as if you have assumed $ x\\equal{}f(x)$ and then showed that it worked. (Specifically, you have $ f(x) \\equal{} 2x\\minus{}f(f(x))$ and then $ f(f(x))\\plus{}2x\\minus{}f(f(x))\\equal{}2(2x\\minus{}f(f(x)))$ which implies $ f(f(x)) \\plus{} f(x) \\equal{} 2f(x)$, not the original equation).\r\n\r\nI don't think it is possible to manipulate just notation with $ f$'s to solve for the function unless we know more about the function, but perhaps different types of functions that f could be - the more general, the better. For instance, if we were given $ f$ is a polynomial, then $ f(x)\\equal{}x$ or $ f(x)\\equal{}\\minus{}2x\\plus{}b$ iff $ b>0$ and the domain and range restrictions I previously showed are the only possiblities.", "Solution_14": "If there are more solutions besides the polynomial ones, my guess is that they are not continuous anywhere. I'm going to keep working on this problem.", "Solution_15": "MellowMelon that wouldn't surprise me much either, though I would only go as far to say not differentiable. I guess the thing to do would be to stick this in a calculus thread with a link here and ask what they think about my Taylor series idea. Taylor series require infinite differentiability (and must converge), so if no Taylor series is possible, it might (not quite sure what the convergence would mean) rule out infinitely differentiable functions. Anyway, I'll post a link in one of those forums.\r\n\r\nedit: Hope someone comes up with something: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1115411#1115411[/url]", "Solution_16": "In fact $ f(x) \\equal{} x$ is the only solution. \r\n\r\n[hide]\nFix $ x$ and just consider the sequence \n$ x,f(x),f(f(x)),f(f(f(x))),\\ldots$ of positive numbers. \nWhat does the functional equation tell you about this sequence?\n[/hide]", "Solution_17": "@solaris: I think you are missing something. For instance, consider $ y \\equal{} \\minus{} 2x \\plus{} b$ for all $ b > 0$ with domain $ (0,b/2)$ and range $ (0,b)$. $ f(x) \\equal{} x$ is certainly a solution, but not the only solution. All the sequence with $ f$'s shows is that we have a sequence of positive numbers. It is not necessary for this sequence to always be increasing, so I don't really see what's special here.\r\n\r\n\r\nAt the least, I'm going to attempt rational functions.\r\n\r\nLet $ f(x) \\equal{} p(x)/q(x)$ where $ p$ and $ q$ are polynomials with degrees $ m$ and $ n$ respectively. As $ x$ appoachs $ \\pm \\infty$, $ f(x)$ approaches a 'polynomial' with degree $ m \\minus{} n$ or a function that behaves like some multiple of $ x^{m \\minus{} n}$ ($ n > m$). Therefore, $ f(f(x)) \\equal{} p(p/q)/q(p/q)$ approaches a degree $ (m \\minus{} n)^2$ polynomial. As long as $ m \\ge n$ we can only have leading terms of $ f(f(x)) \\plus{} f(x)$ cancel to give a linear term if $ (m \\minus{} n)^2 \\equal{} (m \\minus{} n)$ for which $ m \\minus{} n \\equal{} 1$ or $ m \\minus{} n \\equal{} 0$ (no linear term here). So we must consider rational functions such that $ m \\equal{} n \\plus{} 1$. In the cases when $ n < m$ there cannot be a linear term in $ f(f(x)) \\plus{} f(x)$ so we only need to worry about $ m \\equal{} n \\plus{} 1$. (The linear cases I proved earlier are given by $ n \\equal{} 0$).", "Solution_18": "Let's define $ g(x) \\equal{} \\frac {f(x)}x$ . It's easy to see that $ g(x)\\equal{}\\frac 2{1\\plus{}g(f(x))}$ . Let's define the inferior & superior bounds : $ I \\equal{}\\inf_{x>0}(g(x)) \\leq S \\equal{}\\sup_{x>0}(g(x))$\r\n$ I \\leq g(f(x))\\leq S\\ \\Longrightarrow \\ \\frac 2{1\\plus{}I}\\geq \\frac 2{1\\plus{}g(f(x))}\\equal{}g(x) \\geq \\frac 2{1\\plus{}S}$ $ \\ \\Longrightarrow\\ \\frac 2{1\\plus{}I}\\geq S\\geq I\\geq \\frac 2{1\\plus{}S}$ $ \\ \\Longrightarrow\\ I\\plus{}IS \\geq 2\\geq S\\plus{}IS$\r\nFinally $ I\\equal{}S\\equal{}1$ and $ f(x)\\equal{}x$ ... do you follow me ? :cool:", "Solution_19": "What the heck? Stop making bogus solutions that claim $ f(x)$ must be $ x$. There are other possibilities for $ f$, like $ \\minus{} 2x$ plus any constant.\r\n\r\nEDIT: Sorry, that was too harsh; the other posts in this topic had already irritated me. You have made some good progress there, but $ I \\equal{} S \\equal{} \\minus{} 2$ is another solution to the last inequality. There are others too, since $ f(x) \\equal{} \\minus{}2x$ isn't the only other solution.", "Solution_20": "[quote=\"MellowMelon\"]What the heck? Stop making bogus solutions that claim $ f(x)$ must be $ x$. There are other possibilities for $ f$, like $ \\minus{} 2x$ plus any constant.\n\nEDIT: Sorry, that was too harsh; the other posts in this topic had already irritated me. You have made some good progress there, but $ I \\equal{} S \\equal{} \\minus{} 2$ is another solution to the last inequality. There are others too, since $ f(x) \\equal{} \\minus{} 2x$ isn't the only other solution.[/quote]\r\n\r\nThat's not possible because of $ f: R^\\plus{}\\rightarrow R^\\plus{}$ :rotfl:", "Solution_21": "Oh geez, I probably could have solved it if I had seen that. :blush: Well, do you have any ideas if we extend it to $ \\mathbb{R}$?", "Solution_22": "@facis: well yes, but the given domain is $ (0,\\infty)$ and the given range is $ (0,\\infty)$, so you \r\nare considering a different problem (which may be more interesting).\r\n\r\n@Diogene: I guess my solution goes along similar lines: Fix $ x$ and consider the sequence \r\n$ a_n : \\equal{} f^n(x) : \\equal{} f(f( \\ldots f(x) \\ldots ))$, then the functional equation gives a linear recurrence \r\nfor $ a_n$, which can be solved explicitly (in terms of $ a_0$ and $ a_1$), and from the explicit \r\nformula it is evident that the sequence has to contain negative reals if $ a_0 \\neq a_1$.\r\n\r\n@Mellowmelon: If you consider the same problem on the whole set of reals, you get much more \r\nsolutions: Consider a base $ B$ of the reals as a vector space over the rationals, and define $ f$ \r\non all 1-dimensional subspaces spanned by a $ b \\in B$ separately (e.g. as $ f(x) \\equal{} x$ or as \r\n$ f(x) \\equal{} \\minus{}2x$). Then the resulting $ f$ satisfies the equation. But somehow I would guess that there \r\nare a lot more solutions than these, so I am slightly pessimistic that one can find all solutions \r\nto your problem if there are no further conditions on $ f$ like continuity or monotonicity. \r\nHowever, if you have additional conditions a lot depends on what these conditions are. \r\nE.g. if you restrict your attention to functions that are increasing, it is obivous that $ f(x) \\equal{} x$ \r\nis the only solution. If you would rather like to focus on continuous functions \r\nit is getting much harder, I guess ...", "Solution_23": "[quote=\"gagarin\"]Find f:R+->R+ such that\nf(f(x))+f(x)=2x[/quote]\r\n\r\nLet's define $ a_0 \\equal{} x$, $ a_1 \\equal{} f(x)$, $ a_n \\equal{} f(f(...f(x)...))$(n times).\r\nThan the recursive relation holds: $ a_n \\plus{} a_{n \\minus{} 1} \\equal{} 2a_{n \\minus{} 2}$. The characteristic equation is $ \\alpha^2 \\plus{} \\alpha \\minus{} 2 \\equal{} 0$, with solutions $ 1$ and $ \\minus{} 2$. So we have $ a_n \\equal{} c_1 \\plus{} c_2*( \\minus{} 2)^n$, with $ c_1,c_2$ being real contstants. But since the co-domain of our function is $ \\mathbb{R}^ \\plus{}$ obviously $ c_2 \\equal{} 0$. So $ a_n \\equal{} c_1, \\forall n \\in \\mathbb{N}_0$. And since $ a_0 \\equal{} x$, we get $ c_1 \\equal{} x$, so $ f(x) \\equal{} x$ is the only solution.", "Solution_24": "[quote=\"solyaris\"]@facis: well yes, but the given domain is $ (0,\\infty)$ and the given range is $ (0,\\infty)$, so you \nare considering a different problem (which may be more interesting).\n\n@Diogene: I guess my solution goes along similar lines: Fix $ x$ and consider the sequence \n$ a_n : \\equal{} f^n(x) : \\equal{} f(f( \\ldots f(x) \\ldots ))$, then the functional equation gives a linear recurrence \nfor $ a_n$, which can be solved explicitly (in terms of $ a_0$ and $ a_1$), and from the explicit \nformula it is evident that the sequence has to contain negative reals if $ a_0 \\neq a_1$.\n[/quote]\r\n\r\nSolyaris, I interpreted f:R+->R+ as meaning for all real positive numbers in the range we get a real positive number, not that the domain and range are required to include all real positives. \r\n\r\nBehemont, how do you go from $ a_n \\plus{} a_{n \\minus{} 1} \\equal{} 2a_{n \\minus{} 2}$ to $ a^2 \\plus{} a \\minus{} 2 \\equal{} 0$? I'm not very good with finding generating functions. On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.\r\n\r\nDiogene, I do not follow. Perhaps I just can't think well right now but I do not see where you get your formula for $ g(x)$ based on $ g(f(x))$.", "Solution_25": "[quote=\"facis\"]\nSolyaris, I interpreted f:R+->R+ as meaning for all real positive numbers in the range we get a real positive number, not that the domain and range are required to include all real positives. \n[/quote]\n\nI see. Well, the usual interpretation of $ f: A \\to B$ is that $ f$ has to be defined on all of $ A$ and \nall values have to be in $ B$. (This is what I meant in my post.)\n\n[quote=\"facis\"]\nBehemont, how do you go from $ a_n \\plus{} a_{n \\minus{} 1} \\equal{} 2a_{n \\minus{} 2}$ to $ a^2 \\plus{} a \\minus{} 2 \\equal{} 0$? I'm not very good with finding generating functions. On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.\n[/quote]\r\n\r\nIf you want to avoid the theory of recurrence equations, you can simply rewrite the equation as \r\n$ a_n \\minus{} a_{n \\minus{} 1} \\equal{} \\minus{} 2(a_{n \\minus{} 1} \\minus{} a_{n \\minus{} 2})$, and writing $ b_n \\equal{} a_n \\minus{} a_{n \\minus{} 1}$ this gives \r\n$ b_n \\equal{} \\minus{} 2 b_{n \\minus{} 1}$, i.e. $ b_n \\equal{} ( \\minus{} 2)^{n \\minus{} 1} b_1$, so \r\n$ a_n \\minus{} a_0 \\equal{} b_n \\plus{} \\ldots \\plus{} b_1 \\equal{} (( \\minus{} 2)^{n \\minus{} 1} \\plus{} \\ldots \\plus{} 2^1 \\plus{} 2^0) b_1 \\equal{} \\frac {1 \\minus{} ( \\minus{} 2)^n}{3}(a_1 \\minus{} a_0)$. \r\nAs $ a_n > 0$ for all $ n$ and $ \\frac {1 \\minus{} ( \\minus{} 2)^n}{3}$ can take arbitrarily large positive and negative values, this implies $ 0 \\equal{} a_1 \\minus{} a_0 \\equal{} f(x) \\minus{} x$.\r\n\r\nThere is no need to go \"from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$\". It is rather \r\nthat from a general property of the sequence (positive terms) you conclude something about the \r\nstarting values of the sequence $ a_0,a_1$.", "Solution_26": "solyaris, very well explained.\r\n\r\nAs for 'going from $ f^n$ back to $ f$' I meant that if we ignore the f:R+->R+ condition (as I had done, since I had never seen this specific notation before) one way to find all possibilities for $ f$ would be to find all functions such that $ f^n(x) \\equal{} \\frac {1 \\minus{} ( \\minus{} 2)^n}{3}(f(x) \\minus{} x) \\plus{} x$. I see that the format of the answer requires that $ a_1\\minus{}a_0\\equal{}f(x)\\minus{}x\\equal{}0$ because the other factor diverges and would guaruntee a negative value for some sufficiently large $ x$ otherwise.\r\n\r\nAs for \"avoiding the theory of recurrence equations\" I'm not sure it's necessary, I just haven't heard of it before and I am not very good at finding generating functions (which I believe is what you are referring to) - probably more a lack of experience than inability. My best guess is that $ a^2\\plus{}a\\minus{}2\\equal{}0$ is being solved to find a partial fraction decomposition. And then the decomposition is changed into an infinite geometric series, and the coefficients of each term equated to the values of the sequence. But I have no clue how to get to the expression to do the decomposition on.", "Solution_27": "[quote=\"facis\"]\n\nBehemont, how do you go from $ a_n \\plus{} a_{n \\minus{} 1} \\equal{} 2a_{n \\minus{} 2}$ to $ a^2 \\plus{} a \\minus{} 2 \\equal{} 0$? I'm not very good with finding generating functions. On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.\n\n[/quote]\r\n\r\nits a known method for solving recurrence equations..for the second question, read my solution once again.. :)", "Solution_28": "@Behemont: As I said right above your post...\r\n\r\n[quote=\"facis\"]\nAs for 'going from $ f^n$ back to $ f$' I meant that if we ignore the f:R+->R+ condition (as I had done, since I had never seen this specific notation before) one way to find all possibilities for $ f$ would be to find all functions such that $ f^n(x) \\equal{} \\frac {1 \\minus{} ( \\minus{} 2)^n}{3}(f(x) \\minus{} x) \\plus{} x$.\n[/quote]\r\n\r\nI completely understand how this problem works out with the condition $ f: R\\minus{}>R$ but what I meant was assuming we ignore this condition, how could we find all possible solutions? Thoughts?" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Given sequence u_{n} determined by \r\n$ u_{0} \\equal{} 6;u_{1} \\equal{} 42$ and $ u_{n \\plus{} 2} \\equal{} u_{n \\plus{} 1} \\plus{} 6u_{n} \\plus{} 6n$\r\nFind the general formula of $ u_n$", "Solution_1": "[u]xref4search [/u]: closed general form of sequence u_{n + 2} = u_{n + 1} + 6u_{n} + 6n[quote=\"nkht-tk14\"]Given sequence u_{n} determined by \n$ u_{0} \\equal{} 6;u_{1} \\equal{} 42$ and $ u_{n \\plus{} 2} \\equal{} u_{n \\plus{} 1} \\plus{} 6u_{n} \\plus{} 6n$\nFind the general formula of $ u_n$[/quote]\r\n\r\nSo $ (u_{n\\plus{}2}\\plus{}n\\plus{}2\\plus{}\\frac 16)\\equal{}(u_{n\\plus{}1}\\plus{}n\\plus{}1\\plus{}\\frac 16)\\plus{}6(u_n\\plus{}n\\plus{}\\frac 16)$\r\n\r\nSo $ u_n\\plus{}n\\plus{}\\frac 16\\equal{}\\frac{111}{10}3^n\\minus{}\\frac{74}{15}(\\minus{}2)^n$\r\n\r\nSo $ u_n\\equal{}\\frac{111}{10}3^n\\minus{}\\frac{74}{15}(\\minus{}2)^n\\minus{}n\\minus{}\\frac 16$", "Solution_2": "Your answer is incorrect.\r\n\r\nWhy don't you make a confirmation for $ n\\equal{}0$?", "Solution_3": "[quote=\"kunny\"]Your answer is incorrect.\n\nWhy don't you make a confirmation for $ n \\equal{} 0$?[/quote]\r\n\r\nMy anwer is correct.\r\n\r\nWhy dont you make a confirmation of your counter-examples ?\r\n\r\n$ u_n \\equal{} \\frac {111}{10}3^n \\minus{} \\frac {74}{15}( \\minus{} 2)^n \\minus{} n \\minus{} \\frac 16$\r\n\r\nSo $ u_0 \\equal{} \\frac {111}{10} \\minus{} \\frac {74}{15} \\minus{} \\frac 16$ $ \\equal{} \\frac {333}{30} \\minus{} \\frac {148}{30} \\minus{} \\frac {5}{30}$ $ \\equal{} \\frac {333 \\minus{} 148 \\minus{} 5}{30}$ $ \\equal{} \\frac {180}{30} \\equal{} 6$\r\n\r\nI made my confirmations only for $ n\\equal{}0$, $ n\\equal{}1$, $ n\\equal{}2$, $ n\\equal{}3$ and $ n\\equal{}4$ and considered that is was enough.", "Solution_4": "Sorry, my calculation iwas incorrrect. :blush:" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "geometry proposed" ], "Problem": "$\\triangle ABC$ is acute with altitudes $h_a,h_b,h_c.$ $R,r,r_0$ denote the radii of its circumcircle, incircle and incircle of its orthic triangle. Prove synthetically the relation\n\\[h_a \\plus{} h_b \\plus{} h_c \\equal{} 2R \\plus{} 4r \\plus{} r_0 \\plus{} \\frac{r^2}{R} \\]", "Solution_1": "Hint: If $H$ is the orthocenter, show that $HA+HB+HC=2(R+r).$ Also, keep in mind that $\\odot(HBC),$ $\\odot(HCA),$ $\\odot(HAB)$ are congruent to $\\odot(ABC).$\n\nHint for a trigonometric solution: Use the identity $r_0=2R \\cos A \\cos B \\cos C.$", "Solution_2": "$ \\|\\ \\sum h_a = 2R + 4r + r_o + \\frac {r^2}{R}\\ \\|\\ \\odot\\ 2R$ $ \\Longleftrightarrow$ $ \\|\\ \\begin{array}{c} \\sum bc = 4R^2 + 8Rr + 2Rr_0 + 2r^2 \\\\\n \\\\\n(\\ \\begin{array}{c} bc + ca + ab = p^2 + r^2 + 4Rr \\\\\n \\\\\nr_0 = 2R\\cos A\\cos B\\cos C\\end{array}\\ )\\end{array}\\ \\|$ $ \\Longleftrightarrow$ \n\n$ \\boxed {\\ p^2 = (2R + r)^2 + 2Rr_0\\ }$ $ \\Longleftrightarrow$ $ \\cos A\\cos B\\cos C=\\frac {p^2-(2R+r)^2}{4R^2}\\ .$ [b]Nice relation ![/b]", "Solution_3": "$ \\|\\ \\sum h_a = 2R + 4r + r_o + \\frac {r^2}{R}\\ \\|\\ \\odot\\ 2R$ $ \\Longleftrightarrow$ $ \\|\\ \\begin{array}{c} \\sum bc = 4R^2 + 8Rr + 2Rr_0 + 2r^2 \\\\\n \\\\\n(\\ \\begin{array}{c} bc + ca + ab = p^2 + r^2 + 4Rr \\\\\n \\\\\nr_0 = 2R\\cos A\\cos B\\cos C\\end{array}\\ )\\end{array}\\ \\|$ $ \\Longleftrightarrow$ \n\n$ \\boxed {\\ p^2 = (2R + r)^2 + 2Rr_0\\ }$ $ \\Longleftrightarrow$ $ \\cos A\\cos B\\cos C = \\frac {p^2 - (2R + r)^2}{4R^2}$ , what is truly.\n\n[b][u]Remark[/u].[/b] A triangle $ ABC$ is nonobtuse $ \\Longleftrightarrow\\ \\sum\\sin A\\ \\ge\\ 1+\\sum\\cos A\\ \\Longleftrightarrow\\ p\\ \\ge\\ 2R+r\\ .$", "Solution_4": "Dear Mathlinkers,\n\n[url= http://jl.ayme.pagesperso-orange.fr/Docs/Trigonometrie.pdf] here [/url] P. 23...\n\nSincerely\nJean-Louis" } { "Tag": [ "percent" ], "Problem": "Fifteen is twenty percent of what number?", "Solution_1": "if 15 is twenty percent of a number then we can write the equation 15 = $ \\frac{20x}{100}$ you may simplify this to x/5 but either way you still arrive at x=75", "Solution_2": "Generally, to solve percent problems, we use this equation:\r\n\r\n$ \\text{PART} \\equal{} \\text{PERCENT} \\times \\text{WHOLE}$\r\n\r\nTo solve percents, we substitute the numbers we know. So let's use the problem as an example. $ 15$ is the part, and $ 20$ is the percent. We're solving for the whole.\r\n\r\n$ 15 \\equal{} 0.20 \\times \\text{WHOLE}$\r\n\r\n(Remember that $ 20\\%$ is $ 0.20$ or $ 0.2$.)\r\n\r\n$ \\frac{15}{0.20} \\equal{} \\text{WHOLE}$\r\n\r\n$ 75 \\equal{} \\text{WHOLE}$\r\n\r\nSo our answer is $ \\boxed{75}$." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that there exists a infinite number of integers $ m$ satisfying $ m! \\equal{} 2^p(2k \\plus{} 1)$ $ (m,p,k\\in \\mathbb N)$ and $ m \\minus{} p \\equal{} 1^2 \\plus{} 2^2 \\plus{} \\cdots \\plus{} 2005^2$", "Solution_1": "Your titles are completely useless. Use titles that describe the problem! (No, \"m,p,k\" doesn't, thats why it got that new title by me; same for the other two such topics :wink: )", "Solution_2": "Let $ c>0$ be an integer and $ m\\equal{}\\sum_{i\\equal{}1}^c2^{n_i}$, where $ n_1>n_2>\\cdots>n_c>0$ are integers.\r\nWe can show \r\n$ \\sum_{j\\equal{}1}^{\\infty}[\\frac{m}{2^j}]\\equal{}\\sum_{i\\equal{}1}^c(2^{n_i}\\minus{}1)\\equal{}m\\minus{}c$.\r\nObviously we can let $ c\\equal{}1^2\\plus{}2^2\\plus{}\\cdots\\plus{}2005^2$ and $ p\\equal{}m\\minus{}c$.", "Solution_3": "[quote=\"ZetaX\"]Your titles are completely useless. Use titles that describe the problem! (No, \"m,p,k\" doesn't, thats why it got that new title by me; same for the other two such topics :wink: )[/quote]\r\nThe current title is really funny. :P :rotfl: . Do you want to spread consciousness for good naming by it? :D" } { "Tag": [], "Problem": "If a six sided dice is rolled, how many times would we expect to roll it until all numbers appear at least once?What if we extend this to an n sided dice?", "Solution_1": "[hide=\"Hint\"]\nThis can be broken into pieces. After rolling one number, on average how many turns will it take to roll a different number? After rolling two different numbers, how long on average will it take to roll a third different number? Continue this.[/hide]\n[hide=\"Answer\"]I get $ 6(\\frac {1}{6} \\plus{} \\frac {1}{5} \\plus{} \\frac {1}{4} \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{2} \\plus{} \\frac {1}{1}) \\equal{} 14.7$ The pattern is obvious for the extended problem.[/hide]", "Solution_2": "how did you get the last equation?", "Solution_3": "Let's say, for example, you've already chosen four different numbers. You still need to choose one of 2, which you have a 2/6=1/3 chance of. Thus, it should take 3=6/2 turns to do this (the rigorous way to prove this is using an infinite series). So the answer is 6/6+6/5+6/4+6/3+6/2+6/1.", "Solution_4": "can you show the inifinite series proof?", "Solution_5": "[quote=\"kl2836\"]can you show the inifinite series proof?[/quote]\r\n\r\nNo, the rigorous way does not have to use infinite series. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228627]here[/url].\r\n\r\nIn fact, the infinite series method is inelegant - who would want to read it? The preferred method is recursion.", "Solution_6": "Here is the infinite series method anyways, after reading yongyi's post.\r\nOk, in the example I gave, there is a 2/6 chance it will take one turn, $ 2/6 \\cdot 4/6$ chance it will take two, $ 2/6 \\cdot (4/6)^2$ chance it will take 3, etc. So it should take $ 1(2/6) \\plus{} 2(4/6)(2/6) \\plus{} 3(4/6)^2(2/6) \\plus{} 4(4/6)^3(2/6)....$ turns on average. Let this equal x. It is clear $ x \\minus{} (4/6)x \\equal{} x$, from which we find $ x \\equal{} 6/2 \\equal{} 3$." } { "Tag": [ "email" ], "Problem": "Excuse me, anybody knows how to subscribe for Mathematics and Informatics Quarterly magazine, because I've already contacted 2 email address at the website http://olympiads.win.tue.nl but they haven't replied my email until now. If somebody have subscribe this magazine could you tell me how to subscribe that? Thanks", "Solution_1": "is m&ic a really good magazine? what's in it?", "Solution_2": "I don't know it, but this is a math magazine for high school student, of course, there's many fun and nice problem. Actually, Bulgaria is a good country in mathematics." } { "Tag": [ "search", "floor function", "arithmetic sequence", "number theory proposed", "number theory" ], "Problem": "For real positive numbers $x,$ the set $A(x)$ is defined by \r\n\\[A(x)=\\{[nx]\\ | \\ n\\in{ \\mathbb{N}}\\},\\] where $[r]$ denotes the greatest integer not exceeding real numbers $r.$Find all irrational numbers $\\alpha >1$ satisfying the following condition.\r\nCondition: If positive real number $\\beta$ satisfies $A(\\alpha)\\supset A(\\beta),$ then $\\frac{\\beta}{\\alpha}$ is integer.", "Solution_1": "I get a contradictory thing:\r\n\r\nLet $\\alpha$ be such number. There exist $x\\in R$ such that\r\n$\\frac{1}{\\alpha}+\\frac{1}{x}=1$.\r\nTake also irrational numbers $\\beta_{n}$ and $\\frac{x}{n}$ such that\r\n$\\frac{1}{\\beta_{n}}+\\frac{1}{x/n}=1$.\r\nFor any $n \\in N$ we have sequence $[kx]$ is in sequence $[k\\cdot x/n]$.\r\nFrom Beatty Theorem both sequences cover $N$ with no overlapping,\r\ntherefore sequence $[k\\beta_{n}]$ is in sequence $[k\\alpha]$.\r\nThus $\\frac{\\beta_{n}}{\\alpha}$ is an integer for any $n, \\text{ }\\Rightarrow$\r\n$\\frac{x-1}{x}\\cdot \\frac{x}{x-n}=\\frac{x-1}{x-n}$ is an integer for any $n$. The only possibility for that $x=1$, it follows that there does not exist\r\nsuch $\\alpha$. Maybe I misunderstood something.", "Solution_2": "You have to ensure that $\\frac{x}{n}>1$! (which is seldom a case, for any $\\alpha >2$ and $n>1$ we have $\\frac{x}{n}<1$). But your proof implies that if $\\alpha$ exists at all then $\\alpha >2$.", "Solution_3": "You are right Megus, let me try again.\r\n\r\nFirst step: Let $\\alpha$ be such number. There exist $x\\in R$ such that $\\frac{1}{\\alpha}+\\frac{1}{x}=1$.\r\nTake also irrational numbers $\\beta_{n}$ and $\\frac{x}{n}$ such that $\\frac{1}{\\beta_{n}}+\\frac{1}{x/n}=1$.\r\nFor any $n \\in N$ we have sequence $[kx]$ is in sequence $[k\\cdot x/n]$.\r\n\r\nFrom Beatty Theorem both sequences cover $N$ with no overlapping, therefore sequence $[k\\beta_{n}]$ is in sequence $[k\\alpha]$.\r\nThus $\\frac{\\beta_{n}}{\\alpha}$ is an integer for any $n, \\text{ }\\Rightarrow$ $\\frac{x-1}{x}\\cdot \\frac{x}{x-n}=\\frac{x-1}{x-n}$ is an integer for any $n \\geq 2$. \r\nContradiction, it follows that $\\frac{x}{n}\\leq 1$, for $n \\geq 2$, thus $x \\leq 2$ and therefore $\\alpha > 2$.\r\n\r\nSecond step: We search all $\\alpha >2$, for which there exist $\\beta$ such that $A(\\beta) \\subset A(\\alpha)$ and $\\frac{\\beta}{\\alpha}$ is not an integer.\r\nLet $\\beta=\\alpha(k+x)$ where $k \\in \\mathbb{Z}$ and $x\\in (0,1)$. Observe that for all $n$, $n\\beta=kn\\alpha+xn\\alpha$\r\n\r\nThus $kn\\alpha+[xn]\\alpha\\leq kn\\alpha+xn\\alpha \\leq kn\\alpha+([xn]+1)\\alpha$ $\\text{ }\\text{ }\\text{ (1)}\\text{ }$ \r\n\r\nand $n\\beta$ is between these two consecutive multiples of $\\alpha$.\r\n\r\nThe main condition is not satisfied when we can find $n$ such that\r\n\r\n$xn\\alpha-[xn]\\alpha \\geq 1$ and $([xn]+1)\\alpha-xn\\alpha \\geq 1$.\r\n\r\nIf $x$ is irrational then $\\{xn\\}$ is dence in $(0,1)$ and there exist $n$ such that $\\frac{1}{2}\\geq xn-[xn]\\geq \\frac{1}{\\alpha}$\r\nand it satisfies both conditions, as $\\alpha > 2$.\r\n\r\nThus $x$ is rational and therefore if there exist such $\\beta$, then $\\beta/\\alpha$ is rational number.\r\n\r\nThird step: Let $x=\\frac{p}{q}$, where $(p,q)=1$. We want to do not be captured by interval $[\\frac{1}{\\alpha},\\frac{1}{2}]$.\r\n\r\nIf $q$ is even, $q =2q_{1}$, then $xn-[xn]=\\frac{p}{2q_{1}}-[n\\frac{p}{2q_{1}}]$. \r\nTaking $n=q_{1}$ we have thatt $\\frac{1}{2}= n\\frac{p}{2q_{1}}-[n\\frac{p}{2q_{1}}]=\\frac{p}{2}-[\\frac{p}{2}]$, contradiction.\r\n\r\nIf $q$ is odd $q =2q_{1}+1$ then we can find $n$ such that $\\{\\frac{np}{q}\\}= \\frac{q_{1}}{2q_{1}+1}$.\r\nAs we do not want to be captured by the interval $[\\frac{1}{2},\\frac{1}{\\alpha}]$ we need $\\frac{1}{\\alpha}> \\frac{q_{1}}{2q_{1}+1}$, thus $\\frac{2q_{1}+1}{q_{1}}>\\alpha$. \r\nBecause we need the existance of such $q_{1}$ and $q_{1}\\geq 1$, there might exist a construction for $\\beta$ iff $\\alpha < 3$.\r\n\r\nFourth step: $\\frac{\\beta}{\\alpha}=\\frac{u}{v}$, where $(u,v)=1$ and we know $v$ is odd, $\\Rightarrow$ $v\\geq 3 > \\alpha$.\r\n\r\nConsider $l$, inverse of $u$ modulo $v$, $0< l < v$ and $lu=mv+1$, where $m$ is fixed.\r\nConsider numbers $n$ of the form $n=kv+l$, then\r\n$n\\beta=(kv+l)\\frac{u}{v}\\alpha=(k+m)\\alpha+\\frac{1}{v}\\alpha$\r\nAs $\\alpha$ is irrational and $\\{x\\alpha\\}$, for $x \\in \\mathbb{Z}$ is dense in $(0,1)$ there exist $k$ such that $1-\\frac{\\alpha}{v}\\leq \\{(k+m)\\alpha\\}\\leq 1$, then\r\n$[(k+m)\\alpha] < [(k+m)\\alpha+\\frac{1}{v}\\alpha]=[n\\beta]$\r\nAlso $[n\\beta] < [(k+m+1)\\alpha]$ because $(k+m+1)\\alpha-\\left((k+m)\\alpha+\\frac{1}{v}\\alpha\\right)=\\frac{v-1}{v}\\alpha \\geq 1$, ($v \\geq 2, \\alpha > 2$).\r\n\r\nIt follows that all irrational numbers $\\alpha > 2$ are good. It means that if there exist $\\beta$ such that $A(\\beta) \\subset A(\\alpha)$ then $\\frac{\\beta}{\\alpha}$ is an integer.", "Solution_4": "I think there is a simpler way to show that for $\\alpha>2$, $A(\\beta)\\subseteq A(\\alpha)\\implies\\beta/\\alpha\\in\\mathbb{Z}$.\n\nAssume for the sake of contradiction that $\\beta/\\alpha\\notin\\mathbb{Z}$ but $A(\\beta)\\subseteq A(\\alpha)$; then for every $n\\in\\mathbb{N}$, there exists $m$ such that $\\lfloor{m\\alpha}\\rfloor=\\lfloor{n\\beta}\\rfloor$, or equivalently, $[\\lfloor{n\\beta}\\rfloor/\\alpha,\\lfloor{n\\beta+1}\\rfloor/\\alpha)$ contains an integer for every $n$. (*)\n\nWeakening (*) a bit, this means $((n\\beta-1)/\\alpha,(n\\beta+1)/\\alpha)$ has an integer for all $n\\ge1$. If $\\beta/\\alpha$ is irrational, then since $2/\\alpha<1$, Kronecker's density theorem yields a contradiction.\n\nOn the other hand, if $\\beta/\\alpha=p/q$ for some coprime positive integers $p,q$ with $q>1$, then $\\beta$ is irrational, then by B\u00e9zout's identity we can find some arithmetic progression $P=\\{qm+r\\}_{m\\ge0}$ such that for every $n\\in P$, we have $\\{n(\\beta/\\alpha)\\}=1/q\\le1/2<1-1/\\alpha$. But $q\\beta$ is irrational, so by Kronecker's density theorem we can get $\\{qm\\beta\\}\\in[\\{-r\\beta\\},\\{-r\\beta\\}+\\epsilon)$ for any $\\epsilon>0$ and thus $\\{n\\beta\\}\\in[0,\\epsilon)$, whence there exists $n\\in P$ such that $0<\\{n(\\beta/\\alpha)-\\{n\\beta\\}/\\alpha\\}\\le1-1/\\alpha$, contradicting (*).", "Solution_5": "Does anyone know which Bezout's identity math154 is referring to, or provide a link? Thanks.", "Solution_6": "Also, does anyone know any motivations as to math154 other than \"experience\", because I do not know how understanding the solution to this problem would help with solving problems on another math olympiad; i.e., I do not get the reasons for math154 to use Kronecker's density theorem and Bezout's theorem which I have yet to understand. To me, his solution just looks like a variable bash, and I do not see any pattern to his reasoning. Can someone explain any possible motivation for his solution? Thanks!" } { "Tag": [ "inequalities" ], "Problem": "Let a,b,c be nonnegative real numbers. Prove that\r\nab+bc+ca :ge: :sqrt:(3abc(a+b+c)).\r\n\r\nLet a,b,c be nonnegative real numbers such that a+b+c = 1. Prove that\r\na:^2:+b:^2:+c:^2:+:sqrt:(12abc):le:1.", "Solution_1": "[quote=\"TripleM\"]Let a,b,c be nonnegative real numbers. Prove that\nab+bc+ca :ge: :sqrt:(3abc(a+b+c)).\n\nLet a,b,c be nonnegative real numbers such that a+b+c = 1. Prove that\na:^2:+b:^2:+c:^2:+:sqrt:(12abc):le:1.[/quote]\r\n\r\nthe first one looked intimidating but was really not that bad...\r\nworking backwards, we square it to get a:^2:b:^2:+a:^2:c:^2:+b:^2:c:^2:+2ab:^2:c+2abc:^2:+2a:^2:bc :ge:3ab:^2:c+3abc:^2:+3a:^2:bc , which works out to be a:^2:b:^2:+a:^2:c:^2:+b:^2:c:^2:-ab:^2:c-abc:^2:-a:^2:bc:ge:0, which is true because it is in fact weaker than expanding the trivial inequality (ab+bc-ca):^2:+(ab-bc+ca):^2:+(-ab+bc+ca):^2::ge:0", "Solution_2": "For number2, using AM-GM, we get that max of abc is 1/27, attained when a=b=c=1/3.\r\n\r\nThis means that a^2+b^2+c^2<=1/3.\r\nSince the max of the LHS is attained when max(abc) is attained, the max of a^2+b^2+c^2=1/3.\r\nThus, the original one holds.", "Solution_3": "[quote=\"darkquantum\"]For number2, using AM-GM, we get that max of abc is 1/27, attained when a=b=c=1/3.\n\nThis means that a^2+b^2+c^2<=1/3.\nSince the max of the LHS is attained when max(abc) is attained, the max of a^2+b^2+c^2=1/3.\nThus, the original one holds.[/quote]\r\n\r\nwhy must a^2+b^2+c^2:le:1/3? If a=1 and b=c=0, then the LHS is 1 which is bigger than 1/3.", "Solution_4": "it holds for max abc", "Solution_5": "....so? you're not trying to prove it for max abc, you're trying to prove it for all (a,b,c) as long as a+b+c=1. \r\n\r\n\r\nYou have now successfully confused me.", "Solution_6": "errr\r\nperhaps some revision is in order...", "Solution_7": "for the 2nd one :\n\n\n\n[hide]\n\n\n\nsquaring a+b+c=1 and using first inequality doesn't seem to be a bad idea \n\n\n\n[/hide]", "Solution_8": "yup..you're right..that solves it. On the second one, I had gotten up to ab+bc+ca:ge::sqrt:(3abc) and then I sorta forgot all about the first inequality (which solves it since a+b+c=1).", "Solution_9": "lol \r\nyea\r\ni forgot about taht 2\r\n\r\nso" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose $ G$ is a finite group and H is a subgroup of G and [G:H]=n>1. Prove that either G has a non-trivial normal subgroup whose index divides $ n!$ or G is isomorphic to a subgroup of $ S_n$.", "Solution_1": "I think you want: ... whose index divides $ n!$ and is not equal to $ n!$ ...", "Solution_2": "consider the natural map $ G \\to Sym(G/H)$.", "Solution_3": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=19724]here[/url] for a proof that $ K\\equal{}\\bigcap_{g\\in G}(gHg^{\\minus{}1})$ is a normal subgroup, and $ G/K$ can be identified with a subgroup of $ S_n$.", "Solution_4": "Thank you! That is a nice solution." } { "Tag": [], "Problem": "How many three-digit odd integers are greater than 299?", "Solution_1": "the first three-digit odd integer greater than 299 is 301\r\nthe last three-digit odd integer greater than 299 is 999\r\n\r\ntherefore, we have a list of odd integers: 301,303,305...995, 997, 999\r\n\r\nhow do we count these???\r\nwell, we know that the interval between each of these numbers is two, so let's divide by two\r\nhowever, if we do that, we will deal with decimals, and i don't like those in counting problems\r\nso let's add one to each of them first\r\n\r\nnow we have the list 302,304,306....996,998,1000\r\nnow we divide by 2 to get the list 151, 152, 153....498, 499, 500\r\n\r\nwe can subtract 150 from each term to get: 1, 2, 3......348, 349, 350\r\n\r\ntherefore, there are $ \\boxed{350}$ 3-digit integers greater than 299" } { "Tag": [ "\\/closed" ], "Problem": "Are there gift certificates/cards for the Aops bookstore? nice present for math teachers...", "Solution_1": "Currently, you can only order these by phone. Eventually, we'll have a system that allows you to order them online." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all prime numbers $p,q$ and even $n>2$ such that :\r\n\r\n$p^n+p^{n-1}+...+p+1 = q^2+q+1$ .", "Solution_1": "See the topic Primes http://www.mathlinks.ro/Forum/viewtopic.php?t=29570" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $a_{i},b_{i}\\in \\mathbb{R}\\ \\ ,\\ 1\\le i\\le n$ be constant numbers find Max and Min of :\r\n\r\n\r\n$y=b_{i}|x-{a}_{i}|+b_{2}|x-{a}_{2}|+\\dots+b_{n}|x-{a}_{n}|$ that $x,y \\in \\mathbb{R}$", "Solution_1": "Where is $y$ in the inequality? :maybe:", "Solution_2": "I said find Max and Min of the $y$", "Solution_3": "Oh sorry, I did not pay close attention. :blush:", "Solution_4": "no one???? :maybe: :maybe:", "Solution_5": "It seems as far as the maximum gose, you could make it as large as you wanted...\r\nAs for the minimum, it seems to be at the weighted arithmetic mean of the $a_{i}$, ie $\\frac{a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}}{b_{1}+b_{2}+...+b_{n}}$? Maybe I'm not understanding it...or I'm just wrong.", "Solution_6": "[quote=\"K81o7\"]It seems as far as the maximum gose, you could make it as large as you wanted...\nAs for the minimum, it seems to be at the weighted arithmetic mean of the $a_{i}$, ie $\\frac{a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}}{b_{1}+b_{2}+...+b_{n}}$? Maybe I'm not understanding it...or I'm just wrong.[/quote]\r\n\r\n\r\nYou are wrong Consider $b_{i}<0\\ ,\\ i=1,2,...,n$ and you can not make it as large as you want.", "Solution_7": "let $f(x)=b_{1}|x-a_{1}|+...+b_{n}|x-a_{n}|$ \r\nand S={$f(-\\infty),f(a_{1}),...,f(a_{n}),f(\\infty)$} then\r\n Max(f)=Max(S) ,Min(f)=Min(S)", "Solution_8": "[quote=\"ali666\"]let $f(x)=b_{1}|x-a_{1}|+...+b_{n}|x-a_{n}|$ \nand S={$f(-\\infty),f(a_{1}),...,f(a_{n}),f(\\infty)$} then\n Max(f)=Max(S) ,Min(f)=Min(S)[/quote]\r\n\r\ncan you prove it please?", "Solution_9": "suppose that $a_{1}\\leq{a_{2}}\\leq...\\leq{a_{n}}$and$-\\infty=a_{0},\\infty=a_{n+1}$\r\n (1):$\\exists{A_{1},A_{2},...,A_{n},A_{n+1},B_{1},B_{2},...,B_{n},B_{n+1}\\in{R}}$ such that:\r\n$f(x)=A_{1}x+B_{1}$ for $x\\in[a_{0},a_{1}]$\r\n$f(x)=A_{2}x+B_{2}$ for $x\\in[a_{1},a_{2}]$\r\n .\r\n .\r\n .\r\n$f(x)=A_{n}x+B_{n}$ for $x\\in[a_{n-1},a_{n}]$ \r\n$f(x)=A_{n+1}x+B_{n+1}$ for $x\\in[a_{n},a_{n+1}]$\r\n(2):if $g: [a,b]\\rightarrow\\mathbb{R},g(x)=cx+d$ then max(g)=max{g(a),g(b)},min(g)=min{g(a),g(b)}(because if c>0 max(g)=g(b),min(g)=g(a). if c<0 max(g)=g(a),min(g)=g(b).if c=o max(g)=min(g)=g(a)=g(b))\r\n\r\nif $t\\in[a_{i},a_{i+1}]$ then $f(t)\\leq$max{$f(a_{i}),f(a_{i+1})$}\r\ntherefore if $x\\in\\mathbb{R}$then $f(x)\\leq$max{$f(a_{0}),...,f(a_{n+1})$}\r\n and ..." } { "Tag": [ "geometry", "perimeter", "ratio", "trigonometry", "quadratics", "function", "algebra" ], "Problem": "At some point or another, we were all beginning problem solvers. Yet almost every post (even in this forum) contains questions that are inaccessible, at least for the time being, to true beginning problem solvers. Therefore I have decided to post some problems (in approximate increasing order of difficulty) that I feel will be of reasonable difficulty to a student of this level. I encourage all true beginners (especially those who may only frequent the non-math forums on this site) to attempt some of these.\r\n\r\n1. If [tex]30x-18=72[/tex], [tex]xy+6y=72[/tex], and [tex]xyz=72[/tex], find [tex]x+y+z[/tex].\r\n\r\n2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square.\r\n\r\n3. What is the positive difference between the 2004th positive odd integer and the 2004th negative odd integer? \r\n\r\n4. A side of a regular hexagon has length 1. Find the area of the hexagon. \r\n\r\n5. (a) If [tex](x+1)(x-1)=80[/tex], find [tex]x^2[/tex].\r\n(b) If [tex](x+1)(x-1)=22[/tex], find [tex]x^2[/tex].\r\n\r\n6. Find the sum of the first 15 positive odd integers.\r\n\r\n7. (more challenging) The ratio of a circle's circumference to its area is 1:7. What is the radius of the circle?", "Solution_1": "[quote=\"JSRosen3\"]At some point or another, we were all beginning problem solvers.[/quote]\r\n\r\nAnd I mostly still am a beginning problem solver, because I don't get in enough practice time. Thanks for sharing problems at a good level for beginners.", "Solution_2": "Anyone want to give these a shot? Even if you can't completely solve them, you can post your work so far and someone can give you a hint to help you finish it.", "Solution_3": "JSRosen3 wrote:2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square.\n\n\n\n\n2. My solution will be in spoiler, of course. \n\n\n\n[hide]A problem-solver learns to think \"Pythagorean theorem\" as soon as the term \"right triangle\" is mentioned. This problem is asking about a triangle with one right angle (that is, 90 degree angle) in it. The two sides next to that angle are called \"legs.\" The longest side of the triangle, which is opposite the right angle, is called the hypotenuse (Greek for \"thing hanging below\"). The famous theorem of Pythagoras says that the area of the squares on the two legs adds up to the area of the square on the hypotenuse. \n\n\n\nOne leg has length 13. A square constructed on that leg will have area 13 times 13, or 169. The other leg has length 15, so the square has area 15 times 15, or 225. Adding 169 to 225, we find the area of the square on the hypotenuse is 394, which is our answer for part (b) of the question. The length of the hypotenuse, then, is the square root of 394, which is approximately equal to 19.8494 units. [/hide] \n\n\n\nIs that process of solution clear? This is definitely within the scope of Getting Started Forum. \n\n\n\nI'll try to insert a diagram here.", "Solution_4": "Yes, and it's also very thorough. I think I'll start doing that too so that beginners looking at solutions can understand them. Another thing to note:\n\n\n\n[hide]An immediate result of the Pythagorean Theorem is that if a and b represent the lengths of the legs, and c represents the length of the hypotenuse, then a^2 + b^2 = c^2. Thus, we do not have to think strictly geometrically. If you look at the left of the screen below the navigation, there will sometimes be an animated geometric proof of the Pythagorean Theorem[/hide]", "Solution_5": "As a beginner, I'll give a shot. Just one easy one:\n\n\n\n [hide]6. Find the sum of the first 15 positive odd integers. Basically, it's 1+3+5+..... but rather than doing all these, we can simply write this as x+x+2+x+4+x+6 so, just find 2+4+6+ to +28 which is 210. Then we have 11x's where x is 1. So, 11+210 = 221. I hope this is right because it's kinda shame if I get this easy one wrong. Plus, tell me if there's easier way.\n\n[/hide]", "Solution_6": "You've got the right idea, but you made some calculation errors. BTW, there's no shame whatsoever involved here. I get Getting Started stuff wrong all the time due to small errors. If you want an easier way, you can look at the little animated figure under the navigation on your left right here on AoPS. Occasionally it will show a formula for the sum of the first so many odd numbers.", "Solution_7": "Good work so far, Silverfalcon. You're almost there. The only mistake you made was that [hide]you said there were 11 x's rather than 15.[/hide] Fix that and you should be good to go.\n\n\n\nAnother way to do this is [hide]to start small, namely by adding the first 2 odd integers, then the first 3 odd integers, etc. and looking for a pattern. You'll find one pretty quickly[/hide].", "Solution_8": "3.\n\n[hide]\n\nSince all even numbers are multiples of two, we can expect all even number to be in the form 2n. For example, 1*2 = 2, 2*2 = 4, 3*2 = 6, etc. All odds are one less than evens. Thus the odds are 2n-1. Note that we don't use 2n+1, because that would make the first odd number 3, which is wrong.\n\n\n\nThus the 2004th odd number is 4007. The negative of the 2004 odd number is -4007. 4007-(-4007) = 4007+4007=8014\n\n[/hide]", "Solution_9": "probability1.01 wrote:I get Getting Started stuff wrong all the time due to small errors.\n\n\n\nOne of my worst getting started mistakes was saying that 2/2 = 0. \n\n\n\n1.\n\n[hide]\n\n30x - 18 = 72. Adding 18 to both sides and dividing by thirty gives x = 3.\n\nPlugging x = 3 into the next equation gives 9y = 72, or y =8. Putting three and 8 into the last one gives 24z = 72. Dividing by 24 gives z = 3. 3+8+3 = 14.\n\n[/hide]\n\n\n\n5.\n\n[hide]\n\nOne very useful factorization is called the difference of squares. If we have x^2-y^2, it can be factored as (x-y)(x+y) We can expand (x-1)(x+1) and get x^2 - 1.\n\n(a). x^2 - 1 = 80\n\n x^2 = 81\n\n(b). x^2 - 1 = 22\n\n x^2 = 23\n\nRecognizing these factorizations is important because, as you see for part b, guessing and testing doesn't help since x was not a rational number, but x^2 is, and x^2 was what the question asked for.[/hide]", "Solution_10": "I commend JSRosen3 for the wonderful level of these problems. Also, all of the solutions have been very clear and have avoided 'hand wavy' stuff (hand waving is bad, so try not to do it). I did find one problem though. Once you think you have solved a problem, [b]go back and look to see what was being asked[/b]. For #2 that tokenadult solved, he did everything correctly up until the very end. If he had went back and reread the problem, he would have seen that part (a) was asking for the [b]perimeter[/b] of the square, not the sidelength. I'm just trying to stress the importance of answering what was being asked. Good job and keep it up!", "Solution_11": "7.\n\n[hide]\n\nThe circumference of a circle is given by the formula C = d*pi where d is the diameter, or C = 2*pi*r, where r is the radius. The area is pi*r^2. We want the ratio of these two quantities to be 1/7. Let's start with (2r*pi)/(pi*r^2) and start simplifying. The pi's on the numerator and the denominators cancel out. 2r/r^2 becoms 2/r because the r's cancel out. Now we just solve for r in the equation 1/7 = 2/r. Multiplying both sides by 7r gives r = 14.[/hide]", "Solution_12": "I suggest some other people try these problems too. So far, it's been nr1337, Silverfalcon, and tokenadult. Can't think of any others, but BHorseMath agreed to do some problems in a post on the F&G forum :twisted:", "Solution_13": "[quote=\"probability1.01\"]I suggest some other people try these problems too. So far, it's been nr1337, Silverfalcon, and tokenadult. Can't think of any others, but BHorseMath agreed to do some problems in a post on the F&G forum :twisted:[/quote]\r\n\r\nSorry - I had to do something to fight my boredom today!", "Solution_14": "1. [hide]If 30x-18=72, x=3. So xy+6y=72 can be rewritten as 3y+6y=72, or y=8. That means xyz=72 is equal to 3*8z=72, or z=3. x+y+z=3+8+3=14.[/hide]\n\n2. [hide]Using the Pythagorean Theorem, the hypotenuse is (13^2+15^2)^1/2, or :sqrt: 394. The perimeter of the square is 4 :sqrt: 394, and the area is ( :sqrt: 394)^2=394.[/hide]\n\n3. [hide]2004th positive odd integer=2004*2+1=4009\n\n 2004th negative odd integer=-2004*2-1=-4009\n\n 4009-(-4009)=4009+4009=8018[/hide]\n\n5. [hide](x+1)(x-1) can be multiplied out into x^2-1. If x^2-1=80, x^2=81. If x^2-1=22, x^2=23.[/hide]\n\n6. [hide][15(1+31)]/2=240[/hide]\n\n7. [hide]2 :pi: r: :pi: r^2=1:7\n\n 7*2 :pi: r= :pi: r^2\n\n 14=r[/hide]\n\nI don't know why but I cannot get number 4.", "Solution_15": "4.\n\n[hide](Sorry, I don't have Geometer's Sketchpad so I'll explain as clearly as I can with text)\n\n\n\nDivide the hexagon into six equilateral triangles by connecting the center of the hexagon with each vertex. Now, focus on one of the triangles. The central angle of the triangle (the angle containing the center of hexagon) is 60 degrees. (Because there are 6 triangles, and the total number of degrees in the center is 360). \n\n\n\nNow, draw the altitude of the triangle, which is also the apothem of the polygon. The altitude/apothem divides the base of the equilateral triangle in two, forming two right triangles that have a base of 0.5. Notice the right triangles are 30-60-90 triangles, so the apothem will be :rt3: / 2\n\n\n\nThen, following the formula of the area of a regular polygon, 1/2 aP, subsitute in the values. (:rt3: / 2 for apothem, 6 for perimeter). The area of the regular hexagon is (3 :rt3:) / 2.[/hide]", "Solution_16": "LynnelleYe,\n\n\n\nOn #3, the 2004th positive odd integer is not [hide]4009[/hide], because the nth positive odd integer is [hide]2n-1[/hide], not [hide]2n+1[/hide]. Same thing for #6. You had the right idea on both, though.", "Solution_17": "1. [hide]30x=90, x=3\n\n9y=72, y=8\n\n3*8*z=72, z=3[/hide]", "Solution_18": "Revision for tolkenadult's number 2:\n\n\n\n[hide]He has not finished the question because :sqrt: 394 is only the side length. Therefore, correct answer is 4 :sqrt: 394.[/hide]", "Solution_19": "My way of doing number 4:\n\n[hide]If you know the formula for the area of a hexagon, which is 6s^2 :rt3: /4 where s=side of the hexagon, you could just plug 1 in for s and get 3 :rt3: /2. If you didn't know the formula, you could divide the hexagon up into 6 equilateral triangles like one of the posters above me did.[/hide][/hide]", "Solution_20": "Finally somebody posted what my teacher calls the 1-2-3-4 formula. It's for the area of an equilateral triangle, and it's a lot quicker than doing everything with the apothem.\r\n\r\n1-1 variable\r\n2-to the power of 2\r\n3-multiplied by :rt3: \r\n4-divided by 4\r\n\r\nso, the formula is, with s being the side of the eqiuilateral triangle, s:^2:*:rt3: /4\r\n\r\nThen, for a regular hexagon, you just multiply that result by 6 for the 6 triangles that make up a hexagon.", "Solution_21": "Another useful formula is the general formula for the area of a regular n-gon with n sides and side length x:\r\n\r\n[tex]A=\\frac{nx^{2}}{4\\tan \\frac{180}{n}}[/tex]\r\n\r\nSo if you plugged in 6 for n, you'd get that the area of a regular hexagon is \r\n\r\n[tex]\\frac{6x^{2}}{4\\tan \\frac{180}{6}}=\\frac{6x^{2}}{4\\tan 30}=\\frac{6x^{2}}{4(\\frac{1}{\\sqrt{3}})}=\\frac{6x^{2}\\sqrt{3}}{4}=\\frac{3x^{2}\\sqrt{3}}{2}[/tex].\r\n\r\nThe derivation of the general formula is a good exercise in trig identities. Anyone want to give it a shot?", "Solution_22": "Doesn't the trig seem a little non-beginner-ish?", "Solution_23": "The trig might be a little advanced, but if you know the identities it's definitely doable.", "Solution_24": "While we're talking about trig, what's tan[36]? My brother has been wanting me to find the area of a regular pentagon for a while, but I need to know tan[36] (and NOT a deciimal approximation).", "Solution_25": "[tex]\\tan{36}=\\sqrt{5-2\\sqrt{5}}[/tex]\r\n\r\nI'm trying to think of a good way to derive this. Using quintuple-angle formulas is probably not such a good idea :D", "Solution_26": "To find tan (or anything else) 36, consider\r\n\r\n[tex]x^4+x^3+x^2+x+1=0.[/tex]\r\n\r\nOne way to find the roots is to divide by [tex]x^2[/tex] so that you get a quadratic in x+1/x. Another way is to multiply by x-1 to get [tex]x^5-1=0[/tex]. Find the roots each way and equate real and imaginary parts.", "Solution_27": "Okay, I haven't taken trig (I just finished geometry), but i have had a minute exposure to sin cos tan, but i have no idea what anything beyond that is. What are identities and quintuple-angle formulas, and how did you get that for tan 36? :shock:", "Solution_28": "Trig identities are just identities for trig functions; i.e., they are always true. Here are a few examples:\r\n\r\n[tex]\\sin{2\\theta}=2\\sin{\\theta}\\cos{\\theta}[/tex]\r\n\r\n[tex]\\cos{(\\alpha+\\beta)}=\\cos{\\alpha}\\cos{\\beta}-\\sin{\\alpha}\\sin{\\beta}[/tex]\r\n\r\nSo, if you wanted to find [tex]\\cos{105}[/tex], you could write it as [tex]\\cos{(60+45)}[/tex], and then use the identity above to obtain [tex]\\displaystyle \\cos{105} = \\cos{(60+45)} = \\cos{60} \\cos{45} - \\sin{60} \\sin{45}[/tex] = [tex](\\frac{1}{2} \\times \\frac{\\sqrt{2}}{2}) - (\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{2}}{2}) = \\frac{\\sqrt{2} - \\sqrt{6}}{4}[/tex].\r\n\r\nQuintuple-angle formulas are expressions for things like [tex]\\tan{(5\\theta)}[/tex] in terms of only [tex]\\tan{\\theta}[/tex]. Since [tex]\\tan{180}[/tex] can be expressed as [tex]\\tan{(5 \\times 36)}[/tex], we can use a quintuple-angle identity for the tangent function to find [tex]\\tan{36}[/tex] in terms of [tex]\\tan{180}[/tex], which we know to be 0. It would be very messy, though, so I'd look for an alternative (perhaps geometric?) method before I plowed into quintics.", "Solution_29": "Identities are things true for any angle. For example,\r\n[tex]\r\n\\sin^2 \\theta + \\cos^2 \\theta = 1\r\n\r\n\\cos^2 \\theta = \\frac{1 + cos 2\\theta}{2}\r\n[/tex]\r\n\r\nWhen he said quintuple ange formulas, that means a formula for (some trig function) of 5x. For example,\r\n[tex]\r\n\\cos 5\\theta = \\cos^5 \\theta - 10\\cos^3 \\theta \\sin^2 \\theta + 5 \\cos \\theta \\sin^4 \\theta\r\n[/tex]\r\n\r\nAnd now we are WAY beyond getting started.", "Solution_30": "Yeah, we're a little beyond the realm of Getting Started, although I don't think that all trig is necessarily beyond it. Perhaps a mod would be willing to post an introduction to trig, just like the one for arithmetic sequences? The one for arithmetic sequences was excellent.", "Solution_31": "Just for fun,\r\n[tex]\r\n\\displaystyle\\tan 5\\theta = \\frac{5 \\cos^4 \\theta \\sin \\theta - 10 \\cos^2 \\theta \\sin^3 \\theta + \\sin^5 \\theta}{\\cos^5 \\theta - 10 \\cos^3 \\theta \\sin^2 \\theta + 5 \\cos \\theta \\sin^4 \\theta}\r\n[/tex]", "Solution_32": "Gah! :o That's frightening. That's why I'm looking for a geometric way to find [tex]\\tan{36}[/tex].", "Solution_33": "Did you look at my method? You can find it that way.", "Solution_34": "Using ComplexZeta's method:\r\n\r\n[tex]x^4 + x^3 + x^2 + x + 1 = 0[/tex]\r\n\r\n[tex]x^2 + x + 1 + \\frac{1}{x} + \\frac{1}{x^2} = 0[/tex]\r\n\r\nLet [tex]y=x+\\frac{1}{x}[/tex]. Then [tex]y^2 + y - 1 = 0[/tex], so [tex]y = \\frac{-1 \\pm \\sqrt{5}}{2}[/tex].\r\n\r\nRecall that [tex]y=x+\\frac{1}{x}[/tex]. Rearranging:\r\n\r\n[tex]x + \\frac{1}{x} = \\frac{-1 \\pm \\sqrt{5}}{2}[/tex]\r\n\r\n[tex]2x^2 + 2 = -x \\pm x\\sqrt{5}[/tex]\r\n\r\n[tex]2x^2 + (1 \\pm \\sqrt{5})x + 2 = 0[/tex]\r\n\r\nUsing the Quadratic Formula, the four solutions are:\r\n\r\n[tex]\\displaystyle x_1 = \\frac{\\sqrt{5}-1}{4} + \\frac{\\sqrt{2(\\sqrt{5}+5)}}{4} \\times i[/tex]\r\n\r\n[tex]\\displaystyle x_2 = \\frac{\\sqrt{5}-1}{4} - \\frac{\\sqrt{2(\\sqrt{5}+5)}}{4} \\times i[/tex]\r\n\r\n[tex]\\displaystyle x_3 = \\frac{-\\sqrt{5}-1}{4} + \\frac{\\sqrt{-2(\\sqrt{5}-5)}}{4} \\times i[/tex]\r\n\r\n[tex]\\displaystyle x_4 = \\frac{-\\sqrt{5}-1}{4} - \\frac{\\sqrt{-2(\\sqrt{5}-5)}}{4} \\times i[/tex]\r\n\r\nLet [tex]P(x) = x^4 + x^3 + x^2 + x + 1[/tex], and let [tex]Q(x)= (x-1)P(x) = x^5-1[/tex]. The roots of [tex]P(x)[/tex], are the same as the roots of [tex]Q(x)[/tex], disregarding the root [tex]x=1[/tex], since we introduced it when we multiplied [tex]P(x)[/tex] by [tex]x-1[/tex]. The roots of [tex]Q(x)[/tex] are the fifth roots of unity, which are (excluding 1):\r\n\r\n[tex]\\cos{72}+i(\\sin{72})[/tex]\r\n\r\n[tex]\\cos{144}+i(\\sin{144})[/tex]\r\n\r\n[tex]\\cos{216}+i(\\sin{216})[/tex]\r\n\r\n[tex]\\cos{288}+i(\\sin{288})[/tex]\r\n\r\nNow, set one of these roots equal to its corresponding root that was found above. Since we want [tex]\\tan{36}[/tex], choose the second root (since 144 and 36 are supplementary). Since this root is in the second quadrant, we need to choose the root above that is also in the second quadrant, namely [tex]x_3[/tex]. So, evaluating real and imaginary parts seperately, we have:\r\n\r\n[tex]\\cos{144} = \\displaystyle \\frac{-\\sqrt{5}-1}{4}[/tex].\r\n\r\n[tex]\\cos{36} = -\\cos{144} = \\displaystyle \\frac{\\sqrt{5} + 1}{4}[/tex].\r\n\r\n[tex]i \\sin{144} = \\displaystyle \\frac{\\sqrt{-2(\\sqrt{5}-5)}}{4} \\times i[/tex]\r\n\r\n[tex]\\sin{36} = \\sin{144} = \\displaystyle \\frac{\\sqrt{-2(\\sqrt{5}-5)}}{4} = \\displaystyle \\frac{\\sqrt{2(5-\\sqrt{5})}}{4}[/tex]. \r\n\r\nAnd finally:\r\n\r\n[tex]\\tan{36} = \\displaystyle \\frac{\\sin{36}}{\\cos{36}} = \\frac{\\frac{\\sqrt{2(5-\\sqrt{5})}}{4}}{\\frac{\\sqrt{5} + 1}{4}} = \\frac{\\sqrt{2(5-\\sqrt{5})}}{\\sqrt{5}+1}[/tex].\r\n\r\n(also, if a mod wants to move this discussion to intermediate, feel free to do so--but please don't move the whole thread, just the recent discussion of identities, complex numbers, etc.)\r\n\r\n[i]P.S. Thanks for showing me how to [tex]\\TeX[/tex] the plus/minus, ComplexZeta.[/i]", "Solution_35": "[tex]\\pm[/tex]", "Solution_36": "I understand why the method of finding the roots of the polynomial in two different ways and equating them works. But how would we have thought of this in the first place? How can we train ourselves to think [tex]x^4 + x^3 + x^2 + x + 1[/tex] when we hear [tex]\\tan 36[/tex]?", "Solution_37": "You might think about 5th roots of unity when you want to know (insert trig function here)(insert multiple of 18 here). Then you might consider [tex]x^5-1[/tex]. Then you might get rid of the root you already know, and voila!", "Solution_38": "Sadly, this method gets very messy when we have 360/some big number. I tried to find cos 360/17 like Gauss did using this method and didn't get very far very quickly.", "Solution_39": "[spoiler]The ratio of a circle's circumference to its area is 1:7. What is the radius of the circle?\r\n\r\nArea = pi.r^2\r\ncircumference = 2pi.r\r\n\r\n1/7 = circumference/area (using ratios I guess)\r\n1/7 = (2pi.r)/(pi.r^2)\r\n1/7 = 2/r\r\nr = 14?[/spoiler]\r\n\r\nI liked that. Good problem, took some thinking because I haven't seen one like that before.", "Solution_40": "Here are my answers:\n\n[hide]\n\n1.If 30x - 18 = 72 xy + 6y = 72 xyz = 72 what is x+y+z?\n\n30x -18 = 72\n\n30x = 90\n\nx = 3\n\n\n\n3y + 6y = 72\n\n9y = 72\n\ny = 8\n\n\n\n3*8*z = 72\n\n24z = 72\n\nz = 3\n\n\n\nx+y+z = 3+3+8 = 14\n\n\n\n2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square. \n\n\n\na :^2: + b:^2: = c:^2: \n\n13:^2: + 15:^2: = c:^2: \n\n169 + 225 = c:^2: \n\nc:^2: = 394\n\nc = :sqrt: 394\n\n\n\na = :sqrt: 394*4\n\na :approx: 79.40\n\n\n\nb = :sqrt: 394 :^2: \n\nb = 394\n\n\n\n3. What is the positive difference between the 2004th positive odd integer and the 2004th negative odd integer?\n\n \n\nnth positive odd integer = 2(n-1) + 1\n\n2004th positive odd integer = 4007\n\n\n\nnth negative odd integer = -(2(n-1) +1) = -(nth positive odd integer)\n\n2004th negatice odd integer = -4007\n\n\n\n4007 - (-4007) = 4007*2 = 8014\n\n8014\n\n[/hide]\n\n\n\nI'll try the rest in a couple days\n\n\n\nOh yeah, thanks for the easier problems." } { "Tag": [ "inequalities" ], "Problem": "Prove that \\[ a^rb^s\\plus{}b^rc^s\\plus{}c^ra^s>a^sb^r\\plus{}b^sc^r\\plus{}c^sa^r\\] unde \\[ a>b>c>0, r>s>0\\]", "Solution_1": "Let $ a=\\alpha c , b=\\beta c, \\alpha>\\beta>1$,and the inequality becomes after simplyfing with $ c^{r+s}$, \\[ \\alpha^r \\beta^s+ \\beta^r +\\alpha^s > \\alpha^s \\beta^r +\\beta^s +\\alpha^r\\] equivalent to \\[ \\frac{\\alpha^r -1}{\\alpha^s -1}> \\frac{\\beta^r-1}{\\beta^s-1}\\] and if $ \\alpha^s=u and \\beta^s=v, u>v>1, r\\s=t>1$.The inequality becomes \\[ \\frac{u^t-1}{u-1}> \\frac{v^t-1}{v-1}\\] and to prove this one we have to prove that $ f(x)=(x^t-1)/(x-1) \\nearrow$ with $ x>1$.We have \\[ f'(x)=\\frac{tx^{t-1}(x-1)-(x^t-1)}{(x-1)^2}=\\frac{(t-1)x^t-tx^{t-1}+1}{(x-1)^2}\\] To have $ f'(x)>0$ we have to have \\[ g(x)=(t-1)x^t-tx^{t-1}+1>0\\] for any x>1.\r\nBut $ g'(x)=t(t-1)x^{t-2}(x-1)>0$ it results that $ g \\nearrow$ and then $ g(x)>lim_{x \\rightarrow 0,x>0}g(x)=0$ and the inequality is proved." } { "Tag": [ "induction", "algebra proposed", "algebra" ], "Problem": "The sequence $ a_n$ is defined as follows: $ a_0\\equal{}1, a_1\\equal{}1, a_{n\\plus{}1}\\equal{}\\frac{1\\plus{}a_{n}^2}{a_{n\\minus{}1}}$.\r\nProve that all the terms of the sequence are integers.", "Solution_1": "Lets prove by induction that $ a_n=F_{2n-1}$ for $ n\\ge 2$ with $ F_i$ the fibonacci sequence $ F_0=0$ and $ F_1=1$ and $ F_{n+1}=F_n+F_{n-1}$ for $ n>1$.\r\n\r\n$ a_2=F_3=2$ and $ a_3=F_5=5$\r\n \r\ngiven that $ a_{n-1}=F_{2(n-1)-1}$ and $ a_n=F_{2n-1}$ then:\r\n\r\n$ a_{n + 1} = \\frac {1 + a_{n}^2}{a_{n - 1}}=\\frac {1 + F_{2n-1}^2}{F_{2n - 3}}$ \r\n\r\n$ a_{n + 1} = \\frac {1 + (F_{2n-2}+F_{2n-3})^2}{F_{2n - 3}}$ \r\n\r\n$ a_{n + 1} = F_{2n-3}+2F_{2n-2}+{\\frac {1 + F_{2n-2}^2}{F_{2n - 3}}}$ Using Cassiny's identity:\r\n\r\n$ a_{n + 1} = F_{2n-1}+F_{2n-2}+{\\frac {F_{2n-1} F_{2n-3}}{F_{2n - 3}}}$ \r\n\r\n$ a_{n + 1} =F_{2n}+F_{2n-1}$\r\n\r\n$ a_{n + 1} =F_{2n+1}$\r\n\r\n$ a_{n + 1} =F_{2(n+1)-1}$", "Solution_2": "[quote=\"bambaman\"]The sequence $ a_n$ is defined as follows: $ a_0 \\equal{} 1, a_1 \\equal{} 1, a_{n \\plus{} 1} \\equal{} \\frac {1 \\plus{} a_{n}^2}{a_{n \\minus{} 1}}$.\nProve that all the terms of the sequence are integers.[/quote]\r\n[hide=\"Solution\"]\nWe proceed with induction. $ a_{2}\\equal{}2$, so $ a_2$ is an integer (as are $ a_0$ and $ a_1$). Now, assume that $ a_i$ is an integer for $ 1\\le i\\le n$. We have that $ a_{n\\plus{}1}\\equal{}\\frac{1\\plus{}(\\frac{a_{n\\minus{}1}^2\\plus{}1}{a_{n\\minus{}2}})^2}{a_{n\\minus{}1}}$. Upon multiplying out, we have that $ a_{n\\plus{}1}\\equal{}\\frac{a_{n\\minus{}2}^2\\plus{}a_{n\\minus{}1}^4\\plus{}2a_{n\\minus{}1}^2\\plus{}1}{a_{n\\minus{}2}^2a_{n\\minus{}1}}$. Now, $ a_{n\\minus{}1}$ and $ a_{n\\minus{}2}$ are coprime since $ a_{n\\minus{}1}^2\\plus{}1\\equal{}a_{n\\minus{}2}a_n$. Thus, it suffices to show that $ a_{n\\minus{}2}^2|(a_{n\\minus{}2}^2\\plus{}a_{n\\minus{}1}^4\\plus{}2a_{n\\minus{}1}^2\\plus{}1)$ (which is clearly true since $ a_{n\\minus{}2}^2|(a_{n\\minus{}1}^2\\plus{}1)^2$) and $ a_{n\\minus{}1}|(a_{n\\minus{}2}^2\\plus{}a_{n\\minus{}1}^4\\plus{}2a_{n\\minus{}1}^2\\plus{}1)$ (which is clearly true since $ a_{n\\minus{}1}|(a_{n\\minus{}2}^2\\plus{}1)$). Since both are proven, $ a_{n\\plus{}1}$ is an integer, so induction is complete. [/hide]", "Solution_3": "Together,\r\n\\[ a_{n\\plus{}1}a_{n\\minus{}1}\\equal{}1\\plus{}a_n^2\\wedge a_{n\\plus{}2}a_{n}\\equal{}1\\plus{}a_{n\\plus{}1}^2\\implies \\frac{a_{n\\plus{}1}}{a_n}\\equal{}\\frac{a_{n\\plus{}2}\\plus{}a_n}{a_{n\\plus{}1}\\plus{}a_{n\\minus{}1}}\\]\r\n\r\nTelescoping, we find that \\[ a_{n\\plus{}1}\\equal{}\\frac{a_{n\\plus{}2}\\plus{}a_n}{3}\\iff a_{n\\plus{}2}\\equal{}3a_{n\\plus{}1}\\minus{}a_n\\] hence the conclusion.", "Solution_4": "It suffices to show that $a_{n+1}=3a_n-a_{n-1}$ for $n \\ge 2$. This holds for $n=2$ and assume it holds for $n$. Then\n\n$a_{n+1}a_{n-1}=1+a_n^2$\n$a_{n+1}(3a_n-a_{n+1})=1+a_n^2$\n$3a_{n+1}-a_n=(1+a_{n+1}^2)/a_n$\n$3a_{n+1}-a_n=a_{n+2}$\n\nas desired." } { "Tag": [ "inequalities", "integration", "calculus", "abstract algebra" ], "Problem": "$a,b,c>0$ $abc=1$\r\n\r\n$\\displaystyle \\frac{a}{a^2+1}+\\frac{b}{b^2+1}+\\frac{c}{c^2+1} \\leq \\frac{3}{2}$", "Solution_1": "yes, that's obviously true. But what's the joke? :|", "Solution_2": "the joke is that abc=1 is useless?", "Solution_3": "Yes, the joke is not only that abc = 1 is a useless condition, but also that the inequality is completely trivial:\r\n\r\nFor every real number x, we have $\\displaystyle \\frac{x}{x^2+1} \\leq \\frac12$.\r\n\r\nBut I wonder why Maverick posts this in the \"Unsolved Problems\" section when it's a real joke.\r\n\r\n Darij", "Solution_4": "Even a,b,c > 0 are unnecessary, right?", "Solution_5": "[quote=\"fuzzylogic\"]Even a,b,c > 0 are unnecessary, right?[/quote]\r\n\r\nI think so.", "Solution_6": "[quote=\"siuhochung\"][quote=\"fuzzylogic\"]Even a,b,c > 0 are unnecessary, right?[/quote]\n\nI think so.[/quote]\r\n\r\nYes, because (x-1)^2 implies x/(x^2 + 1) <= 1/2 which obviously holds for any x.", "Solution_7": "This starts getting funny :D\r\n\r\nHey why don't we make a contest \"Weakest inequality of the month\" :maybe:", "Solution_8": "[quote=\"Peter VDD\"]This starts getting funny :D\n\nHey why don't we make a contest \"Weakest inequality of the month\" :maybe:[/quote]\r\n :D \r\nthat's interesting", "Solution_9": "[quote=\"Peter VDD\"]This starts getting funny :D\n\nHey why don't we make a contest \"Weakest inequality of the month\" :maybe:[/quote]\r\n\r\nGiven a >= b\r\n\r\nProve 2a >= 2b", "Solution_10": "yeah, but that one is rather cheap.\r\n\r\nOne that takes you a few seconds before you see it :D", "Solution_11": "In triangle abc, prove that:\r\n\r\na^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.", "Solution_12": "[quote=\"blahblahblah\"]In triangle abc, prove that:\na^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.[/quote]\r\n\r\nLOL?? :shock: :shock: :shock:\r\n\r\nIs that a funny ineq for which conditions are too much?? I was going to send that one in for our national junior olympiad... didn't even know it was well-known or so :? but dEUS already told me the same...", "Solution_13": "[quote=\"blahblahblah\"]In triangle abc, prove that:\n\na^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.[/quote]\r\n\r\nHmm, this actually rewrites as\r\n\r\n$a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+c^{2}a-\\left( a^{3}+b^{3}+c^{3}+2abc\\right) \\geq 0$,\r\n\r\nor as\r\n\r\n$\\left( b+c-a\\right) \\left( c+a-b\\right) \\left( a+b-c\\right) \\geq 0$,\r\n\r\nwhat is pretty clear for a triangle abc. Although I wouldn't call it a joke. Just one thing: If you say that abc is a \"real\" triangle, then you can replace the $\\geq$ sign by a strict > sign.\r\n\r\n Darij", "Solution_14": "Yes, but that rewriting is pretty much impossible for pupils having rarely worked with algebra, so I found some more straightforward way to convert the thing to proving that c/(a+b) < 1, which is obvious. :)\r\n\r\nSurprises me that this is soo all-round known :P I really never heard it anywhere before.", "Solution_15": "Speaking of jokes...\r\nWhat's $\\displaystyle{\\int_1^{cabin}dx/x}$", "Solution_16": "ln cabin? :)", "Solution_17": "Peter VDD wrote:Yes, but that rewriting is pretty much impossible for pupils having rarely worked with algebra, so I found some more straightforward way to convert the thing to proving that c/(a+b) < 1, which is obvious. \n\nSurprises me that this is soo all-round known I really never heard it anywhere before.\n\n\n\nA neat problem would be, in triangle abc, prove:\n\n\n\na^3 + b^3 + c^3 + 3abc >= a^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.\n\n\n\nThe leftmost inequality is handled by [hide]Schur[/hide], the rightmost is handled by the triangle inequality.\n\n\n\nBut there are factorizations for each part:\n\n\n\nClearly x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) >= 0, and clearly (x+y-z)(x+z-y)(y+z-x) >= 0 as well.", "Solution_18": "[quote=\"Peter VDD\"]ln cabin? :)[/quote]\r\n\r\nGrrrr. That wasn't the answer I wanted and you know it! ;)", "Solution_19": "Given: a < b\r\n\r\nProve: b > a", "Solution_20": "[quote=\"JSRosen3\"]Given: a < b\n\nProve: b > a[/quote]\r\n\r\nWTF?! That's what they have Pre-Olympiad and Advanced for. Keep that stuff out of Intermediate!", "Solution_21": "Oh, sorry, I forgot to cite the problem. It's IMO 1983, Problem 6. Sorry about the difficulty, Magnara, but if you try it for a while you may find a very elegant solution!", "Solution_22": "mangara, what did you mean then? :) sorry I'm not getting it :P (me=non-english)", "Solution_23": "log cabin\r\nhahahaha", "Solution_24": "I don't even know that :P\r\n\r\nAs soon as it's not mathematics... :D", "Solution_25": "[quote=\"Peter VDD\"]ln cabin? :)[/quote]\r\nOld joke but the actual answer is \" House Boat\"\r\n\r\n(It is log cabin plus \"sea\")\r\n(Of course you have to change the problem a little to make the integral indefinite)\r\n\r\n(Now Another classic one: \" what is purple and commutes? \")", "Solution_26": "abelian grape (haha)", "Solution_27": "[quote=\"Gyan\"][quote=\"Peter VDD\"]ln cabin? :)[/quote]\nOld joke but the actual answer is \" House Boat\"\n\n(It is log cabin plus \"sea\")\n(Of course you have to change the problem a little to make the integral indefinite)[/quote]\r\n\r\nGrooooaaaaan.", "Solution_28": "Tell me if I'm not allowed to simply do this for the aa proof:\r\nGiven: aa proof:\nGiven: a0$ over $ Q(\\rho)\\equal{}\\{x_j\\in[0,1], \\textstyle{\\sum_j x_j\\ge N\\rho}\\}$. This is non-trivial only for $ \\rho>\\frac 12$ when it reduces to finding the asymptotics of the derivative of the logarithm of the volume of the $ N\\minus{}1$-dimensional section of the cube $ [0,1]^N$ by the hyperplane $ \\sum_j x_j\\equal{} N\\rho$ with respect to $ \\rho$. That, in turn, reduces to finding the saddle point of $ \\left|\\frac{\\sin z}z e^{i\\lambda z}\\right|$ for $ 0<\\lambda<1$ on the imaginary axis, i.e. to the minimization of $ \\frac{e^{y}\\minus{}e^{\\minus{}y}}y e^{\\minus{}\\lambda y}$ over $ y\\in(0,\\plus{}\\infty)$. The answer is the unique root of some transcendental equation. You can easily compute it with arbitrary precision but there is no nice formula for it. \r\n\r\nAs to a reference, I doubt it is written anywhere: it is just a routine exercise in Fourier analysis and the mountain pass lemma with no particularly nice answer.", "Solution_2": "Thanks for you reply, but I'm not sure I got how is it a reformulation of getting the asymptotics of that expression. Also I don't get how it is only non-trivial for $ \\rho > \\frac 12$.", "Solution_3": "Oops, I misread the problem a bit (took $ N \\minus{} 1$ for $ N$ in hurry) :blush:. But the correct version is almost the same. Let's talk about the first quotient.\r\n\r\nIf you have any function $ f$ of one variable, then\r\n\\[ \\begin{aligned} \\sum_{k \\equal{} 0}^N \\binom{N}{k} ( \\minus{} 1)^{k}f(k) & \\equal{} ( \\minus{} 1)^N \\int_{[0,1]^N}f^{(N)}(x_1 \\plus{} \\dots \\plus{} x_N)\\,dx_1\\dots dx_N \\\\\r\n& \\equal{} \\int_0^N f^{(N)}(t)S(t)\\,dt \\end{aligned}\\]\r\nwhere $ S(t)$ is the $ N \\minus{} 1$-dimensional volume of the cross-section of $ [0,1]^N$ by the plane $ x_1 \\plus{} \\dots x_N \\equal{} t$. In the denominator, we have just $ (N \\minus{} 1)!S(N\\rho)$ because the $ N$-th derivative of $ (k \\minus{} \\rho N)_ \\plus{} ^{N \\minus{} 1}$ is just the $ \\delta$-function with that factor. In the numerator, we can wtite $ k \\equal{} (k \\minus{} N\\rho) \\plus{} N\\rho$ and see that the numerator is $ (N \\minus{} 1)! N\\rho F(N\\rho)$, which cancels with the denominator nicely, plus $ N!\\int_{N\\rho}^N F(t)\\,dt$. But, if $ \\rho > \\frac 12$, then $ F(t)\\le F(N\\rho)e^{ \\minus{} c(t \\minus{} N\\rho)}$ with $ c \\equal{} \\minus{} \\frac d{dt}\\log F(t)_{t \\equal{} N\\rho} > 0$, so only an interval of constant length is interesting for the asymptotics, on which $ F(t)$ is almost the same as $ F()e^{ \\minus{} c(t \\minus{} N\\rho)}$ and you only need to know the asymptotics of $ c$ to find the answer (which I showed).\r\n\r\nNow disregard what I said about $ \\rho < \\frac 12$. If $ N \\minus{} 1$ were $ N$, we would be dealing with the upper portion of the cube, which, in this case, would be the entire cube. Unfortunately, with $ N \\minus{} 1$, we need to be more careful. The integral becomes trivial, indeed (just $ 1$) because almost all of the volume comes from $ t\\approx \\frac N2$, but now it dominates the poor cross-section volume term and we need to know the asymtotics of $ F(N\\rho)$ rather than just that of $ \\log F(N\\rho)$, which requires more accurate computations in the mountain pass lemma. Anyway, here the ratio is huge, not approaching a constant, so I do not know if you really need to know it rather than just its logarithm and I told you how to estimate the latter.\r\n\r\nThe analysis of the remaining exceptional case $ \\rho \\equal{} \\frac 12$ is completely covered by the central limit theorem, so it shouldn't present any difficulties.\r\n\r\nThe second quotient can be treated similarly.\r\n\r\nPlease, accept my apologies for misreading the statement. Feel free to ask as many more questions as you need :).", "Solution_4": "thanks again for your reply, but I realized I should have stated the problem differently, since the term I want to get the asymptotic behavior for is\r\n$ \\frac{(N(1\\minus{}2\\rho)\\plus{}1\\plus{}2\\rho)\\sum_{k\\equal{}\\lceil N\\rho \\rceil}^N \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k\\minus{}2\\sum_{k\\equal{}\\lceil N\\rho \\rceil}^N \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k^2}{N^3\\sum_{k\\equal{}\\lceil N\\rho \\rceil}^N \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}}$\r\nwhich I realized can be restated as\r\n$ \\frac{(N(1\\minus{}2\\rho)\\plus{}1\\plus{}2\\rho)\\sum_{k\\equal{}0}^{\\lceil N\\rho \\rceil\\minus{}1} \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k\\minus{}2\\sum_{k\\equal{}0}^{\\lceil N\\rho \\rceil\\minus{}1} \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k^2}{N^3\\sum_{k\\equal{}0}^{\\lceil N\\rho \\rceil\\minus{}1} \\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}}$\r\nfor this, I expect to have a finite constant this term approaches and I will try to use the same idea you have mentioned to work it out.", "Solution_5": "Are you absolutely sure that you have $ N^3$, not $ N^2$ as a factor in the denominator? I'm asking because, from what I said above, it follows that the expression you have now tends to $ 0$ for $ \\rho>\\frac 12$... :? Also, it should blow up exponentially for $ \\rho<\\frac 12$, which you will, probably, not like in the slightest (that is because the the numerator represents some integral over the main portion of the cube but the denominator is still just the cross-section volume). Are you sure you do not have $ N$ instead of $ N\\minus{}1$, i.e., that my misreading wasn't actually the truth?", "Solution_6": "I'm sorry, I forgot an N in the numerator, now it's\r\n$ \\frac{(N^2(1\\minus{}2\\rho)\\plus{}N(1\\plus{}2\\rho))\\sum_{k\\equal{}0}^{\\lceil N\\rho\\rceil\\minus{}1}\\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k\\minus{}2\\sum_{k\\equal{}0}^{\\lceil N\\rho\\rceil\\minus{}1}\\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}k^{2}}{N^{3}\\sum_{k\\equal{}0}^{\\lceil N\\rho\\rceil\\minus{}1}\\binom{N}{k}(\\minus{}1)^{N\\minus{}k}(k\\minus{}N\\rho)^{N\\minus{}1}}$\r\nbut I am sure that I have the power as (N-1) rather than N", "Solution_7": "Are you sure that you didn't forget $ N$ in the sum with $ k^2$ too?\r\n\r\nAnyway, as long as you have $ N\\minus{}1$, it blows up for $ \\rho<\\frac 12$ (but for $ \\rho>\\frac 12$, you have a finite non-zero limit, indeed).", "Solution_8": "I think there are no N's I'm missing now, but are you sure it still blows up even after making the sum from $ 0$ to $ N\\rho \\minus{} 1$ rather than from $ N\\rho$ to $ N$ ?", "Solution_9": "Oops, missed that. Let me think a bit then :)" } { "Tag": [ "geometry", "incenter", "geometric transformation", "reflection", "angle bisector", "geometry solved" ], "Problem": "I once saw this property(possibly from Yufei) but I can not remember how to prove this. I do remember that the proof was very simple and short.\r\nLet M and N be points in the triangle ABC. Show that angle MCB= angle NCA, if angle MAB=NAC and angle MBA= angle NBC.\r\nIs this something to do with incentre?", "Solution_1": "M and N are reflections of each other in both the angle bisector of angle A and the angle bisector of angle B (two distinct lines). the only way this can happen is if M = N = I, the incenter. and the result follows.", "Solution_2": "[quote=\"pleurestique\"]M and N are reflections of each other in both the angle bisector of angle A and the angle bisector of angle B (two distinct lines). the only way this can happen is if M = N = I, the incenter. and the result follows.[/quote]\r\nM and N is not necessarily reflections. But AN is reflection of AM by bisector of angle A, BN is reflection of BM by bisector of angle B(straight line, not segment).", "Solution_3": "The problem follows easily from the angle version of Ceva's theorem.\r\n\r\nM and N are then said to be isogonal conjugates:\r\n[url=http://mathworld.wolfram.com/IsogonalConjugate.html]mathworld.wolfram.com/IsogonalConjugate.html[/url]", "Solution_4": "Here's another approach:\r\n\r\nLet $M_a,M_b,M_c$ be the projections of $M$ on $BC,CA,AB$ respectively. Let $\\ell_a,\\ell_b,\\ell_c$ be the isogonal conjugates of the lines $AM,BM,CM$ respectively. Show (very simple angle chase) that $\\ell_a\\perp M_bM_c$, and similar relations hold for $\\ell_b,_c$. \r\n\r\nThe perpendiculars from $M_a,M_b,M_c$ to $BC,CA,AB$ respectively are concurrent in $M$, which means that $ABC,M_aM_bM_c$ are orthologic. This means that the perpendiculars from $A,B,C$ to $M_bM_c,M_cM_a,M_aM_b$ respectively are also concurrent. According to the above paragraph, these perpendicularsd are precisely $\\ell_a,_b,_c$, so we have shown that $\\ell_a,_b,_c$ are concurrent, Q.E.D.", "Solution_5": "you're right. sorry." } { "Tag": [ "LaTeX", "trigonometry", "geometry theorems", "geometry" ], "Problem": "ABC is triangle $\\angle ABC=x$. Then altitude of $BC$ from $A$ is $h$ and $h$ intersect $BC$ at $E$\r\n$h^{2}-h.(BC).cotx=BE*EC$", "Solution_1": "excuse me ...\r\n\r\ncould u plz use latex ;)", "Solution_2": "Now I forgot you. Please, learn and write in [b]LaTeX[/b] !\r\nHere is the posted Osmantelli's problem:\r\n\r\n[u]The hypothesis:[/u] $\\triangle ABC$: $E\\in BC$, $AE\\perp BC\\ (AE=h_a).$\r\n\r\n[u]The conclusion:[/u] $EB\\cdot EC=h^2_a-ah_a\\cot B.$\r\n\r\n[i]Osmantelli[/i], is it correctly ?. I am waiting your .... and the answer ! $\\blacksquare .$\r\n\r\n[b]Remark. [/b] $(1)\\ \\ h^2_a+EB\\cdot EC=ah_a\\cdot cot B,\\ \\underline {E\\in (BC)}\\Longrightarrow B=45^{\\circ}.$\r\n\r\n[u]A solution.[/u] From the relations $h_a=c\\sin B$, $a=b\\cos C +c\\cos B$,\r\n\r\n$EB=c\\cos B$, $EC=b\\cos C$ results:\r\n\r\n$(1)\\Longleftrightarrow c^2\\sin ^2B+bc\\cos B\\cos C=ac\\sin B\\cdot \\frac{\\cos B}{\\sin B}\\Longleftrightarrow$\r\n\r\n$c\\sin ^2 B+b\\cos B\\cos C=a\\cos B\\Longrightarrow$\r\n\r\n$c\\sin ^2B=\\cos B(a-b\\cos C)\\Longrightarrow c\\sin ^2B=c\\cos^2 B\\Longrightarrow B=45^{\\circ}.$", "Solution_3": "but i didnt forget u dear... :rotfl:", "Solution_4": "Osmantelli's problem cannot be correct as it is written now, since one of the equality members is symmetrical in $B$ and $C$, while the other isn't ($cot B \\neq cot C$)\r\n\r\n\r\nEDIT : it is in fact correct if we replace $cot B$ by $cot A$, giving : \r\n\r\n\\[ EB . EC = h_a^2 - a h_a cot A \\]" } { "Tag": [], "Problem": "What is the maximum number of non-intersecting diagonals in a convex $n$-gon?", "Solution_1": "[hide=\"hint\"] Induction/recursion works well. For a convex $n$-gon, draw any diagonal and you have reduced it to a case of two smaller polygons. [/hide]", "Solution_2": "Alternatively, is it not just clearly $n-3$? This is achievable, and using the fact that one cannot draw a diagonal within a triangle, it must be maximal." } { "Tag": [ "function", "induction", "algorithm", "algebra", "domain", "number theory", "relatively prime" ], "Problem": "Find all function $f:Q_+ \\rightarrow Q_+$ such that :\r\n\r\n1) $f(x) = f( \\frac{1}{x} ) $ for all $x \\in Q_+$.\r\n2) $( \\frac{1}{x} +1)f(x)=f(x+1)$ for all $x \\in Q_+$.", "Solution_1": "Let f(1) = y.\r\n\r\n$f:Q_+ \\rightarrow Q_+ => y \\in Q_+$.\r\n\r\n$( \\frac{1}{x} +1)f(x)=f(x+1) => f(n) = ny$ for all $n \\in N_$. \r\n(This is an induction. $( \\frac{1}{1} +1)f(1)=f(2) => f(2)=2y$ \r\nand we are already given that $( \\frac{1+x}{x} )f(x)=f(x+1)$ so $( \\frac{1+n}{n} )n*y=f(n+1)$.)\r\n\r\n$f(x) = f( \\frac{1}{x} ) => f( \\frac{1}{n} )=ny$ for all $n \\in N_$.\r\n\r\nNote that 2) may be rewritten as $f(x)=( \\frac{x}{x-1} )*f(x-1)$ for x>1.\r\n\r\nIf n > m*k for $k \\in N_$ then $f( \\frac{n}{m} )= \\frac{n}{n-m*k} *f( \\frac{n-m*k}{m} )$. \r\n(This is an induction as well. $f( \\frac{n}{m} )= \\frac{n/m}{(n-m)/m} *f( \\frac{n}{m} -1)=\\frac{n}{n-m} *f( \\frac{n-m}{m} )$ \r\nand $\\frac{n}{n-m*(p-1)} *f( \\frac{n-m*(p-1)}{m} )$ \r\n=$ \\frac{n}{n-m*(p-1)} *( \\frac{ \\frac{n-m*(p-1)}{m} }{ \\frac{n-m*(p-1)}{m} -1} )*f( \\frac{n-m*(p-1)}{m} -1)$ \r\n=$ \\frac{n}{n-m*(p-1)} *( \\frac{ \\frac{n-m*(p-1)}{m} }{ \\frac{n-m*p}{m} } )*f( \\frac{n-m*p}{m} )$ \r\n=$ \\frac{n}{n-m*p}*f( \\frac{n-m*p}{m} )$.)\r\n\r\nThere exist $m,n \\in N_$ for all $x \\in Q_+$ such that $x = \\frac{m}{n}$ and m and n are relatively prime.\r\n\r\nIf m\u0395\u03bc\u03b2\u03b1\u03b4\u03cc(\u03a1\u0391\u0392) [\u03b1\u03c6\u03bf\u03cd \u039c\u0391*\u039c\u0392=\u0395\u03bc\u03b2\u03b1\u03b4\u03cc(\u039c\u0391\u0392)/\u03b7\u03bc\u03c6 \u03ba\u03b1\u03b9 \u03a1\u0391*\u03a1\u0392=\u0395\u03bc\u03b2\u03b1\u03b4\u03cc(\u03a1\u0391\u0392)/\u03b7\u03bc\u03c6]. \u0391\u03bb\u03bb\u03ac \u0395\u03bc\u03b2\u03b1\u03b4\u03cc(\u039c\u0391\u0392)>\u0395\u03bc\u03b2\u03b1\u03b4\u03cc(\u03a1\u0391\u0392)<=>d(M,AB)>d(P,AB), \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9, \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf \u039c \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 \u03c4\u03cc\u03be\u03bf\u03c5 \u0391\u0392", "Solution_2": "K\u03ce\u03c3\u03c4\u03b1 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03bd\u03c5\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7\r\n\u03a1\u0391.\u03a1\u0392=(\u03b4.\u03a1\u0391.\u03b4.\u03a1\u0392)/\u03c3\u03c5\u03bd\u03a1=[(\u03a1\u039a)^2-(\u039a\u0392)^2]/\u03c3\u03c5\u03bd\u03a1<[(\u039c\u039a)^2-(\u039a\u0392)^2]/\u03c3\u03c5\u03bd\u039c=(\u03b4.\u039c\u0391)(\u03b4.\u039c\u0392)/\u03c3\u03c5\u03bd\u039c=(\u039c\u0391)(\u039c\u0392) \u03cc\u03c0\u03bf\u03c5 \u039a \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 \u0391\u0392", "Solution_3": "\u0397\u03bb\u03af\u03b1 \u03ba\u03b1\u03b9 \u03a7\u03c1\u03ae\u03c3\u03c4\u03bf \u03c3\u03b1\u03c2 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce. \r\n\r\n\u0397 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03c5\u03c0\u03cc\u03c8\u03b7 \u03bc\u03bf\u03c5, \u03b2\u03b1\u03c3\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b9\u03b4\u03ad\u03b1 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ P\\neq M,$ \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03cc\u03c4\u03b9 $ PA \\plus{} PB < MA \\plus{} MB \\equal{} 2\\kappa$ $ ,(1)$\r\n\r\n\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03ce\u03c1\u03b1, \u03c4\u03bf \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03b4\u03cd\u03bf \u03ac\u03bd\u03b9\u03c3\u03c9\u03bd \u03b5\u03c5\u03b8\u03cd\u03b3\u03c1\u03b1\u03bc\u03bc\u03c9\u03bd \u03c4\u03bc\u03b7\u03bc\u03ac\u03c4\u03c9\u03bd, \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf \u03c4\u03bf\u03c5 \u03b7\u03bc\u03b9\u03b1\u03b8\u03c1\u03bf\u03af\u03c3\u03bc\u03b1\u03c4\u03cc\u03c2 \u03c4\u03bf\u03c5\u03c2 ( \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03b5\u03c4\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b1\u03bb\u03b3\u03b5\u03b2\u03c1\u03b9\u03ba\u03ac \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac ), \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03ac\u03bc\u03b5\u03c3\u03b1 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf.\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03b7\u03c2 $ (1),$ \u03ad\u03c7\u03c9 \u03c5\u03c0\u03cc\u03c8\u03b7 \u03bc\u03bf\u03c5 \u03bc\u03af\u03b1 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c0\u03bf\u03bb\u03cd \u03b8\u03b5\u03c9\u03c1\u03af\u03b1, \u03b1\u03bb\u03bb\u03ac \u03af\u03c3\u03c9\u03c2 \u03bc\u03b5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 ( \u03cc\u03c0\u03bf\u03c5 \u03b4\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c0\u03bf\u03bb\u03cd \u03bb\u03af\u03b3\u03b1 \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 ), \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ac \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1. \r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_4": "Esto $ x$ i gonia ton $ AB,MB$ kai $ y$ i gonia ton $ AB,BM$ \r\nIsxii oti $ x \\plus{} y$=sta8ero\r\nkai episis $ PA*PB \\equal{} 4R^2 *sinx*siny$\r\nkai epidi $ sinx*siny \\equal{} \\frac {cos(x \\minus{} y) \\minus{} cos(x \\plus{} y)}{2}$,i arxiki posostita ginete megisti otan \r\n$ cos(x \\minus{} y)$ megistopiite (afu cos(x+y)=sta8ero)\r\ndiladi otan $ x \\equal{} y$\r\n\r\nEdit:\r\nMe entelos omia silogistiki mpori na vgi oti k to a8risma ine mikrotero...dld $ PA \\plus{} PB < MA \\plus{} MB \\equal{} 2\\kappa$" } { "Tag": [ "AMC", "AIME", "floor function" ], "Problem": "For any positive integer x, let S(x) be the sum of the digits of x, and let T(x) be abs(S(x+2)-S(x)) How many values T(x) do not exceed 1999?", "Solution_1": "[hide=\"(edited) Solution\"]\n$T(x)$ takes on the value $2$ and values $\\equiv -2 \\bmod 9$ (easy to prove). It's easy to construct examples where it takes on all of them, depending on how many digits need to be carried by the addition $x + 2$. Therefore, it takes on $\\lfloor \\frac{1999 - 7}{9} \\rfloor + 1 = \\boxed{223}$ values. [/hide]", "Solution_2": "[hide]Do you mean $\\equiv 7 \\bmod 9$ ... ? (remember abs)\n\n(and 2 of course)[/hide]\r\nAlso, you seem to have made an arthimetic error in the last step. ;)" } { "Tag": [ "logarithms" ], "Problem": "Solve in $ \\mathbb{R}$ the equation: $ 3^{|x\\plus{}1|}\\minus{}2|3^x\\minus{}1|\\equal{}3^x\\plus{}2$ .", "Solution_1": "Divide in to three case in order to solve it.\r\nvery easy.", "Solution_2": "Let's look at when $ x \\plus{} 1$ and $ 3^x \\minus{} 1$ are nonnegative/negative.\r\n\r\n[hide]$ x \\plus{} 1 \\geq 0 \\rightarrow x \\geq \\minus{} 1$\n\n$ x \\plus{} 1 < 0 \\rightarrow x < \\minus{} 1$\n\n$ 3^x \\minus{} 1 \\geq 0 \\rightarrow x \\geq 0$\n\n$ 3^x \\minus{} 1 < 0 \\rightarrow x < 0$[/hide]\nCase 1 : $ 0 > x \\geq \\minus{} 1$\n\n[hide]$ 3^{x \\plus{} 1} \\plus{} 2(3^x \\minus{} 1) \\equal{} 3^x \\plus{} 2$\n\n$ 4 \\cdot 3^x \\minus{} 4 \\equal{} 0$\n\n$ x \\equal{} 0$[/hide]\n\nCase 2: $ x \\geq 0$\n\n[hide]$ 3^{x \\plus{} 1} \\minus{} 2(3^x \\minus{} 1) \\equal{} 3^x \\plus{} 2$\n\n$ 0 \\equal{} 0$ \nThis means infinite solutions for all nonnegative $ x$.[/hide]\n\nCase 3: $ x < \\minus{} 1$\n\n[hide]$ 3^{ \\minus{} x \\minus{} 1} \\plus{} 2(3^x \\minus{} 1) \\equal{} 3^x \\plus{} 2$\n\nSolving gives $ x \\equal{} \\log_{3} (2 \\pm \\frac {\\sqrt {33}}{3})$[/hide]\r\n\r\n\r\nPS: Tell me if you see any mistkes." } { "Tag": [ "analytic geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $A_1,A_2,...,A_n$ be the sets each consisting of $n$ segments on a given line.Prove that the intersection $A_1\\cap A_2\\cap...\\cap A_n$ contains not more than $n^2-n+1$ disjoint segments.", "Solution_1": "[quote=\"ehsan2004\"]Let $A_1,A_2,...,A_n$ be the sets each consisting of $n$ segments on a given line.Prove that the intersection $A_1\\cap A_2\\cap...\\cap A_n$ contains not more than $n^2-n+1$ disjoint segments.[/quote]\r\nFor $n^2$ is easy. But I don't see how to solve for $n^2-n+1$.", "Solution_2": "for each point calculate the coordinate of it right endpoint, and let $r_12 )/2...\r\nI use vectors to prove that the sums (XA 2n+XB2n+XC2n+XD2n) are constant for each nonnegative integer.\r\nFrom now on xa means the vector XA (not the segment XA).\r\nxa=xo+oa.\r\nApply Newton formula for (a+b)k using the scalar product for vectors.\r\nIn that formula use xa2 =XA2 , and oa+ob+oc+od=0. Because those sums are constant it should easily follow the fact that the sums cosn(XOA)+... are constant." } { "Tag": [ "function", "induction", "algebra proposed", "algebra" ], "Problem": "Find all the continuous functions $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that :\r\n$\\forall(x,y) \\in \\mathbb{R}^2$,\r\n$f(x+y)\\cdot f(x-y) = [f(x)\\cdot f(y)]^2$\r\n\r\nEDIT : Sorry, maybe this post should rather go into the \"proposed and own problems\" section :oops:", "Solution_1": "A variation on the old Cauchy theme.\r\n\r\nIf $f(x)=0$ for any $x$ then $f$ must be uniformly 0.\r\n\r\nOtherwise, it's easier to coinsider $g(x)=\\log|f(x)|$. Clearly $g(x+y)+g(x-y)=2g(x)+2g(y)$. By induction $g(nx)=n^2g(x)$ for all integers $n$, and so $g(\\frac{n}{m}x)=n^2g(\\frac{1}{m}x)=(\\frac{n}{m})^2g(x)$ for all rational numbers $\\frac{n}{m}$. By continuity, then, $g(x)=kx^2$, so for any $x$, $f(x)=\\pm a^{x^2}$, where $a$ is some positive constant.\r\n\r\nSubstituting $x=r+s$, $y=r-s$, we see that $f(r)f(s)=[f(r+s)f(r-s)]^2>0$ for all $r$, $s$, so either $f(x)=a^{x^2}$ for all $x$ or $f(x)=-a^{x^2}$ for all $x$." } { "Tag": [ "inequalities" ], "Problem": "The numerator of a fraction is $ 6x \\plus{} 1$, then denominator is $ 7 \\minus{} 4x$, and $ x$ can have any value between $ \\minus{}2$ and $ 2$, both included. The values of $ x$ for which the numerator is greater than the denominator are:\r\n\r\n$ \\textbf{(A)}\\ \\frac{3}{5} < x \\le 2\\qquad \r\n\\textbf{(B)}\\ \\frac{3}{5} \\le x \\le 2\\qquad \r\n\\textbf{(C)}\\ 0 < x \\le 2\\qquad \\\\\r\n\\textbf{(D)}\\ 0 \\le x \\le 2\\qquad \r\n\\textbf{(E)}\\ \\minus{}2 \\le x \\le 2$", "Solution_1": "We just get $ 6x\\plus{}1>7\\minus{}4x$. Solving:\r\n\r\n$ 6x\\plus{}1>7\\minus{}4x$\r\n$ 10x\\plus{}1>7$\r\n$ 10x>6$\r\n$ x>\\frac{3}{5}$\r\n\r\nBut we also have a limit of 2. So:\r\n\r\n$ \\frac{3}{5} 1$, the value of $f(2)$ so that $|f - f_2| > 1$, etc. In order for this process to not terminate, you need to have $f$ defined on some infinite set -- otherwise you'd run out of degrees of freedom for your function $f$.\r\n\r\nTo prove that $\\mathcal{C}(X,\\mathbf{R})$ is separable when $X$ is finite, just use that $\\mathbf{R}$ is separable.", "Solution_4": "Thank you. I realized it later on and its pretty obvious now. It's very similar to the diagonal construction in the proof of the uncountability of $\\mathbb{R}$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Given a,b,c$\\ge{0}$ Prove that:\r\n$(a^2+b^2+c^2)^2\\geq\\cdot{3(a^3b+b^3c+c^3a)}$", "Solution_1": "It was posted at least one time:\r\n$(a^2+b^2+c^2)^2\\geq\\3(a^3b+b^3c+c^3a) \\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\\ge3(a^3b+b^3c+c^3a) \\Leftrightarrow\\frac{1}{2}[(a^2-2ab+bc-c^2+ca)^2+(b^2-2bc+ca-a^2+ab)^2+(c^2-2ca+ab-b^2+bc)^2]\\ge0$" } { "Tag": [], "Problem": "calculate the amount in M of HCO-3 that is added to blood for the blood PH to rise to 8.0\r\n\r\nand then calculate the amount of H+ in M that is removed when this amount HCO-3 of is added. \r\n\r\nWhat equation can i use, i dont want the answer i just want an equation or a certain way i can do this problem", "Solution_1": "In this problem we are considering that the pH of blood is controled solely by the buffer system represented by the equilibrium\r\n\r\n$H_{2}CO_{3}\\rightleftharpoons H^{+}+HCO_{3}^{-}$\r\n\r\nwhere the carbonic acid arises from the equilibrium\r\n\r\n$CO_{2}+H_{2}O \\rightleftharpoons H_{2}CO_{3}$.\r\n\r\nI don't know what is data that you are given to solves this problem, but perhaps the Henderson-Hasselbach equation may help you.\r\n\r\nCarcul" } { "Tag": [ "algebra", "binomial theorem" ], "Problem": "Can 4x^2-9 be factored?", "Solution_1": "[hide]since 4x^2-9 is the difference of two squares, it can be factored down to (2x+3)(2x-3)[/hide]", "Solution_2": "More generally,\r\n\r\n$x^{2}-y^{2}= (x+y)(x-y)$.", "Solution_3": "Also,\r\n\r\n$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$\r\nand\r\n$x^{4}-y^{4}=(x-y)(x+y)(x^{2}+y^{2})$\r\n\r\nThe pattern continues.", "Solution_4": "In general, the binomial theorem can be used.\r\n$(x+y)^{n}=\\binom{n}{0}x^{n}+\\binom{n}{1}x^{n-1}y+...+\\binom{n}{n-1}x^{1}y^{n-1}+\\binom{n}{n}y^{n}$", "Solution_5": "Better yet,\r\n\r\n$(x+y)^{n}= \\sum^{4}_{r=0}\\binom{4}{r}x^{4-r}y^{r}$", "Solution_6": "[quote=\"Temperal\"]Better yet,\n\n$(x+y)^{n}= \\sum^{4}_{r=0}\\binom{4}{r}x^{4-r}y^{r}$[/quote]\r\n\r\nShould not it be $(x+y)^{4}$?", "Solution_7": "[quote=\"AstroPhys\"][quote=\"Temperal\"]Better yet,\n\n$(x+y)^{n}= \\sum^{4}_{r=0}\\binom{4}{r}x^{4-r}y^{r}$[/quote]\n\nShould not it be $(x+y)^{4}$?[/quote]\r\n\r\nOr $(x+y)^{n}= \\sum^{n}_{r=0}\\binom{n}{r}x^{n-r}y^{r}$.", "Solution_8": "Yeah. Oops.", "Solution_9": "Why are we discussing the binomial theorem?\r\n\r\n[hide=\"anyways\"]\n$4x^{2}-9$ can be factored as $(2x-3)(2x+3)$.\n[/hide]\r\n\r\nThe question is pretty much answered and the thread is getting off topic.", "Solution_10": "[quote=\"alanchou\"]Why are we discussing the binomial theorem?\n\n[hide=\"anyways\"]\n$4x^{2}-9$ can be factored as $(2x-3)(2x+3)$.\n[/hide]\n\nThe question is pretty much answered and the thread is getting off topic.[/quote]\r\n\r\nWell the thread isn't really \"off topic\" because we're presenting generalizations that jamesstine will probably find very helpful later.", "Solution_11": "[quote=\"Temperal\"]Also,\n\n$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$\nand\n$x^{4}-y^{4}=(x-y)(x+y)(x^{2}+y^{2})$\n\nThe pattern continues.[/quote]\r\n\r\nAccording to Temperal, this is more general: $x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{2}...+x^{2}y^{n-3}+xy^{n-2}+y^{n-1})$" } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove that if $ AB \\minus{} BA \\equal{} A$, then $ \\det(A) \\equal{} 0$.", "Solution_1": "Suppose $ \\det(A)\\neq 0$, then $ A$ is invertible, so\r\n\r\n\\begin{eqnarray*}\r\nAB & = & A+BA\\\\\r\n & = & (I+B)A \\\\\r\n\\det(AB) & = & \\det((I+B)A) \\\\\r\n \\det(A)\\det(B) & = & \\det(I+B)\\det(A) \\\\\r\n \\det(B) & = & \\det(I+B) \\\\\r\n\\end{eqnarray*}\r\n\r\nA contradiction.", "Solution_2": "[quote=\"Beginner\"]Prove that if $ AB \\minus{} BA \\equal{} A$, then $ \\det(A) \\equal{} 0$.[/quote]\r\nActually , having these conditions we can prove that $ A^n\\equal{}O_n$", "Solution_3": "if $ \\det (A) \\not\\equal{}0$ and $ AB\\minus{}BA\\equal{}A$ then $ B\\minus{}A^{\\minus{}1}BA\\equal{}I$ which is impossible since $ \\text{tr}(B)\\equal{}\\text{tr}(A^{\\minus{}1}BA)$.\r\n\r\n@Svejk\r\nNice result, do you mean that $ \\text{Rank}(A)\\equal{}1$ ?", "Solution_4": "@HilbertThm90: \r\n$ \\det(B) \\equal{} \\det(I \\plus{} B)$ is not a contradiction. There exist (a lot of) polynomials $ P$ such that $ P(0) \\equal{} P(1)$\r\n\r\n@Beginner: \r\n$ A^n \\equal{} 0$ does not imply that $ Rank(A) \\equal{} 1$ (google: Jordan block)", "Solution_5": "Beginner's argument works over a field of characteristic zero. Over a field of finite characteristic, the original result may not hold.\r\n\r\nOver a field of characteristic $ 2,$ consider $ A\\equal{}\\begin{bmatrix}0&1\\\\1&0\\end{bmatrix}$ and $ B\\equal{}\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix}.$", "Solution_6": "Proof of Svejk's claim, assuming we're working over a field of characteristic zero.\r\n\r\nWe have $ A \\equal{} AB \\minus{} BA,$ which we can also write as $ AB\\equal{}A \\plus{} BA.$\r\n\r\n$ A^2 \\equal{} A(AB \\minus{} BA) \\equal{} A^2B \\minus{} ABA \\equal{} A^2B \\minus{} (A \\plus{} BA)A \\equal{} A^2B \\minus{} A \\minus{} BA^2.$ \r\n\r\nHence, $ 2A^2 \\equal{} A^2B \\minus{} BA^2$\r\n\r\nAssume $ kA^k \\equal{} A^kB \\minus{} BA^k.$ Then\r\n\r\n$ kA^{k \\plus{} 1} \\equal{} A^{k \\plus{} 1}B \\minus{} ABA^k \\equal{} A^{k \\plus{} 1}B \\minus{} (A \\plus{} BA)A^k$ yielding\r\n\r\n$ (k \\plus{} 1)A^{k \\plus{} 1} \\equal{} A^{k \\plus{} 1}B \\minus{} BA^{k \\plus{} 1}.$\r\n\r\nThus $ kA^k \\equal{} A^kB \\minus{} BA^k$ for all $ K\\in\\mathbb{N}.$\r\n\r\nTake the trace and divide by $ k$ to get that $ \\text{Tr}(A^k) \\equal{} 0\\ \\forall\\,k\\in\\mathbb{N}.$\r\n\r\nNow, all we need is a proof of the fact that if all of the powers of a matrix have trace zero, then the matrix is nilpotent.", "Solution_7": "[quote=\"Kent Merryfield\"]\n\nNow, all we need is a proof of the fact that if all of the powers of a matrix have trace zero, then the matrix is nilpotent.[/quote]\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=205969]Here[/url] it is :wink:", "Solution_8": "An alternate version of the same argument:\r\n\r\nLet's try this. Let $ p(x)$ be the characteristic polynomial of $ A.$ By Cayley-Hamilton, $ P(A)\\equal{}0.$ Take the trace of $ P(A)\\equal{}0$ and we get that $ \\text{Tr}(a_0I)\\equal{}0,$ from which $ a_0,$ the constant term of the characteristic polynomial, must be 0.\r\n\r\nLet $ x$ be an eigenvector such that $ Ax\\equal{}0.$ Build an invertible matrix $ P$ whose first column is $ x.$ Then $ P^{\\minus{}1}AP\\equal{}\\begin{bmatrix}0&*\\\\0&A'\\end{bmatrix}.$ But if we take powers of this, we see that $ \\text{Tr}(A'^k)\\equal{}0$ for all $ k\\in\\mathbb{N}.$ So the same process can be applied to $ A'.$\r\n\r\nAfter $ j$ steps of this process we have that $ A$ is similar to $ \\begin{bmatrix}N&*\\\\0&B\\end{bmatrix},$ where $ N$ is strictly upper triangular (upper triangular with zeros on the main diagonal). But then $ \\text{Tr}(A^k)\\equal{}\\text{Tr}(B^k)\\equal{}0,$ so the argument applies to $ B$ and we can take another zero off. This recursive process can only end when $ B$ vanishes altogether. We conclude that $ A$ is similar to a strictly upper triangular matrix, and hence $ A$ is nilpotent.\r\n\r\n(Essentially, this is just a minor modification of the proof of Schur's Lemma.)" } { "Tag": [ "calculus", "email", "\\/closed" ], "Problem": "(I'd PMed this to you, Mr. Rusczyk, if you were wondering about an email notification, but I thought it might help others as a forum topic)\r\n\r\n\r\nI'd enrolled in the Calculus class, but today's Math Jam has left me with a lot of doubt as to how useful it will be for me, since I've already taken Calculus.\r\n\r\nI'd like to drop the class and enroll in WOOT instead. Is it too late? I noticed the \"real\" problem solving sessions haven't started yet.\r\n\r\n[quote=\"My Classes\"]Click the 'Drop Class' button if you wish to drop the course. A full refund will be issued within 30 days.[/quote]\r\nIf I drop the Calculus class, how will it be refunded? Will I receive a check, or a money transfer directly to my dad's bank account? Also, exactly how fast will the process be? My mom seems worried that we might be a bit short on money this month if we don't receive the refund before I pay for and enroll in WOOT.\r\n\r\nI hope my late enrollment won't cause too much trouble for you guys. Thanks!", "Solution_1": "It is definitely not too late to enroll in WOOT. This week's classes are just \"orientation\"; the mathematical content of WOOT doesn't start until next week.\r\n\r\nNormally, if you drop a class, we issue the refund via the original form of payment (for example, if you paid by credit card, we'll issue a refund to your credit card). On occasion, though, we're unable to do this (for reasons I don't want to go into), in which case we'll mail you a check.\r\n\r\nHowever, if you want to drop one class and immediately enroll in another, we can instead issue you a non-refundable AoPS gift certificate for the dropped class that you can immediately apply to the new class. \r\n\r\nEmail us at classes@artofproblemsolving.com with your information (essentially, with what you've said in this post), and we can set up the details.\r\n\r\nBy the way, PM'ing an individual is usually the least-effective way to contact us. Emailing us at one the addresses on the \"Contact Us\" page (there's a link at the bottom of every web page) is by far the better way to get in touch with us.", "Solution_2": "Thank you for the info, Dr. Patrick! I will use the classes email address." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c,d$ be non-nagative real numbers such that: $ a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}$. Find the min value of the following expression:\r\n\\[ N=abc+bcd+cda+dab \\]", "Solution_1": "? $ a=1$ and $ b=c=d=0$?", "Solution_2": "to my brother, you should try it :P Don't ask me :P Actually, it is very easy.", "Solution_3": "Hi my younger brother :lol: I am not spamming ! I want to ask if you have an typo in \"minimum\" instead of \"maximum\". The minimum is clearly $ 0$ as in my example.", "Solution_4": "[quote=\"hungkhtn\"]Hi my younger brother :lol: I am not spamming ! I want to ask if you have an typo in \"minimum\" instead of \"maximum\". The minimum is clearly $ 0$ as in my example.[/quote]\r\nokie, I am always understand you :D But the minimum value of $ N$ is not equal $ 0$. It means you are wrong :D I haven't ever tried to find out the maximum value but I will try now :P", "Solution_5": "I have 2 solutions for above problem. The proof is using Cauchy-Schawrz what I like best :D", "Solution_6": "[quote=\"zaizai-hoang\"][quote=\"hungkhtn\"]Hi my younger brother :lol: I am not spamming ! I want to ask if you have an typo in \"minimum\" instead of \"maximum\". The minimum is clearly $ 0$ as in my example.[/quote]\nBut the minimum value of $ N$ is not equal $ 0$. [/quote]\r\n\r\nWhy the minimum is not equal to $ 0$? For $ a=1,b=c=d=0$ then $ a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}$ and\r\n$ abc+bcd+cda+dab=0$. Even if the problem is given for positive real numbers, the infimum is $ 0$, too." } { "Tag": [ "absolute value" ], "Problem": "Hi there!\r\n\r\nLet $x_{1}\\, , \\, x_{2}\\, \\, \\ldots \\, , \\, x_{N}$ be $N \\geqslant 2$ (real) numbers and consider the range of them, that is, the greatest possible absolute value of the differences between any 2 of them: \\[range = \\,\\,\\mathop{\\max }_{1\\, \\leqslant \\,u,v\\, \\leqslant \\,N}\\,\\left|{x_{u}-x_{v}}\\right|\\]\r\n\r\nIf $\\mu$ is the average of the $N$ given numbers, is it possible to have $\\left|{x_{j}-\\mu }\\right| > \\,\\frac{{range}}{2}$ , where $j \\in \\left\\{{1,2, \\ldots ,N}\\right\\}\\, \\,$ ?", "Solution_1": "It is certainly possible, consider the data set $\\{1,99,99,99,99,99,99\\}$. The range is $98$ and $\\mu = 85$. And $|1-85|> \\frac{98}{2}$.\r\n\r\nEDIT: Yes sorry, it is fixed.", "Solution_2": "Correct, illcrowflu. You certainly mean that the range is 98 and $\\left|{1-85}\\right| > \\,\\,\\frac{{98}}{2}$\r\n\r\nOn \"the same subject\", considering the standard deviation, defined as:\r\n\r\n\\[\\sigma = \\sqrt{\\frac{1}{N}\\cdot \\sum_{j = 1}^{N}{\\left({x_{j}-\\mu }\\right)^{2}}}\\]\r\n\r\nis it possible to have $\\, \\, \\, \\sigma > \\frac{{range}}{2}$ ?", "Solution_3": "I\u00b4ve posted another (related-) problem at my last message. It\u00b4s a nice one, I guess! Have a try! \r\n\r\nSmall technical hint:\r\n[hide]Without loss of generality, you may consider $x_{1}\\leqslant x_{2}\\leqslant \\ldots \\leqslant x_{N}$ and, therefore, $range = x_{N}-x_{1}$[/hide]" } { "Tag": [], "Problem": "In a circle with center M, two perpendicular chords AC and BD are given. The circle with diameter AM intersects the circle with diameter BM in M and P. The circle with diameter BM intersects the circle with diameter CM in M and Q. The circle with diameter CM intersects the circle with diameter DM in M and R. The circle with diameter DM intersects the circle with diameter AM in M and S.\r\nShow that PQRS is a rectangle.", "Solution_1": "Because angles MQB and MQC are right we can see that Q is, in fact, the projection of M on BC, so it's the midpt of BC in the same way we find that P, R, S are the midpts of AB, CD, DA respectively. This means that PQ and RS are parallel to AC and QR and PS are parallel to BD, and from here we derive the conclusion." } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Let M ={ 1,2,....,40} .Find the smallest positive integer n for which it is possible to partition M into n disjoint subsets such that whenever a, b, and c ( not necessarily distinct ) are in the same subset ,then a is not equal to (b+c)", "Solution_1": "This is related to the so-called Schur's numbers.\r\n\r\nSchur's original construction proves that the set $\\{1, \\cdots , \\frac {3^n - 1} 2 \\}$ can be splitted into $n$ pairwise disjoint subsets such that none of them contains a solution of the equation $x+y=z$ (in non necessarly distincts integers).\r\nTherefore if $ 40 \\leq \\frac {3^n - 1} 2$ that is $n \\geq 4$ such a partition does exist.\r\n\r\nIn another hand it is a classical problem to prove that $\\{ 1, 2 , \\cdots, 14 \\}$ cannot be partitioned into three subsets such that none of them contains a solution of the equation $x+y=z.$\r\n\r\nThus the desired minimum is $n = 4$.\r\n\r\nPierre.", "Solution_2": "If n =4 , how can we divide the set into 4 subsets that satisfy the condition ?", "Solution_3": "you can prove by induction that (3^n-1)/2 numbers can be divided into such n subsets." } { "Tag": [], "Problem": "There's a fair, handing out 218 prizes. 9 groups try to win the prizes.\r\n3 groups do better compared to the others;\r\nGroup A is 15% of the population and gets 46 prizes\r\nGroup B is 19%, and gets 47\r\nGroup C is 6% and gets 19\r\nWhat are the odds that someone in each group can win a prize?", "Solution_1": "[hide]Group A: 86/23\n\nGroup B: 171/47\n\nGroup C: 199/19\n\nOther : 53/56[/hide]" } { "Tag": [ "geometry", "circumcircle", "incenter", "geometry unsolved" ], "Problem": "The circumcircle of triangle ABC has center O and diameter AD . Let I be the incenter of triangle ABC . The lines AI, DI cut again the circumcircle at H, K respectively. Draw the line IJ perpendicular to BC at J . Prove that H, K, J are collinear .", "Solution_1": "Let K' be the intersection of HJ and (O). K'I meets (O) at K' and D'.\r\n$ \\hat{BK'H}\\equal{}\\hat{HBJ} \\equal{}> BHJ\\sim K'HB \\equal{}> HJ.HK'\\equal{}HB^2\\equal{}HI^2 \\equal{}> IHJ\\sim K'HI$\r\n$ \\equal{}> \\hat{JIH}\\equal{}\\hat{HK'I}\\equal{}\\hat{HK'D'}\\equal{}\\hat{HAD'}$\r\nIJ parallel to OH => $ \\hat{JIH}\\equal{}\\hat{IHO}\\equal{}\\hat{HAO}$\r\nHence, $ \\hat{HAO}\\equal{}\\hat{HAD'}$, so $ A,O,D'$ are collinear.\r\n$ \\equal{}> D'\\equiv D \\equal{}> K'\\equiv K$, hence $ H,K,J$ are collinear.", "Solution_2": "Dear Mathlinkers,\r\nlet 0 be the circumcircle of ABC, U the orthogonal projection of I on the A-altitude of ABC.\r\nIt is knowed that U, J and H are collinear.\r\nLet Th the tangent to 0 at H and 1 the circle with diameter AI ; it goes through K and E .\r\nWe have Th // UI\r\nApplying the Reim's theorem to 0 and 1, it follows that K, U and H are collinear...\r\nSincerely\r\nJean-Louis", "Solution_3": "[quote=\"conan_naruto236\"] [color=darkblue]Let $ [AD]$ be the diameter of the circumcircle $ w \\equal{} C(O,R)$ of $ \\triangle ABC$ and let $ C(I,r)$ be the incircle of $ \\triangle ABC$ . The \n\nlines $ AI$ , $ DI$ cut again the circle $ w$ at $ H$ , $ K$ respectively. Denote $ J\\in BC$ for which $ IJ\\perp BC$ Prove that $ J\\in HK$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof.[/u][/b] This nice problem is one from the celebrated old properties of a triangle. Observe easily that \n\n$ \\widehat {IAD}\\equiv\\widehat {JIH}$ and $ \\frac {AI}{AD} \\equal{} \\frac {IJ}{IH}$ because the power of $ I$ w.r.t. $ w$ is $ p_w(I)\\equiv\\overline {IA}\\cdot \\overline {IH} \\stackrel {(*)}{\\ \\equal{}\\ } \\minus{} 2Rr$ and \n\n$ AD\\cdot IJ \\equal{} 2Rr$ . Therefore $ \\triangle IAD\\sim\\triangle JIH$ $ \\implies$ $ \\widehat {JHI}\\equiv\\widehat {IDA}$ $ \\implies$ $ \\widehat {JHI}\\equiv\\widehat {KDA}$ , i.e. $ J\\in HK$ . [/color]\r\n\r\n========================================================================\r\n\r\n$ (*)\\ \\blacktriangleright$ Denote $ R\\in AC$ so that $ IR\\perp AC$ and the diameter $ [NS]$ of $ w$ so that $ NS\\perp BC$ \r\n\r\nand the line $ BC$ separates $ A$ , $ S$ . Observe easily that $ \\triangle AIR\\sim\\triangle NSC$ $ \\implies$ $ \\frac {AI}{NS}\\equal{}\\frac {IR}{SC}$ $ \\implies$ \r\n\r\n$ IA\\cdot SC\\equal{}2Rr$ $ \\stackrel {(**)}{\\ \\implies\\ }$ $ IA\\cdot IS\\equal{}2Rr$ , i.e. the power of $ I$ w.r.t. $ w$ is $ p_w(I)\\equal{}\\minus{}2Rr$ .\r\n\r\n========================================================================\r\n\r\n$ (**)\\ \\blacktriangleright\\ m(\\widehat {SIC})\\equal{}m(\\widehat {SCI})\\equal{}\\frac {A\\plus{}C}{2}\\equal{}90^{\\circ}\\minus{}\\frac B2$ $ \\implies$ $ SB\\equal{}SC\\equal{}SI$ ." } { "Tag": [ "analytic geometry", "geometry", "3D geometry" ], "Problem": "Let $ S$ be the set of all points with coordinates $ (x,y,z)$, where $ x, y,$ and $ z$ are each chosen from the set $ \\{ 0, 1, 2\\}$. How many equilateral triangles have all their vertices in $ S$?\r\n\r\n$ \\textbf{(A)}\\ 72 \\qquad \\textbf{(B)}\\ 76 \\qquad \\textbf{(C)}\\ 80 \\qquad \\textbf{(D)}\\ 84 \\qquad \\textbf{(E)}\\ 88$", "Solution_1": "I don't think I missed any...\r\n\r\n[hide=\"Answer\"]There are $9$ distinct cubes with edges parallel to the axes. Each of these cubes has $8$ distinct equilateral triangles formed by three face diagonals of the cube, so the answer is $8\\cdot 9=72\\Rightarrow \\boxed{A}$.[/hide]", "Solution_2": "hmm..\r\ni dont see how you got 9 distinct cubes. ;)", "Solution_3": "One 2x2x2 cube and eight 1x1x1.", "Solution_4": "How can we be sure (or prove) there are no more equilateral triangles?", "Solution_5": "There can't be any more equilateral triangles per cube. If there were, it wouldn't fit any of the answer choices. ;) The only thing I'm not entirely sure of is that there aren't any more cubes, but I am [b]fairly[/b] sure that there aren't any others.\r\n\r\nAnyway, to be a little more thorough, there can only be eight equilateral triangles per cube because I covered all where there are two points on top and two on bottom, and that automatically covers the sides, as well. It's a little hard to explain, but I'm sure there aren't any more. ;)", "Solution_6": "your statement about the equalateral triangles is correct.", "Solution_7": "There is also a cube with verticies on the points $(1,1,0)$ ,$(0,0,1)$, $(2,0,1)$, $(0,2,1)$, $(2,2,1)$, and $(1,1,2)$. This cube has eight new triangles, so the final answer is $80$.", "Solution_8": "Cubes have eight vertices. :huh:", "Solution_9": "[quote=\"JesusFreak197\"]Cubes have eight vertices. :huh:[/quote]\r\nlooks 6 to me. ;)", "Solution_10": "That's an octahedron... :rotfl:", "Solution_11": "hmm..\r\nweird, Problem 25 on the \r\n[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2005]Contest Resorces[/url]\r\npage is different from this one.! ;)\r\n\r\n(its actually a duplicate of the problem 24)", "Solution_12": "[quote=\"amirhtlusa\"]looks 6 to me. ;)[/quote]\r\n\r\nExactly my point. ;)\r\n\r\nI made that same mistake at first as well, but I corrected it.", "Solution_13": "[quote=\"JesusFreak197\"]Cubes have eight vertices. :huh:[/quote]\r\n\r\nOops sorry about that. :)", "Solution_14": "I checked the official answer and it's 80, but I can't think of any triangles we missed...\r\n\r\nhttp://www.unl.edu/amc/mathclub/problems/H-problems/H-web/2005web/ha05-25-ia.html", "Solution_15": "That would mean that we missed a cube, but I don't see how you can possibly make another cube with the given points. :huh:", "Solution_16": "what about the $\\sqrt{2}$ cube?", "Solution_17": "Check the math jam solution.", "Solution_18": "Oh, wow, I completely didn't think of using other points in the larger cube. :oops:", "Solution_19": "i got 80\r\ni think the ones you are missing are on the planes in the middle of the 3x3x3 cube\r\nthere are 2 planes and each plane has 4 triangles on it", "Solution_20": "Could someone maybe post a rigorous solution for this since the transcripts to the mathjam are not available and everyone else in this thread were not able to?", "Solution_21": "The case that was missed is when the side length is $\\sqrt{1^2+1^2+2^2}=\\sqrt{6}$, which gives 8 more triangles. This happens when we take 3 midpoints of the edges of the 2x2x2 cube (that are also non-parallel and nonadjacent). The first edge can be any of 12, WLOG let it be parallel to the x-axis. Then, of the 4 edges parallel the the y-axis, we can choose 2 (the other 2 are adjacent). The edge parallel to the z-axis is then fixed. Since we have counted each triangle 3 times, we divide by 3 to get 12*2/3 = 8 more triangles.", "Solution_22": "I dont rrly get the solutions on the wiki, or here, so can someone please explain it to me if they get how to do it?\n" } { "Tag": [ "inequalities", "number theory unsolved", "number theory" ], "Problem": "I found this problem in some old papers, from some class\r\n\r\nPROBLEM: Let $1=d_1= \\sqrt[k]{d_1d_2...d_k}$ (AM-GM)\r\n\r\nNow, I put $d_i$ with $d_{k-i}$ because $d_id_{k-i}=n$\r\n\r\nFIRST CASE: n is even:\r\n\r\n$\\frac{d_1+d_2+...+d_k}{k}>= \\sqrt[k]{n^{\\frac{n}{2}-1}*n/2} >= \\sqrt[k]{n^{\\frac{k}{2}}}$ because k is the number of divisor.\r\nand we are done\r\n\r\nSECOND CASE : n is odd:\r\n\r\n$\\frac{d_1+d_2+...+d_k}{k}>= \\sqrt[k]{n^{\\frac{n}{2}-1}} >= \\sqrt[k]{n^{\\frac{k}{2}}}$ because k is the number of divisor.\r\nand we are done", "Solution_3": "I forgot to add, in the text it goes that $n$ is natural and $n>1$ so equaltiy doesn't hold at all", "Solution_4": "Thank you for the solution libillau but that part wasn't relly my problem, look at my questions :D :D D \r\n\r\nP.S. You have typo, it isn't $d_id_{k-i}=n$, its $d_id_{k-i+1}=n$ because $d_1$ goes with $d_k$, $d_2$ goes with $d_{k-1}$ and so on...\r\nAnd you really don't need cases but never mind your solution is in basics correct and almost same as mine", "Solution_5": "I will post now second part of that problem so if you are interestef look at it, but it $wierder$ then this one" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Define $ n \\in \\mathbb{N}$ as the minimum positive integer such that in every set of $ n$ consecutive positive integers there is at least one element with digital sum multiple of 11. Show that for all $ m \\in \\mathbb{N}$ there exist a infinite number of sets of $ n\\plus{}7$ consecutive positive integers such that all begin with $ m$ (in the usual decimal representation) and exactly one element has its digital sum multiple of 11.\r\n\r\n\r\n(i.e. with \"begin\" i mean that 676498 begins with 67, or also with 6764..)\r\n\r\nI hope you'll like it :lol:", "Solution_1": "[b]1) We claim that the minimum positive integer $ n$ as described in the problem post equals $ 39$.[/b]\r\n\r\n[i]Proof[/i]: Consideration of the series $ \\{999981,999991,\\ldots,1000018\\}$ yields $ n > 38$.\r\n\r\nFor any positive integer $ N = \\sum_{k = 0}^Ma_k10^k$ let $ S(N)$ denote the set of digital sums corresponding to the numbers $ N,N + 1,\\ldots,N + 9$. If $ 10|N\\Longleftrightarrow a_0 = 0$ we have $ S(N) = \\{\\sum_{k = 1}^Ma_k,1 + \\sum_{k = 1}^Ma_k,\\ldots,9 + \\sum_{k = 1}^Ma_k\\}$. Therefore, for such $ N$ we have $ S(N)\\cap 11\\cdot\\mathbb{Z} = \\emptyset$, iff $ \\sum_{k = 1}^Ma_k\\equiv 1\\bmod{11}$ holds. Moreover, if additionally $ S(N + 10)\\cap 11\\cdot\\mathbb{Z} = \\emptyset$ holds, we must have $ a_1 = 9$ or equivalently $ N\\equiv 90\\bmod{100}$. In particular, if $ 10|N$ holds, at least one of the sets $ S(N),S(N + 10),S(N + 20)$ contains a multiple of $ 11$. \r\nNow consider an arbitrary series $ N,N + 1,\\ldots,N + 38$ of $ 39$ consecutive numbers. Let $ N\\equiv r\\bmod{10}$ with $ 01$. You don't have the formula already so you can't induct. This goes well with generating functions and, in this case, linear algebra." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c>0$. Prove that:\r\n\r\n$ \\sqrt{\\dfrac{a\\plus{}b}{c}\\plus{}\\dfrac{b\\plus{}c}{a}\\plus{}\\dfrac{c\\plus{}a}{b}}\\plus{}2\\cdot \\sqrt{\\dfrac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}}\\ge \\sqrt{6}\\plus{}2$", "Solution_1": "[quote=\"Inequalities Master\"]Let $ a,b,c > 0$. Prove that:\n\n$ \\sqrt {\\dfrac{a \\plus{} b}{c} \\plus{} \\dfrac{b \\plus{} c}{a} \\plus{} \\dfrac{c \\plus{} a}{b}} \\plus{} 2\\cdot \\sqrt {\\dfrac{ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}}\\ge \\sqrt {6} \\plus{} 2$[/quote]\r\nIt follows from $ \\frac{ab\\plus{}ac\\plus{}bc}{a^2\\plus{}b^2\\plus{}c^2}\\cdot\\sum_{cyc}\\frac{a\\plus{}b}{c}\\geq6.$ :wink:", "Solution_2": "The soln had alreadyy been posted. i didn't see.", "Solution_3": "[quote=\"arqady\"][quote=\"Inequalities Master\"]Let $ a,b,c > 0$. Prove that:\n\n$ \\sqrt {\\dfrac{a \\plus{} b}{c} \\plus{} \\dfrac{b \\plus{} c}{a} \\plus{} \\dfrac{c \\plus{} a}{b}} \\plus{} 2\\cdot \\sqrt {\\dfrac{ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}}\\ge \\sqrt {6} \\plus{} 2$[/quote]\nIt follows from $ \\frac {ab \\plus{} ac \\plus{} bc}{a^2 \\plus{} b^2 \\plus{} c^2}\\cdot\\sum_{cyc}\\frac {a \\plus{} b}{c}\\geq6.$ :wink:[/quote]\r\nMy argument is true because the inequality $ \\frac {ab \\plus{} ac \\plus{} bc}{a^2 \\plus{} b^2 \\plus{} c^2}\\cdot\\sum_{cyc}\\frac {a \\plus{} b}{c}\\geq6$ is wrong.\r\nI'll try to prove it more.", "Solution_4": "It's an old problem of Nguyen Anh Cuong.\r\nI tried to solve it but I think it's not easy to find a nice solution :wink:", "Solution_5": "[img]http://s15.sinaimg.cn/middle/0018zOAxgy70ADG7Aaq9e&690[/img]", "Solution_6": "[quote=Inequalities Master]Let $ a,b,c>0$. Prove that:\n\n$ \\sqrt{\\dfrac{a\\plus{}b}{c}\\plus{}\\dfrac{b\\plus{}c}{a}\\plus{}\\dfrac{c\\plus{}a}{b}}\\plus{}2\\cdot \\sqrt{\\dfrac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}}\\ge \\sqrt{6}\\plus{}2$[/quote]\nBy [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] it's enough to check one case only $b=c=1$, which is easy.\n\n" } { "Tag": [ "analytic geometry", "trigonometry", "geometry", "circumcircle", "rotation", "graphing lines", "slope" ], "Problem": "Let $ k$ be a circle with radius $ r$, let $ A$ be any point on $ k$, and let $ t$ be the tangent line to $ k$ at $ A$. Let $ B$ and $ C$ be points on $ t$ on opposite sides of $ A$ such that $ AB=mr$ and \r\n$ AC=nr$ for some positive real numbers $ m$ and $ n$. Let $ P$ be any point of $ k$ different from $ A$. Show that $ \\cot\\angle APB+\\cot\\angle APC$ is a constant for all such points $ P$, and determine this constant value in terms of $ m$ and $ n$.", "Solution_1": "We apply a system of coordinates with $ A$ as origin and $ O(r,0)$\r\n\r\nLet $ (A)$ be the circle centered at $ A$ with radius $ 2r$\r\n$ B$ is at positive $ y$-axis and $ C$ at negative\r\n\r\nLet $ F(2r,0)$. Let $ \\theta = \\angle FAP$ such that $ \\theta \\in \\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$. The ray $ AP$ intersects the line $ \\boxed{FF: x=2r}$ at $ P_{2}$. Then $ P_{2}(2r,2r\\cdot \\tan\\theta)$\r\n\r\nThe circle $ (B)$ centered at $ B$ and passing through $ A$ intersects $ (A)$ at $ L(x_{L},y_{L})$\r\n\r\n$ L\\in(B)\\cap k \\Rightarrow \\left\\|\\begin{array}{l}x_{L}^{2}+(y_{L}-mr)^{2}= m^{2}r^{2}\\\\ (x_{L}-r)^{2}+y_{L}^{2}= r^{2}\\end{array}\\right\\}\\Rightarrow x_{L}=m\\cdot y_{L}$\r\n\r\nSetting $ M=\\frac{1}{m}$ we get $ y_{L}= Mx\\ \\ (1)$\r\n\r\nThe line $ AL: \\ y=Mx$ intersects $ FF$ at $ B_{0}$. Then $ B_{0}\\left(2r,2rM\\right)$\r\n\r\n[img]10006[/img]\r\n\r\nNow let $ B_{1}$ be the 2nd intersection point between $ BP$ and $ k$. Clearly $ B_{1},P$ are inverse wrt $ (B)$\r\nLet $ B_{2}$ be the inverse of $ B_{1}$ wrt $ (A)$. Of course $ B_{2}\\in FF$ (because $ FF$ and $ k$ are inverse wrt $ (A)$)\r\nWe have $ AB_{1}\\cdot AB_{2}= (2r)^{2}= AP \\cdot AP_{2}\\Rightarrow PB_{1}B_{2}P_{2}$ is cyclic, let $ W_{B}$ be its circumcenter\r\n$ \\hline$\r\n\r\nLet $ \\mathbb{P}_{q}X$ denote the power of the point $ X$ wrt the circle $ q$\r\n\r\n$ \\mathbb{P}_{(W_{B})}A=AP\\cdot AP_{2}= AF^{2}\\Rightarrow (W_{B})\\perp (A)$\r\n$ \\mathbb{P}_{(W_{B})}B=BB_{1}\\cdot BP = BA^{2}\\Rightarrow (W_{B})\\perp (B)$\r\n\r\nSo we have \r\n\r\n$ \\mathbb{P}_{(A)}W_{B}= \\mathbb{P}_{(B)}W_{B}\\Rightarrow W_{B}$ is on the radical axis of $ (A)$ and $ (B)$\r\nBut the circles $ (A),(B)$ are fixed, so their radical axis is also fixed. We can easily find that, the radical axis is the line $ y=\\frac{2r}{m}=2rM$\r\n\r\nSo $ B_{0}W_{B}\\perp B_{2}P_{2}$ and this means that $ B_{0}$ is the midpoint of $ B_{2}P_{2}$\r\n$ \\hline$\r\n\r\nSo we have \r\n$ P(2r,\\ 2r\\cdot\\tan\\theta)$\r\n$ B_{0}(2r,2rM)$ and consequently\r\n$ B_{2}(2r,\\ 2r(2M-\\tan\\theta))$\r\n\r\n$ AB$ is tangent to $ k$, so $ \\angle APB = \\angle B_{2}AB \\Rightarrow \\cot(\\angle APB) = \\cot(\\angle B_{2}AB)$ for any position of $ P$\r\n\r\nIf $ 2M>\\tan\\theta$ then the $ y$-coordinate of $ B_{2}$ is positive and $ \\cot(\\angle B_{2}AB) = \\tan(\\angle B_{2}AF) = \\frac{B_{2}F}{2r}= 2M-\\tan\\theta$\r\n\r\nIf $ 2M<\\tan\\theta$ then $ y<0$ and $ \\angle B_{2}AB= \\angle B_{2}AF+90^{o}\\Rightarrow$\r\n\r\n$ \\cot(\\angle B_{2}AB) =-\\tan(\\angle B_{2}AF)=-\\frac{|B_{2}F|}{2r}= 2M-\\tan\\theta$\r\n\r\nSo it is always $ \\boxed{\\cot(\\angle B_{2}AB) = 2M-\\tan\\theta}$\r\n\r\n[color=red](* We can see the diagram to be rotated and inversed such that $ A$ remains at the origin, $ F$ goes to $ (0,2r)$ and $ B$ to the positive x-axis, and then it is clear, that the line $ FF$ is the axis of the cotangents)[/color]\r\n$ \\hline$\r\n\r\nWe define the respective points $ C_{1},C_{2},W_{C}$ for the point $ C(0,-nr)$\r\nNow the circle $ (C)=(C,nr)$ intersects $ k$ at $ Q$ which satisfies the equation $ y=-Nx$, where $ N=\\frac{1}{n}$ \r\n\r\nThen $ W_{C}$ is the point $ (2r,-2rN)$. Again we show that $ W_{C}$ is the midpoint of $ P_{2}C_{2}$\r\n\r\nSo it is $ C_{2}(2r,-2r(2N+\\tan\\theta))$\r\n\r\n[img]10007[/img]\r\n\r\nAgain $ \\cot(\\angle APC) = \\cot(\\angle CAC_{2})$ but it is $ \\frac{-C_{2,y}}{AF}$, where $ C_{2,y}$ is the y-coordinate of $ C_{2}$ (rotate the diagram 90 degrees left to see it)\r\n\r\nSo we have:\r\n\r\n$ \\cot(\\angle CAC_{2}) = 2r\\frac{2N+\\tan\\theta}{2r}= 2N+\\tan\\theta\\Rightarrow \\boxed{\\cot(\\angle CAC_{2})=2N+\\tan\\theta}$\r\n\r\nand finally $ \\cot\\angle APB+\\cot\\angle APC = 2M+2N = 2\\left(\\frac{1}{m}+\\frac{1}{n}\\right)=$\r\n[color=red]Correction[/color]$ = 2\\frac{m+n}{mn}$", "Solution_2": "It took you a lot of time to writ these and thank you very much for this solution.", "Solution_3": "[quote=\"delegat\"][color=darkred]Let $ A$ be a fixed point on the given circle $ k=\\mathcal C(r)$. Let $ B$ and $ C$ be two fixed points on the tangent $ t=AA$ such that $ A\\in (BC)$ and $ AB=mr$ and $ AC=nr$, where $ m$, $ n$ are given positive real numbers. Let $ P\\ne A$ be a mobile point of $ k$. Show that $ \\cot\\widehat{APB}+\\cot\\widehat{ APC}$ is constant.[/color][/quote] [color=darkblue][b]Proof I (synthetic).[/b] Define the second intersections $ U$, $ V$ of the circle $ k$ with the lines $ PB$, $ PC$ respectively. Denote $ \\{\\begin{array}{c}m(\\widehat{APB})=x\\ ,\\ m(\\widehat{APC})=y\\\\\\ m(\\widehat{AUB})=\\phi\\ ,\\ m(\\widehat{AVC})=\\psi\\end{array}$.\n\nObserve that $ \\phi+\\psi=180^{\\circ}$, i.e. $ \\sin\\phi=\\sin\\psi$. Suppose w.l.o.g. that $ r=\\frac{1}{2}$, $ m=2b$, $ n=2c$. Thus, $ AB=b$, $ AC=c$, $ AU=\\sin x$, $ AV=\\sin y$.\n\nApply the [b]Sinus' theorem[/b] : $ \\{\\begin{array}{ccc}\\triangle ABU\\ : & \\frac{BU}{\\sin x}=\\frac{AB}{\\sin \\phi}\\\\\\\\ \\triangle ACV\\ : & \\frac{CV}{\\sin y}=\\frac{AC}{\\sin\\psi}\\end{array}$ $ \\implies$ $ AB\\cdot CV\\cdot\\sin x=AC\\cdot BU\\cdot \\sin y\\ \\ (1)$.\n\nApply the [b]generalized Pythagoras' theorem[/b] : $ \\{\\begin{array}{ccc}\\triangle ABU\\ : & BU^{2}=b^{2}+\\sin^{2}x-2b\\sin x\\cos x=\\sin^{2}x\\cdot[b^{2}+(b\\cdot\\cot x-1)^{2}]\\\\\\\\ \\triangle ACV\\: & CV^{2}=c^{2}+\\sin^{2}y-2c\\sin y\\cos y=\\sin^{2}y\\cdot[c^{2}+(c\\cdot\\cot y-1)^{2}]\\end{array}$.\n\nTherefore, the relation $ (1)$ becomes : $ b^{2}\\cdot\\sin^{2}x\\cdot\\sin^{2}y\\cdot[c^{2}+(c\\cdot\\cot y-1)^{2}]=c^{2}\\cdot\\sin^{2}y\\cdot\\sin^{2}x\\cdot[b^{2}+(b\\cdot\\cot x-1)^{2}]$ $ \\Longleftrightarrow$\n\n$ |b\\cdot (c\\cdot\\cot y-1)|=|c\\cdot (b\\cdot\\cot x-1)|$ $ \\Longleftrightarrow$ $ \\{\\begin{array}{c}\\cot x-\\cot y=\\frac{c-b}{bc}\\\\\\\\ \\mathrm{OR}\\\\\\\\ \\cot x+\\cot y=\\frac{b+c}{bc}\\end{array}$, i.e. $ \\boxed{\\cot x-\\cot y=\\frac{2(n-m)}{mn}\\ \\ \\vee\\ \\ \\cot x+\\cot y=\\frac{2(m+n)}{mn}}$.\n\n[b]Proof II (analytic).[/b] Suppose w.l.o.g. that the circle is $ \\mathcal C(O,1)$, where the center $ O$ is the origin of the analytical system and $ \\overline{BAC}\\parallel Ox$.\n\nTherefore, $ \\{\\begin{array}{c}P(\\cos\\phi ,\\sin \\phi )\\\\\\\\ A(0,-1)\\ ,\\ B(-m,-1)\\ ,\\ C(n,-1)\\end{array}$ and the slopes of the lines $ PA$, $ PB$, $ PC$ are the following :\n\n $ \\{\\begin{array}{c}s_{PA}=\\tan\\alpha =\\frac{\\sin\\phi+1}{\\cos\\phi}\\\\\\\\ s_{PB}=\\tan\\beta =\\frac{\\sin\\phi+1}{\\cos\\phi+m}\\\\\\\\ s_{PC}=\\tan\\gamma=\\frac{\\sin\\phi}{\\cos\\phi-n}\\end{array}$. Since $ \\{\\begin{array}{c}x=\\alpha-\\beta\\\\\\ y=\\gamma-\\alpha\\end{array}$ obtain : $ \\{\\begin{array}{c}\\tan x=\\frac{\\tan\\alpha-\\tan\\beta}{1+\\tan\\alpha \\tan\\beta}=\\frac{m(1+\\sin\\phi )}{2+m\\cos\\phi+2\\sin\\phi}\\\\\\\\ \\tan y=\\frac{\\tan\\gamma-\\tan \\alpha}{1+\\tan\\gamma\\tan\\alpha}=\\frac{n(1+\\sin\\phi)}{2-n\\cos\\phi+2\\sin\\phi}\\end{array}\\|$ $ \\implies$\n\n$ \\cot x+\\cot y=$ $ \\frac{2+m\\cos\\phi+2\\sin \\phi}{m(1+\\sin\\phi )}+\\frac{2-n\\cos\\phi+2\\sin\\phi}{n(1+\\sin\\phi)}=$ $ \\frac{2(m+n)(1+\\sin\\phi )}{mn(1+\\sin\\phi )}=$ $ \\frac{2(m+n)}{mn}$ $ \\implies$ $ \\boxed{\\cot x+\\cot y=\\frac{2(m+n)}{mn}}$. \n\n[b]Remark.[/b] $ \\frac{UV\\cdot 2r}{AU\\cdot AV}=\\frac{2r\\sin (x+y)\\cdot 2r}{2r\\sin x\\cdot 2r\\sin y}=\\cot x+\\cot y$. Therefore, the conclusion is equivalently with \"[b]the ratio $ \\boxed{\\frac{UV}{AU\\cdot AV}}$ is constant[/b]\". [/color]", "Solution_4": "Most probably problem is not well defined.\r\nMaybe result depends on position of point $ P$ so sometimes it is $ +$ and sometimes $ -$ sign :?:" } { "Tag": [ "Asymptote", "LaTeX" ], "Problem": "Hi I'm trying to set up Aysmptote on my computer, but I'm having some problems. I managed to get it to work, but now I'm trying to get it to work with the TeXnicCenter, which is where I'm stuck. I'm using this guide: http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode\r\n\r\nI did everything it said but when I run the Asympote tool, I get the DOS-like screen giving me an error. It says\r\n\r\n\r\n[code]\nCannot execute 'latex' \\scrollmode\nPlease put in config.asy:\n\nimport settings;\ntextpath=\"PATH\";\n\nwhere PATH denotes the correct path to the directory containing your latex engine (latex).\n\nAlternatively, set the environment variable ASYMPTOTE_TEXPATH or use the command line option -textpath=\"PATH\"\n[/code]\r\n\r\nI've looked for this config.asy, but I couldn't find it, and even if I did, I'm not quite sure exactly what I need to put in it. Can anybody help me?", "Solution_1": "I'm assuming you are using MikTeX on Windows XP as you don't say.\r\n\r\n1. Is TexnicCenter able to compile normal latex documents?\r\n\r\n2. Go to Start, Run then type cmd and press OK. This will give you a DOS box. Type [i]path[/i] then press Enter. In amongst the mess is there something like c:\\program files\\miktex 2.5\\bin? If not, that means miktex hasn't told windows where it should be found.\r\n\r\nThis can be changed but I am reluctant to advise this as it is too easy to make a mistake. So we'll do as Asymptote says and alter config.asy. It can be found in C:\\Documents and Settings\\[i]your_name[/i]\\.asy. Make a copy of it and then open it in Notepad. Add the line\r\ntexpath=\"c:\\program files\\miktex 2.5\\bin\";\r\nusing whatever your path is.", "Solution_2": "Okay I did that; then when I tried it again it began to output a bunch of stuff like it was working. But then it closed without showing anything. Then I couldn't even get it to do that again and it seems to be unable to execute something called \"dvips\". I think it's unable to find that, but I don't know where it is either...", "Solution_3": "Can you tell me if\r\n1. Texniccenter compiles ordinary tex files correctly?\r\n2. Go to Start, Run then type cmd and press OK. This will give you a DOS box. Type path then press Enter. Is there anything about miktex?\r\n3. Is Ghostscript installed?", "Solution_4": "1. Regular LaTeX works fine\r\n2. No, there is nothing about MiKTeX. I added the path to config.asy but now there's another error.\r\n3. Yes I have Ghostscript.", "Solution_5": "[quote=\"tjhance\"]2. No, there is nothing about MiKTeX.[/quote]That's the source of all your problems. You can add it to the path if you look up how to do it (it's called Environmental Variable) but I will give you an alternative risk-free method which will add this info when you run asymptote.\r\nFollow the instructions at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]Showing Asymptote error messages in TeXnicCenter[/url]. But in asymptote.bat after the line [code]echo Asymptote is working ... please wait[/code]add the lines\n[code]set opath=%path%\nset path=%path%;\"c:\\program files\\miktex 2.5\\bin\"[/code]\nAnd on the line after [code]:end [/code]add\n[code]set path=%opath%\nset opath=[/code]I hope this will work.", "Solution_6": "Hm well okay so I did that, and now I've seemed to get rid of this dvips error, but there's still nothing. When I run the Asymptote tool from TeXnicCenter, the black DOS screen comes up and instantly disappears without displaying anything. I do have the path in config.asy and stuff and I'm not sure what's wrong now.\r\n\r\nI really have no clue what I'm doing.", "Solution_7": "You are, I hope, using asymptote on file called somefile.asy NOT somefile.tex I hope?\r\nPut \r\npause\r\nat the end of the aymptote.bat file and the DOS window shouldn't close until you press a key.\r\nIf that doesn't help delete the first line @echo off from asymptote.bat then see what errors it gives you.", "Solution_8": "My file is called \"test.asy\"\r\n\r\nOkay I'm getting output now... but still nothing that looks like an error. Here's a screenshot of my output.\r\n\r\nI still don't quite understand how this tool is supposed to work...", "Solution_9": "Believe it or not that's looking good!\r\n1. Is there a test_asy.log and is there anything in it (I suspect it's empty)?\r\n2. In asymptote.bat can you make a couple of changes (my fault :oops: )\r\na) remove the quotes from c:\\program files\\miktex 2.5\\bin (are you absolutely sure this is the correct path to MiKTeX - it should contain lot of files including latex.exe, please double-check)\r\nb) Move \r\nset path=%opath% \r\nset opath=\r\nto after :end\r\n\r\nIf still no joy try saving your test.asy into a different directory which has no spaces in it and open it in TeXnicCenter from there.", "Solution_10": "Okay I made your changes; first of all, no, i wasn't using the correct path; it's actually c:\\program files\\MiKTex 2.5\\miktex\\bin. I've been aware of that for a while, but I must've forgotten to change it when I made the .bat file. I made the other changes, with the quotes and stuff. I also tried a different directory.\r\n\r\nIt appears to be working but it still isn't producing an image :(", "Solution_11": "Did you save test.asy inside TexnicCenter? If so, are you sure it hasn't been saved as test.asy.tex?\r\nIs there a test_asy.log file?", "Solution_12": "Actually, I don't think test.asy was a pure asy file... the file appears with the TeXNicCenter icon. I created a new file, called supertest.asy, outside of TeXNicCenter to make sure it was just .asy, but it still doesn't work :wallbash:\r\n\r\nAs for the test_asy.log file (or supertest_asy.log), where is it supposed to be? If it is supposed to appear in the same folder as test.asy and supertest.asy, then I see files called \"test_asy\" and \"supertest_asy\" there, but I don't see the .log extension. They appear to be plain text files, and are both empty.", "Solution_13": "Looks like it's all Microsoft's fault - you haven't got extensions turned on so you can't see what they are or are fooled by double extensions. \r\n\r\nIn My Computer, Tools, Folder Options, View, Untick the box that says Hide extensions for known file types, OK out and look again. I bet you find supertest_asy.log and also supertest.asy.tex which you will need to rename to supertest.asy.", "Solution_14": "wow i can see extensions now... that was bugging me for so long before i even started this..... :lol: \r\n\r\nokay well there's test.asy.tex and supertest.asy. Then there's test_asy.log and supertest_asy.log, which are both empty.", "Solution_15": "Should all work now and using test.asy you should get test.pdf. You should also be able to use asymptote code inode a tex file as well.\r\n\r\nOnce it's working add @echo off back on the first line of asymptote.bat and get rid of pause at the end. Then when you call asymptote from TeXnicCenter you'll get a window with a message which will close automatically and any errors will be found in the test_asy.log file. If there are any errors, test_asy.log will open automatically in TexnicCenter.", "Solution_16": "When I open the pdf it says \"the file was damaged and could not be repaired\". And I'm not getting and pdf from the supertest one. I know I'm just missing something stupid now, but I can't figure out what.......\r\n\r\nHere is the Asymptote code I'm using; maybe there's something wrong with it?\r\n\r\n[code]\ndraw((0,0)--(100,100));\ndraw((0,100)--(100,0));\ndot((50,50));\nlabel(\"$P$\",(50,50),S);[/code]\r\n\r\nEDIT- also, while you say I should get a PDF, the guid http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode makes it sound like running the tool should just make the image appear on the screen.", "Solution_17": "What happens if you remove the dollar signs from the last line?\r\nIs there a test.pdf file created? If so, it will be opened which is what \"appear on the screen\" means.", "Solution_18": "removing the dollar signs did nothing. And no, there is no pdf generated.", "Solution_19": "There appears to be a problem with asymptote. Does it work outside TexnicCenter?", "Solution_20": "Yes I did get it to work outside TeXNicCenter", "Solution_21": "Can you double-check that asy.exe is in C:\\Program Files\\Asymptote?", "Solution_22": "If asy.exe is there then try this:\r\n1. In asymptote.bat remove the first part of\r\n\"C:\\Program Files\\Asymptote\\redir.exe\" -e %1 \"C:\\progra~1\\Asymptote\\asy.exe\" -batchView -tex \"pdflatex\" -align C -f \"pdf\" %2\r\nso it reads\r\n\"C:\\progra~1\\Asymptote\\asy.exe\" -batchView -tex \"pdflatex\" -align C -f \"pdf\" %2\r\n2. From your screen shots, on the line \r\nfind /i \".asy\" %3 >nul\r\nthere seems to be an extra 1 before the >nul which needs removing.\r\n\r\nNow try again and show me the screen which will tell me if asy.exe is really running properly.", "Solution_23": "Hmm yeah asy.exe is there. I fixed that one line, but I have no idea where the extra 1 is coming from. I don't see it asymptote.bat. Here's the code for asymptote.bat:\r\n\r\n[code]echo Asymptote is working ... please wait \nset opath=%path% \nset path=%path%;C:\\Program Files\\MiKTex 2.5\\miktex\\bin\n\"C:\\progra~1\\Asymptote\\asy.exe\" -batchView -tex \"pdflatex\" -align C -f \"pdf\" %2 \nfind /i \".asy\" %3 >nul \nIf errorlevel 1 goto end \n\"C:\\Program Files\\TeXnicCenter\\TEXCNTR.EXE\" /ddecmd \"[open('%3')]\" \npause\n:end\nset path=%opath% \nset opath=[/code]\r\n\r\nDo I have the settings for the Asymptote tool set correctly?\r\n\r\nCommand: \"C:\\Program Files\\Asymptote\\asymptote.bat\"\r\nArguments: %tc_asy.log %tc \"%dc\\%tc_asy.log\" \r\nInitial Directory: %dc", "Solution_24": "Everything seems fine and asymptote is being found but isn't producing anything which is odd. (It turns out that the extra 1 is normal, it's just that normally you don't see it when @echo off is there).\r\n\r\nCan you try re-installing Asymptote over the top of the one you have just in case there's a problem there?", "Solution_25": "What happens if I can't find config.asy in the path you've given?", "Solution_26": "Nothing. Under normal circumstances you do not need config.asy at all. It is just one of the files Asymptote executes before reading your source file. You can always create it if it is absent and put whatever you want in there but it is by no means necessary for the proper functioning of Asymptote.", "Solution_27": "I see. I use Foxit Reader for pdf files, and had to create a new config.asy in the .asy directory to change the pdfviewer path; it works now. Yes, it did not change Asymptote's proper functioning. Thank you for this board; it's really helpful." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "Let Triangle ABC be an isosceles triangle(AB=AC) with Angle BAC=20 degrees. Point D is on side AC such that Angle DBC=60 degrees. Point E is on side AB such that Angle ECB=50 degrees. Find, with proof, the measure of Angle EDB.\r\n\r\nCan you please try to find a solution not involving trigonometry? Thanks. :roll: \r\n\r\nIt'll be very helpful to draw it out.", "Solution_1": "See here: http://www.cut-the-knot.org/triangle/80-80-20/index.shtml\r\n\r\n(and here: http://www.mathlinks.ro/viewtopic.php?t=310088 )" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that for given $ a,b,c \\ge0$ we have \r\n\r\n$ (a^3\\plus{}b^3\\plus{}c^3)^2\\ge (a^4\\plus{}b^4\\plus{}c^4)(ab\\plus{}bc\\plus{}ca)$", "Solution_1": "EDITED : Sorry,It seemed to me that the sign was reversed !", "Solution_2": "If you take $ b$, $ c \\to 0$, then the right-hand-side gets arbitrarily close to zero, but the left-hand-side not. So the ineuality holds true in this case. :wink:", "Solution_3": "Yes , the inequality is true . See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106975]here[/url]", "Solution_4": "It is equivalent to $ \\frac {1}{2}[(a \\minus{} b)^6 \\plus{} (b \\minus{} c)^6 \\plus{} (c \\minus{} a)^6] \\plus{} 2[ab(a \\minus{} b)^4 \\plus{} bc(b \\minus{} c)^4 \\plus{} ca(c \\minus{} a)^4] \\plus{} \\frac {1}{2}(a^4(b\\minus{}c)^2\\plus{}b^4(c\\minus{}a)^2\\plus{}c^4(a\\minus{}b)^2) \\ge 0$.\r\n\r\nCan someone tell me how to post cyclic sum and symmetric sum?\r\nEDIT: Thank you, alex2008", "Solution_5": "[quote=\"dgreenb801\"]\nCan someone tell me how to post cyclic sum and symmetric sum?[/quote]\r\n\r\ncyclic sum - $ \\sum_{cyc}$\r\n\r\n[code]\\sum_{cyc}[/code]\n\nsymmetric sum - $ \\sum_{sym}$\n\n[code]\n\\sum_{sym}[/code]\r\n\r\ni hope you was reffering to those ." } { "Tag": [ "inequalities", "ratio", "function" ], "Problem": "Please explain why us humans cannot obtain 100% efficiency with energy. I would prefer an explanation that is so simple that even my 10-year-old cousin could understand. Thanks", "Solution_1": "In principle, the only thing that theoretically limits efficiency of heat machines is Carnot's inequality. But I don't know how to explain it to a 10 year old boy :?", "Solution_2": "[quote=\"aahmadi\"]Please explain why us humans cannot obtain 100% efficiency with energy. I would prefer an explanation that is so simple that even my 10-year-old cousin could understand. Thanks[/quote]\r\n\r\nI'll give it a try. First, there's the 1st law of thermodynamics: the amount of energy that a thermodynamic system can store is the difference between the amount of heat added to the system and the amount of work the system does on its surroundings.\r\n\r\nSecond, the amount of energy the system can store is equal to the greatest amount of heat that the system can add to its surroundings. Further, heat can only flow from \"hot\" to \"cold.\"\r\n\r\nThis uni-directionality of heat flow indicates that the system draws heat from a hot reservoir, and rejects heat to a cold reservoir. Written as an equation, the first law of thermodynamics says: $ Q_C \\equal{} Q_H \\minus{} W$.\r\n\r\nFor a reversible process, which is the most efficient kind, $ \\frac{Q_H}{T_H} \\equal{} \\frac{Q_C}{T_C}$ This implies $ \\frac{T_C}{T_H} \\equal{} \\frac{Q_C}{Q_H}$\r\n\r\nThe efficiency $ \\eta \\doteq \\frac{W}{Q_H}$.\r\n\r\nIf we then substitute $ W$ from the above equation into the definition of $ \\eta$ we get:\r\n\r\n$ \\eta \\equal{}\\frac{Q_H \\minus{} Q_C}{Q_H} \\equal{} 1 \\minus{} \\frac{Q_C}{Q_H}$\r\n\r\nWe can now substitute $ \\frac{T_C}{T_H}$ for $ \\frac{Q_C}{Q_H}$ in the above expression.\r\n\r\n$ \\eta \\equal{} 1 \\minus{} \\frac{T_C}{T_H}$\r\n\r\nNow let's suppose that $ \\eta\\geq 100\\%$.\r\n\r\nThen $ 1 \\minus{}\\frac{T_C}{T_H} \\geq 1$.\r\n\r\nSubtracting 1 from both sides of the inequality,\r\n\r\n$ \\minus{}\\frac{T_C}{T_H} \\geq 0 \\Rightarrow \\frac{T_C}{T_H}\\leq 0$\r\n\r\nThis is impossible according to the third law of thermodynamics; therefore, our supposition is false.\r\n\r\nErgo, $ \\eta < 100\\%$\r\n\r\nI don't know if a ten year old would comprehend the algebra involved, but the conclusion is clear enough: One simply cannot get as much work out of a thermodynamic system as the amount of energy one puts into it.", "Solution_3": "I would just emphasize that, even if a human intends to do work,he can not consentrate 100% to it.So from this,it can be understood that there can't be 100% eficiency.", "Solution_4": "Yes, essentially the explanation of Carnot's inequality is exactly that\r\n\r\na) No matter how cunning we try to be, we cannot create a process the only result of which is passing heat from a cold body to a hot one.\r\n\r\nb) We can be cunning enough to pass heat $ Q$ from the cold body to the hot one using work $ W$ much less than $ Q$ (namely, $ [1\\minus{}(T_{\\text{low}}/T_{\\text{hot}})]Q$)\r\n\r\nErgo, we cannot make the heat machine that passes heat $ Q$ to produce work larger than $ W$ (otherwise run this machine, and use the resulting work to transfer more heat back than was transferred forward) \r\n\r\nUnfortunately, neither (a), nor (b) is obvious. (a) has to be taken for granted, (b) is a bit easier to explain, but it still requires an explicit construction of such a machine and computation of the mechanical work done and the heat transferred. :(", "Solution_5": "[quote=\"Dr. No\"]One simply cannot get as much work out of a thermodynamic system as the amount of energy one puts into it.[/quote]\r\n\r\nWrong again: yes we can. The Kelvin-Planck statement of the second law simply states that such a conversion is not possible [i]for a cyclic process[/i]. There is no problem in converting completely in work a given amount of heat by a non-cyclic process.", "Solution_6": "[quote=\"Carcul\"][quote=\"Dr. No\"]One simply cannot get as much work out of a thermodynamic system as the amount of energy one puts into it.[/quote]\n\nWrong again: yes we can. The Kelvin-Planck statement of the second law simply states that such a conversion is not possible [i]for a cyclic process[/i]. There is no problem in converting completely in work a given amount of heat by a non-cyclic process.[/quote]\r\n\r\nI wish you would stop finding fault with what I write merely for the sake of finding fault.\r\n\r\nYour objection is pointless. I didn't appeal to the Kelvin-Planck statement of 2LoT, or any other statement of it for that matter, in making my argument. Starting with a reversible process, the Carnot inequality guarantees that $ \\frac{Q_H}{T_H} \\equal{} \\frac{Q_C}{T_C}$; that said process is or is not cyclic notwithstanding.\r\n\r\nFollowing which I proved that an efficiency greater than or equal 100% requires a ratio of absolute temperatures that is either negative or zero. By the 3LoT this circumstance is impossible. Thus the efficiency of [b]any[/b] thermodynamic process is less than 100%, cyclic or not.", "Solution_7": "Carnot's inequality applies only to cyclic processes: that's the point of using it to demonstrate the existence of a state function that we call Entropy. And an example of a thermodynamic process with 100% efficiency is the isothermal expansion of an ideal gas.\r\n\r\n[quote=\"Dr. No\"]I wish you would stop finding fault with what I write merely for the sake of finding fault.[/quote]\r\n\r\nIf you don't like people to make a critic about what you write, then think twice before posting. If you want to keep writing wrong things, be me guest. That's your problem and a problem for those who read your posts. And don't worry, I will not make any further comments about your posts.", "Solution_8": "The Loss of Heat\r\nThe reversible system of heat\r\nThe inreversible system of heat", "Solution_9": "[quote=\"Carcul\"]Carnot's inequality applies only to cyclic processes: that's the point of using it to demonstrate the existence of a state function that we call Entropy. And an example of a thermodynamic process with 100% efficiency is the isothermal expansion of an ideal gas.\n\n[quote=\"Dr. No\"]I wish you would stop finding fault with what I write merely for the sake of finding fault.[/quote][/quote]\n\n[i]Oh, good grief....[/i] $ dS\\geq \\frac{\\delta Q}{T}$ has nothing to do with cyclic processes necessarily. The case I cited was the equality, which occurs when $ dS \\equal{} \\frac{\\delta Q}{T}$, i.e. a reversible process.\n\nFor an isothermal expansion of an ideal gas, the efficiency is given by:\n\n$ \\eta \\equal{}\\frac{W}{Q} \\equal{} \\frac{W}{\\Delta U \\plus{} W} < 1$\n\n[quote=\"Carcul\"]If you don't like people to make a critic about what you write, then think twice before posting. If you want to keep writing wrong things, be me guest. That's your problem and a problem for those who read your posts. And don't worry, I will not make any further comments about your posts.[/quote]\r\n\r\nI don't mind criticism, especially when it's legitimate. But you have an annoying habit of trying to pull at loose threads in my arguments that don't exist. I don't intentionally post things that are wrong and I don't come here to get help so much as to give it. I don't have any need to prove myself to be the smartest kid in the room. I know that there are many here, including probably you Carcul, who are brighter than I. But that shouldn't matter since I don't come here with the intention of putting anyone down. I know the subject of physics, I know a fair bit about several fields of mathematics, and I enjoy figuring things out. If I can help anyone in the process of learning things myself then good. If not, then I'll leave but in the meantime I don't like being attacked by 'know-it-all' pups with ego malfunctions.", "Solution_10": "[quote=\"Dr. No\"] I don't like being attacked by 'know-it-all' pups with ego malfunctions.[/quote]\r\n\r\nOuch. Carcul is a teacher. I think it's part of his job to pick out small things. So yeah, let's just stick to the academic stuff.", "Solution_11": "[quote=\"aahmadi\"]I would prefer an explanation that is so simple that even my 10-year-old cousin could understand.[/quote]\r\n\r\nI think a really simple, understandable explanation is that in any machine, parts rub together and release some of the use[i]ful[/i] energy put into them as use[i]less[/i] heat due to friction, or heat is released through other processes (e.g., a wire heating due to the electric current passing through it). Because some of the energy expended is always released as heat, the machine cannot be 100% efficient.\r\n\r\nThat's probably not the most \"scientific\" explanation and not easily derived mathematically and whatnot, but I do think a 10-year-old would be able to understand it.", "Solution_12": "A simple explanation is not an accurate one. Pick what you value more." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "With $ a,b,c > 0$ such that $ ab,ca,bc \\in[0;\\sqrt {2}]$ and $ (ab)^2 \\plus{} (bc)^2 \\plus{} (ca)^2 \\plus{} (abc)^2 \\equal{} 4$\r\nProve that $ a \\plus{} b \\plus{} c \\geq 3$", "Solution_1": "[quote=\"2424\"]With $ a,b,c > 0$ such that $ ab,ca,bc \\in[0;\\sqrt {2}]$ and $ (ab)^2 \\plus{} (bc)^2 \\plus{} (ca)^2 \\plus{} (abc)^2 \\equal{} 4$\nProve that $ a \\plus{} b \\plus{} c \\geq 3$[/quote]\r\nYou can see the inequality than stronger:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=289839" } { "Tag": [ "geometry", "perpendicular bisector", "geometry proposed" ], "Problem": "Let $ABCD$ be a convex quadrilateral with $AB=CD$ and not a parallelogram. $M,N$ are midpoints of diagonals $AC,BD$ respectively. Prove that the orthogonal projections of $AB$ and $CD$ onto $MN$ both have length equal to $MN$.", "Solution_1": "Let $U, V$ be the midpoints of $AD$ and $BC$. Prove the fact the $UV$ is the perpendicular bisector of $MN$. Thus, if $G$ is the centroid of $ABCD$ and $A', B', C', D'$ are the projections of $A, B, C, D$ on $MN$, then $G$ is the common midpoint of $MN, A'D', B'C'$. The result immediately follows.", "Solution_2": "[quote=\"treegoner\"] then $G$ is the common midpoint of $MN, A'D', B'C'$. The result immediately follows.[/quote]\r\n\r\nWhy? I see only $A'D'=B'C'$, not $=MN$...", "Solution_3": "Yeah, I don't see that either...", "Solution_4": "Can somebody continue?", "Solution_5": "Let U be the midpoint of AD. We have 2*UM=CD=AB=2*MN so triangle MUN is isosceles. \r\nLet MN meet AB at K, and PQ is the orthogonal projection of AB onto MN\r\nWe have PQ=AB*cosAKN=2*UN*cosUNM=MN (because triangle MUN is isosceles)\r\nSimilarly with CD we have the result above q.e.d", "Solution_6": "[quote=\"Huy\u1ec1n V\u0169\"]Let U be the midpoint of AD. We have $2 \\cdot UM=CD=AB=2 \\cdot UN$ so triangle $MUN$ is isosceles. \nLet $MN$ meet $AB$ at $K$, and $PQ$ is the orthogonal projection of $AB$ onto $MN$\nWe have $PQ=AB \\cdot \\cos AKN=2 \\cdot UN \\cdot \\cos UNM=MN$ (because triangle $MUN$ is isosceles)\nSimilarly with $CD$ we have the result above q.e.d[/quote]\r\n\r\n Please see the figure. (There was a typo in the above solution. I have corrected it.)" } { "Tag": [ "calculus", "derivative", "algebra", "polynomial", "function", "domain", "complex numbers" ], "Problem": "Solve it:\r\n\r\n ${\\begin{array}{l}x^5-y^5=992\\\\x-y=2\\end{array}}$ \r\n\r\n\r\n\r\n\r\nPlease if somebody could help me with this :mad:", "Solution_1": "Well, \r\nit is easy to see that\r\n$4^5-2^5=992$.\r\nAnd also\r\n$(y+2)^2-y^2$ grows when $y$ is positive because the first derivative of $x^5$ is $5x^4$ and is growing. So this is only solution for positive $x$ and $y$ $(4,2)$. also $x$ and $y$ must have the same sign because otherwise $x^5-y^5\\leq 2.2^5=64<992$.\r\nAnd for negative $x,y$ we find the solution $(-2,-4)$.\r\n\r\nFinally $(4,2)$ and $(-2,-4)$ are the solutions!", "Solution_2": "there isn't more solutions? Are you sure?", "Solution_3": "You mean aren't there more solutions?\r\n\r\nWell, it sounds like you're quizzing instead of asking.\r\n\r\nActually, to the poster, imortal it is a throwaway statement \"it is easy to see that...\"\r\n\r\nIt certainly didn't occur to me to use those numbers! From now on. please don't use that phrasing; it's counterproductive.\r\n\r\nAnd imortal, because the problem never stated to find only integer solutions in x and y, your solutions are incomplete.\r\n\r\nAfter getting (y+2)^5 - y^5 = 992 and subtracting off the y^5 terms\r\nfrom each side, you have a 4th degree equation in y. After setting\r\nit to zero and dividing both sides by 10, you have \r\n\r\ny^4 + 4y^3 + 8y^2 + 8y - 96 = 0\r\n\r\nAfter dividing out the factors (y-2) and (y+4) to go along with the known solutions of y = 2 and y = - 4 from the earlier post, you are left with\r\n\r\ny^2 + 2y + 12 = 0 which has the roots of \r\n\r\ny = -1 - i*sqrt(11) and -1 + i*sqrt(11)\r\n\r\n\r\nOther solutions to x^5 - y^5 = 992 and x - y = 2 are then\r\n\r\n(1 - i*sqrt(11), -1 - i*sqrt(11)) and\r\n\r\n(1 + i*sqrt(11), -1 + i*sqrt(11))", "Solution_4": "thank you, now i see that it is good solution of the problem", "Solution_5": "[hide=\"Here is a solution only with the polynominal of the second degree.\"] $S=x+y$, $P=xy$ $\\Longrightarrow$ $S_2\\equiv x^2+y^2=S^2-2P$ $\\Longrightarrow$ $S_4\\equiv x^4+y^4=$ $\\left(x^2+y^2\\right)^2-2(xy)^2=$ $S^2_2-2P^2=$ $\\left(S^2-2P\\right)^2-2P^2=$ $S^4-4PS^2+2P^2$ $\\Longrightarrow$ $\\boxed {S_2=S^2-2P\\ ,\\ S_4=S_4-4PS^2+2P^2}\\ .$\n$x-y=2,\\ x+y=S\\Longrightarrow \\boxed {x=\\frac{S+2}{2}\\ ,\\ y=\\frac{S-2}{2}}$ $\\Longrightarrow S^2-4=4P\\ \\ (P\\ge -1)\\ .$\n$x^5-y^5=992$ $\\Longleftrightarrow$ $(x-y)\\left[ \\left(x^4+y^4\\right)+xy\\left(x^2+y^2\\right)+(xy)^2\\right]=998$ $\\Longleftrightarrow$ $S_4+PS_2+P^2=496$ $\\Longleftrightarrow$ $S^4-4PS^2+2P^2+P\\left(S^2-2P\\right)+2P^2=496$ $\\Longleftrightarrow$ $\\boxed {S^2=4(P+1)\\ge 0\\ ,\\ S^2\\left(S^2-3P\\right)+P^2=496}$ $\\Longrightarrow$ $4(P+1)(P+4)+P^2-496=0$ $\\Longleftrightarrow$ $P^2+4P-96=0$ $\\Longrightarrow$ $P=8$ $\\Longrightarrow$ $S^2=36$ $\\Longrightarrow$ $S\\in \\{6,-6\\}$ $\\Longrightarrow$ $\\boxed {x=4\\ ,\\ y=2\\ \\ \\vee\\ \\ x=-2\\ ,\\ y=-4}\\ .$\n\n[b]Remark.[/b] For $P=-12$ and $S^2=-44$, i.e. $S\\in \\{\\pm 2i\\sqrt{11}\\}$ we obtain and the another solutions, but in the set of the complex numbers:\n$t^2-2i\\sqrt{11}\\cdot t-12=0\\Longrightarrow \\boxed {x=1+i\\sqrt{11}\\ ;\\ y=-1+i\\sqrt{11}}\\ .$\n$t^2+2i\\sqrt{11}\\cdot t-12=0\\Longrightarrow \\boxed {x=1-i\\sqrt{11}\\ ;\\ y=-1-i\\sqrt{11}}\\ .$\n\n[color=darkred][b]Remark.[/b] I don't know the set in which we find the solutions of the proposed system !\n[b]Convention.[/b] [u]In the case when the domain of the existence isn't specified, then it is the largest domain of the existence, i.e. in the our case, the set of the complex numbers.[/u][/color][/hide]", "Solution_6": "thanks a lot :D", "Solution_7": "Hi!\r\n\r\nOf course, when I saw the problem, I had no idea what kind of numbers x and y are so I decided that they are real. If I knew that x and y may be complex I would try to find them but it was not stated in the problem so I overlooked it :)\r\nBut if you read carefully my solution you will see that I solve the equation in real numbers (NOT ONLY IN INTEGERS) and my solution is full (for real numbers).\r\n\r\nActually, A cute angle, it is not really hard to see that (4,2) is a solution.\r\nWhen I saw the problem I firstly thought: wow, the number 992 sounds very strange, maybe the reason it appers in this problem is that INTEGERS are solutions.\r\n.\r\n\r\nThen I started trying with small integers if they satisfy the condition. Because the differnce between the perfect fifth degrees grows very fast I knew that can check if the equation have integer solutions or not. So I tried it and was lucky to find (4,2).\r\n\r\nSo, maybe it may sound as throwaway \"it is easy to see that...\" to you, but for me it was really easy to see. :D", "Solution_8": "imortal, \r\n\r\nyou do not have the leisure to decide that x and y are exclusively \r\nreal.\r\n\r\nJust because the problem did not state that x and y may be complex does NOT mean you can ignore them as a possibility.\r\n\r\nYour solution in real numbers is not complete enough.\r\n\r\nNo, your reasoning is again off, as pertains 992. It only means\r\nthat perhaps integers CAN be the solutions, not ARE the sole solutions.\r\n\r\nYou'll have to get your story straight; \"it is easy to see that...\"\r\nsounds like the solution jumps off the screen with little effort, BUT you disclosed that you tried (4,2) and were lucky, which is altogether a different meaning. You cannot have it both ways logically.\r\n\r\nThe main point is this: the problem has to have accompanying restrictions on the variable(s) if it stands alone. If there are none, then you solve for as large an encompassing set of solutions that you can, taking into account, for example, real world restrictions.\r\n\r\nYou were wrong to discard complex solutions, for example, with this problem, because you stated you felt like overlooking it.\r\nYou simply don't have a choice in the matter.", "Solution_9": "Hi,\r\nA cute angle, you are correct that I should have solved the problem for complex, not just for real. I agree with that. I won't argue about that.\r\n\r\nBut you are not correct when you are saying that my solution is not correct if we have the restriction $x,y \\in R$. It is correct just like your solution is correct but it is different. Still for a problem there can be more than one solution, right?", "Solution_10": "You are bringing up a specific objection for you that I did not raise.\r\nLook back over my sentences.\r\n\r\nAs long as you, the poster imortal, are having the restriction for x and y\r\nbelonging to the real numbers, then yes, your solutions are correct and consistent with that. My past words are also consistent with that.", "Solution_11": "Read all, please, the my last reply ! I added there a theoretical comment. Finish this exchange of words, please ! Thanks." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "In a graph with even number of vertices, each vertex has degree $ 3$. This graph has a Hamiltonian cycle. Show that it has at least two.", "Solution_1": "[quote=\"nguyentrang\"]In a graph with even number of vertices, each vertex has degree $ 3$. This graph has a Hamiltonian cycle. Show that it has at least two.[/quote]\r\n\r\nPosted before (not quite answered) at http://www.mathlinks.ro/Forum/viewtopic.php?t=80281 .\r\n\r\n darij" } { "Tag": [ "trigonometry", "quadratics", "rational numbers" ], "Problem": "Suppose that $\\tan \\alpha =\\frac{p}{q}$, where $p$ and $q$ are integers and $q \\neq 0$. Prove the number $\\tan \\beta$ for which $\\tan 2\\beta =\\tan 3\\alpha$ is rational only when $p^2 +q^2$ is the square of an integer.", "Solution_1": "If $ \\tan\\alpha \\equal{}\\frac{p}{q}$ then the equation reads $ \\frac{2\\tan\\beta}{1\\minus{}\\tan^{2}\\beta}\\equal{}\\frac{p(p^{2}\\minus{}3q^{2})}{q(3p^{2}\\minus{}q^{2})}$, solving to $ \\tan\\beta$ gives a quadratic with discriminant $ (p^{2}\\plus{}q^{2})^{3}$, so $ \\tan\\beta\\in\\mathbb{Q}$ iff $ p^{2}\\plus{}q^{2}$ is a rational square. But $ p^{2}\\plus{}q^{2}$ is an integer, so is the square of an integer, QED." } { "Tag": [ "trigonometry" ], "Problem": "A homogeneous ball with decelerates during $4.0\\,s$ to rest on a smooth,\r\nhorizontal plane. The braking distance is $s=1,5\\,dm$. What inclination, $\\theta$, should the\r\nplane have if the ball is to roll with constant velocity?\r\n\r\n[color=red]EDITED[/color]", "Solution_1": "assuming you made a mistake. because the plane is horizontal. so the angle is zero.\r\n\r\nassuming misspelling.\r\n\r\nthe ball is rolling down to a inclined plane with constant velocity.\r\n\r\nSo there is no force on it. so the aceleration is zero.\r\n\r\nSo the component of the gravitational force is equal to the aceleration.\r\n\r\nsince you didnt say if there is friction or another thing and the problem and your statement.\r\n\r\nI will be happy if you say it more clear :)", "Solution_2": "lol, right yagaron :D", "Solution_3": "The only thing i have forgot to say is that it decelerates uniformly.\r\n\r\nYes of course friction must be taken into account, otherwise the ball\r\nwont decelerate on a horizontal plane.", "Solution_4": "acctually, there is a lot of ways a particle decelerate without friction.\r\n\r\nwe can use eletromagnetic fields, of others fields.\r\n\r\ni just wanted the cause of deceleration. thanks\r\n\r\nbut this problem is simple. we can use\r\n\r\n$s'' = a$ and $s'=v$\r\n\r\ni put in diferential forme to become more simple, but is the law of movement on newtonian mechanics", "Solution_5": "I have already mentioned the cause: friction\r\n\r\nand of course you have to rely\r\non Newtonian mechanics.\r\n\r\nIf you find it simple, then please post a solution.", "Solution_6": "the solutions is analogous of solution of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151971\r\n\r\nI cannot lose time to make a lot of analogous problem dude.", "Solution_7": "They are not that analogous.\r\n\r\nYou don't have to solve it if you don't want to. :P", "Solution_8": "color it would be helpful if you restate the entire problem . actually it is difficult to get you.", "Solution_9": "Please explain what is difficult. I am sorry for that, but the problem\r\ntext will not reveal more information, that is the point. It gets\r\nmore difficult to solve that way.", "Solution_10": "Well then the problem text is probably wrong. it happens sometimes :wink:", "Solution_11": "I am sure it is not, specify please.", "Solution_12": "[quote=\"color\"]They are not that analogous.\n\nYou don't have to solve it if you don't want to. :P[/quote]\r\n\r\ner?? ahm?!!?\r\n\r\nnot!? are you sure?\r\n\r\nsince, like i said, there is no force acting on the particle. so $s = v t+d$ $a=0$\r\n\r\nand your problem has wrongs things, but you dont want to correct it.\r\n\r\nI just think you thought you made a good problem, and everybody is saying it is wrong and it's simples etc.. and you get angry. dont be mad man..\r\n\r\nthats happen like taakisfox said.\r\n\r\nif you'd be able to rewrite the problem correctly, we can try to solve it.", "Solution_13": "I cant see why the problem is wrong.\r\nAlso I have said that friction acts on the ball, so count that as a force.\r\nAnd I am not mad :lol: \r\n\r\nI will post my solution later, but I cant see why the problem should be changed.\r\n---------------------------------\r\n\r\nA ball with mass $m$ and radius $R$ decelerates uniformly during $4.0$s to rest\r\non a smoth horizontal table due to friction.\r\nThe breaking distance is $l$ (1.5 dm). If one instead want the ball to roll with constant velocity\r\nwhat should the inclination $\\theta$ be?", "Solution_14": "no i assume roll her means roll without slipping but if it has to roll without slipping and with a constant speed .that implies the friction(which must act because it has to roll without slipping to produce the net torque) must balance the component of the gravity along the surface.But then this friction gives ise to torque which in turn provides angular accelaration and since the speed of c.m has to be constant the ball no longer rolls.\r\n\r\nif you meant simple rolling not considering about sliding then \r\n\r\n$mg \\sin \\theta$ $=$ $mg \\cos \\theta \\frac{2s}{t^{2}g}$\r\n\r\n$\\ tan \\theta$ $=$ $\\frac{2s}{t^{2}g}$\r\nthis means there is no use of teh so given variables like $R$...", "Solution_15": "Sorry for giving extra information like $R\\ldots$\r\nThat is edited.\r\n\r\n[hide=\"Here is a solution\"]\nThe force which causes the ball to decelerate, is the resulting force from the normal force\n$N$ and friction force $f$. Assume the lever arm of the normal force $N$ is $d$ relative the center of mass, and $f$ has the arm $R$.\n\nLet the braking distance be $s$ and braking time be $t$, then we have\n$f=ma$\n$s=\\frac{at^{2}}{2}$ or $a=\\frac{2s}{t^{2}}$\n$Nd-fR=I\\alpha=\\frac{2}{5}mR^{2}\\cdot \\frac{a}{R}=\\frac{2}{5}mRa$\nall these give $s=\\frac{14Rs}{5gt^{2}}$\n\nIf the plane is inclined with $\\theta$ in braking is compensated. Braking torque will be zero as well as acceleration. (This means the resultant of $N$ and $f$ goes through the center of the ball).\nThis gives:\n$f-mg\\sin{\\theta}=0$\n$Nd-fR=0$ or $mg\\cos{\\theta}\\cdot d-fR=0$\ngives\n$\\tan{\\theta}=\\frac{d}{R}=\\frac{14Rs}{5gt^{2}}=\\boxed{\\frac{14s}{5gt^{2}}}$\n\nso $\\theta\\approx 0.15^{o}$\n[/hide]", "Solution_16": "Come on man, The normal force gets exerted at the point where the ball touches the floor, and since the ball is homogenous, its c.m. is above this point so the normal force cant have torque on c.m. :wink:", "Solution_17": "[b]@Thaakisfox[/b]\r\nSo how would you solve it?\r\n\r\nPlease, explain more detailed than [b]pardesi[/b]s solution above\r\n\r\n[b]@pardesi[/b]\r\nCould you develop your conditions and the solution a bit more, please?\r\nAlso what do you think about my solution?", "Solution_18": "I would do it the same way Pardesi did:\r\nThe coefficient of friction from the given data:\r\n$\\mu =\\frac{2s}{gt^{2}}$\r\nsince the ball will roll with constant velocity:\r\n$mg\\sin \\theta =\\mu m g \\cos \\theta$\r\n\r\nSo the needed angle:\r\n$\\theta =\\arctan \\frac{2s}{gt^{2}}$" } { "Tag": [ "probability", "complementary counting" ], "Problem": "Complementary Probability Part 3\n\nWe use complementary probability to find the probability that a geyser goes off in 150 minutes, given that it is 2/3 likely to go off at least once during any 50 minutes.", "Solution_1": "Umm...this is late at night, and I have to go to bed right after this\r\n\r\nIf it is 2/3 likely to go off during any 50 minutes, then it is 1/3 likely that it will not go off.\r\nWe can divide 150 minutes into 3 50-minute segments.\r\nThus the probablility is $ \\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}*\\equal{}\\frac{1}{27}$\r\n\r\nNope, that's not complementary counting, it's multiplying probabilities. Oh well.", "Solution_2": "aren't we supposed to find the probability that it [i]does [/i]go off?\r\n\r\ntherefore, the probability that it doesn't go off is $ \\frac{1}{27}$, so the probability that it [i]will[/i] go off is $ \\frac{26}{27}$\r\n\r\nsince we found the part not wanted to find the part wanted, we used complementary counting", "Solution_3": "i don't know how to solve the question mentioned at the end of the video. \r\n\"If the probability that you will see it go off is 2/3 in 50 minutes, then what is the probability you will see in 20minutes\"\r\n :| :| :| ~~ \r\nThanks, for anyone who replied", "Solution_4": "I have the same question as Fersolve. Can someone please answer it?" } { "Tag": [ "number theory open", "number theory" ], "Problem": "Solve the folowing diophantine equation:\r\n\r\n$n!+1=m^{2}$", "Solution_1": "posted before. Known 4 solution.", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=103178" } { "Tag": [ "analytic geometry" ], "Problem": "Let $p={(2,7), (-1,\\pi), (4, \\sqrt{3}), (0,7)}$.\r\nand let $r={(-2, 1), (\\pi, 4), (\\sqrt{3}, 2), (6, 0), (2,9)}$. Find the sum of all distinct $x$ such that $p(r(x))=7$.\r\n\r\nCan you please give me an example as to how the problem works? Thanks!", "Solution_1": "[quote=\"jeez123\"]Let $p={(2,7), (-1,\\pi), (4, \\sqrt{3}), (0,7)}$.\nand let $r={(-2, 1), (\\pi, 4), (\\sqrt{3}, 2), (6, 0), (2,9)}$. Find the sum of all distinct $x$ such that $p(r(x))=7$.\n\nCan you please give me an example as to how the problem works? Thanks![/quote]\r\n\r\n[hide]If $p(r(x))=7$, then you see where $p(\\text{something})=7$ so you look in the set of points of $p$ to see where the y-coordinate is 7, which is at $(2,7)$ and $(0,7)$. Then you know that $r(x)=2$ or $r(x)=0$ so you look at the set of points in $r$ to see where the y-coordinate is $2$ or $0$ and you get $(\\sqrt{3},2)$ and $(6,0)$ so the sum of the $x$ where $p(r(x))=7$ is $\\sqrt{3}+6$[/hide]\r\n\r\nHope this helped...", "Solution_2": "It clears things up. Thanks :D" } { "Tag": [ "analytic geometry", "MATHCOUNTS", "Pythagorean Theorem", "geometry" ], "Problem": "mathnerd314 is at the center of a circular building of radius 127. He decides to start walking north (somehow he happens to know the compass directions even though he has no compass). Halfway to the wall, he stops and decides to go east. Again he only gets halfway to the wall before he changes his mind yet another time and walks south. Then, he walks halfway to the wall and turns to walk west. After going halfway to that wall, he stops, utterly lost. How far is he from the center?\r\n\r\nI didn't understand the explanation on mathcounts.saab.org (where I got this problem).\r\n\r\nPlease tell me if my wording is too unclear.", "Solution_1": "[quote=\"mathnerd314\"]mathnerd314 is at the center of a circular building of radius 127. He decides to start walking north (somehow he happens to know the compass directions even though he has no compass). Halfway to the wall, he stops and decides to go east. Again he only gets halfway to the wall before he changes his mind yet another time and walks south. Then, he walks halfway to the wall and turns to walk west. After going halfway to that wall, he stops, utterly lost. How far is he from the center?\n\nI didn't understand the explanation on mathcounts.saab.org (where I got this problem).\n\nPlease tell me if my wording is too unclear.[/quote]\r\n\r\nwhen you say west or south do you mean mathnerd314's :lol: west and south or west and south in general?", "Solution_2": "Let the equation of this circle be $x^2+y^2=127^2$. Mathnerd314 starts at $(0,0)$ and walks North half of the radius to $(0, \\frac{127}{2})$. To find how far away the circle is to the left, let $y=\\frac{127}{2}$. $x^2+(\\frac{127}{2})^2 = 127^2 \\Rightarrow x = \\pm \\sqrt{127^2 - (\\frac{127}{2})^2}$. Choose the positive because Mathnerd314 is walking east into the first quadrant. The new point is half of this distance, $x= \\frac{\\sqrt{127^2 - (\\frac{127}{2})^2}}{2}$ So now with the point $(\\frac{\\sqrt{127^2 - (\\frac{127}{2})^2}}{2}, \\frac{127}{2})$ find the distance to the circle coordinate with the x-coordinate being $\\frac{\\sqrt{127^2 - (\\frac{127}{2})^2}}{2}$ and the y-coordinate being negative, because it's in the fourth quadrant because Mathnerd314 is walking South. The numbers get increasingly ugly, but this will eventually give you the point where Mathnerd314 ends, and by using the distance formula on that from the origin you get the total distance Mathnerd314 is from the origin.\r\nHope this helped.", "Solution_3": "[quote=\"ch1n353ch3s54a1l\"]when you say west or south do you mean mathnerd314's :lol: west and south or west and south in general?[/quote]I think it means west and south in general. Aren't west and south always the same anyway? ;) I think you're confusing this with left and right.", "Solution_4": "[quote=\"catcurio\"][quote=\"ch1n353ch3s54a1l\"]when you say west or south do you mean mathnerd314's :lol: west and south or west and south in general?[/quote]I think it means west and south in general. Aren't west and south always the same anyway? ;) I think you're confusing this with left and right.[/quote]\r\n\r\noh. i should have known..oops :oops: :? :roll: \r\n\r\ni got ........[hide]42? [/hide]", "Solution_5": "[quote=\"ch1n353ch3s54a1l\"][hide]42? [/hide][/quote]\r\n\r\nAh, such an inexplicably beautiful number... :P\r\n\r\nHow did you arrive at that answer?\r\n\r\n-BLt", "Solution_6": "[quote=\"thebrokenlight\"][quote=\"ch1n353ch3s54a1l\"][hide]42? [/hide][/quote]\n\nAh, such an inexplicably beautiful number... :P\n\nHow did you arrive at that answer?\n\n-BLt[/quote]\n\nwell ACTUALLY, i was estimating half the time. :D \n\nso he goes up $\\frac{127}{2}$. Forming a parallel line that goes through where mathnerd314 is and the diameter of the circle, you can find the length of the line, divide it by two, and then divide it by two again, and i found that to be $\\approx54.9926$. this is how much he goes east. then the \"halfway south\" part i found to be $\\frac{127}{2}+\\frac{127}{4}$. then \"halfway west\" i found to be 88.9799, u take away 54.9926 and get 33.9873, use pythagorean theorem and get...........................................................46.51. hm, different answer............. I don't think 42's right anymore.. hehe\nso [hide][size=200]46.51[/size][/hide]", "Solution_7": "But 42 is the answer to life, the universe, and everything! How could it not be right?", "Solution_8": "[quote=\"Hokkage\"]But 42 is the answer to life, the universe, and everything! How could it not be right?[/quote]\r\n\r\nwait huh? so the answer's 42?? :lol: :P :P", "Solution_9": "You need to look at my diagram for my solution to make sense.\r\n[hide]\nWe start out at the center of the circle then walk along the radius $OE$ until mathnerd stops halfway between the center and the wall. He has walked a distance of $\\frac{127}{2}$. So $OA = \\frac{127}{2}.$\n\nThe next distance we need to find is $AB$ You notice that $AB = \\frac{1}{2}(AG)$.\n\n$AG$ can be determined by using the Pythagareon Theorem for $\\triangle AOG$.\n\n$AG = \\sqrt{(127)^2 - (\\frac{127}{2})^2}$ so $AB = \\frac{1}{2} \\sqrt{(127)^2 - (\\frac{127}{2})^2}$.\n\nThe next distance to be determined is $BC$. $BC = \\frac{1}{2}(BJ)$\n\n$BJ = BL + LJ$\n\nUsing $\\triangle LOJ$ we can find $LJ$\n\n$LJ = \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]$ since $OL = AB$\n\n$BL = AO = \\frac{127}{2}$ so $BJ = \\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]$\n\nand $BC =\\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}])$\n\n$BL = AO = \\frac{127}{2}$; $LC = BC - BL = \\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}$\n\nWe can now find distance $CD$ where $CD = \\frac{1}{2}KC$\n\nFirst of all $KC = KM + MC$; $MC = AB = \\frac{1}{2} \\sqrt{(127)^2 - (\\frac{127}{2})^2}$\n\n$KM$ can be found by once again using the Pythagereon Theorem and $\\triangle KOM$\n\n$OM = LC = \\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}$\n\n$KM = \\sqrt{(127)^2 - [\\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}]^2}$\n\n$CD = \\frac{1}{2} \\sqrt{(127)^2 - [\\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}]^2}$\n\n$DM = CD - MC = \\frac{1}{2} \\sqrt{(127)^2 - [\\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}]^2} - \\frac{1}{2} \\sqrt{(127)^2 - (\\frac{127}{2})^2}$\n\nWe can now find $OD$, which is the distance that mathnerd is from the center of the circle.\n\n$(OD)^2 = (DM)^2 + (OM)^2$ so $OD = \\sqrt{(DM)^2 + (OM)^2}$\n\n$DM = \\frac{1}{2} \\sqrt{(127)^2 - [\\frac{1}{2} (\\frac{127}{2} + \\sqrt{(127)^2 - \\frac{1}{4}[(127)^2 - (\\frac{127}{2})^2}]) - \\frac{127}{2}]^2} - \\frac{1}{2} \\sqrt{(127)^2 - (\\frac{127}{2})^2}$\n\nCalculating everything out:\n\n$DM = 62.208 - 54.99 = 7.218$; $(DM)^2 = 52.10$\n\n$OM = 88.988 - 63.5 = 25.488$; $(OM)^2 = 649.64$\n\n$OD = \\sqrt{52.10 + 649.64} = 26.49$\n\n[/hide]", "Solution_10": "[quote=\"Hokkage\"]But 42 is the answer to life, the universe, and everything! How could it not be right?[/quote]\r\nSo long, so long so long, and thanks for all the fish! :rotfl:", "Solution_11": "Please stop spamming, everyone.\r\n\r\n[quote=\"Hokkage\"]The numbers get increasingly ugly,[/quote]\r\n\r\nThat was the problem with the website's solution. I'm trying to find an easy solution to this problem.", "Solution_12": "Maybe it's just me.....but don't you guys think this problem belongs under the intermediate forum and not the getting started one. Don't get me wrong the problem sounds fairly simple, but I have tried it and came up with some very ugly numbers.....and unless there's an easier way to do it, things like this should be posted somewhere else. :?:", "Solution_13": "[quote=\"gmslinger\"]Maybe it's just me.....but don't you guys think this problem belongs under the intermediate forum and not the getting started one. Don't get me wrong the problem sounds fairly simple, but I have tried it and came up with some very ugly numbers.....and unless there's an easier way to do it, things like this should be posted somewhere else. :?:[/quote]\r\n\r\nAt this point I agree with this (and it's been moved anyway). I just thought that since it was from mathcounts.saab.org, it would be around Mathcounts level.\r\n\r\nI had hoped for a less ugly, fairly simple solution too.", "Solution_14": "Thanks guys,\r\n Sorry I'm new at this and some of this seems very overwhelming. I'm only a freshman in geometry.\r\n\r\n [b]THANKS[/b]", "Solution_15": "One simple question:\r\n\r\nIf the initial point is in somewhere on the first quadrant, and then go south, then go west, go north, go east, and so on....\r\n\r\nWhat is the initial point, if at the end of each step, the distance from the origin is constant?\r\n\r\nAngel \"Java\" Lopez", "Solution_16": "No no no... the initial point is the center of the circle, and the distance from the origin does not remain constant. Unless I've totally misunderstood your idea...", "Solution_17": "btw it's a mathcounts target round problem... IMO it should be in the MathCounts forum.", "Solution_18": "And BTW that's where I got it (I posted it in the Getting Started forum), but it was moved because of all the hard solutions submitted. Do you have the Mathcounts-level solution, Treething?", "Solution_19": "walks up half radius, then right rsin(60)/2\r\nthen its kinda good to know how far it is up to the wall from that point.\r\nThat would be sqrt(r^2 - (r^2)(sin^2(60))/4) which happens to be:\r\nsqrt(r^2 - (3/16)r^2)=sqrt((9/16)r^2)=(3/4)r\r\nWhich means that the distance to the other wall is (3/4)r-(1/2)r+(3/4)r=r\r\nHe now walks r/2 from the coordinates where he first was, and end up at \r\nthe coordinates (relative the middle) (rsin(60)/2 ; 0)\r\nthe way to the (our) left wall is now (r+rsin(60)/2)\r\nk divide by two..\r\n\r\nr/2+rsin(60)/4... thats the way he walks\r\n\r\nlets subtract that from the x-coordinate\r\naand get a better value for sin(60) again.. sqrt(3)/2 yess\r\nrsqrt(3)/4 - r/2 + rsqrt(3)/8 = r(3sqrt(3)/8-1,5).. according to me that should be negative, so i guess he passed origo.. yei\r\nso take |127(3sqrt(3)/8 - 1,5)| for an answer. thats like 108 and something.\r\n\r\nI may have done some counting mistakes, but the rest should due. ;P\r\nExcuse me cant handle latex.. and btw sqrt(x) is supposed to mean squareroot" } { "Tag": [ "analytic geometry", "pigeonhole principle", "number theory unsolved", "number theory" ], "Problem": "Find the largest integer $n$ for which there exist $n$ different integers such that none of them are divisible by either of $7,11$ or $13$, but the sum of any two of them is divisible by at least one of $7,11$ and $13$.", "Solution_1": "the answer is actually 8. \r\n\r\nwe may represent each integer by a triple $ (a, b, c)$, where $ a, b, c$ are the remainders upon division by 7, 11, 13, respectively. for each remainder $ a$, let $ a'$ denote the complementary remainder; e.g., the complementary remainder modulo 7 of 3 is 4. for two $ n$-tuples $ (a_1, ..., a_n)$, $ (b_1, ..., b_n)$ of remainders, we say that they \"match\" in the $ i$-th position if $ a_i \\equal{} b_i'$. a set of $ n$-tuples will be called \"matching\" if any two of them match in at least one position; otherwise, we call them \"mismatched.\" so the statement we want to prove is, that a set of matching triples must have size at most 8. first we prove the following lemma.\r\n\r\nLemma: a set of matching pairs can have size at most 4. \r\n\r\nProof: assume, for a contradiction, that we have a set of 5 matching pairs, and let $ a, b, c, d, e$ denote these pairs. assume for the moment that $ a$ matches 3 of the other pairs - WLOG we may assume they are $ b, c, d$ - in the first position. then $ b, c, d$ coincide in the first position, so any two of them must match each other in the second position - but this is impossible. so $ a$ can match at most 2 of the other pairs in the first position, and similarly it can match at most 2 of the other pairs in the second position. since there are four other pairs, a must match *exactly* 2 in the first position and 2 in the second position. now let $ a \\equal{} (x,y)$. by what we've just proved, the other pairs must be of the form $ (x, a), (x, a'), (b, y), (b', y)$. now, for $ (x, a')$ to match $ (b, y)$, we must have either $ x \\equal{} b'$ or $ y \\equal{} a$. in both cases, we have 3 pairs which coincide in one position, which as before, is impossible. this completes the proof of the lemma. \r\n\r\nnow to the result we want to prove. assume, for a contradiction, that we have a set of 9 matching triples. take the 9 numbers in the first coordinates of these triples and partition them into two sets so that no two numbers in the same set are complementary. (this is possible because every number has a unique complement and no number is its own complement because the moduli are odd.) by the pigeonhole principle, one of these sets contains 5 numbers, and these numbers are pairwise non-complementary. thus there are 5 triples which are mutually mismatched in their first coordinates. their second and third coordinates then give us 5 matching pairs, contrary to the lemma.\r\n\r\nto see that 8 is indeed possible, take 8 triples of the form $ (a, b, c), (a', b, d), (a, b', e), (a', b', f), (a', b', f'), (a, b, c'), (a', b, d'), (a, b', e')$.", "Solution_2": "thank you.\r\n\r\nI realized 120 was too high, as I thought that every pair added up to a multiple of 7, 11, 13 in my set of 120, but I was wrong.\r\n\r\nNice solution." } { "Tag": [ "function", "limit", "inequalities open", "inequalities" ], "Problem": "Hello everybody.\r\n\r\nI would like to know if there exists a linear function f so that for all positive integers n : $ (x\\plus{}y)^n \\le f(n) (x^n \\plus{} y^n)$.\r\n\r\nThanks.", "Solution_1": "Take $ x\\equal{}y$. Hence $ f(n) \\ge 2^{n\\minus{}1}$ for all $ n$. $ f(n)\\equal{}An\\plus{}B \\ge 2^{n\\minus{}1} \\ge n^2$ for $ n \\ge 7$. But $ \\lim_{n \\to infty}g(n)\\equal{}\\lim_{n \\to \\infty}\\minus{}n^2\\plus{}An\\plus{}B \\to \\minus{} \\infty$. Hence there exists no such function." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "problem1: let ABC be square triangle ($c>a \\geq b$) \r\n find minimal of expression P \r\n$P= \\frac{a^2(b+c)+b^2(c+a)}{abc}$\r\nand proble m 2 :\r\nfind maximum of expression P \r\n$P= \\frac{a}{4+bc} + \\frac{b}{4+ac} + \\frac{c}{4+ba}$ where $a,b,c \\in$ [$0;2$]", "Solution_1": "[quote=\"tranthanhnam\"]let ABC be square triangle ($c>a \\geq b$) \n find minimal of expression P \n$P= \\frac{a^2(b+c)+b^2(c+a)}{abc}$\nand problem 2 :\nfind minimal of expression P \n$P= \\frac{a}{4+bc} + \\frac{b}{4+ac} + \\frac{c}{4+ba}$ where $a,b,c \\in$ [$0;2$][/quote]\r\nWhat is a square triangle? :?", "Solution_2": "I guess he means \"right triangle\".\r\nhttp://mathworld.wolfram.com/SquareTrianglePicking.html for \"square triangle\"", "Solution_3": "$P= (1-1/\\sqrt{2})(a^2+b^2)/ab +\\frac{1}{\\sqrt{2}}.(a^2+b^2)/ab +(a+b)/c$\r\n\r\nwe have :$a^2+b^2 \\geq 2ab$ and $a+b \\geq 2\\sqrt{ab}$\r\n hence, $min P=2+\\sqrt{2}$", "Solution_4": "[quote=\"tranthanhnam\"]problem1: let ABC be square triangle ($c>a \\geq b$) \n find minimal of expression P \n$P= \\frac{a^2(b+c)+b^2(c+a)}{abc}$\nand proble m 2 :\nfind maximum of expression P \n$P= \\frac{a}{4+bc} + \\frac{b}{4+ac} + \\frac{c}{4+ba}$ where $a,b,c \\in$ [$0;2$][/quote]\r\nsolution for problem 2:\r\nlet $f(a)=\\frac{a}{4+bc} + \\frac{b}{4+ac} + \\frac{c}{4+ba}$\r\nwe have $\\frac{d^2f}{da^2}\\geq 0$\r\nso P reach the maximum when $a,b,c$ is $0$ or$2$\r\nthen the maximum is 1" } { "Tag": [ "function", "limit", "algebra", "domain", "real analysis", "real analysis unsolved" ], "Problem": "It's well-known that a function $f: E\\subset\\mathbb{R}^{n}\\rightarrow F\\subset\\mathbb{R}^{m}$ continuous at almost every points in $E$ is measurable.\r\nCan you find a function measurable on $E$ that is not continuous at almost every points in $E$?", "Solution_1": "take a non-trivial dense mesurable subset $S$, $\\mathbb{Q}^{n}$ for example, and the function $1_{S}$.", "Solution_2": "Thanks :) And how can you say that $1_{S}$ is not continuous on $\\mathbb{R}^{n}\\backslash S$, it takes the constant value 0. So $\\forall y \\in \\mathbb{R}^{n}\\backslash S,\\hspace{0,2cm}\\lim_{\\{x\\rightarrow y; x\\in\\mathbb{R}^{n}\\backslash S\\}}1_{S}(x)=0=1_{S}(y)$. I'm maybe confused by the definition of \"$f$ is continuous at point $y$\" for $y$ in a strange subset :p...", "Solution_3": "Continuity of $f$ depends on what is understood by the domain of $f$. In alekk's post the domain ($E$ in your question) is the entire space $\\mathbb R^{n}$" } { "Tag": [ "inequalities", "geometry", "circumcircle", "analytic geometry", "ratio", "absolute value", "inequalities proposed" ], "Problem": "Let ABC be a triangle. M and N are arbitrary points inside ABC. D, E, F are points that belong to BC, CA, AB, respectively such that: MD is parallel to NA, ME is parallel to NB, and MF is parallel to NC. Prove that:\r\n\r\narea of DEF =< 1/4 area of ABC.\r\n\r\n\r\nYou may know the problem in some more specific cases :\r\n\r\n1) AN is orthogonal to BC, so on...\r\n\r\nor\r\n\r\n2) M and N are the same points\r\n\r\nThis problem is little more general. ;)", "Solution_1": "This hangs around unsolved since June?\r\n\r\nWe begin with a lemma:\r\n\r\n[b]Lemma 1.[/b] If M is an arbitrary point inside triangle ABC, then the area of the pedal triangle of the point M with respect to triangle ABC is less or equal to $\\frac14$ of the area of triangle ABC.\r\n\r\n[i]Note.[/i] The pedal triangle of a point M with respect to triangle ABC is the triangle formed by the orthogonal projections of the point M on the sides BC, CA, AB of triangle ABC.\r\n\r\n[i]Proof of Lemma 1.[/i] After what I proved in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5658[/url], the area of T the pedal triangle of the point M with respect to triangle ABC equals $T=\\frac{Sp}{4R^2}$, where S is the area and R is the circumradius of triangle ABC, and p is the absolute value of the power of the point M with respect to the circumcircle of triangle ABC. We have to prove $T \\leq \\frac14 S$. This simplifies to $\\frac{Sp}{4R^2} \\leq \\frac14 S$, i. e. to $\\frac{p}{R^2} \\leq 1$. In other words, we have to show $p \\leq R^2$. But this is immediately clear: The point M lies inside the triangle ABC, hence also inside the circumcircle of triangle ABC, and thus the power of the point M with respect to the circumcircle is negative. On the other hand, this power must be $\\geq -R^2$ (in fact, if O is the center of the circumcircle of triangle ABC, the power of M with respect to the circumcircle equals $OM^2-R^2$, what is clearly $\\geq -R^2$ since $OM^2 \\geq 0$). Hence, the absolute value p of this power is $\\leq R^2$, and Lemma 1 is proven.\r\n\r\nNow I cite a lemma which I won't prove since the proof is straightforward:\r\n\r\n[b]Lemma 2.[/b] If ABC is a triangle with the sidelengths a, b, c, then the homogeneous barycentric coordinates of the orthocenter of triangle ABC with respect to triangle ABC are $\\left(\\frac{1}{b^2+c^2-a^2}:\\frac{1}{c^2+a^2-b^2}:\\frac{1}{a^2+b^2-c^2}\\right)$.\r\n\r\nNow a more important lemma:\r\n\r\n[b]Lemma 3.[/b] For every triangle ABC and every point N inside triangle ABC, there exists an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'.\r\n\r\n[i]Proof of Lemma 3.[/i] Let $\\left(x:y:z\\right)$ be the homogeneous barycentric coordinates of the point N with respect to triangle ABC, such that all the numbers x, y, z are positive (this is possible, since the point N lies inside triangle ABC). Let A'B'C' be a triangle lying somewhere in the plane and having the sidelengths\r\n\r\n$a^{\\prime} = B^{\\prime}C^{\\prime} = \\sqrt{\\frac{1}{y}+\\frac{1}{z}}$;\r\n$b^{\\prime} = C^{\\prime}A^{\\prime} = \\sqrt{\\frac{1}{z}+\\frac{1}{x}}$;\r\n$c^{\\prime} = A^{\\prime}B^{\\prime} = \\sqrt{\\frac{1}{x}+\\frac{1}{y}}$.\r\n\r\nThen,\r\n\r\n$b^{\\prime}\\ ^2 + c^{\\prime}\\ ^2 - a^{\\prime}\\ ^2 = \\left(\\frac{1}{z}+\\frac{1}{x}\\right) + \\left(\\frac{1}{x}+\\frac{1}{y}\\right) - \\left(\\frac{1}{y}+\\frac{1}{z}\\right) = 2\\frac{1}{x} = \\frac{2}{x}$,\r\n\r\nand thus\r\n\r\n$\\frac{1}{b^{\\prime}\\ ^2 + c^{\\prime}\\ ^2 - a^{\\prime}\\ ^2} = \\frac{x}{2}$.\r\n\r\nSimilarly,\r\n\r\n$\\frac{1}{c^{\\prime}\\ ^2 + a^{\\prime}\\ ^2 - b^{\\prime}\\ ^2} = \\frac{y}{2}$;\r\n$\\frac{1}{a^{\\prime}\\ ^2 + b^{\\prime}\\ ^2 - c^{\\prime}\\ ^2} = \\frac{z}{2}$.\r\n\r\nHence, after Lemma 2, the homogeneous barycentric coordinates of the orthocenter of triangle A'B'C' with respect to triangle A'B'C' are\r\n\r\n$\\left(\\frac{1}{b^{\\prime}\\ ^2 + c^{\\prime}\\ ^2 - a^{\\prime}\\ ^2}:\\frac{1}{c^{\\prime}\\ ^2 + a^{\\prime}\\ ^2 - b^{\\prime}\\ ^2}:\\frac{1}{a^{\\prime}\\ ^2 + b^{\\prime}\\ ^2 - c^{\\prime}\\ ^2}\\right)$\r\n$=\\left(\\frac{x}{2}:\\frac{y}{2}:\\frac{z}{2}\\right)=\\left(x:y:z\\right)$.\r\n\r\nNow, it is well-known that any triangle can be mapped to any other triangle by an affine transformation. Consider the affine transformation mapping the triangle ABC to the triangle A'B'C'. Let N' be the image of the point N under this transformation. Since affine transformations preserve homogeneous barycentric coordinates, it follows that the homogeneous barycentric coordinates of the point N' with respect to triangle A'B'C' are the same as the homogeneous barycentric coordinates of the point N with respect to triangle ABC. In other words, the homogeneous barycentric coordinates of the point N' with respect to triangle A'B'C' are $\\left(x:y:z\\right)$. But we know that $\\left(x:y:z\\right)$ are the homogeneous barycentric coordinates of the orthocenter of triangle A'B'C' with respect to triangle A'B'C'. Hence, the point N' is the orthocenter of triangle A'B'C'. And Lemma 3 is proven.\r\n\r\nNow, in order to solve the problem, we apply Lemma 3 and conclude that there is an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'. Let this affine transformation map the points D, E, F, M to the points D', E', F', M', respectively; of course, then, the points M' and N' lie inside the triangle A'B'C', and the points D', E', F' lie on the lines B'C', C'A', A'B', respectively. And since affine transformations preserve parallelisms, from MD || NA it follows that M'D' || N'A'. But since the point N' is the orthocenter of triangle A'B'C', we have $N^{\\prime}A^{\\prime} \\perp B^{\\prime}C^{\\prime}$. Hence, $M^{\\prime}D^{\\prime} \\perp B^{\\prime}C^{\\prime}$. In other words, the point D' is the orthogonal projection of the point M' on the side B'C' of triangle A'B'C'. Similarly, the points E' and F' are the orthogonal projections of the point M' on the sides C'A' and A'B' of triangle A'B'C'. Hence, the triangle D'E'F' is the pedal triangle of the point M' with respect to triangle A'B'C'. Now, since the point M' lies inside the triangle A'B'C', Lemma 1 tells us that the area of the pedal triangle of the point M' with respect to triangle A'B'C' is less or equal to $\\frac14$ of the area of triangle A'B'C'. In other words, the area of triangle D'E'F' is less or equal to $\\frac14$ of the area of triangle A'B'C'. But since affine transformations preserve ratios of areas, the ratio between the areas of triangles D'E'F' and A'B'C' equals the ratio between the areas of triangles DEF and ABC. Hence, we see that the area of triangle DEF is less or equal to $\\frac14$ of the area of triangle ABC. Problem solved.\r\n\r\nNote that, as a special case of your inequality for M = N, we get the following famous inequality: If N is a point inside a triangle ABC, and DEF is the cevian triangle of this point N with respect to triangle ABC (in other words, the points D, E, F are the points of intersection of the lines AN, BN, CN with the lines BC, CA, AB, respectively), then the area of triangle DEF is less or equal to $\\frac14$ of the area of triangle ABC.\r\n\r\n Darij" } { "Tag": [ "geometry", "circumcircle", "inequalities unsolved", "inequalities" ], "Problem": "[i]First Day - Budapest, April 9[/i]\r\n\r\nLet $ABC$ be an acute-angled triangle. The tangents to its circumcircle at $A,B,C$ form a triangle $PQR$ with $C\\in PQ$ and $B\\in PR$. Let $C_1$ be the foot of the altitude from $C$ in $\\triangle ABC$. Prove that $CC_1$ bisects $\\angle QC_1P$.", "Solution_1": "Excuse me. There is a mistake. This shouldn`t be in this section :( .", "Solution_2": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=55918 or http://www.mathlinks.ro/Forum/viewtopic.php?t=59664 .\r\n\r\n darij" } { "Tag": [], "Problem": "George is planning a dinner meeting for 150 people. Each table will be set for nine people. If there are enough tables to accommodate everyone, what is the fewest possible number of unused table settings?", "Solution_1": "Each table seats nine people, so 17 tables will seat 17*9 or 153 people.\r\nHowever, there are only 150 people at the dinner meeting, so there will be 153-150=$ 3$ unused table settings." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ x_{1},x_{2}...x_{3n}\\ge 0$, prove that\r\n\\[ 2^{n}\\cdot\\prod_{k=1}^{3n}\\frac{1+x_{k}^{2}}{1+x_{k}}\\ge(1+\\prod_{k=1}^{3n}x_{k}^{\\frac{1}{n}}) \\]\r\nAn especial case $ x,y,z\\ge 0$ prove that\r\n\\[ \\frac{1+x^{2}}{1+x}\\cdot\\frac{1+y^{2}}{1+y}\\cdot\\frac{1+z^{2}}{1+z}\\ge\\frac{1+xyz}{2}\\]", "Solution_1": "[quote=\"gemath\"]Given $ x_{1},x_{2}...x_{3n}\\ge 0$, prove that\n\\[ 2^{n}\\cdot\\prod_{k=1}^{3n}\\frac{1+x_{k}^{2}}{1+x_{k}}\\ge(1+\\prod_{k=1}^{3n}x_{k}^{\\frac{1}{n}}) \\]\n[/quote]\r\nI think you missed $ n$.. so that what to prove is $ 2^{n}\\cdot\\prod_{k=1}^{3n}\\frac{1+x_{k}^{2}}{1+x_{k}}\\ge(1+\\prod_{k=1}^{3n}x_{k}^{\\frac{1}{n}})^{n}$.", "Solution_2": "Anyhow, this is my proof. $ (\\frac{1+x^{2}}{1+x})^{3}\\geq \\frac{1+x^{3}}{2}$ is equivalent to $ (x-1)^{4}(x^{2}+x+1) \\geq 0$. Hence $ \\prod_{k=1}^{3n}\\frac{1+x_{k}^{2}}{1+x_{k}}\\geq \\prod_{k=1}^{3n}(\\frac{1+x_{k}^{3}}{2})^{\\frac{1}{3}}$\r\n$ \\geq (\\frac{1+(x_{1}\\cdots x_{3n})^{\\frac{3}{3n}}}{2})^{\\frac{3n}{3}}=(\\frac{1+(x_{1}\\cdots x_{3n})^{\\frac{1}{n}}}{2})^{n}$ by Helder's inequality.", "Solution_3": "Oh you are right I missed $ n$ and I am suprised at your solution, here is the solution I known." } { "Tag": [], "Problem": "Write the value of $ 6600 \\div .006$ in scientific notation.", "Solution_1": "This is just $ 6.6\\times 10^{3}\\div 6\\times 10^{\\minus{}3}\\equal{}\\frac{6.6\\times 10^{3}}{6\\times 10^{\\minus{}3}}\\equal{}\\frac{(6.6\\times 10^{3})\\times 10^{3}}{6}$. Now just re-write it." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "Putnam", "calculus", "algebra", "polynomial" ], "Problem": "Take the first n squares (where n is a variable) and insert plus or minus signs in between them\r\nprove you can get any number\r\nfor example\r\n1=1^2\r\n2=-1^2-2^2-3^2+4^2\r\n3=-1^2+2^2\r\n4=1^2-2^2-3^2+4^2", "Solution_1": "[quote=\"Scrambled\"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them\nprove you can get any number\nfor example\n1=1^2\n2=-1^2-2^2-3^2+4^2\n3=-1^2+2^2\n4=1^2-2^2-3^2+4^2[/quote]\r\n\r\nI may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting!", "Solution_2": "[quote=\"djimenez\"][quote=\"Scrambled\"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them\nprove you can get any number\nfor example\n1=1^2\n2=-1^2-2^2-3^2+4^2\n3=-1^2+2^2\n4=1^2-2^2-3^2+4^2[/quote]\n\nI may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting![/quote]\r\ni got this from a book, which is a collection of good problems from various sources (including, but not limited to USAMO and putnam, but theres no need for calculus)", "Solution_3": "I know this is in Zeitz, is that where you got it from? Also, I started a discussion about this question a while back on the forum. If you are asking about the question because you need help I'll find the link...if you are just posing a challenge I'll let people try it before giving the linik ;)", "Solution_4": "[quote=\"djimenez\"][quote=\"Scrambled\"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them\nprove you can get any number\nfor example\n1=1^2\n2=-1^2-2^2-3^2+4^2\n3=-1^2+2^2\n4=1^2-2^2-3^2+4^2[/quote]\n\nI may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting![/quote]\r\nBut it is really trivial, just apply well known trick $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ !!! Therefore it sufficies to obtain representation for 1, 2, 3 and 4 as showed above ;)", "Solution_5": "[quote=\"Myth\"][quote=\"djimenez\"][quote=\"Scrambled\"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them\nprove you can get any number\nfor example\n1=1^2\n2=-1^2-2^2-3^2+4^2\n3=-1^2+2^2\n4=1^2-2^2-3^2+4^2[/quote]\n\nI may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting![/quote]\nBut it is really trivial, just apply well known trick $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ !!! Therefore it sufficies to obtain representation for 1, 2, 3 and 4 as showed above ;)[/quote]\r\n\r\n :first: :first: :cool:", "Solution_6": "[quote=\"Myth\"][quote=\"djimenez\"]I may be wrong, and just loosing the whole point, but ... [/quote]\nBut it is really trivial, just apply well known trick $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ !!! Therefore it sufficies to obtain representation for 1, 2, 3 and 4 as showed above ;)[/quote]\r\n\r\nUpss! :blush: :blush: I didn't see that!", "Solution_7": "It's also true for changing the square to another power, or even for a special type of polynomial:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=30998" } { "Tag": [ "USAMTS" ], "Problem": "It says we can't use outside resources.\r\nAre we allowed:\r\n1) Calculators?\r\n2) To code computer programs to solve stuff?\r\n3) Printed reference material (formula books, etc.)", "Solution_1": "1) Definitely\r\n2) This I would say no. In USAMTS, they dont mind if you use programming, but I think programming would fall under an \"outside resource.\" That, and they probably wouldn't give you much credit if you said \"Plugging this into Mathematica gives me...\"\r\n3) Not sure about\r\n\r\n\r\nman, the cmor people need to be more clear with their rules." } { "Tag": [ "geometry" ], "Problem": "[geogebra]a5d1d921ed4d1930f0538e3cdbaa82aadaaa9ed3[/geogebra]\r\n\r\nExplain why the figure above is impossible given that ...\r\n\r\nA' on the same side of BC as A, A' in the interior of triangle ABC, AB = A'B, AC = A'C. \r\n\r\nNote that BC is the common segment, and A and A' are the points that are above (it doesn't really matter which you pick to be A and A' (I didn't know how to label them on the Geogebra that this post allows for...)\r\n\r\nNOTE: In that book Plane Euclidean Geometry: Theory and Problems, we have not \"learned\" the SSS Theorem yet, but we do know the SAS Postulate (i.e. prove the above assertion of impossibility using the SAS Postulate).\r\n\r\nAny help in this matter is greatly appreciated. I thank you in advance.", "Solution_1": "Because a circle centered at B having radius AB and a circle centered C having radius AC can't intersect in 2 points on the same side of BC to each other (A and A'), since their intersections must be symmetrical about the line BC.\r\n\r\nCheers,\r\n\r\nRofler", "Solution_2": "okay.....\r\n\r\nBut I want to have a more \"solid\" proof (I totally understand that your intuitive proof works - and if we were to set up the formal geometry, it would totally be a great solution) in the sense that we start with the SAS Postulate (which then proves the Isosceles Triangle Theorem, which are about the only two things that the book has proved before that problem) and then prove the impossibility of this result. \r\n\r\nIs anyone willing to take the challenge?\r\n\r\n:)", "Solution_3": "I guess not... :( \r\n\r\nBut luckily I found the solution in a footnote in the dover publication of euclid's elements (book one - footnote on proposition 8)\r\n\r\n[hide=\"Hint\"]\nIntroduce three auxiliary lines\n[/hide]" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Let p \\geq 3 be a prime and n \\in N. Written in the 'regular' base 10 p^n has 20 digits.\r\n\r\nProve that there is a digit which appears more than twice !", "Solution_1": "Since there are 10 digits (0 through 9) and p^n has 20 digits, if there is no digit which appears more than twice then all the digits must appear exactly twice. If p>3, p^n has digit sum =2(1+2+..+9), which is divisible by 9, so 9|p^n, which is impossible. I guess this only leaves the case p=3." } { "Tag": [ "email" ], "Problem": "We have just revamped our [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]directory of contests[/url]. I will add NOML and AHSMIC within the next couple of days, but otherwise please email me at crawford@artofproblemsolving.com if you recognize any omissions from this list.", "Solution_1": "Kinda unrelated, but I would recommend adding SIMUW (http://www.math.washington.edu/~simuw/) to the list of summer programs. Yes it's only restricted to students from WA but adding it won't flood the list either. :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$a,b,c,d >0$ satisfy $abcd=1$.Prove that:\r\n $\\frac{1+ab}{1+a}+\\frac{1+bc}{1+b}+\\frac{1+cd}{1+c}+\\frac{1+da}{1+d} \\geq 4$", "Solution_1": "hope my solution is correct:\r\nlet:$a=\\frac y x,b=\\frac z y,c=\\frac w z,d=\\frac x w$\r\nwe will find the inequality is equal to:\r\n$\\frac{x+z}{x+y}+\\frac{y+w}{y+z}+\\frac{z+x}{z+w}+\\frac{w+y}{w+x} \\ge 4$\r\nbut use caughy's,we have:\r\n$\\left((x+y)(x+z)+(y+z)(y+w)+(z+w)(z+x)+(w+x)(w+y)\\right)\\left( \\frac{x+z}{x+y}+\\frac{y+w}{y+z}+\\frac{z+x}{z+w}+\\frac{w+y}{w+x}\\right) \\ge 4(x+y+z+w)^2$\r\nbut $(x+y+z+w)^2=(x+y)(x+z)+(y+z)(y+w)+(z+w)(z+x)+(w+x)(w+y)$\r\n :)", "Solution_2": "Wonderful solution :)", "Solution_3": "Posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=85389", "Solution_4": "[quote=\"nhat\"]$a,b,c,d >0$ satisfy $abcd=1$.Prove that:\n $\\frac{1+ab}{1+a}+\\frac{1+bc}{1+b}+\\frac{1+cd}{1+c}+\\frac{1+da}{1+d} \\geq 4$[/quote]\r\nit is equivalent to :\r\n$\\sum \\frac{2+a+ad}{1+a}\\geq 8 \\leftrightarrow 2\\sum \\frac{1}{1+a}+\\sum \\frac{a(1+b)}{1+a}\\geq 8$ \r\nbecause of wellknown $\\sum \\frac{1}{1+x_i}\\geq \\frac{n}{2}$ when $\\prod x_i=1$\r\nand AM-GM", "Solution_5": "[quote=\"Albanian Eagle\"]\n...because of wellknown $\\sum \\frac{1}{1+x_i}\\geq \\frac{n}{2}$ when $\\prod x_i=1...$\n[/quote]\r\nThis is wrong. Try $n=3,$ $x_1=x_2=2,$ $x_3=\\frac{1}{4}.$ ;)" } { "Tag": [ "\\/closed" ], "Problem": "please help me in attaching pdf and img files to the post ???? :?: :(", "Solution_1": "When you're posting an attachment, you can click the \"Add an Attachment\" next to the \"Message body\" tab." } { "Tag": [ "AMC" ], "Problem": "I found a past complete AMC 12 exam on this website, but cannot find the solutions to them. I don't need complete worked-out solutions; I'd just like the letter-answers to the multiple choice questions. Does anyone know where I can find this (specifically for the 2004 exam)?\r\n\r\nThank you.", "Solution_1": "look at the amc website under archives", "Solution_2": "Here is for 2004:\r\n\r\nhttp://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2004-10a/04AMC10Aanswers.html\r\n\r\nThere are no \"official\" key or full solutions for AMC in this site yet. We hope to have one in future, however.\r\n\r\nPlease ask this type of question in AMC forum. Moderators, can you move this to AMC forum?", "Solution_3": "[quote=\"Silverfalcon\"]Here is for 2004:\n\nhttp://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2004-10a/04AMC10Aanswers.html\n\nThere are no \"official\" key or full solutions for AMC in this site yet. We hope to have one in future, however.\n\nPlease ask this type of question in AMC forum. Moderators, can you move this to AMC forum?[/quote]Woah! I can check my answers now! :wow:\r\n\r\nShouldn't this be in the AMC forum? ;)" } { "Tag": [], "Problem": "Se considera in plan familia de cercuri $ x^{2}+y^{2}-4x-2\\alpha y+3=0$ unde $ \\alpha$ este un parametru real.\r\ni) Sa se arate ca exista doua puncte prin care trec toate cercurile familiei.\r\nii) Sa se determine cercul de raza minima din familie.\r\niii) Sa se scrie ecuatiile cercurilor din familie care sunt tangente axei $ Oy$.", "Solution_1": "1) Pentru $ y=0$ rezulta ecuatia $ x^{2}-4x+3=0$ cu solutiile $ x_{1}=1,x_{2}=3$. Deci punctele prin care trec toate cercurile din familie sunt $ A(1,0),B(3,0)$.\r\n2) Ecuatia cercului se poate scrie\r\n$ (x-2)^{2}+(y-\\alpha )^{2}=\\alpha^{2}+1$ Rezulta ca $ r^{2}=\\alpha^{2}+1$, unde $ r$ este raza cercului. Atunci raza este minima pentru $ \\alpha =0$.\r\n3) Centrul cercului este $ C(2,\\alpha )$. Cercul este tangent axei $ Oy$ daca distanta de la $ C$ la $ Oy$ este egala cu raza cercului. Atunci $ \\alpha^{2}+1=4\\Rightarrow \\alpha =\\pm\\sqrt{3}$." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "suppose $ R$ be a commutative ring with unity and $ f: M \\longrightarrow M^{\\prime}$ be a R-homomorphism .assume $ I$ be an ideal of $ R$ so that $ f^{*} : \\dfrac{M}{IM} \\longrightarrow \\dfrac{M^{\\prime}}{{IM}^{\\prime}}$ is a R-homomorphism induced by $ f$ is onto\r\nproove that if $ M^{\\prime}$ is finitely generated and $ I \\subseteq J(R)$ then $ f$ is onto.", "Solution_1": "$ U: \\equal{}im(f)$ is a submodule of $ M'$ such that $ I*M'/U \\equal{} M'/U$. nakayama's lemma implies $ M'/U\\equal{}0$, i.e. $ f$ is onto.", "Solution_2": "it is not simple to prove $ I \\dfrac{M^{\\prime}}{U}\\equal{}\\dfrac{M^{\\prime}}{U}$\r\nhow I can do this" } { "Tag": [], "Problem": "Please find all non-negative integers n, a, b, n doesnt equal to 0, and (n^b-1) | (n^a+1).", "Solution_1": "[quote=\"scatteredclouds\"]Please find all non-negative integers n, a, b, n doesnt equal to 0, and (n^b-1) | (n^a+1).[/quote]\r\nSupposition $ \\rightarrow a \\ge b$.We have, $ (n^b\\minus{}1) | (n^a\\plus{}1) \\leftrightarrow n^a\\plus{}n^b \\vdots n^b\\minus{}1 \\rightarrow n^{a\\minus{}b}\\plus{}1 \\vdots n^b\\minus{}1 (gcd(n^b,n^b\\minus{}1)\\equal{}1)$.With $ a\\equal{}kb\\plus{}p (0 \\le p r$.\r\n\r\nHence $ 2\\ge n^b\\minus{}n^r\\ge 1$ and $ n^b\\minus{}n^r\\equal{}1,2$\r\n\r\nIf $ r\\equal{}0$ and $ n^b\\minus{}n^r\\equal{}1$, then $ n\\equal{}2$, $ b\\equal{}1$.\r\nIf $ r\\equal{}0$ and $ n^b\\minus{}n^r\\equal{}2$, then $ n\\equal{}3$, $ b\\equal{}1$.\r\n\r\nIf $ r>0$, $ n|1,2$, $ n\\equal{}1$ is nonsense, so $ n\\equal{}2$. \r\n$ 2^{b\\minus{}1}\\minus{}2^{r\\minus{}1}\\equal{}1$, so $ r\\minus{}1\\equal{}0$ and $ b\\minus{}1\\equal{}1$.\r\n\r\nOur solutions are $ (n,b,r)\\equal{}(2,1,0), (3,1,0), (2,2,1)$ which correspond to\r\n\r\n$ (n,a,b)\\equal{}(2,q,1), (3,q,1), (2,2q\\plus{}1,2)$ where $ q$ is a non-negative integer.\r\n\r\n[one can easily check that these solutions work]." } { "Tag": [ "trigonometry", "integration", "calculus", "calculus computations" ], "Problem": "messed up on previous post the real problem is \r\n\u222b\t(x^4/(1+x^2)^2) dx", "Solution_1": "Again, you should try to use the same methods on those you previously got responded and see if it works.", "Solution_2": "having trouble with the long division because equal powers on top and bottom", "Solution_3": "$\\frac{x^4}{(1+x^2)^2}=\\left(\\frac{x^2}{1+x^2}\\right)^2$", "Solution_4": "Remember that $\\frac{dx}{1+x^2}$ has *something* to do with $\\tan^{-1}x$. :)", "Solution_5": "Some algebra, starting from kunny's expression:\r\n$\\left(\\frac{x^2}{1+x^2}\\right)^2=\\left(1-\\frac1{1+x^2}\\right)^2= 1-\\frac2{1+x^2}+\\frac1{(1+x^2)^2}.$\r\n\r\nYou should be able to integrate that.", "Solution_6": "okay im having trouble integrating that i keep coming up with 1/ sec^2 x please help", "Solution_7": "$\\int \\frac{1}{sec^2 x}dx$? It is $\\int cos^2 xdx = \\int \\frac{1+\\cos 2x}{2}dx$...\r\n\r\nAnd \r\n$\\frac{1}{(1+x^2)^2} = \\frac{1}{1+x^2} - \\frac{x^2}{(1+x^2)^2}$.\r\n\r\nFirst we have $\\int \\frac{x}{(1+x^2)^2}dx = \\frac{1}{2}\\frac{d(x^2)}{(1+x^2)^2} = -\\frac{1}{2(1+x^2)}+c$\r\n$\\int \\frac{x^2}{(1+x^2)^2}dx = -\\frac{1}{2}\\int xd(\\frac{1}{1+x^2}) = -\\frac{x}{2(1+x^2)} + \\frac{1}{2} \\int\\frac{dx}{1+x^2}$. Now you can easily work out the integral following jmerry.", "Solution_8": "please help me get this to a point i can solve. from the begining i may be making a mistake somewhere\r\n\r\n\r\ni get \r\n \r\n\r\nx-1/2 arctan x + 1/2 x + 1/4 sin x + c \r\n\r\n\r\nis that right for the whole problem\r\n\r\n\r\nplease help", "Solution_9": "Your answer is absolutely wrong... There would not be $\\sin x$.\r\n\r\n$\\int \\frac{x^4}{(1+x^2)^2}dx = \\int \\left(\\frac{x^2}{1+x^2}\\right)^2 dx = \\int \\left(1-\\frac{1}{1+x^2}\\right)^2 dx = \\int 1 dx - 2\\int \\frac{dx}{1+x^2} + \\int\\frac{dx}{(1+x^2)^2}$.\r\n\r\nFrom my post,\r\n$\\int\\frac{dx}{(1+x^2)^2} = \\int \\frac{1}{1+x^2}dx - \\int \\frac{x^2}{(1+x^2)^2}dx = \\int \\frac{1}{1+x^2}dx + \\frac{x}{2(1+x^2)} - \\frac{1}{2} \\frac{dx}{1+x^2}$.\r\n\r\nCombining these results, \r\n\r\n$\\int \\frac{x^4}{(1+x^2)^2}dx = x + \\frac{x}{2(1+x^2)} - \\frac{3}{2}\\int \\frac{dx}{1+x^2} = x + \\frac{x}{2(1+x^2)} - \\frac{3\\arctan x}{2} + c$", "Solution_10": "thanks to all" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "if all the sylow subgroups of a finite group G, are normal in G, then G is soluble?", "Solution_1": "It's even nilpotent. $G$ would be the direct product of its sylow subgroups, and since each sylow subgroup is a $p$-group (for various primes $p$) and thus nilpotent, $G$ itself must be nilpotent.", "Solution_2": "i think it is only true for groups that are abelian", "Solution_3": "No, it's also true in other cases as shown above. And not to agree with it is no reason to open a new topic, especially in Solved Problems!", "Solution_4": "i really can't see why it works for general groups\r\ncan someone help me?", "Solution_5": "As grobber proved and I proved again in http://www.mathlinks.ro/Forum/viewtopic.php?t=62847 , it's true for all groups of the type you gave.\r\nWhere exactly is your problem\u00bf Please be more specific at which point you have problems to understand.", "Solution_6": "i do understand that there is only one sylow p group of G for every prime p dividing the order of G, and clearly the intersection is empty(a part the identity)\r\nbut i cant find a prof that A x B is solvable iff A and B are solvable", "Solution_7": "Let $\\{e\\} = A_0 \\subset A_1 \\subset ... \\subset A_m=A$ and $\\{e\\} = B_0 \\subset B_1 \\subset ... \\subset B_n=B$ be normal series with $A_{k+1}/A_k$, respectivly $B_{k+1}/B_k$, being abelian.\r\nThen the normal series\r\n$\\{e\\} \\times \\{e\\} = A_0 \\times B_0 \\subset A_1 \\times B_0 \\subset ... \\subset A_m \\times B_0 =A \\times B_0 \\subset A \\times B_1 \\subset ... \\subset A \\times B_n= A \\times B$\r\nshould be a normal series with abelian factors.\r\nThe other direction isn't needed here, but can be shown in a similar way.\r\n\r\nAnother way to prove such facts is to consider the $i$-th commutators.", "Solution_8": "yes but the problem is the quotient group, how can I be sure it is abelian?", "Solution_9": "Which ones\u00bf $(A_{k+1} \\times B_0)/(A_k \\times B_0)$ is isomorphic to $A_{k+1}/A_k$, and the last one is given to be abelian (we want to conclude from \"$A,B$ are solvable\" that \"$A \\times B$ is solvable\"). Similar then for the other quotients.", "Solution_10": "now i am getting really lost.\r\n\r\nthe only thing i know by hypothesis is that every sylow p-group is normal in G\r\ni cant follow your proof", "Solution_11": "All what I gave is the prove for $A \\times B$ to be solvable iff $A,B$ are. Not more. It has at first nothing to do with the given $G$.\r\n\r\nSo what is left to prove now is that every Sylow subgroup is solvable. That follows more or less directly from the following: the center of a nontrivial $p$-group is not trivial.", "Solution_12": "that s clear now.\r\nthanks" } { "Tag": [ "logarithms" ], "Problem": "If $ a,b,c>1$ and $ a\\plus{}b\\plus{}c\\equal{}6$ , then prove that : $ \\log_a(\\sqrt[3]{bc}\\plus{}a)\\plus{}\\log_b(\\sqrt[3]{ca}\\plus{}b)\\plus{}\\log_c(\\sqrt[3]{ab}\\plus{}c)\\ge \\frac{11}{2}$ .", "Solution_1": "Please don't post a solution until 10 a.m. This problem is proposed in an online competition.", "Solution_2": "What competition is that? Just curious.", "Solution_3": "send me a private message for details" } { "Tag": [ "quadratics", "group theory", "abstract algebra", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that the unique quadratic subfield of Q(zeta) where zeta is a pth root of unity is\r\n\r\nK= Q(sqrt(p)) if p=1(mod4)\r\nK= Q(sqrt(-p)) if p=3(mod4)", "Solution_1": "$ \\mathbb{Q}(\\zeta_p)/\\mathbb{Q}$ is galois with galois group $ \\cong (\\mathbb{Z}/p\\mathbb{Z})^\\times$, which is cyclic of even order. (as $ \\mathbb{F}_p\\equal{}\\mathbb{Z}/p\\mathbb{Z}$ is a field)\r\nSo it has a unique subgroup of index $ 2$, and it's fixed field is the unique quadratic subfield of $ \\mathbb{Q}(\\zeta_p)$.\r\nThe rest I proved in the lemma [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199531]here[/url]." } { "Tag": [ "logarithms" ], "Problem": "Evaluate $log_{256}(0.0625)+log(125)+log(0.8)$.", "Solution_1": "[hide] Since .0625=1/16, we have that $log_{256}(.0625)=log_{16^{2}}(\\frac{1}{16})=\\frac{-1}{2}.$ Since $logn+logm=logmn$, we have that $log125+log.8=log100$. Note that whenever we say logx, we mean $log_{10}(x)$, so $log100=log_{10}(100)=2$. Adding our two values up, we get $\\frac{-1}{2}+2=\\frac{3}{2}$, which is our answer. [/hide]", "Solution_2": "[quote=\"Ignite168\"]Evaluate $log_{256}(0.0625)+log(125)+log(0.8)$.[/quote]\r\n\r\n[hide]\n$=\\log_{256}\\left(\\frac{1}{16}\\right)+\\log 125+\\log \\left(\\frac{4}{5}\\right)$\n\n$=-\\frac{1}{2}+\\log\\left(125\\cdot \\frac{4}{5}\\right)=\\boxed{\\frac{3}{2}}$.[/hide]" } { "Tag": [], "Problem": "If $\\sqrt{x+2}=2$, then $(x+2)^2$ equals \r\n\r\n$\\text{(A)}\\ \\sqrt{2} \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 4 \\qquad \\text{(D)}\\ 8 \\qquad \\text{(E)}\\ 16$", "Solution_1": "i think i dont understand the question\r\n\r\nif $2=(x+2)^\\displaystyle\\frac{1}{2}$ then $(x+2)^2=2^4$\r\n\r\nReply by mod: Don't worry, that's correct. :)\r\nReply by sol: because a strange reason i wasnt able to see the options", "Solution_2": "[hide=\"Solution\"]Since $\\sqrt{x+2}=2$, $(\\sqrt{x+2})^2=x+2=4 \\Rightarrow x=2$.\n\n$(x+2)^2=4^2=16$ The answer is therefore $E$.[/hide]", "Solution_3": "[hide]\nx has to be 2.\n$2^4=16$\nSo $E$[/hide]", "Solution_4": "HAHAHAHAHAHAHAHAHAHAHA!!!!!!! \r\n\r\n :rotfl: :rotfl: :rotfl: ....\r\n\r\n\r\n[hide]\nx=2 \n\n2+2 =4 and the square root of 4 is 2\n2x2= 4^2= 16\n\n[/hide]", "Solution_5": "[hide]solve the equation and x = 2 so (2+2)(2+2)=4(4)=16=E[/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Determine all function $f : \\mathbb{N}\\rightarrow \\mathbb{R}$ so that for all $k,m,n$ we have $f(km)+f(kn)-f(k) \\cdot f(mn) \\geq 1$", "Solution_1": "This was in a book recommended on this website...", "Solution_2": "Can u tell me where the book is ? Actually, I just want to know the answer.. :maybe:", "Solution_3": "Let $f(1)=x$, then (k=m=n=1) $x^{2}-2x+1\\le 0 \\Longrightarrow x=1$.\r\nm=n=1 give $f(k)\\ge 1$. From $m=n=k$ we get $f(k^{2})(2-f(k))\\ge 1$. Therefore $1\\le f(k)<2$.\r\nLet $a_{n}=f(k^{2^{n}})$, then $a_{n+1}\\le \\frac{1}{2-a_{n}}$. If $1 1$ then there exist $n$ so that $a_{n}> 2$ ?\r\nI think it's not that trivial.. :maybe: \r\n\r\nHow about this,\r\nIf $a_{0}> 1$,\r\n$a_{n+1}\\geq \\frac{1}{2-a_{n}}> a_{n}$\r\nThough I can prove the sequence $a_{n}$ is increasing, is it enough ? Maybe there's a particular value of $a_{0}$ so that the sequence is bounded by 2..", "Solution_5": "[quote=\"RDeepMath91\"]There's a typo, it must be $a_{n+1}\\geq \\frac{1}{2-a_{n}}$..\nBut how can you prove if $a_{0}> 1$ then there exist $n$ so that $a_{n}> 2$ ?\nI think it's not that trivial.. :maybe: \n\nHow about this,\nIf $a_{0}> 1$,\n$a_{n+1}\\geq \\frac{1}{2-a_{n}}> a_{n}$\nThough I can prove the sequence $a_{n}$ is increasing, is it enough ? Maybe there's a particular value of $a_{0}$ so that the sequence is bounded by 2..[/quote]\r\nYes $a_{n+1}\\ge \\frac{1}{2-a_{n}}$. It give contradition with $a_{n}<2$ if $a_{0}>1$.\r\nExactly, if $a_{0}>1+\\frac{1}{k}$, then $a_{1}>1+\\frac{1}{k-1}$,...,$a_{k-1}>2$.", "Solution_6": "I see.. Thx a lot, Rust :)" } { "Tag": [ "algebra", "system of equations" ], "Problem": "In solving the system of equations $ y \\equal{} 7$ and $ x^2\\plus{} y^2\\equal{} 100$, what is the sum of the solutions for $ x$?", "Solution_1": "[quote=\"GameBot\"]In solving the system of equations $ y \\equal{} 7$ and $ x^2 \\plus{} y^2 \\equal{} 100$, what is the sum of the solutions for $ x$?[/quote]\r\n\r\nsum of roots $ \\ \\equal{}0$", "Solution_2": "[quote=\"makar\"]sum of roots $ \\ \\equal{}0$[/quote]\nwhy??", "Solution_3": "hello, plugging $y=7$ in the given equation we get $x^2=100-49$ thus $x=\\pm \\sqrt{51}$ therefore the searched sum is zero.\nSonnhard." } { "Tag": [ "calculus", "integration", "logarithms", "function", "calculus computations" ], "Problem": "Evaluate\r\n\r\n\\[\\int^e_1(x+1)e^x\\ln x dx\\]", "Solution_1": "It's done by parts; substitute $u=e^x$ and $v=(x+1)\\ln x$", "Solution_2": "Well,what's your answer? :D", "Solution_3": "I don't want to spoil it for someone who could get something out of it. The answer is:\r\n\r\n[hide]$e+(e-1)e^e$[/hide]", "Solution_4": "[quote=\"blahblahblah\"]It's done by parts; substitute $u=e^x$ and $v=(x+1)\\ln x$[/quote]\r\n\r\nhum :D", "Solution_5": "Well, they still have to do the integration themselves. With a product of 3 functions, that's a lot of integrals to calculate.", "Solution_6": "$(xe^x\\ln x)'=e^x\\ln x+xe^x\\ln x+xe^x\\frac{1}{x}=(x+1)e^x\\ln x+e^x$ :D", "Solution_7": "Very good, kunny :lol: You have many interesting integral exercises ;)" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "What is the formula to find the number of faces/edges/vertices based on information given for example:\r\n\r\nIf the measure of an interior angle of a regular polygon is 160 degrees, and this polygon is the base of a prism, the prism has how many edges?\r\n\r\nI know there's a formula relating faces/edges/vertices, i just forgot, and google isn't too helpful", "Solution_1": "Ok so the formula for the number of degrees in each angle of a regular polygon is\r\n180(n-2)/n=160\r\n180n-360=160n\r\n20n=360\r\nn=18\r\n\r\nso it has 18 sides.\r\n\r\nIf it is a prism,\r\nthen it should have\r\n1+1+18 faces\r\n18*2 vertices \r\n\r\nI think you are thinking of \r\nV+F=E+2\r\nwhere v,f and e are vertices, faces, and edges.\r\n\r\nso \r\n36+20=E+2\r\n\r\n56=E+2\r\nE=54\r\nIt has 54 edges", "Solution_2": "Alternatively, for the first one, b/c the angle is 160 degrees, it's external angle is 20. External angles sum to 360 degrees, if each external angle is 20, there are 360/20=18 angles, therefore 18 sides." } { "Tag": [ "MATHCOUNTS", "percent", "ratio", "geometry", "probability", "analytic geometry", "trapezoid" ], "Problem": "I wrote up a Mock MathCounts for the Math Nerds of New Jersey. (Since I noticed that the Illinois Forum has one.) Anyone can participate in the Chapter competition. You can private message your answers to me.\r\n\r\nSprint Round:\r\n\r\n1. Triangle ABC is an obtuse, isosceles triangle. Angle A measures 20 degrees. What is the measure of the\r\nlargest interior angle of triangle ABC?\r\n\r\n2. In Clara county, 25% of households earn less than 30,000 per year, and 65% of households earn less than 80,000 per year. What is the largest possible percent of households that could\r\nearn between 30,000 and 80,000 per year?\r\n\r\n3. The mean of four distinct positive integers is 5. If the largest integer is 13, what is the smallest integer?\r\n\r\n4. What is the sum of the odd integers from 11 through 39, inclusive?\r\n\r\n5. A line contains the points (-1, 6), (6, k) and (20, 3). What is the value of k?\r\n\r\n6. A particular convex polygon with seven sides has exactly one right angle. How many diagonals does this seven-sided polygon have?\r\n\r\n7. The product of three consecutive odd integers is 1287. What is the sum of the three integers?\r\n\r\n8. In square ABCD, point M is the midpoint of side AB and point N is the midpoint of side BC. What is the ratio of the area of triangle AMN to the area of square ABCD? Express your answer as a common fraction.\r\n\r\n9. How many non-congruent triangles are there with sides of integer length having at least one side of length 5 units and having no side longer than 5 units?\r\n\r\n10. A customer ordered 15 pieces of gourmet chocolate. The order can be packaged in small boxes that contain 1, 2 or 4 pieces of chocolate. Any box that is used must be full. How many different combinations of boxes can be used for the customer\u2019s 15 chocolate pieces? One such combination to be included is to use seven 2-piece boxes and one 1-piece box.\r\n\r\n11. The summary of a survey of 100 students listed the following totals:\r\n59 students did math homework\r\n49 students did English homework\r\n42 students did science homework\r\n20 students did English and science homework\r\n29 students did science and math homework\r\n31 students did math and English homework\r\n12 students did math, science and English homework\r\nHow many students did no math, no English and no science homework?\r\n\r\n12. Grady rides his bike 60% faster than his little brother Noah. If Grady rides 12 miles further than Noah in two hours, how fast does Noah ride?\r\n\r\n13. Either increasing the radius or the height of a cylinder by six inches will result in the same volume. The original height of the cylinder is two inches. What is the original radius?\r\n\r\n14. Four couples are at a party. Four people of the eight are randomly selected to win a prize. No person can win more than one prize. What is the probability that both members of at least one couple win a prize? Express your answer as a common fraction.\r\n\r\n15. New York and Denver are in different time zones. When it is noon in New York, it is 10 a.m. in Denver. A train leaves New York at 2 p.m. (New York time) and arrives in Denver 45 hours later. What time is it in Denver when the train arrives?\r\n\r\n16. One type of cat food recommends that a cat have a daily serving of 13 ounce of dry cat food per pound of body weight. If a cat is fed 233 ounces of dry food a day according to these recommendations, how many pounds does the cat weigh?\r\n\r\n17. If one quart of paint is exactly enough for two coats of paint on a 9-foot by 10-foot wall, how many quarts of paint are needed to apply one coat of paint to a 10-foot by 12-foot wall? Express your answer as a common fraction.\r\n\r\n18. A stock loses 10% of its value on Monday. On Tuesday it loses 20% of the value it had at the end of the day on Monday. What is the overall percent loss in value from the beginning of Monday to the end of Tuesday?\r\n\r\n19. Five balls are numbered 1 through 5 and placed in a bowl. Josh will randomly choose a ball from the bowl, look at its number and then put it back into the bowl. Then Josh will again randomly choose a ball from the bowl and look at its number. What is the probability that the product of the two numbers will be even and greater than 10? Express your answer as a common fraction.\r\n\r\n20. If a certain negative number is multiplied by six, the result is the same as 20 less than the original number. What is the value of the original number?\r\n\r\n21. If 2004 is split after the third digit into a three-digit integer and a one-digit integer, then the two integers, 200 and 4, have a common factor greater than one. The years 2005 and 2006 each have this same property, too. What is the first odd-numbered year after 2006 that has this property?\r\n\r\n22. Two consecutive even numbers are each squared. The difference of the squares is 60. What is the sum of the original two numbers?\r\n\r\n23. At the beginning of my bike ride I feel good, so I can travel 20 miles per hour. Later, I get tired and travel only 12 miles per hour. If I travel a total of 122 miles in a total time of 8 hours, for how many hours did I feel good? Express your answer as a common fraction.\r\n\r\n24. When reading a book, Charlie made a list by writing down the page number of the last page he finished reading at the end of each day. (He always finished reading a page that he started.) His mom thought his list indicated the amount of pages he had read on each day. At the end of the 8th day of reading, she added the numbers on his list and thought Charlie had read 432 pages. If Charlie started reading the book on page one, and he read the same amount of pages each day of this eight-day period, how many pages did he actually read by the end of the 8th day?\r\n\r\n25. Suppose that each distinct letter in the equation MATH = COU + NTS is replaced by a different digit chosen from 1 through 9 in such a way that the resulting equation is true. If H = 4, what is the value of the greater of C and N?\r\n\r\n26. One gear turns $ 33\\frac{1}{3}$ times in a minute. Another gear turns 45 times in a minute. Initially, a mark on each gear is pointing due north. After how many seconds will the two gears next have both their marks pointing due north?\r\n\r\n27. It takes 24 minutes for Jana to walk one mile. At that rate, how far will she walk in 10 minutes? Express your answer as a decimal to the nearest tenth.\r\n\r\n28. When plotted in the standard rectangular coordinate system, trapezoid ABCD has vertices A(1, \u22122), B(1, 1), C(5, 7) and D(5, 1). What is the area of trapezoid ABCD?\r\n\r\n29. What is the sum of all the distinct positive two-digit factors of 144?\r\n\r\n30. The positive difference of the cube of an integer and the square of the same integer is 100. What is the integer?\r\n\r\nGood luck!", "Solution_1": "What about your number theory problems? Stop procrastinating.", "Solution_2": "what are you talking about, thats in the Math Olympics......\r\nI have noticed that some of the problems you didn't make up.....can I do it?", "Solution_3": "Sure, anyone can.", "Solution_4": "Where did the target round go?", "Solution_5": "No one showed any interest at all. :(", "Solution_6": "How do you start #8?", "Solution_7": "Revive much? :P\n\nSubtract areas from the square.", "Solution_8": "what is answer to #8\n", "Solution_9": "REVIVE. $$$$", "Solution_10": "#8 is 1/8th! because I'm pretty sure that i did this problem on an aops book because this problem is so popular", "Solution_11": "What is number 11?\n\n", "Solution_12": "what is the answer to #23? NEED IMMEDIATELY.", "Solution_13": "here's a hint.\nN" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that\r\n\\[ 3\\sqrt[9]{\\frac{a^9\\plus{}b^9\\plus{}c^9}{3}}\\geq\\sqrt[10]{\\frac{a^{10}\\plus{}b^{10}}{2}}\\plus{}\\sqrt[10]{\\frac{a^{10}\\plus{}c^{10}}{2}}\\plus{}\\sqrt[10]{\\frac{b^{10}\\plus{}c^{10}}{2}}\\]", "Solution_1": "By the EV-Theorem, it suffices to prove the inequality for $ b\\equal{}c\\equal{}1$.", "Solution_2": "Thank you, Vasc! But I still didn't read a proof of this theorem. :(", "Solution_3": "[quote=\"arqady\"]Thank you, Vasc! But I still didn't read a proof of this theorem. :([/quote]\r\nSee http://jipam.vu.edu.au/article.php?sid=828" } { "Tag": [], "Problem": "Hi, I play both the piano and violin, but have always had trouble reading rhythms and playing to the beat of a metronome. Anyone have any ideas on how I can become better at this? My teacher tell me to count, but that's not really helping, because sometimes I have no idea what the song sounds like until someone else plays it. And I just go along with that until i've memorized the rhythm and mastered the notes in the piece and it has really slowed my progress. Just looking for some suggestions, thanks!", "Solution_1": "Acidentally posted twice, sorry." } { "Tag": [ "trigonometry" ], "Problem": "Let $ x$ be a real number such that the five numbers $ \\cos(2 \\pi x)$, $ \\cos(4 \\pi x)$, $ \\cos(8 \\pi x)$, $ \\cos(16 \\pi x)$, and $ \\cos(32 \\pi x)$ are all nonpositive. What is the smallest possible positive value of $ x$?", "Solution_1": "[hide]For $ \\cos(2\\pi u)$ to be nonpositive, the fractional part of $ u$ must lie between $ \\frac{1}{4}$ and $ \\frac{3}{4}$.\n\nWe will now consider all numbers mod 1 (or the fractional part of all numbers). Call a number valid if (its fractional part) lies in $ [\\frac{1}{4},\\frac{3}{4}]$ We must have $ x,2x,4x,8x,16x$ be valid. Note that the range $ [\\frac{1}{4},\\frac{1}{2}]$, when doubled, maps to $ [\\frac{1}{2},1]$. The valid part of this range is $ [\\frac{1}{2},\\frac{3}{4}]$, meaning that only the first half of the range $ [\\frac{1}{4},\\frac{1}{2}]$ will go to a valid number when doubled, Similarly, only the second half of the range $ [\\frac{1}{2},\\frac{3}{4}]$ will go to a valid number (in $ [\\frac{1}{4},\\frac{1}{2}]$) when doubled.\n\nSince we want $ x$ minimized, assume that it lies in $ [\\frac{1}{4},\\frac{1}{2}]$. For $ 2x$ to be valid, $ x$ must lie within the first half of this range, namely $ [\\frac{1}{4},\\frac{3}{8}]$. Now, $ 2x$ lies in $ [\\frac{1}{2},\\frac{3}{4}]$, so for $ 4x$ to be valid, $ 2x$ must lie within the second half of its range, meaning that $ x$ lies within in the second half of $ [\\frac{1}{4},\\frac{3}{8}]$, which is $ [\\frac{5}{16},\\frac{3}{8}]$. Repeating this process, we have that $ x$ must lie in $ [\\frac{5}{16},\\frac{11}{32}]$ for $ 8x$ to be valid, and that $ x$ must lie in $ [\\frac{21}{64},\\frac{11}{32}$ for $ 16x$ to be valid. Thus, the smallest possible value of $ x$ is $ \\boxed{21}{64}$.[/hide]", "Solution_2": "Right. Your final box came out a little off. You mean $ \\boxed{\\frac{21}{64}}$." } { "Tag": [ "percent", "probability" ], "Problem": "What is the percent probability that a two-digit number selected at random will have its tens digit at least two more than its units digit?", "Solution_1": "We can start to list the different possibilities from smallest to largest:\r\n\r\n\\[ 20 \\]\r\n\r\n\\[ 30, 31 \\]\r\n\r\n\\[ 40, 41, 42 \\]\r\n\r\n\\[ 50, 51, 52, 53 \\]\r\n\r\n\\[ ... \\]\r\n\r\nThere's a pattern! Our answer is just 1 + 2 + 3 + 4 + ... + 8 = $ \\boxed{36}$.", "Solution_2": "Thus 36 numbers out of a possible 90 work, so 36/90=6/15=2/5.", "Solution_3": "[quote=\"Mewto55555\"]Thus 36 numbers out of a possible 90 work, so 36/90=6/15=2/5.[/quote]\r\n\r\nGasp! You read incorrectly?\r\n\r\nSo $ \\frac25\\implies\\boxed{40\\%}$. :P" } { "Tag": [ "function", "algebra", "polynomial", "logarithms" ], "Problem": "Inverse of f(x) = e^x + x. \r\n\r\nQuestion : \r\n\r\nI have heard (many times before) that the function f(x) = e^x + x\r\n\r\nhas no closed form inverse, in terms of elementary functions like: finite sums, products, exponentiations and compositions of polynomials, rational functions, trigonometric functions, exponential functions, hyperbolic functions and all their inverses. \r\n\r\nHowever, I've never seen a proof for the case at hand.\r\nI was wondering whether there is such a proof and whether or not it's simple enough for me to understand. \r\n\r\nThis is not a class-assigned problem; it is just for my own curiosity.\r\n\r\nThank you", "Solution_1": "i just think that it is the fact that you cannot solve for x in this expression", "Solution_2": "On a related note,\r\n\r\nIs it impossible to solve for x in the general form\r\n\r\n$f(x)=n^x+x, n \\neq 1, n \\neq 0 \\text{ (ha ha ha)}$\r\n\r\n?", "Solution_3": "Thank you for the tip.", "Solution_4": "Well, think about how an inverse is found. You're basically expressing the equation as $x=N$ (then replacing $x$ with $y$ and vice-versa) instead of $y=N$. However, with an $x$ in an exponent and an $x$... not as an exponent :P, you can't do that. All equations of that type have no inverse that can be expressed by an equation. For example, you can say that the inverse of $y=e^x \\Rightarrow x=\\ln y$ is $y=\\ln x$. The inverse of $y=3x\\Rightarrow x=\\frac{1}{3}y$ is $y=\\frac{1}{3}x$. But the way the equation is laid out, that's impossible to do.", "Solution_5": "Thank you for your response.\r\n\r\nClearly, if I could solve for x in the equation:\r\n\r\nf(x) = e^x + x\r\n\r\nI would have found the inverse function. But this again raises the question: How do I know that it can't be done? I could try again and again to solve for x in that equation and fail each time; but how can I prove that this is because it really is impossible, and not because I just haven't come up with an ingenious enough approach?\r\n\r\nThank you again." } { "Tag": [ "limit", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $f_{n} ;[-1, 1]\\to\\mathbb{R}$ , $f_{n}(x)=\\frac{1+x^{n}}{1+|x|}$.\r\n Find \\[ \\lim_{n\\to\\infty} \\int_{[-1, 1]} f_{n} d\\lambda. \\]", "Solution_1": "$(f_{n}(x))$ converges a.e to $\\frac{1}{1+|x|}$, and $|f_{n}(x)| \\leq 2$. Applying Lebesgue's Bounded Convergence Theorem gives the result $2ln(2)$." } { "Tag": [ "probability", "geometry", "MATHCOUNTS" ], "Problem": "A square ABCD is drawn such that AB=37453 cm. A point P is randomly chosen inside the square. What is the probability that APB is acute? Express your answer to the nearest whole percent.", "Solution_1": "Let AB=2x. Consider the geometric locus consisting in all Points such that $Q$ is midpoint.\n" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "rotation", "congruent triangles", "cyclic quadrilateral" ], "Problem": "Given triangle $ABC$, squares $ABEF, BCGH, CAIJ$ are constructed externally on side $AB, BC, CA$, respectively. Let $AH \\cap BJ = P_1$, $BJ \\cap CF = Q_1$, $CF \\cap AH = R_1$, $AG \\cap CE = P_2$, $BI \\cap AG = Q_2$, $CE \\cap BI = R_2$. Prove that triangle $P_1 Q_1 R_1$ is congruent to triangle $P_2 Q_2 R_2$.", "Solution_1": "The triangle pairs $\\triangle ABI \\cong \\triangle AFC,\\ \\triangle BCE \\cong \\triangle BHA,\\ \\triangle CAG \\cong \\triangle CJB$ are congruent by SAS, for example, the 1st pair has the sides AB = AF and AI = AC equal, because ABEF and CAIJ are both squares and the angles $\\angle BAI = \\angle FAC = \\angle A + 90^o$ are also equal. Hence, the other corresponding angles of these congruent triangle pairs are also equal. Moreover, the triangle $\\triangle ABI$ is obtained by rotating the triangle $\\triangle AFC$ around the vertex A by the angle $\\angle FAB = 90^o$ and similarly for the other 2 triangle pairs. Hence, the line pairs $IB \\perp FC,\\ EC \\perp AH,\\ GA \\perp BJ$ are perpendicular to each other. The angles $\\angle P_1Q_1R_1 = \\angle P_2Q_2R_2$ are equal, because\r\n\r\n$\\angle P_1Q_1R_1 = \\widehat{(P_1Q_1, R_1Q_1)} \\equiv \\widehat{(BJ, FC)} =$\r\n\r\n$= \\widehat{(GA, FC)} \\equiv \\widehat{(P_2Q_2, R_2Q_2)} = \\angle P_2Q_2R_2$\r\n\r\nand similarly, the angles $\\angle Q_1R_1P_1 = \\angle Q_2R_2P_2,\\ \\angle R_1P_1Q_1 = \\angle R_2P_2Q_2$ are also equal. Thus we established that the triangles $\\triangle P_1Q_1R_1 \\sim \\triangle P_2Q_2R_2$ are similar.\r\n\r\nLet K, L, M be the diagonal intersections of the squares ABFE, BCGH, CAIJ and X, Y, Z the intersections of the perpendicular line pairs $IB \\perp FC,\\ EC \\perp AH,\\ GA \\perp BJ$. The quadrilateral AXCI is cyclic, because the angles $\\angle AIX \\equiv \\angle AIB = \\angle ACF \\equiv \\angle ACX$ of the congruent triangles $\\triangle ABI \\cong \\triangle AFC$ are equal. Since the angle $\\angle CXI = 90^o$ is right, the circumcenter of this cyclic quadrilateral is the midpoint of the chord CI, the diagonal of the square CAIJ. Hence, the circumcircle of the quadrilateral AXCI is identical with the circumcircle $(M)$ of the square CAIJ and the square vertex J also lies on this circle. Since AJ is a diameter of the circle $(K)$ and the angle $\\angle AZJ = 90^o$ is also right, the point Z also lies of this circle. The power of the triangle vertex B to the circle $(K)$ is equal to\r\n\r\n$BX \\cdot BI = BZ \\cdot BJ$, which implies $\\frac{BI}{BJ} = \\frac{BZ}{BX}$\r\n\r\nThe right angle triangles $\\triangle BXQ_1 \\sim \\triangle BZQ_2$, because the angles $\\angle BXQ_1 = \\angle BZQ_2 = 90^o$ are both right and the angles $\\angle BQ_1X \\equiv \\angle P_1Q_1R_1 = \\angle P_2Q_2R_2 \\equiv \\angle ZQ_2B$ of similar triangles $\\triangle P_1Q_1R_1 \\sim \\triangle P_2Q_2R_2$ are equal. Consequently,\r\n\r\n$\\frac{BQ_2}{BQ_1} = \\frac{BZ}{BX} = \\frac{BI}{BJ}$\r\n\r\nBut this means that the triangles $\\triangle BQ_2Q_1 \\sim \\triangle BIJ$ are centrally similar with the similarity center B and thus the lines $Q_1Q_2 \\parallel JI \\parallel CA$ are parallel. In a similar way, we could show that the lines $P_1P_2 \\parallel HG \\parallel BC$ and the lines $R_1R_2 \\parallel FE \\parallel AB$ are also parallel.\r\n\r\nLet S be the concurrency point of the lines AL, BM, CK, lying (together with the centroid and the 1st Fermat point of the triangle $\\triangle ABC$) on Kiepert's hyperbola. Since the angles $\\angle AMC = \\angle AYC = 90^o$ are both right, the quadrilateral AYCM is cyclic. Since N is the midpoint of the are AC of its circumcircle opposite to the vertex Y, the line YC bisects the right angle $\\angle AYC$. Let $(M_1), (M_2)$ be 2 centrally similar circles on centrally similar diameters $IJ \\sim Q_1Q_2$ with the similarity center B. Since the right angle triangle $\\triangle IMJ$ with the circumcircle $(M_1)$ is isosceles with IM = JM, the line BM cuts the corresponding arcs $JI, Q_1Q_2$ of the circles $(M_1), (M_2)$ at their midpoints M, S', the right angle triangle $\\triangle Q_1S'Q_2$ is also isosceles with $Q_1S' = Q_2S'$ and the angles $\\angle Q_2Q_1S = \\angle Q_1Q_2S' = 45^o$. Since the angles $\\angle Q_1XQ_2 = \\angle Q_1ZQ_2 = 90^o$ formed by the perpendicular lines $Q_1X \\equiv FC \\perp IB \\equiv Q_2X$ and $Q_1Z \\equiv JB \\perp AG \\equiv Q_2Z$ are both right, the points $X, Z$ both lie on the circle $(M_2)$. Since the angle $\\angle AZK \\equiv \\angle Q_2ZS = 45^o$, this angle spans the same arc $Q_2S'$ as the angle $\\angle Q_2Q_1S' = 45^o$ and as a result, the points $S' \\equiv S$ are identical. Thus we proved that the triangle $\\triangle Q_1SQ_2$ is isosceles with the angle $\\angle Q_1SQ_2 = 90^o$ and $Q_1S = Q_2S$. In a similar way, we could prove that the triangles $\\triangle P_1SP_2,\\ \\triangle R_1SR_2$ are also isosceles with the angles $\\angle P_1SP_2 = \\angle R_1SR_2 = 90^o$ and $P_1S = P_2S,\\ R_1S = R_2S$. As a result, the triangle $\\triangle P_2Q_2R_2$ is obtained from the triangle $\\triangle P_1Q_1R_1$ by a rotation around the center S by the angle $\\angle P_1SP_2 = \\angle Q_1SQ_2 = \\angle R_1SR_2 = 90^o$, which means that these 2 triangles are congruent.\r\n\r\nYetti", "Solution_2": "thats crazy, can't we use analytic?", "Solution_3": "Dear Mathlinkers,\nan article concerning the Vecten's figure and its developpement can be seeing on my site\nhttp://perso.orange.fr/jl.ayme , la figure de Vecten vol. 5 p. 40\nSincerely\nJean-Louis" } { "Tag": [ "trigonometry", "geometry proposed", "geometry" ], "Problem": "Let $ D$ be a point on the side $ AC$ of $ \\triangle{ABC}$.Let $ E$ and $ F$ be points on the segment $ BD$ and $ BC$,respectively,such that $ \\hat{BAE}\\equal{}\\hat{CAF}$.Let $ P$ and $ Q$ be points on $ BC$ and $ BD$,respectively,such that $ EP \\parallel{} DC$ and $ FQ \\parallel{} CD$.Prove that $ \\hat{BAP}\\equal{}\\hat{CAQ}$", "Solution_1": "Apply C\u00e9va's theorem for: $ \\triangle ABI$ with $ BE,IE,AD$ concurrent we obtain:\r\n$ \\frac{\\sin{\\widehat{DAB}}}{\\sin{\\widehat{DAC}}}.\\frac{\\sin{\\widehat{EIA}}}{\\sin{\\widehat{EIB}}}.\\frac{\\sin{\\widehat{EBI}}}{\\sin{\\widehat{EBA}}}\\equal{}1$\r\n\r\nApply C\u00e9va's theorem for $ \\triangle ABF$ with $ AD,BJ,FJ$ concurrent, we obtain:\r\n$ \\frac{\\sin{\\widehat{DAB}}}{\\sin{\\widehat{DAC}}}.\\frac{\\sin{\\widehat{JFA}}}{\\sin{\\widehat{JFB}}}.\\frac{\\sin{\\widehat{JBF}}}{\\sin{\\widehat{JBA}}}\\equal{}1$\r\n\r\nCompare these two expressions above, with notice that $ \\widehat{EIA}\\equal{}\\widehat{JFA}\\equal{}\\widehat{ACB}$ and $ \\widehat{EBA}\\equal{}\\widehat{JFB}$, we obtain:\r\n$ \\frac{\\sin{\\widehat{EBI}}}{\\sin{\\widehat{EIB}}}\\equal{}\\frac{\\sin{\\widehat{JBF}}}{\\sin{\\widehat{JBA}}}$\r\n\r\nOn the other hand: $ \\widehat{EBI}\\plus{}\\widehat{EIB}\\equal{}\\widehat{EBI}\\plus{}\\widehat{CBI}\\equal{}\\widehat{EBC}\\equal{}\\widehat{JBA}\\equal{}\\widehat{JBF}\\plus{}\\widehat{JBA}$\r\n\r\nHence, $ \\widehat{EBI}\\equal{}\\widehat{JBF}$ and $ \\widehat{EIB}\\equal{}\\widehat{JBA}$, which ends our proof." } { "Tag": [ "geometry", "incenter", "inradius", "trigonometry", "symmetry", "ratio", "angle bisector" ], "Problem": "$ I_A$ is incenter of $ AB'C'$ where $ A'$ is feet of height from A.Let $ N_A$ is the point where incircle touches $ BC$.\r\nProve that $ N_AI_CN_BI_AN_CI_B$ is a hexagon with all sides equal lengths.", "Solution_1": "[hide]Let $ E$ and $ F$ be the feet of perpendicular from $ B$ and $ C$ to $ AC$ and $ AB$ respectively. Let the incircle of $ \\triangle AEF$ touches the side $ AE$ at $ G$. Also let $ \\angle BAC=2\\alpha$. Note that $ A,I_A,I$ lie on the angle bisector of $ \\angle BAC$, and hence are colllinear. It is well known that $ \\triangle AEF\\sim\\triangle ABC$. Hence $ \\frac{AI_A}{AI}=\\frac{AG}{AN_B}=\\frac{AE+AF-EF}{AC+AB-BC}=\\frac{AE}{AB}=\\cos2\\alpha$. Consider the right triangle $ N_BIA$ and a cevian $ N_BI_A$. Let $ r$ be the inradius of $ \\triangle ABC$. We have $ IN_B=r$, $ IA=\\frac{r}{\\sin\\alpha}$, $ AN_B=\\frac{r}{\\tan\\alpha}$, $ AI_A=\\frac{r\\cos2\\alpha}{\\sin\\alpha}$, $ I_AI=\\frac{r(1-\\cos2\\alpha)}{\\sin\\alpha}=2r\\sin\\alpha$. By Stewart's Theorem, we have \\[ IN_B^2\\cdot AI_A+AN_B^2\\cdot II_A=AI(I_AN_B^2+AI_A\\cdot II_A)\\]\n\n\\[ r^2\\frac{r\\cos2\\alpha}{\\sin\\alpha}+\\frac{r^2}{\\tan^2\\alpha}2r\\sin\\alpha=\\frac{r}{\\sin\\alpha}(I_AN_B^2+\\frac{r\\cos2\\alpha}{\\sin\\alpha}2r\\sin\\alpha)\\]\n\n\\[ \\frac{r^3\\cos2\\alpha}{\\sin\\alpha}+\\frac{2r^3\\cos^2\\alpha}{\\sin\\alpha}=\\frac{r}{\\sin\\alpha}(I_AN_B^2+2r^2\\cos2\\alpha)\\]\n\n\\[ r^2\\cos2\\alpha+2r^2\\cos^2\\alpha=I_AN_B^2+2r^2\\cos2\\alpha\\]\n\n\\[ r^2(2\\cos^2\\alpha-\\cos2\\alpha)=I_AN_B^2\\]\n\n\\[ r^2=I_AN_B^2\\]\n\nHence $ I_AN_B=r$. By symmetry, we also have $ N_AI_C=I_CN_B=N_BI_A=I_AN_C=N_CI_B=I_BN_A=r$. Q.E.D.[/hide]", "Solution_2": "If $ AA'$ and $ BB'$ are altitudes ($ A'$ and $ B'$ on $ BC$, $ AC$ respectively), it is well known that $ \\frac{A'B'}{AB}\\equal{} cos(\\angle C)$, and this is the similitude ratio of the triangles $ CA'B'$ and $ CAB$, hence we have to prove that $ \\frac{CI_C}{CI}\\equal{}cos(\\angle C)$ ( 1 ), $ I_C$ being the symmetrical of $ I$ about $ N_AN_B$.\r\nLet $ P\\equal{}CI\\cap N_AN_B$. If $ IN_A\\equal{}r$ (inradius), from the right-angled $ \\triangle IN_AC$ we have $ r^2\\equal{}IP\\cdot IC$ ( 2 ).\r\nBut $ CI_C\\equal{}IC\\minus{}2\\cdot IP$ ( 3 ) and, with (2), $ \\frac{CI_C}{IC}\\equal{}\\frac {IC\\minus{}\\frac {2\\cdot r^2}{IC}}{IC}$ $ \\equal{}1\\minus{}2\\cdot \\frac {r^2}{IC^2}\\equal{}1\\minus{}2\\cdot cos^2 \\left( \\frac {\\angle C}{2} \\right)\\equal{}cos (\\angle C)$, hence $ I_C$ IS the incenter of $ \\triangle CA'B'$ and $ N_AI_C\\equal{}N_BI_C\\equal{}r$, and similarly for the other vertices.\r\n\r\nBest regards,\r\nsunken rock" } { "Tag": [], "Problem": "1) Find the lowest integer the can be written as the sum of 3 consecutive integers, 4 OTHER consecutive integers, and 5 other consecutive integers.\r\n\r\n2) A number that can be written as the product of 2 consecs and 3 consecs...\r\nEither find the lowest one or prove that there are none.", "Solution_1": "1.\r\n[hide]if by \"OTHER\" you mean non-overlapping, then it would be 90. Otherwise it would be 30.[/hide]", "Solution_2": "Then, since it is integers it would be $-90$", "Solution_3": "The product of four consecutive integers can't be negative though.\r\nClosest it can get is 0. ;)", "Solution_4": "Were talking sums here :)", "Solution_5": "2. [hide=\"Click here for solution\"]\n1 x 2 x 3 = 2 x 3 = 6\n\nx=a(a+1)\nx=b(b+1)(b+2)\n\nIt was kind of a guess n' check process, since, for example, 12 = 2x6 = 2x2x3 where 2x3=6, since it is high likely that wither a=b, a=b+1, or a=b+2.\nThrough some figuring out, only a=b+2 works out to be an integer, where b=1 and a=2, thus 1x2x3=2x3=6.\n[/hide]", "Solution_6": "Oh, hehe. :blush: I guess I mixed up the two parts on accident.\r\n\r\nBut anyways, for #1, I think positive is implied, or any negative multiple of 30 would work once you get past -90.\r\n\r\nAnd same for #2, since 0 can be expressed as the product (and sum) of a whole bunch of consecutive numbers." } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Prove that in any triangle $ ABC$ we have\r\n\r\n$ sin^22A\\plus{}sin^22B\\plus{}sin^22C \\ge 36(r/R)^4$", "Solution_1": "[quote=\"Nebraska boy\"]Prove that in any triangle $ ABC$ we have\n\n$ sin^22A \\plus{} sin^22B \\plus{} sin^22C \\ge 36(r/R)^4$[/quote]\r\n\r\n$ sin^22A \\plus{} sin^22B \\plus{} sin^22C \\ge \\frac{1}{3}(sin 2A \\plus{} sin 2B \\plus{} sin 2C)^2\\equal{}\\frac{4s^2r^2}{3R^4} \\ge$ $ 18 (\\frac{r}{R})^3 \\ge 36(\\frac{r}{R})^4$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a;b;c$ be positive real numbers.Prove that:\r\n$\\sqrt[n]{\\frac{a}{b+c}}+\\sqrt[n]{\\frac{b}{a+c}}+\\sqrt[n]{\\frac{c}{a+b}}\\geq 2$\r\n$(n \\geq 2)$", "Solution_1": "I'm sorry but this inequality is not your. It's too old and I remember that I posted once longtime ago!", "Solution_2": "$n_{min}=\\frac{ln2}{ln3-ln2}$\r\nI have a problem more difficult than yours:\r\nLet a,b,c,d$\\geq 0 and n \\geq n_{min}=\\frac{ln2}{ln3-ln2}$.Prove that:\r\n $(\\frac{a}{b+c+d})^{\\frac{1}{n}}+(\\frac{b}{c+d+a})^{\\frac{1}{n}}+(\\frac{c}{d+a+b})^{\\frac{1}{n}}+(\\frac{d}{a+b+c})^{\\frac{1}{n}}\\geq 2$\r\nEven,I have a similar problem which has n variables!", "Solution_3": "Yes, not you :lol: First, not be \"$\\geq$\". It's must be \">\"\r\nSuppose $a\\geq b\\geq c>0$ we prove:\r\n$\\sqrt[n]{\\frac{a}{b+c}}+\\sqrt[n]{\\frac{b}{c+a}}+\\sqrt[n]{\\frac{c}{a+b}}>\\sqrt[n]{\\frac{a+c}{b+c}}+\\sqrt[n]{\\frac{b+c}{c+a}}$ (*) \r\n (*) Equivalent:\r\n$\\sqrt[n]{\\frac{c}{a+b}}>\\frac{\\sqrt[n]{a+c}-\\sqrt[n]{a}}{\\sqrt[n]{b+c}}+\\frac{\\sqrt[n]{b+c}-\\sqrt[n]{a}}{\\sqrt[n]{c+a}}$(**)\r\nWe have:\r\n$\\sqrt[n]{a+c}-\\sqrt[n]{a}=\\frac{a+c-a}{(\\sqrt[n]{a+c})^{n-1}+(\\sqrt[n]{a+c})^{n-2}\\sqrt[n]{a}+...+(\\sqrt[n]{a})^{n-1}}<\\frac{c}{n\\sqrt[n]{a^{n-1}}}$\r\n$\\Rightarrow \\frac{\\sqrt[n]{a+c}-\\sqrt[n]{a}}{\\sqrt[n]{b+c}}<\\frac{c}{n\\sqrt[n]{a^{n-1}}}.\\frac{1}{\\sqrt[n]{b+c}}=\\frac{c}{n\\sqrt[n]{a^{n-1}(b+c)}$\r\nBecause $a^{n-1}(b+c)=a^{n-2}(b+c)=a^{n-2}a(b+c)\\geq c^{n-2}c(a+b)=c^{n-1}(a+b)$ so $\\frac{\\sqrt[n]{a+c}-\\sqrt[n]{a}}{\\sqrt[n]{b+c}}<\\frac{c}{n\\sqrt[n]{a^{n-1}(b+c)}}\\leq \\frac{c}{n\\sqrt[n]{c^{n-1}(a+b)}}=\\frac{1}{n}.\\sqrt[n]{\\frac{c}{a+b}}$\r\nSimilar $\\frac{\\sqrt[n]{b+c}-\\sqrt[n]{a}}{\\sqrt[n]{c+a}}\\leq \\frac{1}{n}.\\sqrt[n]{\\frac{c}{a+b}}$\r\nFrom 2 inequality we have:\r\n$\\frac{\\sqrt[n]{a+c}-\\sqrt[n]{a}}{\\sqrt[n]{b+c}}+\\frac{\\sqrt[n]{b+c}-\\sqrt[n]{a}}{\\sqrt[n]{c+a}}<\\frac{2}{n}\\sqrt[n]{\\frac{c}{a+b}}\\leq \\sqrt[n]{\\frac{c}{a+b}}$\r\nSo (**) is right." } { "Tag": [ "conics", "ellipse" ], "Problem": "The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=5$ is:\r\n\r\n$\\text{(A)} \\ 0 \\qquad \\text{(B)} \\ 1 \\qquad \\text{(C)} \\ 2 \\qquad \\text{(D)} \\ 3 \\qquad \\text{(E)} \\ 4$", "Solution_1": "I have a hint, cause I like this problem. :)\r\n\r\n[hide]Try adding to two equations. You get a nice circle.\n\nWhat's special about it?[/hide]", "Solution_2": "Is it that they their center is on the origin?", "Solution_3": "it is not that hard to just solve the system...", "Solution_4": "[quote=\"Altheman\"]it is not that hard to just solve the system...[/quote]Yeah, but where's the fun it that? :D", "Solution_5": "what does it mean by \"distinct points\"\r\na circle results from manipulating equations but no answer is one of the choices", "Solution_6": "$x^2+4y^2-1=0$\r\nMultiplying the second equation by 4 gives\r\n$16x^2+4y^2-20=0$\r\n\r\nSubtracting the first from the second:\r\n$x^2=\\frac{19}{15}$\r\nSo $x$ can have 2 values\r\nAnswer [b]$2$[/b]\r\n\r\n[b]BUT[/b] this of course means $y$ has to be complex. So on the real axes, there are [b]NO[/b] distinct points.", "Solution_7": "[hide]so we add them, and get $x^2+y^2=\\frac{6}{5}$. if some $x$ and $y$ satisfied those two original equations, they satisfy the circle equation too. \n\nbut if you subtract this equation from $x^2+4y^2=1$, you get $3y^2=\\frac{-1}{5}$, so $y$ is imaginary. so no real solutions.[/hide]", "Solution_8": "[quote=\"4everwise\"][quote=\"Altheman\"]it is not that hard to just solve the system...[/quote]Yeah, but where's the fun it that? :D[/quote]\r\n\r\nI would do the same. But you are right: there is no fun.\r\nBut what is the logic to add the two equations? We get the expression of a circle.", "Solution_9": "[quote=\"4everwise\"][quote=\"Altheman\"]it is not that hard to just solve the system...[/quote]Yeah, but where's the fun it that? :D[/quote]\r\n\r\nMath is never fun :P \r\n\r\nwhat's so special about that circle you mentioned in your hint?", "Solution_10": "Sorry to post on this old topic,\r\n\r\nbut according to [b]Contest Problem Book III[/b] published by the MAA, the question should be as follows:\r\n\r\n[hide=\"original\"]\n\nThe number of distinct points common to the cruve $ x^2\\plus{}4y^2\\equal{}1$ and $ 4x^2\\plus{}y^2\\equal{}4$ is:\n\nA. 0\nB. 1\nC. 2\nD. 3\nE. 4\n\n[/hide][/b]", "Solution_11": "in that case, you can just graph the eclipses and its pretty clear that there are two solutions at x=$ \\pm 1$ and y=0", "Solution_12": "You mean ellipses.\r\n\r\nAn eclipse is when the Earth passes between the moon and the Sun (Lunar), or the moon passes between the Earth and the Sun (Solar).", "Solution_13": "[quote=\"fishythefish\"]You mean ellipses.\n\nAn eclipse is when the Earth passes between the moon and the Sun (Lunar), or the moon passes between the Earth and the Sun (Solar).[/quote]\r\n\r\nthank you so much" } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "$\\sum (tg(A/2))^4tg(B/2)+1/3\\sum tg(A/2)\\ge 2/\\sqrt{3}$.", "Solution_1": "what's conditions of $A$ and $B$ ? \r\n$A=B=0$\r\n$0\\geq\\frac{2}{\\sqrt{3}}$", "Solution_2": "Sorry! $ABC$ is triangle" } { "Tag": [], "Problem": "By placing a 2 at both ends of a number its value increases by 2137.what is the number", "Solution_1": "[hide]I don't really know how to formally prove this first part, but you know that we're looking for a 2-digit number. So when a two is added to the front and back, the number will be 2_ _2 and will increase by around $ 2000$. That said, call the original number $ X$\n\n$ 2000\\plus{}10X\\plus{}2\\equal{}2137\\plus{}X \\Longrightarrow\n9X\\equal{}135 \\Longrightarrow\nX\\equal{}15$[/hide]" } { "Tag": [ "inequalities", "logarithms", "number theory unsolved", "number theory" ], "Problem": "Show that there exists an absolute constant $a>1$ with the property that for any $n$ and any system of natural numbers $ x_1 a^n$.", "Solution_1": "can $n$ be equal to $2$ ? ( $x_1=1$ and $x_2=2$ implies $a=1$)", "Solution_2": "What about $x_1=1,x_2=2,x_3=3$? The estimate needs to be more complicated than you stated.", "Solution_3": "Sorry, indeed, we must assume that $x_1>1$.", "Solution_4": "But then we can choose $x_1=2,x_2=4,x_3=6$.Are you sure that we don't need any additon? :?", "Solution_5": "So what, in this case $ a=\\sqrt[10]{2}$ works very well. I don't understand your question.", "Solution_6": "I'm sorry,I thought $a\\in\\mathbb N$ :blush: .Nice looking-problem!", "Solution_7": "Here's a solution:\r\n\r\nStep 1: Since $x_k-x_1|2x_1$ for all $k$, $x_n\\le3x_1.$\r\nStep 2: For each integer $m,$ at most two of the $x_k$ can be divisible by $3^m$ but not by $3^{m+1}.$ If two of the $x_k$ are congruent mod $3^{m+1},$ both must be divisible by $3^{m+1}.$\r\n\r\nIf $n=2m+1$, there is at least one $x_k$ which is divisible by $3^m$ and hence greater than or equal to $3^m.$ If $n=2m+2,$ there are at least two $x_k$ divisible by $3^m$ and therefore one is at least $2\\cdot3^m.$ Combining these two cases, $x_n\\ge3^{\\frac{n-1}2}.$ Using the inequality in the first step, $x_1\\ge3^{\\frac{n-3}2}.$\r\nSince we restrict to cases where $x_1\\ge2$, we ignore all $n\\le3.$ For $n\\ge4,$ $n-3\\ge\\frac n4,$ so $x_1\\ge\\sqrt[8]{3}^n.$ Equality is only possible for $n=4$; since $2>\\sqrt{3}$ this does not occur.\r\n\r\nThe maximal $a$ for which $x_1\\ge a^n$ is in fact $\\sqrt[4]{2}$. We can show that for $n\\ge5$, $x_1\\ge3^{\\frac n5}$, and $\\sqrt[5]{3}>\\sqrt[4]{2}.$ For $n=4,$ we have the example $2,3,4,6.$", "Solution_8": "Jmerry's nice ideas can be used to prove a stronger lower bound $x_1\\geqslant e^{cn\\ln n}$ for some $c>0$. To prove it, one has to run jmerry's argument for all primes $p$ between $3$ and $n/2$. The result is that there are not more than $p-1$ numbers not divisible by $p$, not more than $p-1$ numbers divisible by $p$ but not by $p^2$ and so on. So, the average power of $p$ in $x_j$ is at least $c\\frac n{p}$. Thus, there exists at least one $x_j$ that is not less than\r\n$\\prod_{p\\text{ prime},3\\leqslant p\\leqslant n/2}p^{c n/p} =\\exp\\left\\{cn\\sum_{p\\text{ prime},3\\leqslant p\\leqslant n/2}\\frac{\\ln p}p\\right\\}$ but the last sum is known to be approximately $\\ln n$. Since all $x_j$ must be the same up to a factor not exceeding $3$, the inequality follows for large $n$. I don't know whether this new lower bound is the best possible one." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all pairs of prime numbers $ (p,q)$ such that the number $ (p\\plus{}1)^q$ is a perfect square.", "Solution_1": "If $ q\\equal{}2$ then the number $ (p\\plus{}1)^q$ is always a square for every $ p$. If $ q>2$ then $ (p\\plus{}1)^q$ is a perfect square only if $ p\\plus{}1$ is a perfect square (remenber that $ q$ is odd): $ p\\plus{}1\\equal{}n^2$, namely $ p\\equal{}n^2\\minus{}1$, $ p\\equal{}(n\\plus{}1)(n\\minus{}1)$. Since $ p$ must be prime, $ n\\minus{}1\\equal{}1$ and $ n\\equal{}2$ and $ p\\equal{}3$. Thus, the solutions are $ (\\forall p,2)$ and $ (3,\\forall q)$." } { "Tag": [], "Problem": "Is there a fast way to do this problem without actually graphing the lines?\r\n(If not please still give a solution.) \r\n\r\nPlanet Nowt and Sonemenid have two-dimensional orbits, described by the equations x squared + y squared= 1296 and x squared +y squared= 4225, respectively. How many more units does planet Sonemenid travel than plant Nowt in one orbit.", "Solution_1": "[hide=\"Hint\"]\nRemember, $x^2+y^2=r^2$ describes a circle centered at the origin with radius $r$.\n[/hide]", "Solution_2": "[hide]They're equations of circle. $x^2+y^2=a^2$ where a is radius.\n\nradius of Nowt: $\\sqrt{1296}=36$\n\nradius of Sonemenid: $\\sqrt{4225}=65$\n\ncircumference of Nowt orbit$=72\\pi$\n\ncircumference of Sonemenid orbit$=130\\pi$\n\n$130\\pi-72\\pi=58\\pi$[/hide]", "Solution_3": "Haha this did some \"WAKE UP\" to me. I thought i was messing with an linear equation" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ a,b,c>0$ and $ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\equal{}1$ prove: \r\n\r\n$ \\frac{a}{1\\minus{}a^{3}}\\plus{}\\frac{b}{1\\minus{}b^{3}}\\plus{}\\frac{c}{1\\minus{}c^{3}}\\geq\\frac{4\\sqrt[3]{4}}{3}$", "Solution_1": "I wonder when the equality occurs. It seems that you use analytic tools to get the inequality." } { "Tag": [ "trigonometry", "inequalities", "algebra", "polynomial", "function", "combinatorics proposed", "combinatorics" ], "Problem": "Prove that\r\n$ \\sum_{k\\geq 0}\\binom{2n}{k}f_{2k}\\equal{}5^nf_{2n}$\r\nWhere $ f_{0}\\equal{}f_1\\equal{}1$ and $ f_{n\\plus{}1}\\equal{}f_n\\plus{}f_{n\\minus{}1}$", "Solution_1": "\\begin{eqnarray*}\r\n(\\cos \\theta + 1 + i \\sin \\theta)^n\r\n& = & (1 + e^{i\\theta})^n \\\\\r\n& = & \\sum_{k=0}^n\\binom{n}{k}e^{ik\\theta} \\\\\r\n& = & \\sum_{k=0}^n\\binom{n}{k}\\cos k\\theta + i\\sum_{k=0}^n\\binom{n}{k} \\sin k\\theta\r\n\\end{eqnarray*}\r\n\r\nLet, $ x=2\\cos \\theta$,\r\n\\[ (x+2+\\sqrt{x^2-4})^n = 2^{n-1}\\left( \\sum_{k=0}^n\\binom{n}{k} t_k(x) + \\sqrt{x^2-4}\\sum_{k=0}^n\\binom{n}{k}u_{k-1}(x)\\right).\\]\r\n\r\n$ \\left(\\begin{array}{l}\r\nt_n(x)=2T_n\\left( \\frac{x}{2} \\right), u_n(x)=U_n\\left( \\frac{x}{2} \\right) \\\\\r\n\\text{and} \\\\\r\nT_n(x): \\text{Chebyshev Polynomial of the First Kind}, \\\\\r\nU_n(x): \\text{Chebyshev Polynomial of the Second Kind}\r\n\\end{array}\\right)$\r\n\r\nAnd let $ x=3$,\r\n\\begin{eqnarray*}\r\n2^{n-1}\\left( \\sum_{k=0}^n\\binom{n}{k} L_{2k} + \\sqrt{5}\\sum_{k=0}^n\\binom{n}{k} F_{2k} \\right)\r\n& = & (5+\\sqrt{5})^n \\\\\r\n& = & \\left(2\\sqrt{5} \\cdot \\frac{1+\\sqrt{5}}{2}\\right) \\\\\r\n& = & 2^n \\cdot (\\sqrt{5})^n \\cdot \\left(\\frac{1+\\sqrt{5}}{2}\\right)^n \\\\\r\n& = & 2^n \\cdot (\\sqrt{5})^n \\cdot \\frac{L_{n} + \\sqrt{5}F_{n}}{2}\r\n\\end{eqnarray*}\r\n\r\nThus,\r\n\\[ \\sum_{k=0}^n\\binom{n}{k} L_{2k}\r\n=\\left\\{\\begin{array}{ll}\r\n5^{n/2}L_{n} & \\text{(n: even)} \\\\\r\n5^{(n+1)/2}F_{n} & \\text{(n: odd)}\r\n\\end{array}\\right.\\]\r\n\r\n\\[ \\sum_{k=0}^n\\binom{n}{k} F_{2k}\r\n=\\left\\{\\begin{array}{ll}\r\n5^{n/2}F_{n} & \\text{(n: even)} \\\\\r\n5^{(n-1)/2}L_{n} & \\text{(n: odd)}\r\n\\end{array}\\right.\\]\r\nAnd,\r\n\\[ \\sum_{k=0}^n\\binom{n}{k} L_{k^2}\r\n=\\left\\{\\begin{array}{ll}\r\n5^{n/2}L_{n} & \\text{(n: even)} \\\\\r\n5^{(n+1)/2}F_{n} & \\text{(n: odd)}\r\n\\end{array}\\right. \\ \\left(\\because \\ L_{2n} = L_{n}^2+(-1)^{n-1}2, \\ \\sum_{k=0}^n\\binom{n}{k}(-1)^k = (1-1)^k = 0\\right)\\]\r\n\r\n\r\n\\[ \\sum_{k=0}^n\\binom{n}{k} F_{k}^2\r\n=\\left\\{\\begin{array}{ll}\r\n5^{(n-2)/2}L_{n} & \\text{(n: even)} \\\\\r\n5^{(n-1)/2}F_{n} & \\text{(n: odd)}\r\n\\end{array}\\right. \\ \\left(\\because \\ L_{n}^2 = 5F_{n}^2+(-1)^n 4\\right)\\]", "Solution_2": "Let $ F(x) \\equal{} \\sum f_{2k} \\frac {x^k}{k!} \\equal{} \\frac {e^{\\phi^2 x} \\minus{} e^{\\varphi^2 x}}{\\phi \\minus{} \\varphi}$. The generating function of the desired sequence is (see the lemma [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=634]here[/url])\r\n\r\n$ e^x F(x) \\equal{} \\frac {e^{ (\\phi^2 \\plus{} 1)x} \\minus{} e^{(\\varphi^2 \\plus{} 1)x} }{\\phi \\minus{} \\varphi}$.\r\n\r\nNow note that $ \\phi^2 \\plus{} 1 \\equal{} \\phi \\plus{} 2 \\equal{} \\sqrt {5} \\phi$ and the same for $ \\varphi$ and we have our result.\r\n\r\nEdit: I'd be interested in a combinatorial proof using tilings, though.", "Solution_3": "There exists one! :wink:" } { "Tag": [ "geometry", "trapezoid", "combinatorial geometry", "circles", "IMO Shortlist" ], "Problem": "In the plane we are given two circles intersecting at $ X$ and $ Y$. Prove that there exist four points with the following property:\r\n\r\n(P) For every circle touching the two given circles at $ A$ and $ B$, and meeting the line $ XY$ at $ C$ and $ D$, each of the lines $ AC$, $ AD$, $ BC$, $ BD$ passes through one of these points.", "Solution_1": "- Let $ \\omega_1$ and $ \\omega_2$ be the given circles and $ w$ is tangent to $ \\omega_1$,$ \\omega_2$ at $ B,A$,respectively.Note that $ w$ is tangent either internally,or externally to both of $ \\omega_1$ and $ \\omega_2$,otherwise it can not be intersected by $ XY$.Suppose that $ w$ is externally tangent to $ \\omega_1$ and $ \\omega_2$. Denote intersection points of $ XY$ with $ w$,as $ C$ and $ D$,such that $ D$ lies between $ l_1$ and $ l_2$ and $ C$ doesn't lie there and distance between $ C$ and $ l_2$ is greater than distance between $ C$ and $ l_1$.Let $ K\\in DB\\cap\\omega_1$,$ N\\in DA\\cap\\omega_2$,$ M\\in CA\\cap\\omega_2$ and $ L\\in CB\\cap\\omega_1$.Now if we prove that $ KN$ and $ LM$ are common tangent lines to $ \\omega_1$ and $ \\omega_2$,we will be done.\r\n-Consider the homothety with center at $ B$,which turns $ w$ to $ \\omega_1$.Then obviously $ D\\rightarrow K$ and $ C\\rightarrow L$,thus $ KL\\| CD$.As we know $ CB\\cdot CL \\equal{} CX\\cdot CY \\equal{} CA\\cdot CM$,hence quadrilateral $ CBLM$ is cyclic,so $ \\widehat{MLB} \\equal{} \\widehat{BAC} \\equal{} \\widehat{LKB}$.From where we conclude that $ LM$ is tangent to $ \\omega_1$.In similar way we prove that $ LM$ is also tangent to $ \\omega_2$,so $ LM$ and $ KN$ are common tangent lines to $ \\omega_1$ and $ \\omega_2$ and obviously $ K,L,M,N$ are fixed points.So we are done.\r\n-The case when $ w$ is internally tangent to $ \\omega_1$ and $ \\omega_2$ is totally similar to the previous case.", "Solution_2": "I want to understand your solution and I don't understand who's $ l_1$ and $ l_2$. Please answer me !", "Solution_3": "Here's a fix for the typos (I think this solution is the same?)\r\nLet CA intersect the circle tangent at A again at E, CB intersect the other circle again at F. Let BA intersect the circle containing E again at B'. EB'A ~ CBA, so mABC=mEB'A. C is on the radical axis of the 2 circles, so EABF is cyclic, and we have mABC=mFEA, and EF is a tangent to the circle containing E.\r\n\r\nApplying the same argument 4 times over, or extending DA, DB and taking note of the chords being parallel to the radical axis gives us that the given lines intersect the 2 circles at the points of tangency of the common tangents.", "Solution_4": "Too many circles. Invert!\r\n\r\nInvert about a circle centered at X. Then the the two circles intersecting at X become lines intersecting the radical axis at Y.\r\n\r\nEasy to show that AC always makes the same angle with the radical axis since A,B,C,D move uniformly with the circle, so then we can easily construct one of our points P so as to make a trapezoid XPAC, where P lies on what is now the inverted first circle. Then XPAC is cyclic, and is always cyclic, corresponding to the desired condition in the original diagram. Repeat with the other 3 lines/circles in the inverted diagram, and we have the result.", "Solution_5": "Hm... 90% perspiration (drawing the diagram :P), 10% inspiration (oh yeah, the four points obviously are the common tangents' tangency points :D)\n\nLet $MN$ be a common tangent of the two given circles $\\omega_1$ and $\\omega_2$, and call $\\omega$ the third circle. Homotheties taking $\\omega_1$, $\\omega_2$ to $\\omega$ take $M, N$ to $D_1, D_2$ respectively and $MN$ to a tangent of $\\omega$ parallel to $MN$, $l$. Because there is only one such tangent $l$, we have $D_1 = D_2$.\n\nIf $F$ is a point on $l$ such that $\\angle MND = \\angle NDF$, then incidentally $\\angle D_1 AB = \\angle ND_1 F$, so $MNBA$ is cyclic. Thus $D_1$ lies on $XY$ by radical axis theorem, so $D_1 = D$ or $C$. Similar result holds for other common tangent.", "Solution_6": "[hide=Solution]\nWe claim that the four points are the tangency points of the two common external tangents with each circle.\n\nLet \\(H=AC\\cap \\omega_1\\), \\(I=BC\\cap \\omega_2\\), \\(F=AD\\cap \\omega_1\\), \\(G=BD\\cap \\omega_2\\). We show that \\(HI\\) and \\(FG\\) are the common external tangents.\n\nFirst note that due to homothety, \\(IG\\), \\(HF\\) and \\(CD\\) are parallel. Now, observe that\n\\begin{align*}\n\\measuredangle GFH &= \\measuredangle AFH + \\measuredangle GFD \\\\\n&= \\measuredangle ADC + \\measuredangle DBA \\\\\n&= \\measuredangle ABC + \\measuredangle DBA \\\\\n&= \\measuredangle DAC = \\measuredangle FAH \\\\\n\\end{align*}\nso that \\(FG\\) is tangent to \\(\\omega_1\\). We can similarly prove the other tangencies, so we are done.\n[/hide]", "Solution_7": "[hide=Solution][quote=ISL 2000 G1]In the plane we are given two circles intersecting at $ X$ and $ Y$. Prove that there exist four points with the following property:\n\n(P) For every circle touching the two given circles at $ A$ and $ B$, and meeting the line $ XY$ at $ C$ and $ D$, each of the lines $ AC$, $ AD$, $ BC$, $ BD$ passes through one of these points.[/quote]\n[b][color=#000]Solution:[/color][/b] Let the given circles be $\\omega_1, \\omega_2$. Define: $AC \\cap \\omega_1$ $=$ $K$, $BC $ $\\cap$ $ \\omega_2$ $=$ $L$, $BD$ $\\cap$ $\\omega_2$ $=$ $M$ and $AD$ $\\cap$ $\\omega_1$ $=$ $N$\n--------------------------------------------------\nBy Homothety at $A,B$, $NK||XY||LM$ and using the fact that $KABL$ is cyclic ($\\because CA \\times CK=CX \\times CY$ $=$ $CB$ $\\times$ $CL$),\n$$\\angle AKL=\\angle CBA=\\angle CDA=\\angle CNK$$\nAnd similarly $\\angle BLK=\\angle BML$ $\\implies$ we get $KL$ is the common tangent of $\\omega_1, \\omega_2$. Similarly, show $MNAB$ is cyclic and using this $\\angle MNA=\\angle AKN$ & $\\angle NMB=\\angle MLB$ $\\implies$ $MN$ is the other common tangent of $\\omega_1,$ $\\omega_2$. Hence, as a result $M,N,K,L$ are fixed $\\qquad \\blacksquare$\n[/hide]\n", "Solution_8": "No one has posted a diagram yet so I'll post mine(my solution uses the same ideas):\n\n[asy]\nsize(10cm);\npath circ1, circ2, dcirc1, dcirc2, circ, radaxis;\nreal r1=5, r2=4.2, h=2, r=3.8, e=.7; \npair O1= (-sqrt(r1^2-h^2),0), O2=(sqrt(r2^2-h^2),0); \ncirc1=circle(O1,r1);\ncirc2=circle(O2,r2);\ndcirc1=circle(O1,r1+r);\ndcirc2=circle(O2,r2+r);\npair [] x=intersectionpoints(circ1, circ2);\nradaxis=(0,10^7) -- (0,-10^7);\npair [] z=intersectionpoints(dcirc1, dcirc2);\ncirc=circle(z[1],r); \npair A=r1*dir(z[1]-O1)+O1, B=r2*dir(z[1]-O2)+O2;\npair [] c=intersectionpoints(radaxis, circ);\npair H=(r1*O2-r2*O1)/(r1-r2); \npair M1=conj((O1-H)^2/(O1+(0,r1)-H))+H, M2=conj((O2-H)^2/(O2+(0,r2)-H))+H;\npair N2=(O2-H)^2/(O2+(0,r2)-H)+conj(H), N1=(O1-H)^2/(O1+(0,r1)-H)+conj(H);\ndraw(circ1); draw(circ2); draw(x[0]+(0,e) -- c[1]-(0,e),mediumred); draw(circ,heavyblue); draw(M1--M2,heavygray); draw(N2--N1,heavygray); \ndraw(M1--c[1]--M2,heavygreen); draw(N1--c[0]--N2,heavygreen);\ndraw(circumcircle(A,B,M1),dashed+heavycyan);\ndot(\"$A$\", A, 1.5*dir(80)); dot(\"$B$\", B, 1.5*dir(110)); dot(\"$C$\", c[0], dir(310)); dot(\"$D$\", c[1], dir(220)); dot(\"$X$\", x[0], dir(165)); dot(\"$Y$\", x[1], dir(190)); dot(\"$O$\", z[1], dir(60)); dot(\"$M_1$\", M1, dir(120)); dot(\"$N_1$\", N1, dir(260)); dot(\"$M_2$\", M2, dir(40)); dot(\"$N_2$\", N2, dir(300));\n[/asy]\n\nCall the two original circles $\\omega_1$ and $\\omega_2$, with $A$ on $\\omega_1$ and $B$ on $\\omega_2$. Let $M_1$ be the second intersection of $\\overline{AD}$ with $\\omega_1$, and let $M_2$ be the second intersection of $\\overline{BD}$ with $\\omega_2$. Define $N_1$ and $N_2$ analogously, with point $D$ replaced with point $C$. \n\nSince $A$ is the center of the homothety with negative scale factor sending $\\omega$ to $\\omega_1$, and this homothety sends $\\overline{CD}$ to $\\overline{N_1M_1}$, $\\angle ACD=\\angle AN_1M_1$. Notice that $\\overline{XY}$ is the radical axis of $\\omega_1$ and $\\omega_2$, so $DA\\cdot DM_1=DB\\cdot DM_2$ and it follows that quadrilateral $ABM_2M_1$ is cyclic. Thus, \n$$\\angle M_2M_1A=\\angle ABD=\\angle ACD=\\angle AN_1M_1,$$\nso $\\overline{M_1M_2}$ is tangent to $\\omega_1$. We can repeat this three times to find $\\overline{M_1M_2}$ and $\\overline{N_2N_2}$ are tangent to $\\omega_1$ and $\\omega_2$, so points $M_1$, $M_2$, $N_1$, and $N_2$ are fixed. These are the desired four points and we are done. (This is another configuration where the circles are internally tangent, but the same proof still holds.)", "Solution_9": "Let $\\omega_1, \\omega_2$ be the names of the two circles with $A$ corresponding to $\\omega_1$ and $B$ corresponding to $\\omega_2$. Let $E$ and $F$ be the intersection of $BD$ and $BC$ respectively with $\\omega_2$. Lastly, let $O_1,O_2,O_3$ be the centers of $\\omega_1,\\omega_2,$ and the circle tangent to both, respectively. We claim that $E$ and $F$ are fixed, this will solve the problem because we can apply a similar thing to the other two points.\n\nNotice that by homothety triangles $CBD$ and $FBE$ are similar, and we get that $CD\\parallel FE$. In addition, by [url=https://artofproblemsolving.com/community/c5h2484038p20872596]2021 AIME I #13[/url], we get that given the two circles $\\omega_1$ and $\\omega_2$, the angle $\\angle CO_3D$ is fixed, which means that by homothety, the angle $\\angle EO_2F$ is fixed. Therefore minor arc $FE$ of $\\omega_2$ is fixed. Combined with the parallel condition and the fact that $E$ and $F$ lie on $\\omega 2$, this proves that $E$ and $F$ are fixed, and we are done.", "Solution_10": "Let our circles be $\\omega_1$ and $\\omega_2$ and their common tangents touch them at point's $S,T$ and $K,M$ respectively. Let common tangents meet at $P$.\n\nStep1 : $SKBA$ is cyclic.\nBy Homothety of circles $\\omega_1$ and $\\omega_2$ we have $SK$ and $AB$ both meet at $P$ and $\\angle PBK = \\angle PSA$.\n\nStep2 : $C,A,S$ and $C,B,K$ are collinear.\nLet tangents to $\\omega_1$ and $\\omega_2$ at $A$ and $B$ meet at $Q$.\n$\\angle QAC = \\angle ABC = \\angle ASK \\implies C,A,S$ are collinear.\n$\\angle QBC = \\angle BAC = \\angle BKS\\implies C,B,K$ are collinear.\n\nSo now we have $S$ and $K$ are two of those 4 points and same way we can prove $T$ and $M$ are other points.\nwe're Done.", "Solution_11": "Thought inversions first, but unforunately don't know how inversions work.\n\n[img width=50]https://media.discordapp.net/attachments/925784397469331477/949510510099988540/Screen_Shot_2022-03-04_at_8.35.30_PM.png?width=1602&height=1170[/img]\n\nLet the two given circles be $S_1,S_2,$ with the new circle $C$ touching $S_1,S_2$ at $A,B$ respectively. Let $C$ be closer than $D$ to $Y.$ Let $CA$ intersect $S_1$ at $E,$ $CB$ intersect $S_2$ at $F,$ $DA$ intersect $S_1$ at $G$ and $DB$ intersect $S_2$ at $H.$ We claim that $EF,GH$ are always the common tangents.Note that since $C,D$ are on the radical axis of $S_1,S_2,$ $ABFE$ and $ABHG$ are cyclic quadrilaterals.\n\nThus $\\angle GEA=\\angle DCA$ (famous lemma, to prove use homothety about $A$) and so $\\angle DCA=\\angle DBA=\\angle HGA,$ so $HG$ tangent to $S_1$ Similarly, $GH$ tangent to $S_2.$\n\nOn the other hand, $\\angle BHF=\\angle CDB=\\angle CAB=\\angle CFB$ so $EF$ tangent to $S_2,$ and similarly to $S_1,$ as desired.", "Solution_12": "Let $\\measuredangle$ denote a directed angle.\nLet $\\omega_1, \\omega_2$ be the original two circles and let $\\omega$ be a circle that is tangent to them at $A, B$ respectively. We claim that the four distinct points $P_1, P_2, P_3, P_4$ such that $P_1P_2$ and $P_3P_4$ are each tangent to both $\\omega_1$ and $\\omega_2$ satisfy the problem statement.\n\nLet $C$ be an intersection of $XY$ and $\\omega$. Let $A'$ be the second intersection of $AC$ with $\\omega_1$ and let $B'$ be the second intersection of $BC$ with $\\omega_2$. (Note that $CA$ cannot be tangent to $\\omega_1$ at $A$, since then it would have to tangent to $\\omega$ as well.) We claim $A'B'$ is tangent to $\\omega_1$ and $\\omega_2$ at $A'$ and $B'$.\n\nBecause $C$ is on the radical axis of $\\omega_1$ and $\\omega_2$, $CA\\cdot CA'=CB\\cdot CB'$. Therefore, $AA'BB'$ is cyclic. So, we have $\\measuredangle B'A'A=\\measuredangle B'BA=\\measuredangle CBA$. Similarly, $\\measuredangle A'B'B=\\measuredangle CAB$.\n\nLet $O$ be the radical center of the three circles. Note $OA$ is tangent to $\\omega_1, \\omega$ and $OB$ is tangent to $\\omega_2, \\omega$. So, we have $\\measuredangle CAO=\\measuredangle CBA=\\measuredangle B'A'A$. Let $P$ be a point on $\\omega_1$ not equal to $A$ or $A'$. Since $OA$ is tangent to $\\omega_1$, we have $\\measuredangle A'PA=\\measuredangle A'AO=\\measuredangle CAO=\\measuredangle B'A'A$. Since $\\measuredangle A'PA=\\measuredangle B'A'A$, we know $A'B'$ is tangent to $\\omega_1$ at $A'$. By the same logic $A'B'$ is tangent to $\\omega_2$ at $B'$. Thus $A', B'\\in\\{P_1, P_2, P_3, P_4\\}$, which completes the proof since $CA$ and $CB$ pass through $A'$ and $B'$, respectively.", "Solution_13": "Does this work? Not sure if I fakesolved.\n\nConsider the diagram as shown below. I claim that the four points $E$, $F$, $G$, and $H$ which we require are the four points which are the intersections of the two external tangents of the circles, which we call $\\omega_1$ and $\\omega_2$. Also, define $M$ to be the intersection of $XY$ and $GH$. Obviously $A$, $B$, $G$, and $H$ are concyclic which means that $$\\measuredangle DCA=\\measuredangle DBA=\\measuredangle HGD$$ which implies that $A$, $C$, $M$, and $G$ are concyclic. By Reim's Theorem, $CD\\parallel EG$, which implies that $$\\measuredangle HGA=\\measuredangle DCA=\\measuredangle GEA$$ so that $GH$ is tangent to $\\omega_1$. The final result immediately follows. $\\blacksquare$", "Solution_14": "this is extremely similar to 2021 aime i 13 and actually trivializes it (angle between external tangent and line connecting centers is 30 degrees)\nvery nice", "Solution_15": "Let $\\odot (AXY)$ meets rays $AC,AD$ again at $P,Q$ and $\\odot (BXY)$ meets $BC,BD$ again at $R,S.$ Since $C$ lies on radical axis of given circles, points $A,B,P,R$ are concyclic, so by Reim's theorem $\\overline{PR}$ is parallel to tangent to $\\odot (ABC)$ at $C.$ Thus by homothety $PR$ is tangent to $\\odot (AXY),\\odot (BXY),$ which yields that points $P,R$ (and similarly $Q,S$) are fixed." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "(x^2-x+3)/3sq.rt X dx from -1 to -2", "Solution_1": "$\\sqrt{x}$ makes no sense for $-2\\le x\\le-1.$", "Solution_2": "[quote=\"Kent Merryfield\"]$\\sqrt{x}$ makes no sense for $-2\\le x\\le-1.$[/quote]\r\n\r\nit is 3 sq.rt of X or 1/3X", "Solution_3": "Maybe s/he means $\\sqrt[3]{x}$? In which case you just divide through and integrate term by term.", "Solution_4": "If that's the case- it's an improper integral. Don't blindly integrate across a singularity.", "Solution_5": "[quote=\"jmerry\"]If that's the case- it's an improper integral. Don't blindly integrate across a singularity.[/quote]\r\n\r\nI don't follow. Where is the singularity in $\\frac{x^{2}-x+3}{\\sqrt[3]{x}}$ for $-2 \\le x \\le-1$?", "Solution_6": "I misread the limits." } { "Tag": [ "function", "analytic geometry", "LaTeX", "factorial", "algorithm", "induction", "floor function" ], "Problem": "Like in the numbertheory section,zetax,mathmanman,etc have made a collection of all theorems and formulas,any such thing here?", "Solution_1": "Calatan numbers:\r\n\r\n$C_n = \\frac{\\binom{2n}{n}}{n+1}$\r\n$C_0 = 1, C_n = C_0C_{n-1}+C_1C_{n-2}+C_2C_{n-3}+...+C_{n-1}C_0$\r\n$C_1 = 1, C_2 = 2, C_3 = 5, C_4 = 14, C_5 = 42, C_6 = 132$\r\n\r\nBijections of catalan: number of ways to draw diagonals of polygon without crossing, number of ways to fully parenthesize expression, number of paths along a line never going behind starting point and ending at starting point.\r\n\r\nGraphs:\r\n\r\nA graph is Eulerian iff every node has in-degree equal to out-degree, or if by drawing one extra edge we can make this so.\r\n\r\nIn an undirected graph, the condition of even degree for each node is equivalent.", "Solution_2": "So can we start a collection of all formulas,like done for number theory [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76610]here[/url]\r\n\r\nI hope i get lots of cooperation,it'll be pretty cool(if its okay with pbornsztein,megus adn the other mods :) )", "Solution_3": "Let $f$ be a function from $S$ finite to $T$ finite.\r\nWe then have :\r\n\\[ |f(S)| \\le \\min{(|S|,|T|)}. \\]\r\n[i]In particular, if $f$ is injective $|S| \\le |T|$; if it's surjective $|T| \\le |S|$ and if it's bijective $|S| = |T|$.[/i]\r\n\r\nThe number of ways to go from $(0, 0)$ to $(m, n)$ through a path where, in each step, one of the coordinates increases from one unity is :\r\n\\[ \\binom {m+n}m = \\binom {m+n}n. \\]\r\n\r\nThe number of sub-sets with $k$ elements of $\\{1, 2, \\ldots, n\\}$ which does not contain consecutive integers is :\r\n\\[ \\binom {n+1-k}k. \\]", "Solution_4": "Elementary ones...\r\n\r\n\r\n[b]Sum Of Numbers Formed by n non-zero digits[/b]\r\n\r\n$=(Sum\\ of\\ digits)(n-1)!(\\frac{10^n-1}{10-1})$\r\n\r\n\r\n[b]Exponent Of PRIME p In n![/b]\r\n[i]Let[/i] $E_p(n)$ [i]denote the exponent of the prime[/i] $p$ [i]in the positive integer[/i] $n$.Then,\r\n\r\n$E_p(n!)=[\\frac{n}{p}]+[\\frac{n}{p^2}]+\\cdots+[\\frac{n}{p^s}]$\r\n\r\nwhere $s$ is the largest positive integer such that $p^s\\leq n \\leq p^{s+1}$\r\n\r\n[b][size=150]SELECTION OF ONE OR MORE ITEMS[/size][/b]\r\n\r\n[b]Selection from different items[/b]\r\n[i]The number of ways of selecting one or more items from a group of[/i] $n$ [i]distinct items is [/i]$2^n-1$\r\n\r\n[b]Selection from identical items[/b]\r\n\r\n[i]The total number of selections of some or all out of[/i] $p+q+r$ \r\n\r\n[i]items where [/i]$p$ [i]are alike of one kind[/i],$q$ [i]are alike of \n\nsecond kind and rest are alike of third kind is[/i] \r\n\r\n$[(p+1)(q+1)(r+1)]-1$\r\n\r\n[b][size=150]DIVISION OF ITEMS INTO GROUPS[/size][/b]\r\n\r\n[b]Division of items into groups of unequal size[/b]\r\n\r\n[i]The number of ways to distribute [/i]$(m+n+p)$ [i]items among \n\n[/i]$N$ [i]persons in the groups containing [/i]$m,n$ and $p$ [i]items is[/i]\r\n\r\n$\\frac{(m+n+p)!}{m!n!p!}*N!$\r\n\r\n[b]Division of items into groups of equal size[/b]\r\n\r\n[i]The number of ways in which[/i] $mn$ [i]different items can be \n\ndivided equally into[/i] $m$ [i]groups,each containing [/i]$n$ [i]\n\nobjects and the order of the groups is not important is[/i]\r\n\r\n$(\\frac{(mn)!}{(n!)^m})\\frac{1}{m!}$\r\n\r\nwhen order of groups is important\r\n\r\n$\\frac{(mn)!}{(n!)^m}$", "Solution_5": "A friend and I wrote up two lecture-type thingies with a whole lot of formulas. Unfortunately most of the formatting is lost here, so if someone wants to tell me how to get the LaTeX code to be interpreted, I shall be very happy.\r\n\r\nHere they are (from http://www.tjhsst.edu/~dschafer/VMT/CombTut/CombTutExpl.pdf and http://activities.tjhsst.edu/vmt/arml-lectures/2006/11a-advancedcombolecture.pdf):\r\n\r\n\\begin{enumerate}\r\n\r\n\\item How many ways can one arrange the elements of a set of size $n$?\r\n\r\n\\par \\fbox{$\\displaystyle n!$}\r\n\\par You can place the first element in any of $n$ places. You can place the second element in any of $n-1$ places. You can place the $k$th element in any of $n-k+1$ places. Thus there are a total of $n \\cdot (n-1) \\cdot (n-2) \\cdot \\ldots \\cdot 3 \\cdot 2 \\cdot 1 = n!$ possibilities.\r\n\r\n\r\n\\item How many ways can one choose $k$ out of $n$ items (order does matter)?\r\n\r\n\\par \\fbox{$\\displaystyle \\frac{n!}{(n-k)!}$}\r\n\\par Use the same argument as before, except stop just before the $k+1$st element. Then you get $n \\cdot (n-1) \\cdot (n-2) \\cdot \\ldots \\cdot (n-k+1) = \\frac{n!}{(n-k)!}$ possibilities.\r\n\r\n\r\n\\item How many ways can one choose $k$ out of $n$ items (order doesn't matter)?\r\n\\par \\fbox{$\\displaystyle \\frac{n!}{k!(n-k)!}$}\r\n\\par This is the definition of combination, so it is clearly ${n \\choose k} = \\frac{n!}{k!(n-k)!}$. This can be easily derived; there are $n$ choices for the first item, $(n-1)$ for the second, until eventually there are $(n-(k-1))$ for the $k$th. However, there are $k!$ orderings of these $k$ items, there must be $\\frac{n(n-1)\\cdots(n-k+1)}{k!} = \\frac{n!}{k!(n-k)!}$ total combinations.\r\n\r\n\r\n\\item How many sequences $a_1,a_2,\\ldots,a_k$ of positive integers exists such that $a_i < a_{i+1} \\leq n$?\r\n\\par \\fbox{$\\displaystyle {n \\choose k}$}\r\n\\par Clearly, each of the $a_i$ is distinct. Also, given any $k$ distinct numbers, there will be exactly $1$ sequence of $a_i$ with those $k$ numbers in it that satisfies the requirement. This creates a bijection\\footnote{An over-simplified explanation of a bijection is that it is a one-to-one relationship between two sets created by a function. Their main use in combinatorics is that the sizes of the two sets must be the same, so if one set's size is known, both sizes are known.} between sequences $a_1,\\ldots,a_k$ and sets of $k$ numbers chosen from the numbers $1$,$2$,\\ldots,$n$. The number of sets for the latter is clearly ${n \\choose k}$, so there are ${n \\choose k}$ such sequences.\r\n\r\n\r\n\\item How many ways can $n$ distinct items be partitioned into $k$ groups?\r\n\\par \\fbox{$\\displaystyle n^k$}\r\n\\par Consider each item individually. For each item, there are $k$ options of what group to put it in. There are $n$ such choices, each of which is independent, so there are $n^k$ ways to perform the partition.\r\n\r\n\r\n\\item How many ways can one travel from $(0,0)$ to $(x,y)$ traveling only to the right and up and only between adjacent lattice points.\r\n\\par \\fbox{$\\displaystyle {{x+y} \\choose x}$}\r\n\\par Consider representing each path with a list of what direction was travelled, either right (R) or up (U). Clearly, any path can be represented by a distinct arrangement of the sequence $RR\\cdots RUU \\cdots U$, where there are $x$ $R$s and $y$ $U$s. This creates a bijection, so the number of paths is the same as the number of arrangements, which is ${{x+y} \\choose x}$.\r\n\r\n\r\n\\item How many ways can $n$ distinct items be partitioned into $k$ groups of sizes $s_1,s_2,\\ldots,s_k$, where $s_1+\\ldots+s_k = n$?\r\n\\par \\fbox{$\\displaystyle \\frac{n!}{s_1!s_2!\\cdots s_k!}$}\r\n\\par Consider selecting each group seperately. For the first group, there are ${n \\choose s_1}$ options, for the second, ${{n-s_1} \\choose s_2}$, etc. Multiplying these out, there is a total of ${n \\choose s_1}{{n-s_1} \\choose s_2}{{n-s_1-s_2} \\choose s_3}\\cdots = \\frac{n!}{s_1!s_2!\\cdots s_k!}$. This value is often expressed as ${n \\choose {s_1,s_2,\\ldots,s_k}} = \\frac{n!}{s_1!s_2!\\cdots s_k!}$, and is refered to as \\textit{multinomial} coefficients. Note that these have a relationship with the term $a_1^{s_1}\\cdots a_k^{s_k}$ in the expansion of $(a_1+\\ldots+a_k)^n$ similar to that of binomial coefficients to the expansion of $(x+y)^n$. Also note that if $s_1+\\ldots+s_k < n$, one can add a pseudo-set $s_{k+1}$ which contains all of the items not in any of the first $k$ sets (so its size will be $n-(s_1+\\ldots+s_k)$), and calculate the coefficient from there. Note that this value is also the number of ways to arrange $n$ objects in $k$ indistinct groups of sizes $s_1, s_2, \\ldots, s_k$, where the sizes again sum to $n$.\r\n\r\n\r\n\\item Simplfify $\\binom{n}{k}+\\binom{n}{k+1}$.\r\n\\par \\fbox{$\\displaystyle \\binom{n+1}{k+1}$}\r\n\\par \\textbf{Method one}. Writing out the factorial notation and simplifying quickly yields the desired reuslt. $\\frac{n!}{k!(n-k)!} + \\frac{n!}{(k+1)!(n-k-1)!} = \\frac{n!(k+1)}{(k+1)!(n-k)!} + \\frac{n!(n-k)}{(k+1)!(n-k)!} = \\frac{n!(n+1)}{{(k+1)!(n-k)!}} = \\binom{n+1}{k+1}$. This is the basis of Pascal's Triangle.\r\n\\par \\textbf{Method two}. Consider selecting $k+1$ items from $n+1$. Either the first item is chosen, in which case there are $\\binom{n}{k}$ ways to select the remaining $k$, or the first item is not chosen, in which case there are $\\binom{n}{k+1}$ ways to choose the remaining $k+1$. Thus, $\\binom{n+1}{k+1} = \\binom{n}{k}+\\binom{n}{k+1}$.\r\n\r\n\r\n\\item Simplfify $\\binom{n}{0}+\\binom{n}{1}+\\ldots+\\binom{n}{n-1}+\\binom{n}{n}$.\r\n\\par \\fbox{$\\displaystyle 2^n$}\r\n\\par This can be derived easily by counting in two ways. This is the total number of subsets of a set of size $n$; for each element of the original set, there are two options: in the set, or not in the set. Thus, there are $2^n$ subsets, and the expression above simplifies to $2^n$.\r\n\r\n\r\n\\item How many positive integer solutions exist to the system $a_1 + a_2 + \\ldots + a_k = n$?\r\n\\par \\fbox{$\\displaystyle {{n-1} \\choose {k-1}}$}\r\n\\par Draw $n$ circles in a row, and consider drawing $k-1$ lines in the $n-1$ gaps between the circles. Let $a_1$ be the number of circles before the first line, $a_2$ be the number of circles between the first and second lines, etc. Clearly, there will be $k$ total $a_i$ (if there are $k-1$ lines), all $a_i$ will be at least $1$ (since there is at least $1$ circle between each gap) and the sum of the $a_i$ will be $n$ (since each circle is counted exactly once). Thus, there is a bijection between the sequence $a_1,\\ldots,a_k$ and the drawing; as the drawing was generated by choosing $k-1$ lines from $n-1$ gaps, there are ${{n-1} \\choose {k-1}}$ drawings, and thus ${{n-1} \\choose {k-1}}$ sequences. This algorithm is sometimes refered to as ``Stars and Bars.''\r\n\r\n\r\n\\item How many non-negative integer solutions exist to the system $a_1 + a_2 + \\ldots + a_k = n$?\r\n\\par \\fbox{$\\displaystyle {{n+k-1} \\choose {k-1}}$}\r\n\\par Draw $n+k-1$ circles in a row, and cross out $k-1$ of them. Let $a_1$ be the number of circles before the first crossed-out one, $a_2$ be the number of circles between the first crossed-out one and the second crossed-out one, etc. Clearly, there will be $k$ total $a_i$ (if there are $k-1$ crossed out circles), all $a_i$ will be at least $0$ (since there may be two consecutive crossed-out circles) and the sum of the $a_i$ will be $n$ (since each remaining circle is counted exactly once, and there are $(n+k-1)-(k-1)=n$ remaining circles). Thus, there is a bijection between the sequence $a_1,\\ldots,a_k$ and the drawing; as the drawing was generated by choosing $k-1$ circles from $n+k-1$ choices, there are ${{n+k-1} \\choose {k-1}}$ drawings, and thus ${{n+k-1} \\choose {k-1}}$ sequences. Note that this algorithm can be used to determine the number of ways to partition $n$ indistinct objects into $k$ sets.\r\n\r\n\r\n\\item How many integer solutions exist to the system $a_1 + a_2 + \\ldots + a_k = n$ if $a_i > m$?\r\n\\par \\fbox{$\\displaystyle \\binom{n-km-1}{k-1}$}\r\n\\par \\textbf{Method one}. Use algebraic manipulation. Subtract $km$ from both sides, so that now we have $b_1 + b_2 + \\ldots + b_k = n-km$ with $b_i > 0$ (we subtracted $m$ from each of the $a_i$'s). Now apply the previous argument and arrive at $\\binom{n-km-1}{k-1}$.\r\n\\par \\textbf{Method two}. Use a bijection. This is the same as dividing up a set of size $n$ into $k$ groups, all of size greater than $m$. But this means we can remove $m$ items from each group and still have a positive number of items in each group, yielding a scenario in which we divide up the set of size $n-km$ into $k$ non-empy groups. Also, for every scenario with this smaller set we can add $m$ items to each group to get back to what we had in the larger set. This means there is a one-to-one correspondence between the number of possibilities in both cases. We count the easier quantity (the set of size $n-km$). To ensure that all sets are non-empty, we simply place $k-1$ dividing lines between elements of the set to divide the set into $k$ subsets. There are $\\binom{n-km-1}{k-1}$ ways to do this, as there are $n-km-1$ spaces between successive elements of the set.\r\n\r\n\r\n\\item How many ways can one arrange $n$ of one object and $k$ of another if none of the $k$ objects may be next\r\nto each other?\r\n\\par \\fbox{$\\displaystyle {{n+1} \\choose k}$}\r\n\\par First, lay out the first objects in a row of length $n$. There are now $n+1$ gaps to place the $k$ other objects in (there are $n-1$ gaps between the layed-out objects and $1$ on both ends of the row). Thus, there are ${{n+1} \\choose k}$ arrangements of these objects to avoid having any of the $k$ objects touch each other.\r\n\r\n\r\n\\item How many ways can one place $k$ indistinct items between $n$ other indistinct items? (Any number of the\r\nfirst type can go between the sucessive items of the second type).\r\n\\par \\fbox{$\\displaystyle {{n+k-2} \\choose {k}}$}\r\n\\par Draw $n+k$ circles in a row, and mark $k$ of them as objects of the second type. Clearly, one cannot mark either the first or the last, as the problem specifies that the second objects go between objects of the first type. Thus, there are $n+k-2$ markable objects, and $k$ marks, so there are ${{n+k-2} \\choose {k}}$ arrangements.\r\n\r\n\\item Simplify $\\binom{n}{0}+\\binom{n+1}{1}+\\ldots+\\binom{n+k}{k}$.\r\n\\par \\fbox{$\\displaystyle \\binom{n+k+1}{k}$}\r\n\\par The proof of this is trivial by induction. The base case is $k=1$, and it is clear that $\\binom{n}{0}+\\binom{n+1}{1}=1+n+1 = \\binom{n+2}{1}$. For the induction, $\\binom{n}{0}+\\binom{n+1}{1}+\\ldots+\\binom{n+k-1}{k-1}+\\binom{n+k}{k} = \\binom{n+k}{k-1}+\\binom{n+k}{k} = \\binom{n+k+1}{k}$. This is known as the \\textit{Hockey-stick Identity}, due to its appearance when drawn on Pascal's Triangle. Note that replacing $\\binom{n}{i}$ with $\\binom{n}{n-i}$ throughout the problem yields an alternative form of the identity.\r\n\r\n\r\n\\item How many paths of length $n$ can be made using only left or right moves of length 1, starting on the left side of a line segment of length $n$?\r\n\\par \\fbox{$\\displaystyle \\binom{n}{\\lfloor\\frac{n}{2}\\rfloor}$}\r\n\\par Consider the number of ways one can return to the origin with a path of length $2n$ (note that there is clearly no way to return to the origin with a path of length $2n+1$); let this value be $g(n)$. Clearly, the first move and the last move are preordained; the first move is invariably a move to the right and the last move a move to the left. Thus, if the path never returns to the origin until the last move, there are $g(n-1)$ options. If the path returns on the second move, then there are $g(n-2)$ options; on the fourth move, $g(1)g(n-3)$, etc. Thus, there are $g(n) = g(n-1)+g(n-2)+g(1)g(n-3)+g(2)g(n-4)+\\ldots+g(n-4)g(2)+g(n-2)g(1)+g(n-2)+g(n-1)$ options. This is a well-known recurrance relation for the Catalan numbers, so $g(n)$ is the $n$th Catalan number, which is $\\frac{(2n)!}{(n)!(n+1)!}$.\r\n\\par Now, consider $f(n)$ to be the number of paths possible after $n$ moves (without the return to the origin restriction placed on $g$.) A recurrance relation can be easily found; any path of length $x-1$ that does not end at the origin has two child paths of length $x$, and any path of length $x-1$ that does end at the origin has one child path of length $x$. Thus, for even numbers ($2n+2$), the recurrance is $f(2n+2)=2f(2n+1)$, as no odd path can end at the origin. For odd numbers ($2n+1$), the recurrance is $f(2n+1)=2f(2n)-g(n)$.\r\n\\par The proof will be completed through induction. Assume the solution is $f(n) = \\binom{n}{\\lfloor\\frac{n}{2}\\rfloor}$. The base cases are $n=1 \\rightarrow f(n)=1$ and $n=2 \\rightarrow f(n)=2$, so the base case is complete. For any odd number $2n+1$, $f(2n+1) = 2f(2n)-g(n)$. Testing this, $2f(2n)-g(n) = 2\\binom{2n}{\\lfloor\\frac{2n}{2}\\rfloor} - \\frac{1}{n+1}\\binom{2n}{n} = 2\\binom{2n}{n} - \\frac{1}{n+1}\\binom{2n}{n} = \\binom{2n}{n}(2-\\frac{1}{n+1}) = \\binom{2n}{n}(\\frac{2n+1}{n+1}) = \\binom{2n+1}{n+1} = \\binom{2n+1}{n} = f(2n+1)$, so the induction works for all $2n+1$. For any even number $2n+2$, $f(2n+2)=2f(2n+1)$. Testing this in the induction, $2f(2n+1) = 2\\binom{2n+1}{\\lfloor\\frac{2n+1}{2}\\rfloor} = 2\\binom{2n+1}{n} = \\binom{2n+1}{n}+\\binom{2n+1}{n+1} = \\binom{2n+2}{n+1} = f(2n+2)$, so the induction works for all $2n+2$. Thus, the induction is complete, and $\\binom{n}{\\lfloor\\frac{n}{2}\\rfloor}$ is the number of paths possible.\r\n\r\n\\item How many ways can one place in order a total of $n$ elements of two different types if one can at no point have placed down more elements of the first type than of the second type?\r\n\r\n\\par \\fbox{$\\displaystyle \\binom{n}{\\lfloor\\frac{n}{2}\\rfloor}$}\r\n\\par If one considers placing items of the second type moves to the right and items of the first type moves to the left, there is a clear bijection to the scenario of the previous problem.\r\n\r\n\r\n\\end{enumerate}\r\n\r\n\\begin{enumerate}\r\n\r\n\\item \\[ \\binom{n}{k} = \\frac{n!}{k!(n-k)!} \\]\r\n\r\n\\item \\[ \\binom{n}{k} = \\binom{n-1}{k-1} + \\binom{n-1}{k} \\]\r\n\\noindent Go through the elements of the $n$-element set in order. You can either take or not take each one.\r\n\r\n\\item Number of positive integer solutions to\r\n\\[ a_1 + a_2 + ... + a_k = n \\]\r\n\\noindent Treat this as having $n$ items that you wish to partition into $k$ groups. Then you must place $k-1$ dividing lines into $n-1$ possible places between items. So there are $\\binom{n-1}{k-1}$ ways to do this.\r\n\r\n\\item \\[ \\binom{n}{0}^2 + \\binom{n}{1}^2 + \\binom{n}{2}^2 + ... + \\binom{n}{n}^2 = \\binom{2n}{n} \\]\r\n\r\nConsider walking from $(0,0)$ to $(n,n)$ on a grid and only being able to follow grid lines. How many ways can this be done? The left-hand side counts this exhaustively by considering possible ways to cross the diagonal. The right-hand side makes a more clever argument by stating that you must go down $n$ times and right $n$ times, so there are $\\binom{2n}{n}$ ways to do this. As the same entity is being counted in both cases, we can place an equals sign between them.\r\n\r\nAn alternate solution with generating functions: note that $(1+x)^{2n} = ((1+x)^n)^2$. Now equate the coefficients of $x^n$ in each equation and you arrive at the same identity.\r\n\r\n\\end{enumerate}\r\n\r\n\\begin{enumerate}\r\n\r\n\\item Sometimes a problem will have a really big number in it but still require recursion. In this case a recursion with linear coefficients can be solved explicitly. Example:\r\n\r\n\\[ f_n = f_{n-1} + 2*f_{n-2} \\]\r\n\r\nSome long-dead mathematician was kind enough to prove for us that such a recurrence will always result in a closed form exponential equation (i.e. $f_n = A*B^{n}+C*D^{n}$ if $f_n$ is based only on $f_{n-1}$ and $f_{n-2}$, with more exponents if there are more terms in the recurrence). We would probably like to know $B$ and $D$, so we solve the equation $x^{2} = x + 2$ (note that this works because we can simply factor out all necesarry constants that appear). We solve the quadratic and have $x = -1,2$, so $B = -1$ and $D = 2$. $A$ and $C$ are determined by the base cases.\r\n\r\n\\item It will also be useful if one ever does Olympiad Combinatorics to know about the Pigeonhole Principle, which states that if I have $n$ pigeons and $k$ holes then there must be at least $\\lceil\\frac{n}{k}\\rceil$ pigeons in one hole and at most $\\lfloor\\frac{n}{k}\\rfloor$ pigeons in another hole.\r\n\r\n\\item It is often easier to find the average value of something and then multiply by however many of that something there are than to count the value outright.\r\n\r\n\\item Sums often lead to binomial coefficients. For example:\r\n\r\n\\[ 1 = \\binom{n}{0} \\]\r\n\r\n\\[ \\sum_{i=1}^n1 = \\binom{n}{1} \\]\r\n\r\n\\[ \\sum_{i=1}^{n-1}\\sum_{j=i+1}^n1 = \\binom{n}{2} \\]\r\n\r\n\\[ \\sum_{i=1}^{n-2}\\sum_{j=i+1}^{n-1}\\sum_{k=j+1}^n1 = \\binom{n}{3} \\]\r\n\r\nand so on. From these and a few other identities one can derive some stranger but still very useful identities:\r\n\r\n\\item \\[ \\binom{n}{k} = \\sum_{i=1}^{n-k+1}\\binom{n-i}{k-1} = \\sum_{i=0}^a\\binom{a}{i}\\binom{n-a}{k-i} \\]\r\n\r\n\\end{enumerate}\r\n\r\nA \\textit{graph} is defined as a collection of points, called \\textit{vertices}, and a collection of \\textit{edges} collecting those points. Examples of a graph are a hexagon, a tetrahedron, and every other shape you know of, with edges and vertices fitting the definition given in Euclidean Geometry. These are known as \\textit{planar graphs}. There are also \\textit{non-planar graphs}, which are, in the most intuitive sense, graphs that cannot be drawn in Euclidean space without edges crossing each other. Edges can have values associated with them, usually referred to as \\textit{weights}, and can also have directions (such a graph is called a \\textit{directed graph}).\r\n\r\nThe \\textit{in-degree} of a vertex is the number of edges pointing to an edge in a directed graph. Similarly the \\textit{out-degree} is the number of edges pointing from an edge in a directed graph.\r\n\r\nAn \\textit{Eulerian cycle} is a path through the graph that visits every edge exactly once. In a directed graph this is possible iff the in-degree equals the out-degree for every edge. In an undirected graph this is possible iff every edge has an even degree (why are these equivalent?).\r\n\r\nA \\textit{Hamiltonian cycle} is a path through the graph that visits every vertex exactly once. There are several theorems stating when a Hamiltonian cycle exists. While they are admittedly useful they will not be discussed here, as they are not biconditional like the theorem for Eulerian cycles.\r\n\r\nA \\textit{complete graph} is one in which all possible edges are drawn. By convention the complete graph with $V$ vertices is referred to as the $K_V$ graph. A \\textit{simple graph} has no more than one edge between any two vertices and no edges from a vertex to itself. By convention all graphs are simple and undirected unless otherwise stated.\r\n\r\nAlso note that the number of paths $N(A,B)$ from $A$ to $B$ is $\\displaystyle\\sum_{v \\in C(A)}N(v,B)$, where $C(X)$ denotes the set of vertices that $X$ has an edge to.", "Solution_6": "those are nice JSteinhardt :lol: thanks a lot\r\n\r\n[b][size=150]SOME IMPORTANT RESULTS[/size][/b]\r\n\r\n[b][u]Result 1[/u][/b] [i]If n items are arranged in a row,then the number of wasy in which they can be rearranged so that no one of them occupies the place assigned to it is[/i]\r\n$n![ 1-\\frac{1}{1!}+\\frac{1}{2!}-\\frac{1}{3!}+\\cdots+(-1)^n\\frac{1}{n!}]$\r\n\r\n[b][u]Result 2[/u] [/b] [i]If there are m items of one kind,n items of another kind and so ion,then the number of ways of choosing ritems out of these items is[/i]\r\n$Coeff.\\ of\\ x^r\\ in\\ (1+x+x^2+x^3+\\cdots+x^m)(1+x+x^2+\\cdots+x^n)\\cdots$\r\n\r\n[b][u]Result 3[/u][/b] [i]If there are m items of one kind,n items of another kind and so on,then the number of ways of choosing r items out of these items such that at least one item of each kind is included in every selection is[/i]\r\n\r\n$Coeff.\\ of\\ x^r\\ in\\ (x+x^2+\\cdots+x^m)(x+x^2+\\cdots+x^n)\\cdots$", "Solution_7": "$\\displaystyle\\binom{n}{x_1,x_2\\cdots x_k}$ is the number of ways of choosing k subsets, each with $x_i$ elements, out of n elements. Define $x_0=0$\r\n\\begin{eqnarray*} \\displaystyle\\binom{n}{x_1,x_2\\cdots x_k} &=& \\displaystyle\\frac{n!}{(n-\\sum x_i)!\\displaystyle\\prod_{i=1}^k x_i!}\\\\ &=& \\displaystle\\prod_{i=1}^k \\binom{n-\\sum_{j=0}^{i-1} x_{j}}{x_i}\\\\ &=& \\binom{n}{\\sum_{i=1}^k x_i} \\binom {\\displaystyle\\sum_{i=1}^k x_i}{x_1,x_2\\cdots x_k} \\end{eqnarray*}\r\n\r\nDoes anyone know a formula for $\\displaystyle\\sum\\binom{n}{x_1,x_2\\cdots x_k}$ for all $\\displaystyle\\sum_{i=1}^k x_i = n$ ?", "Solution_8": "[quote=\"felipeochoa\"]Does anyone know a formula for $\\displaystyle\\sum\\binom{n}{x_1,x_2\\cdots x_k}$ for all $\\displaystyle\\sum_{i=1}^k x_i = n$ ?[/quote]\r\n\r\nFor a given k, it's just $N = k^n$. (Each item goes in one of k subgroups). For all k such that $1 \\leq k \\leq n$, you get the sum $N = \\sum^{n}_{r=1}{r^n}$... I'm not sure how to write that explicitly.", "Solution_9": "[color=blue]Result 1[/color] If n items are arranged in a row,then the number of wasy in which they can be rearranged so that no one of them occupies the place assigned to it is \r\n \r\n$n![1-\\frac{1}{1!}+\\frac{1}{2!}-\\frac{1}{3!}+...+(-1)^{n}\\frac{1}{n!}]$\r\n\r\ni dont get this, can someone help plz", "Solution_10": "cane some one prove this:\r\n\r\nResult 2 If there are m items of one kind,n items of another kind and so ion,then the number of ways of choosing ritems out of these items is\r\n$ Coeff.\\ of\\ x^{r}\\ in\\ (1\\plus{}x\\plus{}x^{2}\\plus{}x^{3}\\plus{}\\cdots\\plus{}x^{m})(1\\plus{}x\\plus{}x^{2}\\plus{}\\cdots\\plus{}x^{n})\\cdots$", "Solution_11": "A few mistakes I have just spotted seem to be really popular:\r\n\r\n[quote=\"JSteinhardt\"]There are also \\textit{non-planar graphs}, which are, in the most intuitive sense, graphs that cannot be drawn in Euclidean space without edges crossing each other.[/quote]\n\nSpace $ \\to$ plane.\n\n[quote=\"JSteinhardt\"]An \\textit{Eulerian cycle} is a path through the graph that visits every edge exactly once. In a directed graph this is possible iff the in-degree equals the out-degree for every edge. In an undirected graph this is possible iff every edge has an even degree (why are these equivalent?).[/quote]\n\nYou forgot to require that the graph is connected. Many people tend to forget this condition. Keep in mind that in some cases - for instance, in the proof of the existence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17342]generalized de Bruijn sequences[/url] - this condition is the harder one of the two to check!\n\n[quote=\"JSteinhardt\"]Also note that the number of paths $ N(A,B)$ from $ A$ to $ B$ is $ \\sum_{v\\in C(A)}N(v,B)$, where $ C(X)$ denotes the set of vertices that $ X$ has an edge to.[/quote]\r\n\r\nDepending on your definition of path, this is either wrong or nonsense... Are paths to pass any vertex only once? Any edge only once? If not, than $ N\\left(A,B\\right)$ will most often happen to be $ \\infty$. If yes, then the above is wrong.\r\n\r\n Darij" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove the inequality for numbers $x_1,x_2,...,x_n \\in [a;b]$ where a is positive:\r\n\r\n$(x_1+x_2+...+x_n)(\\frac{1}{x_1}+\\frac{1}{x_2}+...+\\frac{1}{x_n})\\leq n^2\\cdot \\frac{(a+b)^2}{4ab}$\r\n\r\nTo view all the problems of Uzbekistan Undergraduate Olympiad problems visit\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81099[/url]", "Solution_1": "This is a special case of the Polye--Szegoe inequality." } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "$(F_n)_{-\\infty}^{+\\infty}$: $F_0=0$, $F_1=1$, $F_{n+1}=F_{n}+F_{n-1}$. \r\nProve that for all integer $n$, there only one $i_1,i_2,\\ldots,i_k\\in\\{0,1\\}$ such that:\r\n(i) $i_ji_{j+1}=0$\r\n(ii) $i_1F_1+i_2F_2+\\ldots+i_kF_k=n$", "Solution_1": "I think posted before in Pre-Olympiad (can't search now)." } { "Tag": [ "logarithms" ], "Problem": "Solve:\r\n\r\n$ (log_2 (x))^2 \\plus{} log_2 (x) \\plus{} 1 \\equal{} \\frac{7}{log_2 (.5x)}$", "Solution_1": "[quote=\"Stokes93\"]Solve:\n\n$ (log_2 (x))^2 \\plus{} log_2 (x) \\plus{} 1 \\equal{} \\frac {7}{log_2 (.5x)}$[/quote]\r\n[hide]Set $ log_{2}(x)\\equal{}a$ and note that $ log_{2}(.5x)\\equal{}log_{2}(x)\\minus{}log_{2}(2)\\equal{}log_{2}(x)\\minus{}1$\nNow we have that $ a^{2}\\plus{}a\\plus{}1\\equal{}\\frac{7}{a\\minus{}1}$\nYou can multiply both sides by $ a\\minus{}1$ or note that $ \\frac{a^{3}\\minus{}1}{a\\minus{}1}\\equal{}a^{2}\\plus{}a\\plus{}1$, so setting $ a^{3}\\minus{}1\\equal{}7$, you get that $ a\\equal{}2$, there $ x\\equal{}4$.[/hide]", "Solution_2": "hello, rewriting your equation in the form\r\n$ \\left(\\frac {\\ln(x)}{\\ln(2)}\\right)^2 \\plus{} \\frac {\\ln(x)}{\\ln(2)} \\equal{} \\frac {7}{\\frac {\\ln(1/2x)}{\\ln(2)}}$\r\nusing $ \\ln(1/2x) \\equal{} \\ln(x) \\minus{} \\ln(2)$\r\nwe get after some simplifications\r\n$ \\ln(x)^3 \\minus{} \\ln(x)^2\\ln(2)^2 \\minus{} 7\\ln(2)^3 \\equal{} 0$\r\nsolving this we get\r\n$ x \\equal{} {2}^{ \\minus{} {\\frac {1}{72}}\\,{12}^{2/3} \\left( 63 \\plus{} \\sqrt {3957} \\right) ^{2/3 }\\sqrt {3957} \\plus{} {\\frac {7}{8}}\\,{12}^{2/3} \\left( 63 \\plus{} \\sqrt {3957} \\right) ^{2/3} \\plus{} 1/6\\,\\sqrt [3]{12}\\sqrt [3]{63 \\plus{} \\sqrt {3957}}}$\r\nSonnhard.\r\noh, i have forgotten the number $ 1$ in your equation, sorry.", "Solution_3": "hello, no the original equation\r\n$ \\left(\\frac {\\ln(x)}{\\ln(2)}\\right)^2 \\plus{} \\frac {\\ln(x)}{\\ln(2)} \\plus{} 1 \\equal{} \\frac {7}{\\frac {\\ln(1/2x)}{\\ln(2)}}$\r\nsimplifying this we get\r\n$ \\ln(x)^2\\ln(1/2x) \\plus{} \\ln(x)\\ln(2)\\ln(1/2x) \\plus{} \\ln(2)^2\\ln(1/2x) \\minus{} 7\\ln(2)^3 \\equal{} 0$\r\n$ \\ln(x)^3 \\minus{} 8\\ln(2)^3 \\equal{} 0$\r\n$ \\ln(x) \\equal{} \\ln(4)$\r\n$ x \\equal{} 4$\r\nSonnhard.", "Solution_4": "[quote=\"Stokes93\"]Solve:\n\n$ (log_2 (x))^2 \\plus{} log_2 (x) \\plus{} 1 \\equal{} \\frac {7}{log_2 (.5x)}$[/quote]\r\n\r\n\r\n[i]Denote[/i] $ \\log_{2}(x) \\equal{} y \\implies log_{2}^2 (x) \\equal{} y^2$\r\n\r\n$ log_{2}(\\frac {x}{2}) \\equal{} log_{2}(x) \\minus{} \\log_{2}(2) \\equal{} y \\minus{} 1, \\implies (y \\minus{} 1)(y^2 \\plus{} y \\plus{} 1) \\equal{} 7$\r\n\r\n$ \\implies y \\equal{} 2$ , [i]therefore[/i] $ x \\equal{} 2^y \\implies x \\equal{} 4$" } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "1. A and B play a series of games. Each game is independently won by A with probability p and by B with probability 1-p. They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner. Find the probability that A is the match winner.\r\n\r\n\r\n2. A stock market investor owns shares in stock whose present value is 25. She has decided that she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with probability .55 or down 1 point with probability .45, and the successive changes are independt, what is the probability that the investor retires a winner?", "Solution_1": "You seem to be unable to decide if your questions belong in the intermediate forum or the olympiad forum, which suggests that in fact they are best-suited to [url=http://www.artofproblemsolving.com/Forum/index.php?f=151]Pre-Olympiad[/url].\r\n\r\n[quote=\"wly3298456\"]1. A and B play a series of games. Each game is independently won by A with probability p and by B with probability 1-p. They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner. Find the probability that A is the match winner.[/quote]\r\n\r\nA simple probability state diagram is the best way to go. There are five states: A wins (happy state), A + 1, Tied, B + 1, B wins (unhappy state). With these states we associate probabilities $p_{1}, p_{2}, p_{3}, p_{4}$ and $p_{5}$ that A wins starting in the given state, and we want to calculate $p_{3}$. Then we have $p_{1}= 1$, $p_{2}= p+(1-p)p_{3}$, $p_{3}= p\\cdot p_{2}+(1-p)p_{4}$, $p_{4}= p\\cdot p_{3}$ and $p_{5}= 0$. This is straightforward to solve." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "given f(x), g(x) defined in a neighbourhood of c,\r\nis it true that:\r\n$ o_{c}(f)\\equal{}o_{c}(g)\\iff f \\thicksim_{c}g$ ?\r\n\r\n(I have proved the reverse implication, but I'm not sure about the direct one)", "Solution_1": "True. For example, if $ \\liminf_{x\\to c}f(x)/g(x)\\equal{}0$, then you can construct a function $ h$ that is equal to $ f$ when $ f(x)/g(x)$ is \"small\" and is zero otherwise, so that $ h\\in o_c(g)\\setminus o_c(f)$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "On a plane are given finitely many red and blue lines, no two parallel, such that any intersection point of two lines of the same color also lies on another line of the other color. Prove that all the lines pass through a single point.", "Solution_1": "[b]Solution [/b]We shall start the process of drawing the lines in the following fashion. We first draw two red(or blue) lines, then draw a blue line so that it passes through the intersection. Then draw a blue line if the number of blue line is greater than 1 and so on. In case we have more than 2 red lines and just one blue line then we have draw the red line through the same intersection point and the process continues. Suppose we go on like this till we exhaust all our lines and build a system according to the condition of the problem. We are assured that at least one such system exists : consider all the red and blue points passing through a point.\r\n\r\nLet the last line that was drawn be red wlog. It had to be drawn so that it passed through the intersection point of every pair of red lines drawn previously. For any three non parallel lines we get three distinct intersection points (a triangle) or else all of them pass through a same point. The only option was the latter and we conclude that all the red lines meet at a common point. By the condition of the problem, all the blue lines also pass through this common point.", "Solution_2": "I understand virtually nothing from your proof, such as\r\n\r\n[quote=\"srikanth\"] Then draw a blue line if the number of blue line is greater than 1 and so on. [/quote]\r\n\r\nDraw it where, and what do you mean by \"so on\"?", "Solution_3": "I was actually giving a method to construct a set of finite red and blue lines according to condition of the problem. \r\n\r\nLet $ r$ and $ b$ be the number of red and blue lines respectively. Intent of the solution is to explicitly construct a system ( of red and blue lines with the required conditions) by drawing one line at a time. We use the finiteness property later. Say we started by drawing two red lines wlog. This is not a system yet, we have to draw a blue line through the intersection point of the two red lines. The next line we draw can be either red or blue (your quote from my last post- I said we can draw a blue line at this stage only if $ b$>1) and [b]so on[/b]. We observe that that drawing lines will give rise to more intersections, and we need to draw more lines through the intersection points. So we impose following the condition when the last line is drawn (finiteness property). \r\n\r\nLast line in a construction (either red or blue) has to pass through the intersection of every pair of lines ( previously drawn) of same colour. The required result follows immediately. \r\n\r\nPs: Referring to [i]so on[/i]- If you put pen to paper and start constructing a system (without finiteness property) like this you will find that there different ways in which one can proceed. The solution speaks about the terminating step in the finite case and solution does not intend to describe the possible (infinite) constructions in the case without finiteness condition.\r\n\r\nIs the solution clear ?", "Solution_4": "check this topic so amazing proof:\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=446298&p=2512211#p2512211" } { "Tag": [ "calculus", "derivative", "function", "calculus computations" ], "Problem": "If all the n-th derivatives of a function are bounded on an interval ]-a,a[ then it has a series development.\r\nThis is a sufficient condition , but it seems to me that all $C^\\infty$functions have this property including the classical couter example $e^{-\\frac{1}{{x^{2}}}}$ ; am i missing something ?", "Solution_1": "A sufficient condition for analyticity is that the derivatives are [b]uniformly[/b] bounded, that is, there is $C$ such that $|f^{(n)}(x)|\\le C$ for all $x\\in (-a,a)$ and all $n\\ge 1$.", "Solution_2": "You could weaken what mlok said to \r\n\r\n$|f^{(n)}(x)|\\le\\frac{Cn!}{a^{n}}$ for all $x\\in(-a,a)$ and for all $n\\ge 1.$\r\n\r\nThat would still be a sufficient condition; it would still not be a necessary condition." } { "Tag": [], "Problem": "For $n>2k$, show that \\[2^{k}n!>n^{k}\\left(n-k\\right)!.\\]", "Solution_1": "[hide]\n$2n > 2n-2 > \\cdots > 2n-2k+2 > n$\n\n$(2n)(2n-2) \\cdots (2n-2k+2) > n^{k}$\n\n$2^{k}n(n-1) \\cdots (n-k+1) > n^{k}$\n\n$2^{k}n! > n^{k}(n-k)!$[/hide]" } { "Tag": [ "conics", "parabola", "geometry proposed", "geometry" ], "Problem": "The normals at the three points A, B, C on a parabola are concurrent.\r\n\r\nA line is drawn through each vertex of the triangle ABC , parallel to the opposite side (the line through A is parallel to BC, and so on).\r\n\r\nProve that the three lines formed touch a fixed parabola.", "Solution_1": "[quote=\"Michael Niland\"]The normals at the three points A, B, C on a parabola are concurrent.\n\nA line is drawn through each vertex of the triangle ABC , parallel to the opposite side (the line through A is parallel to BC, and so on).\n\nProve that the three lines formed touch a fixed parabola.[/quote]\r\n\r\nLet $A(at_1 ^2 ,\\,2at_1 ),\\,\\,B(at_2 ^2 ,\\,2at_2 ),\\,\\,A(at_3 ^2 ,\\,2at_3 )$ are three points on the parabola $y^2=4ax$ normals at which are concurrent at (h, k).\r\n\r\nThen there are three well known relations\r\n$t_1 + t_2 + t_3 = 0$, $\\sum {t_1 t_2 } = \\frac{{2a - h}}{a}$, $t_1 t_2 t_3 = \\frac{k}{a}$.\r\n\r\nLine through $A$ and parallel to $BC$ is given by $2x - y(t_2 + t_3 ) = 2at_1 ^2 - 2at_1 (t_2 + t_3 )$\r\nUsing first of the above three relations, we can write this equation as\r\n\r\n$2x + yt_1 = 4at_1 ^2$ \r\n$\\Rightarrow y = - \\frac{2}{{t_1 }}x + 4at_1$\r\n\r\nwhich is the standard equation to the tangent to the parabola $y^2=-8ax$.\r\n\r\nIt can be shown that the other two lines too touch the same parabola." } { "Tag": [ "inequalities", "Digits", "algebra", "binary representation", "combinatorics", "Sequence", "IMO Shortlist" ], "Problem": "Let $ \\alpha(n)$ be the number of digits equal to one in the binary representation of a positive integer $ n.$ Prove that:\r\n\r\n(a) the inequality $ \\alpha(n) (n^2 ) \\leq \\frac{1}{2} \\alpha(n)(\\alpha(n) + 1)$ holds;\r\n(b) the above inequality is an equality for infinitely many positive integers, and\r\n(c) there exists a sequence $ (n_i )^{\\infty}_1$ such that $ \\frac{\\alpha ( n^2_i )}{\\alpha (n_i }$ \r\ngoes to zero as $ i$ goes to $ \\infty.$\r\n\r\n\r\n[i]Alternative problem:[/i] Prove that there exists a sequence a sequence $ (n_i )^{\\infty}_1$ such that $ \\frac{\\alpha ( n^2_i )}{\\alpha (n_i )}$ \r\n\r\n(d) $ \\infty;$\r\n(e) an arbitrary real number $ \\gamma \\in (0,1)$;\r\n(f) an arbitrary real number $ \\gamma \\geq 0$;\r\n\r\nas $ i$ goes to $ \\infty.$", "Solution_1": "a) We prove that $ \\alpha(n^2)\\leq\\alpha(n)(\\alpha(n)\\plus{}1)/2$.\r\nLet $ k\\equal{}\\alpha(n)$; $ n\\equal{}2^{u_1}\\plus{}\\cdots{2^{u_k}}$. Thus $ n^2\\equal{}\\sum_{i\\equal{}1}^{k}{2^{2u_i}}\\plus{}\\sum_{i3$\r\nthen $ \\alpha(n_i)=1+(i)=i+1$\r\n\r\n$ n_i=2^{i+1}+2^{i}-1=3.2^{i}-1$ \r\n\r\n$ n^2_i=(3.2^{i}-1)^2=9.2^{2i}-6.2^i+1$\r\n\r\n$ n^2_i=2^{2i+3}+2^{2i}-3 2^{i+1}+1$\r\n\r\n$ n^2_i=2^{2i+3}+2^{2i}-2^{i+3}+2^{i+1}+1$\r\n\r\n$ n^2_i=2^{2i+3}+2^{i+3} (2^{i-3}-1) +2^{i+1}+1$\r\n\r\nthen $ \\alpha(n^2_i)=1+(i-3)+1+1=i$", "Solution_3": "Hi mszew,\r\n$ \\alpha(n_i^2)/\\alpha(n_i)$ goes to $ 1$.", "Solution_4": "Here's a solution for part (c). \n\nLet $k$ be a sufficiently large positive integer, and $a > 2^k \\cdot 10^{10^{10^{10}}}$ be an even larger positive integer. Let $n= 2^{a-1} - 2^{a-2} - 2^{a-4} - \\cdots - 2^{a-2^k}.$ Observe that $\\alpha (n) = 2^k - k.$ On the other hand, we can easily find that\n\n$$n^2 = 2^{2(a-2^k)} + 2 \\sum_{1\\le i,j \\le k, i \\neq j} 2^{a-2^i} \\cdot 2^{a-2^j},$$\n\nwhich implies that $\\alpha(n^2) \\le 1 + \\binom{k}{2}$ since it's a sum of that many powers of two. From here, we are clearly done since\n\n$$\\lim_{k \\rightarrow \\infty} \\frac{1 + \\binom{k}{2}}{2^k-k} = 0.$$\n\n$\\square$ ", "Solution_5": "The part a as stated here is false and there's a typo. In IMO compendium it stated the correct inequality that is $ \\alpha(n^2) \\leq \\frac{1}{2} \\alpha(n)(\\alpha(n) + 1)$ and not $ \\alpha(n) (n^2 ) \\leq \\frac{1}{2} \\alpha(n)(\\alpha(n) + 1)$" } { "Tag": [], "Problem": "About how long would it take you to play every song you have memorized (that you actually USE often -- like \"real\" songs, not ones that you learned for a one-time thing or whatever . . . i.e. \"solos\") on the instrument you are most skilled on? Please name the instrument, too.", "Solution_1": "i voted for 10 to 20 minutes...\r\ni think the last time i touched my accordian was the summer of 2004....", "Solution_2": "I could probably play for a good two hours with just piano stuff. I'm required to do a twenty piece program for Guild auditions, which would probably constitute about an hour and a half. Adding the tunes I do for fun, jazz stuff, and my own compositions would bring it to about two hours.", "Solution_3": "I [i]think[/i] it'd be around an hour--I'm not sure though.\r\n\r\nOh right, I play piano.", "Solution_4": "I could probably do between an hour and an hour and a half on vocals.", "Solution_5": "Probably 0-10 minutes...I don't have anything memorized on piano right now. :?", "Solution_6": "30 or so minutes on piano. I'm not a big memorizing person. I should memorize more stuff, but there's a lot of pieces I can recall if I play again for a couple days.\r\n\r\nI'd say around 40 minutes on violin. Even though I haven't played for a year, I remember a lot of that stuff really well. Mainly because it isn't too advanced.\r\n\r\nforgot to mention, I also sing. :) Actually, I only sang for one year, and that was more like three months. I just decided to try out for Fiddler on the Roof in 6th grade and ended up getting the role of Tevye. It was pretty unexpected, but pretty fun. I still remember all my songs from that, so that's probably another half an hour or so.", "Solution_7": "Hi~\r\n\r\nI'd say around 2 hours on piano. I'm taking DipABRSM this year, which requires around 30 minutes of performance there already,,\r\n\r\nAnother 2 hours on clarinet. I find it easier to memorize on this instrument because it uses a single stave and flows easier than piano.\r\n\r\nI can play French horn, but not that well... maybe another 30 minutes? All I know is Saint-Saens' Morceau de Concert and the Mozart Horn Concertos -_-;;", "Solution_8": "I could go for one or two hours on the bagpipes, but I HAVE to memorize everything because I [obviously] can't use sheet music while marching :lol: The thing is, I don't remember the names of the tunes, just the actual tune, so if somebody wants me to play a tune, they'll have to hum it :blush: So many of the tunes are similar with just a slight nuance that it gets quite confusing sometimes." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find n max such that the equation:\r\n $2x$ $+$ $4y$ $+$ $5z$ $=$ $n$\r\nhas no solution in $N*$", "Solution_1": "$n=2(x+2y)+5z$. If z=x+2y>2 then exist $x,y\\in N$, suth that t=x+2y.\r\nTherefore if n>2*5-2-5+3*2+5=16 exist solution $n=2t+5z,t>2,z>0$ or $n=x+2y+5z, x,y,z\\in N.$\r\n$16=2*1+4*1+5*2$\r\n$15=2*1+4*2+5*1$\r\n$14=2*x+4*y+5*z,x,y,z\\in N\\Longrightarrow z=2,x+2y=2$ have not solution." } { "Tag": [ "ratio", "geometry", "3D geometry", "calculus", "analytic geometry", "integration" ], "Problem": "We have uniformly volumetrically charged cube. What's the ratio of the electrostatic potential in the middle of the cube and in one of the heads of the cube? (the potential is zero in infinity)", "Solution_1": "By dimensional analysis, we can conclude that the potential in any point of the cube is proportional to $a^{2}\\rho$, where $a$ is the length of a side of the cube and $\\rho$ is the charge density of the cube. Let $\\varphi_{C}(a, \\rho)$ be the potential in the center of the cube and $\\varphi_{V}(a, \\rho)$ the potential in one of the vortices. If we consider a cube of side $2a$, but charged with the same density $\\rho$, from superposition we can conclude that $\\varphi_{C}(2a, \\rho) = 8\\varphi_{V}(a, \\rho)$. Since $\\varphi_{C}(2a, \\rho) = 4\\varphi_{C}(a, \\rho)$, the required ratio is\r\n\\[\\frac{\\varphi_{C}(a, \\rho)}{\\varphi_{V}(a, \\rho)}= 2. \\]", "Solution_2": "Here is a solution using calculus:\r\nlets choose the sides of the cube as 1 unit each, and fit a coordinate system $(x,y,z)$ to one of the heads of the cube. \r\naround the point $\\vec{r}=(x,y,z)$ we choose a small $dV=dxdydz$ volume. the potential by this small volume at the 0 point of the coordinate system is :\r\n$k\\frac{dV}{|\\vec{r}|}$\r\nthe potential of the whole charge at the head:\r\n$\\phi_{head}=k\\int\\frac{dV}{|\\vec{r}|}=k\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{1}\\frac{dxdydz}{\\sqrt{x^{2}+y^{2}+z^{2}}}$\r\nsimilarly the potential in the middle of the cube:\r\n$\\phi_{middle}=k\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{1}\\frac{dxdydz}{\\sqrt{(x-\\frac12)^{2}+(y-\\frac12)^{2}+(z-\\frac12)^{2}}}$\r\nso the ratio:\r\n$\\frac{\\phi_{head}}{\\phi_{middle}}=\\frac{\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{1}\\frac{dxdydz}{\\sqrt{x^{2}+y^{2}+z^{2}}}}{\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{1}\\frac{dxdydz}{\\sqrt{(x-\\frac12)^{2}+(y-\\frac12)^{2}+(z-\\frac12)^{2}}}}=\\frac12$\r\njust the same as you got, though your solution is much more elegant.", "Solution_3": "Your solution is nice, too :)" } { "Tag": [ "topology" ], "Problem": "Prove that the only sets that are both open and closed are $ R$ and the empty set $ \\emptyset$.", "Solution_1": "More generally: a set $ A\\ne\\emptyset$ is connected if and only if the only open and closed of its subsets are $ A$ and $ \\emptyset$ .\r\n[hide=\"Proof\"]\n-- If $ A$ is connected.\nSuppose for the sake of contradiction that there exist $ B\\subset A$ different from $ A$ and $ \\emptyset$ which is both open and closed. \nBecause $ B$ is closed we have that $ C \\equal{} A \\minus{} B$ is open. Then there exits a partition $ \\{B, C\\}$ of $ A$ into two open sets. Then $ A$ is disconnected, contradiction. Hence, the only subsets of $ A$ which are open and closed are $ A$ and $ \\emptyset$.\n\n--If the only subsets of $ A$ which are open and closed are $ A$ and $ \\emptyset$.\nThen for every $ B\\subset A$ different from $ A$ and $ \\emptyset$, $ B$ is only open or closed. Then its $ C \\equal{} A \\minus{} B$ is also only open or closed. This means that all partitions $ \\{B, C\\}$ of $ A$ are compound by two sets which are not simultaneoulsy open. Then $ A$ is connected.\n[/hide]\r\n\r\nSince we know that $ \\mathbb{R}$ is connected by the previous result it follows that the only open and closed of its subsets are $ \\mathbb{R}$ and $ \\emptyset$.\r\n\r\n:)", "Solution_2": "Thank u. \r\nBut if I only know the concepts of open and closed, then how to prove it?\r\nA set $ O \\subset \\mathbb{R}$ is open if for all points $ a \\in O$ there exists an e-neighborhood $ V_{e}{(a)} \\subset O$.\r\nA set $ F \\subset \\mathbb{R}$ is closed if it contains its limit points.", "Solution_3": "Well then this helps (for future reference, a classical proof that a segment is connected). \r\n[hide]\nSuppose for the sake of contradiction that there exists $ A\\subset\\mathbb{R}$ different from $ \\mathbb{R}$ and $ \\emptyset$ which is both, open and closed. Since $ A$ is closed, its complement, say $ B$, is open. Clearly we can find a closed interval $ J \\equal{} [c,d]$ which contains points from both, $ A$ and $ B$. Suppose with out lose of generality that $ d\\in A$.\nLet $ r \\equal{} \\sup\\{B\\cap J\\}$. \nSuppose that $ r\\in A$. Since $ A$ is open it contains a whole open interval with center in $ r$, say $ (b, a)$, with $ b > a$. Then, the fact $ a\\in A$ implies $ a > x,\\;\\forall\\; x\\in B\\cap J$, which is a contradiction because $ r > a$ and $ r$ was supposed to be the least real with that property.\nThen $ r\\in B$, so $ r a$, so $ a r$, which is again a contradiction because we should have $ r > a$.\n\nThen $ r$ is real but, $ r\\not\\in A\\cup B$, which is another contradiction because $ A\\cup B \\equal{} \\mathbb{R}$ by hypothesis. \nHence, there does not exists such set $ A$. This means that the only subsets of $ \\mathbb{R}$ which are both open and closed are $ \\mathbb{R}$ and $ \\emptyset$.\n[/hide]\r\n\r\n:)", "Solution_4": "[quote=\"ElChapin\"]Well then this helps (for future reference, a classical proof that a segment is connected). \n[hide]\nSuppose for the sake of contradiction that there exists $ A\\subset\\mathbb{R}$ different from $ \\mathbb{R}$ and $ \\emptyset$ which is both, open and closed. Since $ A$ is closed, its complement, say $ B$, is open. Clearly we can find a closed interval $ J \\equal{} [c,d]$ which contains points from both, $ A$ and $ B$. Suppose with out lose of generality that $ d\\in A$.\nLet $ r \\equal{} \\sup\\{B\\cap J\\}$. \nSuppose that $ r\\in A$. Since $ A$ is open it contains a whole open interval with center in $ r$, say $ (b, a)$, with $ b > a$. Then, the fact $ a\\in A$ implies $ a > x,\\;\\forall\\; x\\in B\\cap J$, which is a contradiction because $ r > a$ and $ r$ was supposed to be the least real with that property.\nThen $ r\\in B$, so $ r < d$. Because $ B$ is also open it contains a whole open interval with center in $ r$, say $ (b,a)$, with $ b > a$, so $ a < d$ also. Then $ a\\in B\\cap J$ and $ a > r$, which is again a contradiction because we should have $ r > a$.\n\nThen $ r$ is real but, $ r\\not\\in A\\cup B$, which is another contradiction because $ A\\cup B \\equal{} \\mathbb{R}$ by hypothesis. \nHence, there does not exists such set $ A$. This means that the only subsets of $ \\mathbb{R}$ which are both open and closed are $ \\mathbb{R}$ and $ \\emptyset$.\n[/hide]\n\n:)[/quote]\r\nWhat does \"$ J \\equal{} [c,d]$ which contains points from both, $ A$ and $ B$\" mean?\r\nand \"say $ (b, a)$, with $ b > a$\"? Why $ a\\in A$ implies $ a > x,\\;\\forall\\; x\\in B\\cap J$?", "Solution_5": "[quote=\"IORI\"]\nWhat does \"$ J \\equal{} [c,d]$ which contains points from both, $ A$ and $ B$\" mean?\n[/quote]\n\nThere exists $ J \\equal{} [c, d]$ with $ c < d$ such that $ A\\cap J\\ne\\emptyset\\ne B\\cap J$. \n\n[quote=\"IORI\"]\nand \"say $ (b, a)$, with $ b > a$\"? Why $ a\\in A$ implies $ a > x,\\;\\forall\\; x\\in B\\cap J$?\n[/quote]\r\n\r\nYou are right about asking this, because I forgot to add something about $ a$, that it is taken such that $ a\\in A$. After this clarification:\r\n$ \\forall\\; x\\in B\\cap J$ we have $ x\\le r$\r\nSuppose that there exists $ y\\in B\\cap J$ such that $ y\\ge a$. Then $ y\\in [a,r]$ but $ [a,r]\\subset A$ then $ y\\in A$, contradiction. Therefore $ \\forall\\; y\\in B\\cap J, y < a$.\r\n:)", "Solution_6": "e...I still can't get your proof.\r\nU wrote: $ J \\equal{} [c,d]$ with $ c < d$ s.t. $ A \\cap J \\ne \\emptyset$, $ B \\cap J \\ne \\emptyset$. Let $ r \\equal{} sup \\{B \\cap J \\}$\r\nThen why \"$ \\forall x \\in B$ we have $ x \\le r$\"?\r\nAnd why \"If there exists $ y \\in B \\cap J$ s.t. $ y \\ge a \\equal{} > y \\in A$ is a contradiction\"?", "Solution_7": "[quote=\"IORI\"]e...I still can't get your proof.\nU wrote: $ J \\equal{} [c,d]$ with $ c < d$ s.t. $ A \\cap J \\ne \\emptyset$, $ B \\cap J \\ne \\emptyset$. Let $ r \\equal{} sup \\{B \\cap J \\}$\nThen why \"$ \\forall x \\in B$ we have $ x \\le r$\"?\nAnd why \"If there exists $ y \\in B \\cap J$ s.t. $ y \\ge a \\equal{} > y \\in A$ is a contradiction\"?[/quote]\n\nI just had a mistake in writing it. Clearly, the correct statement is $ \\forall\\; x\\in B\\cap J$ we $ x\\le r$. \n\n\n[quote=\"IORI\"]e...I still can't get your proof.\nU wrote: $ J \\equal{} [c,d]$ with $ c < d$ s.t. $ A \\cap J \\ne \\emptyset$, $ B \\cap J \\ne \\emptyset$. Let $ r \\equal{} sup \\{B \\cap J \\}$\nThen why \"$ \\forall x \\in B$ we have $ x \\le r$\"?\nAnd why \"If there exists $ y \\in B \\cap J$ s.t. $ y \\ge a \\equal{} > y \\in A$ is a contradiction\"?[/quote] \r\n\r\nI defined $ B$ as $ \\mathbb{R} \\minus{} A$, so $ A\\cap B \\equal{} \\emptyset$.\r\n\r\n:)" } { "Tag": [ "parameterization", "real analysis", "real analysis theorems" ], "Problem": "Suppose $ x \\in R^n$, \r\nI encountered the sentence:\r\n$ |x| \\rightarrow \\infty \\Leftrightarrow |x_1| \\rightarrow \\infty \\vee |x_2| \\rightarrow \\infty \\vee ... \\vee |x_n| \\rightarrow \\infty$\r\nCould anyone explain why this sentence makes sense? How to write it in a clearer way? \r\nThanks.", "Solution_1": "I definitely wouldn't use so many symbols.\r\n\r\nTranslated to use words:\r\n$ |x|$ goes to $ \\infty$ if and only if ($ |x_1|$ goes to $ \\infty$ or $ |x_2|$ goes to $ \\infty$ or ... or $ |x_n|$ goes to $ \\infty$).\r\nA step farther, we can read that chain of \"or\" as \"at least one\":\r\n$ |x|$ goes to $ \\infty$ if and only if at least one of the $ |x_i|$ ($ i\\in\\{1,2,\\cdots,n\\}$) goes to $ \\infty$.", "Solution_2": "[quote=\"jmerry\"]I definitely wouldn't use so many symbols.\n\nTranslated to use words:\n$ |x|$ goes to $ \\infty$ if and only if ($ |x_1|$ goes to $ \\infty$ or $ |x_2|$ goes to $ \\infty$ or ... or $ |x_n|$ goes to $ \\infty$).\nA step farther, we can read that chain of \"or\" as \"at least one\":\n$ |x|$ goes to $ \\infty$ if and only if at least one of the $ |x_i|$ ($ i\\in\\{1,2,\\cdots,n\\}$) goes to $ \\infty$.[/quote]\r\n\r\nBut what does it mean by\r\n$ |x|$ goes to $ \\infty$ \r\n\r\nI mean how to define \r\n\"$ |x|$ goes to $ \\infty$ \"\r\nOr how to express it in terms of a logic expression", "Solution_3": "For all $ R > 0$, there exists $ N > 0$ such that for all $ [ [$whatever your time parameter is$ ] ]$ larger than $ N$, $ |x| > R$. \"For any arbitrarily large distance from the origin, eventually $ x$ is always that far away.\"\r\n[color=green]Moderator edit: accidental Wiki link removed[/color]", "Solution_4": "[quote=\"JBL\"]For all $ R > 0$, there exists $ N > 0$ such that for all $ [ [$whatever your time parameter is$ ] ]$ larger than $ N$, $ |x| > R$. \"For any arbitrarily large distance from the origin, eventually $ x$ is always that far away.\"\n[color=green]Moderator edit: accidental Wiki link removed[/color][/quote]\r\n\r\nIf this is not nonsenses ,what is.\r\n\r\nBut since you are good in refering people to books,refer me to a book that i can find that definition.\r\n\r\nWhat is the meaning of [b]|x| if $ x\\in R^n$[/b]", "Solution_5": "Usually, although the precise meaning is not for us to deduce, rather for you to provide given the context of the statement, the meaning is the following:\r\n\r\nFor $ x \\in \\mathbb{R}^n$ we have that $ |x|$ also denoted $ \\|x\\| \\equal{} \\sqrt{\\sum\\limits_{i\\equal{}1}^n x_i^2}$, the Euclidean norm.", "Solution_6": "|x| on its own in $ R^n$ it has no meaning ,because it could be any norm ,unless you state clearly what norm is\r\n\r\nIt could be the maxnorm.", "Solution_7": "It doesn't actually matter what norm is meant by $ |x|.$ Why not? Because all norms on a finite-dimensional space are equivalent, so if you prove this for one norm, it holds for all norms.\r\n\r\nIn fact, the original statement is closer to the \"maxnorm\", $ \\|x\\|_{\\infty}\\equal{}\\max_{1\\le j\\le n}(|x_j|)$ than it is to the Euclidean norm.", "Solution_8": "yes, but the proving process is completely different for each norm,on a certain problem.\r\n\r\nSo we have to specify which norm we are using.\r\n\r\nYou specified the maxnorm as more appropriate for the original statement" } { "Tag": [ "integration", "logarithms", "search", "real analysis", "real analysis unsolved" ], "Problem": "Prove this:\r\n\r\n\\[\\int_0^1 {\\frac{dx}{x^x}} = \\sum_{n=1}^{\\infty} {\\frac{1}{n^n}}\\]", "Solution_1": "Hint\r\n[hide]$\\frac{1}{x^x}=\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}x^n{\\ln x}^n$[/hide]", "Solution_2": "It has been posted before.\r\nAnd I gave a proof there...\r\nYou can search it on this forum" } { "Tag": [ "topology", "geometry", "inequalities", "number theory" ], "Problem": "I am not sure which forum to post this in, so i aplogize if this isn't where it should be. Anyways, i was just wondering. This is a very broad question but, what is/do you think the hardest/most difficult/most complicated etc field of mathematics that is currently in existence? I guess what i am asking is, what is the highest math such that there is no form of mathematics higher than it? Please grace me with your insight :)", "Solution_1": "Every advanced field is hard. Different people say different things; its only a matter of opinions. For me, I find number theory hard.", "Solution_2": "I was going to say the same thing, Number Theory. It's so theoretical and expansive at times. But specifically, I've heard people say that K-Theory is very difficult. It blends topology, number theory, tensors, etc. Used in theoretical physics, like string theory. Physicists are waiting for more mathematics (like K-Theory) to develop so that they can continue to build upon string theory.", "Solution_3": "geometry", "Solution_4": "Number Theory, and inequalities", "Solution_5": "@last three people: sorry, but do you really know what you are speaking of\u00bf\r\nThis topic seems not to concern olympiad math but \"real\" math.\r\nWhen you meant geometry as \"algebraic geometry\" or something like that, I could understand it. But for the way geomatry is normally used on that forum: I disagree (not saying that olympiad geometry is easy)." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "[quote=\"Vasc\"]\n\nThe general ineq is true for $a,b,c$ positive reals and $p>-2$ and $q\\ge 2p+1$.\n\n$\\displaystyle \\displaystyle \\frac{2a^2+qbc}{b^2+pbc+c^2}+\\frac{2b^2+qca}{c^2+pca+a^2}+\\frac{2c^2+qab}{a^2+pab+b^2} \\ge \\frac{3(2+q)}{2+p}$\n\n[/quote]\r\n\r\nThis is so nice and Vasc promised to post his solution.\r\n\r\nAs it is again solved I hope Vasc will post his solution.\r\n\r\n\r\nThank you very much , Vasc.", "Solution_1": "manlio, do you mean $q \\le 2p+1$?\r\nI wrote this in my exercise book.", "Solution_2": "Yes, sorry. :blush:" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "In parallelogram $ABCD$, a line from $C$ cuts diagonal $\\overline{BD}$ in $E$ and $\\overline{AB}$ in $F$. If $F$ is the midpoint of $\\overline{AB}$, and the area of $\\bigtriangleup BEC$ is $100$, find the area of quadrilateral $AFED$.", "Solution_1": "[hide]\n\\[\\frac{FC}{EC}= \\frac{3}{2},\\;\\frac{DB}{EB}= 3\\]\n\\[\\frac{[FCB]}{[ECB]}= 3/2 \\implies [FEB] = 50\\]\nSimilarly, \\[[DEC] = 200\\]\n\\[[ABD] = [BCD] = 300\\]\n\\[[AFED] = [ABD]-[FEB] = \\boxed{250}\\]\n[/hide]" } { "Tag": [ "geometry" ], "Problem": "Today we celebrate the victory of the democracy in Romania. The democratic-liberal lider Traian Basescu won with 51,23% of the votes the second round of the Presidential Elections in Romania, making him the first political figure in Romanian post-communist era which wins 3 elections in a row (2000 is elected mayor of Bucharest, in june 2004 is again elected mayor of Bucharest in first round - his counter-candidate obtaining half of his procentage, and in december 2004 is elected president after an on-the-edge presidential run). \r\n\r\nThe exit polls showed a 50-50% never enountered situation yesterday when the voting rooms closed. Most of the voters of Basescu are young, open-minded people, with a high-level of education, living especially in urban areas. The counter-cadidate Adrian Nastase, lider of the neo-communist party in Romania, was in general voted by old people, living in rural areas, and generally with below average education. \r\n\r\nFor the first time, the President is elected for 5 years (instead of 4).", "Solution_1": "[quote]Today we celebrate the victory of the democracy in Romania. The democratic-liberal lider Traian Basescu won with 51,23% of the votes the second round of the Presidential Elections in Romania,[/quote]\r\n\r\nI heard the good news from my parents a little earlier today. My parents voted Basescu in the first round, but inclement weather in Detroit made it difficult to vote in the second round. Did you vote Valentin?", "Solution_2": "[quote=\"tetrahedr0n\"][quote]Today we celebrate the victory of the democracy in Romania. The democratic-liberal lider Traian Basescu won with 51,23% of the votes the second round of the Presidential Elections in Romania,[/quote]\n\nI heard the good news from my parents a little earlier today. My parents voted Basescu in the first round, but inclement weather in Detroit made it difficult to vote in the second round. Did you vote Valentin?[/quote]Of course I voted. You can also figure out with who I voted, looking at the pool of voters of the two candidates. :)", "Solution_3": "[quote=\"Valentin Vornicu\"]You can also figure out with who I voted, looking at the pool of voters of the two candidates. :)[/quote]\r\n\r\nI dunno. I'm kind of puzzled here. :D :P \r\nJust kidding, please don't destroy me with your super-admin powers... :D ;)" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "P is a point out of ABC triangle. perpendiculars from P to BC,CA,AB meet them at D,E,F respectively.PCE,PBD and PAF have the same area.prove that the areas of these triangles are equal to area of ABC trianle.", "Solution_1": "Nobody want to help me?" } { "Tag": [ "function", "algebra", "polynomial", "conics", "algebra unsolved" ], "Problem": "find all continous fuction f:R...R such that f(xf(y)+yf(x) )=2f(x)f(y) $x,y\\in\\ R $", "Solution_1": "I'm new, so I'll plead ignorance for not including a formal proof. It appears that it works for f(x) = mx where m is any real number.\r\n\r\nf [ xf(y) + yf(x) ] =\r\nf [ xmy + ymx ] =\r\nf [ 2mxy ] = 2(m^2)xy\r\n\r\n and\r\n\r\n2f(x)f(y) = 2mxmy = 2(m^2)xy.\r\n\r\nCouldn't find it to be true for any other continuous functions (any other polynomials, the conic sections, trigonometric, exponential, or logarithmic functions) for all x, y in R.\r\n\r\nHope this gets some discussion going." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "prove that (a/b+b/c+c/a)^1/2 + ((ab+bc+ca)/(a^2+b^2+c^2))^1/2>=3^1/2 + 1 holds for all nonnegative real number a,b,c.", "Solution_1": "[quote=\"Lee Sang Hoon\"]prove that $ (a/b\\plus{}b/c\\plus{}c/a)^1/2 \\plus{} ((ab\\plus{}bc\\plus{}ca)/(a^2\\plus{}b^2\\plus{}c^2))^1/2>\\equal{}3^1/2 \\plus{} 1$ holds for all nonnegative real number a,b,c.[/quote] ....\r\nnow lets solve", "Solution_2": "$ \\sqrt{\\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a}}\\plus{}\\sqrt{\\frac{ab\\plus{}bc\\plus{}ca}{a^2\\plus{}b^2\\plus{}c^2}} \\geq \\sqrt{3}\\plus{}1$\r\nWe can use ABC theorem :wink:", "Solution_3": "[quote=\"chien than\"]$ \\sqrt {\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a}} \\plus{} \\sqrt {\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}} \\geq \\sqrt {3} \\plus{} 1$\nWe can use ABC theorem :wink:[/quote]\r\n\r\nCan you give exact solution???\r\n\r\nI want to know how can we apply ABC theorem!", "Solution_4": "This inequality is cyclic not symmetric how can i use ABC theroem? do I need to apply some transformation?", "Solution_5": "i have also tried to solve this problem by using some inequalities involving a^2b+b^2c+c^2a. but all faild. can some body give me some hints?" } { "Tag": [ "vector", "probability", "function", "analytic geometry", "probability and stats" ], "Problem": "Hi all. I was wondering if my work for the following problem was correct:\r\n\r\nSuppose (X,Y) is a random vector with a probability mass function p_(X,Y) of (i, j) = 1/10 for 1 \u2264 i \u2264 j \u2264 4.\r\n\r\nCompute the conditional expectation of Y given X.\r\n\r\nSo for this, I realized that the possibilities of (i,j) were (1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4). \r\n\r\nP(1,1) = .1\r\nP(1,2) = .1\r\nP(1,3) = .1\r\nP(1,4) = .1\r\nP(2,1) = 0\r\nP(2,2) = .1\r\nP(2,3) = .1\r\nP(2,4) = .1\r\nP(3,1) = 0\r\nP(3,2) = 0\r\nP(3,3) = .1\r\nP(3,4) = .1\r\nP(4,1) = 0\r\nP(4,2) = 0\r\nP(4,3) = 0\r\nP(4,4) = .1\r\n\r\nP(X=1) = .1 + .1 + .1 + .1 = .4\r\nP(X=2) = .1 + .1 + .1 = .3\r\nP(X=3) = .1 + .1 = .2\r\nP(X=4) = .1\r\n\r\nBelow I did p(x,y)/P(X=x) for each pair of coordinates listed above that had a probability not equal to 0. \r\nFor (1,1), I got 1/4\r\nFor (1,2), I got 1/4\r\nFor (1,3), I got 1/4\r\nFor (1,4), I got 1/4\r\nFor (2,2), I got 1/3\r\nFor (2,3), I got 1/3\r\nFor (2,4), I got 1/3\r\nFor (3,3), I got 1/2\r\nFor (3,4), I got 1/2\r\nFor (4,4), I got 1.\r\n\r\nMultiplying each of these probabilities by the value of the x coordinate, I got (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/3 * 2) + (1/3 * 2) + (1/3 * 2) + (1/2 * 3) + (1/2 * 3) + (1 * 4) = 10\r\n\r\nSo I got 10 for a conditional expectation of Y given X.", "Solution_1": "Which is obviously wrong because $ Y$ never exceeds $ 4$.\r\n\r\n\"Given X\" means you are told that $ X\\equal{}n$ and your answer will depend on $ n$. For example, if $ X\\equal{}3$, then you have either $ (3,3)$ or $ (3,4)$, hence\r\n\\[ E[Y\\, |\\, X\\equal{}3]\\equal{}\\frac{3P[(3,3)]\\plus{}4P[(3,4)]}{P[(3,3)]\\plus{}P[(3,4)]}\\equal{}3.5\\]", "Solution_2": "[quote=\"mlok\"]Which is obviously wrong because $ Y$ never exceeds $ 4$.\n\n\"Given X\" means you are told that $ X \\equal{} n$ and your answer will depend on $ n$. For example, if $ X \\equal{} 3$, then you have either $ (3,3)$ or $ (3,4)$, hence\n\\[ E[Y\\, |\\, X \\equal{} 3] \\equal{} \\frac {3P[(3,3)] \\plus{} 4P[(3,4)]}{P[(3,3)] \\plus{} P[(3,4)]} \\equal{} 3.5\n\\]\n[/quote]\r\n\r\nThanks!\r\n\r\nQuestion - then for the conditional expectation, would the solution just be the individual expectations for each of x=1,2,3,4 done out as what you demonstrated?\r\n\r\ni.e.:\r\n\r\nE(Y|X=1) = 2.5\r\nE(Y|X=2) = 3\r\nE(Y|X=3) = 3.5\r\nE(Y|X=4) = 4\r\n\r\nOR is the conditional expectation each of the E(Y|X=x) multiplied by the value of x and then added up?", "Solution_3": "It's the first. Conditional expectation is not a number - it's a random variable." } { "Tag": [ "function" ], "Problem": "Find the function $f$ that satisfies \\[ xf(x)+2xf(-x)=-1. \\]", "Solution_1": "[hide]\nWe can write another function using the cyclic property of this one:\n$-xf(-x) -2xf(x) = -1$\n\nThen, using system of linear equations we find that $f(x) = \\frac{1}{x}$[/hide]" } { "Tag": [], "Problem": "At a dance party a group of boys and girls exchange dances as follows: one boy dances with $ 5$ girls, a second boy dances with $ 6$ girls, and so on, the last boy dancing with all the girls. If $ b$ represents the number of boys and $ g$ the number of girls, then:\r\n\r\n$ \\textbf{(A)}\\ b \\equal{} g\\qquad \r\n\\textbf{(B)}\\ b \\equal{} \\frac{g}{5}\\qquad \r\n\\textbf{(C)}\\ b \\equal{} g \\minus{} 4\\qquad \r\n\\textbf{(D)}\\ b \\equal{} g \\minus{} 5\\qquad \\\\\r\n\\textbf{(E)}\\ \\text{It is impossible to determine a relation between }{b}\\text{ and }{g}\\text{ without knowing }{b \\plus{} g.}$", "Solution_1": "I think it is $ b\\equal{}g\\plus{}4$, but that is not an answer choice. Are you sure $ \\mathbf{(C)}$ isn't $ b\\equal{}g\\plus{}4$. \r\n\r\nTo solve the problem, you can make a table. \r\n\r\nB G\r\n1 5\r\n2 6\r\n3 7\r\n4 8\r\n5 9\r\n6 10\r\n...\r\n\r\nSo we find the pattern to be $ b\\equal{}g\\plus{}4$.", "Solution_2": "Actually, your table implies that $ b \\equal{} g \\minus{} 4 \\implies \\fbox{(C)}$.", "Solution_3": "Yeah, stupid mistake. I was thinking the other way around, $ g\\equal{}b\\plus{}4$. My answer is $ \\mathbf{(C)}$.", "Solution_4": "the answer is c.", "Solution_5": "I got (C) also :surf:" } { "Tag": [ "probability", "algorithm", "calculus", "inequalities", "vector", "logarithms", "function" ], "Problem": "Suppose there are 4 ships with probability $ P_i$ for the $ i^{th}$ ship of holding treasure (Each $ P_i$ is a real number between 0 and 1). Furthermore, there are 6 pirates who can each select at most 1 of the 4 ships and plunder its treasure with probability of success $ S_j$ for the $ j^{th}$ pirate (Each $ S_j$ is a real number between 0 and 1). Which ships should each pirate target to plunder in order to maximize the likelihood of the pirates stealing the treasure from the ships? Can you generalize with $ n$ ships and $ m$ pirates?\r\n\r\nAssume that the probability of success for each pirate is independent of its targeted ship. Also assume that all of the pirates will select their target ships independently and will plunder the ships at the same time. There are no restrictions to the number of pirates that can plunder a particular ship.", "Solution_1": "This is somewhat ambiguous. If all you want is to maximize the expected quantity of treasure, the answer is given by the greedy algorithm. The best pirate plunders the richest ship, and so forth.\r\n\r\n[b]Edit:[/b] As stated, shouldn't every pirate plunder the richest ship?", "Solution_2": "Sorry for the ambiguity. Let me restate the problem with more numbers.\r\n\r\nSuppose there are 4 ships with probabilities 50%, 30%, 15%, and 5% for the first, second, third, and fourth ship respectively of holding treasure. It is known that exactly 1 of the ships will have the treasure and the other 3 will have nothing valuable. Furthermore, there are 5 pirates who can each independently select at most 1 of the 4 ships and plunder its treasure with respective probabilities of success 90%, 80%, 60%, 50%, and 30%. Which ships should each pirate target to plunder in order to minimize the likelihood that the 5 pirates walk away with no treasure? (ie, the pirates will have no treasure if they decide to plunder the wrong ships or if the ones that plunder the ship with treasure are unsuccessful in plundering it.)\r\n\r\nIs it possible generalize with n ships and m pirates?", "Solution_3": "I think greedy algorithm wins here.", "Solution_4": "So would applying the greedy algorithm mean that you send all of the pirates to the ship with the largest chance of having the treasure (the 50% one)?\r\n\r\nCorrect me if I'm wrong, but the probability of the treasure not being plundered then would be:\r\n\r\n$ P_{plunder} \\equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...\r\n$ P_{plunder} \\equal{} (50\\%)*(1\\minus{}(1\\minus{}90\\%)(1\\minus{}80\\%)(1\\minus{}60\\%)(1\\minus{}50\\%)(1\\minus{}30\\%)) \\plus{} (30\\%)*(0\\%) \\plus{} (15\\%)*(0\\%) \\plus{} (5\\%)*(0\\%) \\equal{} (50\\%)*(99.72\\%) \\equal{} 49.86\\%$\r\n$ P_{not plunder} \\equal{} 1 \\minus{} P_{plunder} \\equal{} 1 \\minus{} 49.86\\% \\equal{} \\boxed{51.14\\%}$\r\n\r\nBut lets say that all pirates go for the 50% ship except the pirate who has an 80% success rate. The one with 80% success rate decides to plunder the ship that is 30% certain to contain the treasure.\r\n\r\nThe probability of the treasure still not being plundered from this pirate-ship assignment is:\r\n\r\n$ P_{plunder} \\equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...\r\n$ P_{plunder} \\equal{} (50\\%)*(1\\minus{}(1\\minus{}90\\%)(1\\minus{}60\\%)(1\\minus{}50\\%)(1\\minus{}30\\%)) \\plus{} (30\\%)*(80\\%) \\plus{} (15\\%)*(0\\%) \\plus{} (5\\%)*(0\\%) \\equal{} (50\\%)*(98.60\\%) \\plus{} (30\\%)*(80\\%) \\equal{} 73.3\\%$\r\n$ P_{not plunder} \\equal{} 1 \\minus{} P_{plunder} \\equal{} 1 \\minus{} 73.3\\% \\equal{} \\boxed{26.7\\%}$", "Solution_5": "[quote=\"Tojoyk\"]So would applying the greedy algorithm mean that you send all of the pirates to the ship with the largest chance of having the treasure (the 50% one)?\n\nCorrect me if I'm wrong, but the probability of the treasure not being plundered then would be:\n\n$ P_{plunder} \\equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...\n$ P_{plunder} \\equal{} (50\\%)*(1 \\minus{} (1 \\minus{} 90\\%)(1 \\minus{} 80\\%)(1 \\minus{} 60\\%)(1 \\minus{} 50\\%)(1 \\minus{} 30\\%)) \\plus{} (30\\%)*(0\\%) \\plus{} (15\\%)*(0\\%) \\plus{} (5\\%)*(0\\%) \\equal{} (50\\%)*(99.72\\%) \\equal{} 49.86\\%$\n$ P_{not plunder} \\equal{} 1 \\minus{} P_{plunder} \\equal{} 1 \\minus{} 49.86\\% \\equal{} \\boxed{51.14\\%}$\n\nBut lets say that all pirates go for the 50% ship except the pirate who has an 80% success rate. The one with 80% success rate decides to plunder the ship that is 30% certain to contain the treasure.\n\nThe probability of the treasure still not being plundered from this pirate-ship assignment is:\n\n$ P_{plunder} \\equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...\n$ P_{plunder} \\equal{} (50\\%)*(1 \\minus{} (1 \\minus{} 90\\%)(1 \\minus{} 60\\%)(1 \\minus{} 50\\%)(1 \\minus{} 30\\%)) \\plus{} (30\\%)*(80\\%) \\plus{} (15\\%)*(0\\%) \\plus{} (5\\%)*(0\\%) \\equal{} (50\\%)*(98.60\\%) \\plus{} (30\\%)*(80\\%) \\equal{} 73.3\\%$\n$ P_{not plunder} \\equal{} 1 \\minus{} P_{plunder} \\equal{} 1 \\minus{} 73.3\\% \\equal{} \\boxed{26.7\\%}$[/quote]\r\n\r\nNo, each pirate would go to whichever ship had the highest probability of containing the treasure. So if the pirate with 90% success plunders the 50%, then the 50% becomes a 5%.\r\n\r\nAlthough in this case using greedy algorithm needs an order to pick the pirates in. So maybe I should really spend more time trying to solve the problem instead of randomly guessing answers.", "Solution_6": "That seems like a logical method. But it seems like a big recursion loop where you evalute, compare, determine next ship, evalute, compare, etc. Is there a more explicit way to determine the answer?\r\n\r\nWhat if the probabilties of success for every pirate were the same?\r\n\r\nps, did I calculate the probabilties correctly for my scenarios?", "Solution_7": "[quote=\"Tojoyk\"]That seems like a logical method. But it seems like a big recursion loop where you evalute, compare, determine next ship, evalute, compare, etc. Is there a more explicit way to determine the answer?\n\nWhat if the probabilties of success for every pirate were the same?\n\nps, did I calculate the probabilties correctly for my scenarios?[/quote]\r\n\r\nMy method is wrong. We're not trying to find a better answer, we're trying to find an answer at all.\r\n\r\nIf the probabilities of success for every pirate were the same, then my method would work. Same if all the ships had equal chances to contain treasure.\r\n\r\nYour probabilities look correct to me.", "Solution_8": "Why is your method wrong? It seems recursive as in a computer program would be able to follow the alogirhm to reach the optimal answer.", "Solution_9": "As stated, the method would need to run through every permutation of the pirates. This has the potential to actually be [b]less[/b] efficient than running through every possible assignment of pirates to ships (depending on how many more ships than pirates there are), although both methods are woefully slow.\r\n\r\nThe problem is a little more sensible when stated in terms of failure rates, since failure rates are multiplicative, so I'll restate it as follows: there are probabilities $ P_i$ for each ship to have the treasure with $ \\sum P_i \\equal{} 1$, and there are probabilities $ F_i$ that each pirate fails to plunder the treasure. Say that ship $ i$ is targeted by pirates to be labeled $ (i, 1), (i, 2), ...$. Then we want to minimize something like\r\n\r\n$ \\sum_{i \\equal{} 1}^{n} P_i F_{(i, 1)} F_{(i, 2)} ...$\r\n\r\nover all assignments of pirates to ships. The greedy algorithm works fine if at most one pirate is allowed per ship; this is just the rearrangement inequality. Allowing multiple pirates to plunder a ship introduces a nonlinearity which is difficult to deal with; two pirates with failure probabilities $ F_i, F_j$ plundering a ship is equivalent to a single pirate with failure probability $ F_i F_j$.\r\n\r\nIt's not even clear to me how to solve this problem if every ship has equal probability of having the treasure. Then the problem is something like the following: we are given some positive real numbers less than $ 1$ and we want to find the minimum value of an expression made using those numbers, $ \\plus{}$, and $ \\times$.", "Solution_10": "How about an easier case when there are 5 pirates who have the each have the same 50% chance of success? Lets say that there are 3 ships with respective probabiltiies 60%, 30%, and 10% respectively of bearing the treasure. What should be the pirates' plan to make sure that they mimimize their chances of not getting the treasure?\r\n\r\nCorrect me if I'm wrong, but following from what I understand with the greedyness, we have 5 steps:\r\n\r\n1) We should send the first pirate to the first ship (first ship now has a 30% of holding treasure after this step)\r\n2) We send the second pirate to the first ship again (first ship now has a 15% of holding treasure after this step)\r\n3) third pirate to the second ship (second ship has now a 15% of holding treasure after this step)\r\n4) fouth pirate to the first ship (first ship has now a 7.5% of holding treasure after this step)\r\n5) fifth pirate to the second ship (second ship now as a 7.5% of holding treasure after this step).\r\n\r\nAfter the 5 pirates have tried to plunder the ships, the ships have a 7.5%, 7.5%, and 10% chance respectively of still having the treasure. Continuing this logic, a 6th pirate would attack the third ship.\r\n\r\nI still think that there should be a more constructive way to plan out the attacks. Like solving some equations where you get (3, 2, 0) as the answer or something?\r\n\r\nWhat about a more general scenario: The pirates will have equal proababilities of success/failure, but each of the ships still had their individual probabilities $ P_i$ to bear the treasure with $ \\sum P_i \\equal{} 1$. What would be the attack plan?", "Solution_11": "[quote=\"Differ\"]If the probabilities of success for every pirate were the same, then my method would work. Same if all the ships had equal chances to contain treasure.[/quote]\r\n\r\nI take this statement back. Looking at just two ships with 50% probability leads to trouble.", "Solution_12": "I was hoping to get some sort of algebraic solution to this problem, but I guess maybe using calculus might be necerssary to optimize the probability?", "Solution_13": "Alas, it is not even calculus. As stated, it includes the infamous bin packing problem as a partial case. Suppose that there are just 2 ships with probabilities $ P_1$ and $ P_2$ of having the treasure. Suppose we have many pirates and $ f_j$ are the [b]failure[/b] probabilities for them. Then you need to minimize $ P_1\\prod_{j\\in J_1}f_j\\plus{}P_2\\prod_{j\\in J_2}f_j$ over all partitions of the set of pirates into two subsets $ J_1$ and $ J_2$. The AM-GM inequality gives the lower bound $ 2\\sqrt {P_1P_2\\prod_j f_j}$ for this quantity. It is attained if and only if you can find a partition satisfying $ P_1\\prod_{j\\in J_1}f_j\\equal{}P_2\\prod_{j\\in J_2}f_j$. If you think of this equality for a few minutes, you'll realize that every bin packing problem can be restated this way. Pirate's life is not easy (actually, NP-hard), you know... :D", "Solution_14": "From the general case:\r\n\r\nEach of the $ m$ pirates have individual probabilties of [b]failure[/b]: $ F_j$. (Each pirate can only choose 1 ship to plunder)\r\nEach of the $ n$ ships have individual probabilties of bearing treasure: $ P_i$.\r\nExactly one of the $ n$ ships has the treasure: $ \\sum_{i\\equal{} 1}^{n} P_i \\equal{} 1$.\r\n\r\n\r\nLet the group of pirates that will target the first ship be denoted as $ J_1$. Likewise, the group of pirates that will target the second ship be denoted as $ J_2$, and so on for all of the ships. I came up with this as my generalized probability equation representing the probability that the treasure will be successfully plundered by the pirates:\r\n\r\n$ P \\equal{} \\sum_{i \\equal{} 1}^{n}(P_i*(1 \\minus{} \\prod_{j\\in J_i}F_j))$\r\n\r\nHowever, I have no idea how to minimize this with the $ P_i$ and $ F_j$ as parameters. \r\n\r\nWhile trying out different methods, I did realize that we can realize this equation as the dot product of two $ n$-component vectors, one of them is the vector with respective components $ a \\equal{} P_i$ and the other vector is one with respective components $ b \\equal{} 1 \\minus{} \\prod_{j\\in J_i}F_j$. And the question becomes trying to minmize the dot product of $ a$ and $ b$. But I didn't get too far with this approach either ...\r\n\r\nAny help would be appreciated. Maybe Im approaching this the wrong way?", "Solution_15": "Have you read my last remark? [b]Nobody[/b] currently knows how to do it in a much better way than considering all possible assignments and just computing the numbers. If you ever come up with one, it'll be quite a revolution in computer science. :roll:", "Solution_16": "Gotcha, but what about when all of the pirates have equal probabilities? Does there exist a more structured approach to the problem?", "Solution_17": "Then it is easy. You need to minimize the sum $ \\sum P_i F^{j_i}$ in your notation. WLOG $ P_1$ is the largest of $ P_i$. Note that there should be at least one factor $ F$ on $ P_1$ (otherwise, moving it there from anywhere else will decrease the sum). Thus, the assignment for the first pirate is clear. Once it has been made, you have exactly the same problem as before but with the numbers $ FP_1,P_2,\\dots,P_n$. So, again, the next pirate should be assigned to the largest of these numbers and so on. This runs in $ (n\\plus{}m)\\log n$ time (you need to sort the numbers in the beginning and after each step; the initial sorting costs $ n\\log n$ and placing the changed number correctly takes $ \\log n$ operations). Probably, you can improve the time to linear, but I do not see how to do this immediately.", "Solution_18": "Excellent, sounds like a simple program can be written to solve this semi-general case.\r\n\r\nHere was my attempt at trying to calculate the pirate-ship distribution:\r\n\r\nIf each of the $ m$ pirates had an equal probability of failure $ F$ and the $ n$ ships had individual probabilties $ P_j$ of bearing the 1 treasure, then the probability of succesfully collecting the treasure would be:\r\n\r\n$ (*) P_{\\text{success}} \\equal{} P_1*(1 \\minus{} F^{r_1}) \\plus{} P_2*(1 \\minus{} F^{r_2}) \\plus{} ... \\plus{} P_n*(1 \\minus{} F^{r_n}) \\equal{} \\sum_{i \\equal{} 1}^{n}(P_i*(1 \\minus{} F^{r_i}))$\r\n\r\nwhere the $ r_j$ is the number of pirates who plan to attack the jth ship.\r\n\r\nSince there are $ m$ total ships, we have a constraint that:\r\n\r\n$ (**) \\sum_{i \\equal{} 1}^{n}(r_i) \\equal{} m$\r\n\r\nSince we're trying to maximize (*), which can be thought of as a function of $ r_n$ ($ n$ total variables), would it be possible to treat the $ P_i$ and $ F$ as constants (they will be constants in less general cases) and apply Lagrange Multipliers [(*) is minimized and (**) is constaint equation]? I haven't actually tried this out yet, but if it works, we'll probably get non-integer answers for the $ r_n$, so we can just search integer solutions nearby to get the answer that we want? \r\n\r\nPlease let me know if my idea will be an utter failure before I actually try this out.", "Solution_19": "I do not see how it is simpler to implement than the algorithm I suggested (especially the last step: search for the integer solutions near the real solution). Also, if you've read my answer to your post in the Calculus forum, you might notice that the real-valued problem has not really much to do with Lagrange multipliers and requires sorting of $ P_i$ anyway. ;) So, unless $ m\\gg n$, I wouldn't bother with trying to improve the running time (not in your way, at least). If $ m\\gg n$, then we can try to think of something better. What are the actual values you want to work with?\r\n\r\nIn the case when the success probabilities are different, we cannot hope to get the truly optimal solution, as I explained, but we can try to get reasonably close to it. Again, it would help to have some idea of the actual numbers involved.", "Solution_20": "Ok thanks for the info. I was just wondering if there were a more explicit way to calculate the solutions.\r\n\r\nSo how would your algorithm work if we had 10 pirates each with 40% chance of failure and 3 ships (70%, 20%, 10%) respectively? How would we distribute the 10 pirates among the 3 ships to minimize chance of treasure leakage?", "Solution_21": "As I said, we can do better if $ m$ is much larger than $ n$ but for $ m$ smaller than or comparable to $ n$ this runs reasonably fast. \r\n\r\nIn your particular example, we'll have the following steps (at each step you replace the largest of the 3 numbers you have by 0.4 times it and add 1 pirate to the corresponding ship):\r\n\r\n0) P: 70,20,10 assignments: 0,0,0\r\n1) P: 28,20,10 assignments: 1,0,0\r\n2) P: 11.2,20,10 assignments: 2,0,0\r\n3) P: 11.2,8,10 assignments: 2,1,0\r\n4) P: 4.48,8,10 assignments: 3,1,0\r\n5) P: 4.48,8,4 assignments: 3,1,1\r\n6) P: 4.48,3.2,4 assignments: 3,2,1\r\n7) P: 1.792,3.2,4 assignments: 4,2,1\r\n8) P: 1.792,3.2,1.6 assignments: 4,2,2\r\n9) P: 1.792,1.28,1.6 assignments: 4,3,2\r\n10) P: 0.7168,1.28,1.6 assignments: 5,3,2\r\n\r\nwith final probability of failure 0.7168+1.28+1.6=3.5968%", "Solution_22": "Excellent! I think i can make a little program out of this. Although I do agree that if we had say 500 pirates and the same three ship (70, 20, 10) configuration, a more explicit optimization approach might be quicker." } { "Tag": [ "algorithm", "number theory" ], "Problem": "If you know how to play Nim, you don't need to click [hide=\"this.\"] 9 coins are set up in 3 rows the top one has 1 coin, the next has 3, and the last has 5. On your turn, you may remove any number of coins from any [b]one[/b] row. The player who removes the last coin loses. [/hide] There have been many game theory problems based on Nim and variations of it.\n\n[b]Question 1:[/b] It is possible to always win Nim. To always win, should you go first or second? And what algorithm do you use to choose your moves?\n\n[hide=\"Variation 1: Extreme Nim\"] The gameplay is the same as original Nim, but the first row starts with 3 coins, the second with 5, and the last with 7.[/hide]\r\n\r\n[b]Question 2:[/b] Is it possible to always win Extreme Nim?\r\n\r\nMore problems coming later...", "Solution_1": "[hide=\"Variation 2: Nim-25\"]\nNim-25 is played with a 5x5 board of coins. On your turn, you may remove up to 4 coins from any row or column. The player that is forced to remove the last coin loses.\n[/hide]\r\n[b]Question 3:[/b] Is it possible to always win Nim-25?[/b]", "Solution_2": "Is the answer to #1,\r\n\r\n[hide]remove 3 from the 5 coin pile so that the coins are 1,3,2 after this step, no matter what your opponent does, you can reduce the coins to 1,1,1 or 2,2 or just 1, where you win[/hide]\r\n\r\nI have to write a paper on game theory, which i know little about (at the moment) and i stumbled upon this", "Solution_3": "I have a number theory book that explain how to win this with any amount of coins in any amount of columns, though it may not work with Nim-25.\r\nThe explaination they provide invovle looking at each piles' number of coins as binary.\r\nI can post it if you want.", "Solution_4": "[quote=\"Yagmoth0000\"][hide]remove 3 from the 5 coin pile so that the coins are 1,3,2 after this step, no matter what your opponent does, you can reduce the coins to 1,1,1 or 2,2 or just 1, where you win[/hide][/quote]\r\n\r\nYeah, that's right.\r\n\r\nThere's a variation of NIM: Pearls before swine.\r\n\r\nlink: http://www.transience.com.au/pearl3.html\r\n\r\nLevels 2 and beyond are all NIM with more pearls. Level 1 is pointless once you figure out an algorithm.", "Solution_5": "[quote=\"AstroPhys\"]I have a number theory book that explain how to win this with any amount of coins in any amount of columns, though it may not work with Nim-25.\nThe explaination they provide invovle looking at each piles' number of coins as binary.\nI can post it if you want.[/quote]I have seen the solution involving binary before, but I forget it. could you post it?", "Solution_6": "This works best if you go first, but it should still work if you go second.\r\nFirst you put the number of coins of each pile into sums of powers of two.\r\nThen you attempt to \"balance\" the piles so that each power of two is in a pair.\r\n\r\nFor example\r\nYou have three piles with $17 \\ , \\ 28 \\ , \\ 30$\r\nIn sums $17 = 16+1$ , $30 = 16+8+4+2$ , $28 = 16+8+4$\r\nThen to balance, you would take you take $15$ from the pile with $17$.\r\nThis leaves you with pairs for each of the remaining powers of two.\r\n$2 \\ , \\ 16+8+4+2 \\ , \\ 16+8+4$.\r\nThe pattern continues until there are no more coins.\r\n\r\nNow to generalize this method, you would just need to make more than one pair of each for multiple piles.\r\nIf modified it may work for Nim-25.", "Solution_7": "@ Astrophys...\r\n\r\n\r\nWhat do you mean by \"Now to generalize this method, you would just need to make more than one pair of each for multiple piles. \"\r\n\r\nand also, isn't it not possible to do the powers of 2 pairs in some cases?[/quote]", "Solution_8": "By generalizing, I meant to extend the method to games of Nim with more than three piles, then having more than one pair per power of two.\r\n\r\nAccording to my book, the method is applicable to all case regardless of the number of coins per pile.", "Solution_9": "Indeed, it is. The method is actually write all the numbers in binary and bitwise xor them together (so 0+0=0, 0+1=1, 1+0=1, and 1+1=0. This is the same as adding without carrying). You want to move into a position where this \"nim-sum\" is 0.\r\n\r\nExample: Say the piles are\r\n45, 32, 16, and 55\r\n\r\n101101 = 45\r\n100000 = 32\r\n010000 = 16\r\n110111 = 55\r\n-------------------\r\n101010\r\n\r\nSo, we need to toggle 101010 on some of them. The winning moves are:\r\n101101 -> 000111 = 45 -> 7\r\n100000 -> 001010 = 32 -> 10\r\n110111 -> 011101 = 55 -> 29\r\n\r\nAlso, technically, this game should be called misere nim. In actual nim, you want to take the last stone, as the person who can't move loses." } { "Tag": [ "function", "trigonometry", "algebra", "polynomial", "inequalities" ], "Problem": "write cos5 theda as a function of cos theda", "Solution_1": "Do you mean $\\cos^{5}\\theta$ or $\\cos 5\\theta$?", "Solution_2": "[quote=\"Ubemaya\"]Do you mean $\\cos^{5}\\theta$ or $\\cos 5\\theta$?[/quote]\r\nprobably(wow took me 5 tries to type this right) the latter.\r\n[hide]Consider $(\\cos x+i\\sin x)^{5}.$ Now expand that thing using binomial thrm.That thing is equal to $\\cos5x+i\\sin5x$ by DeMoivre's thrm. Set the real parts equal to each other, and boom, we're done. \n\noops i used $x$ instead of $\\theta$[/hide]", "Solution_3": "[hide=\"Hint.\"]Use $\\cos{4\\theta+\\theta}$, $\\cos{2(2\\theta)}$, and $\\cos2\\theta$. :D [/hide]", "Solution_4": "[hide=\"partial solution\"]From DeMoivre, $\\cos5\\theta+i\\sin5\\theta=(\\cos\\theta+i\\sin\\theta)^{5}$\n$=\\cos^{5}\\theta+i5\\cos^{4}\\theta\\sin\\theta-10\\cos^{3}\\theta\\sin^{2}\\theta-i10\\cos^{2}\\theta\\sin^{3}\\theta+5\\cos\\theta\\sin^{4}\\theta+i\\sin^{5}\\theta$.\nEquating real parts, $\\cos5\\theta=\\cos^{5}\\theta-10cos^{3}\\theta\\sin^{2}\\theta+5\\cos\\theta\\sin^{4}\\theta$.\n$=\\cos^{5}\\theta-10\\cos^{3}(1-\\cos^{2}\\theta)+5\\cos\\theta(1-\\cos^{2}\\theta)^{2}$\nI don't fell like doing the rest...[/hide]", "Solution_5": "[quote=\"bigboy82892\"][hide=\"Hint.\"]Use $\\cos{4\\theta+\\theta}$, $\\cos{2(2\\theta)}$, and $\\cos2\\theta$. :D [/hide][/quote]\r\n\r\nWell, it's easier to used DeMoivre's theorem, since what if I asked you to find $\\cos{783\\theta}$?", "Solution_6": "DeMoivre's isn't that easy, since you have annoying $(1-\\cos^{2}x)^{k}$ expressions to expand. It would become quite tedious if I asked you to find, say, $\\cos 10x$ as a polynomial in $\\cos x$. \r\n\r\n[hide=\"I would try this instead...\"] Let $T_{n}(\\cos \\theta) = \\cos n \\theta$ be the Chebyshev Polynomials of the First Kind. then\n\n$T_{0}(x) = 1$\n$T_{1}(x) = x$\n$T_{n+1}(x) = 2x T_{n}(x)-T_{n-1}(x)$ (good lemma! Can you prove it? ;) )\n\nFrom here you can quickly calculate that\n\n$T_{5}(x) = 16x^{5}-20x^{3}+5x$ [/hide]" } { "Tag": [], "Problem": "If $n$ is a sum of squares, prove that $2n$ is also a sum of squares.", "Solution_1": "What a well stated problem :roll: \r\n\r\nIf $n$ is sum of squares, then $2n= \\underbrace{1^{2}+1^{2}+...+1^{2}}_{2n \\text{ times}}$ is so, too ;)", "Solution_2": "You meant sum of two squares, of course. \r\n\r\n[hide=\"Generalization\"] $(a^{2}+b^{2})(c^{2}+d^{2}) = (ac \\pm bd)^{2}+(ad \\mp bc)^{2}$\n\nThis problem is the special case $c = d = 1$. [/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions from reals to reals such that:\r\n$ f(x\\plus{}f(y)) \\equal{} x \\plus{} f(f(y))$ and $ f(2004) \\equal{} 2005$\r\n[hide=\"my approach\"]\nThe way I solved this was that I showed that if $ y\\equal{}2004$ then you see that $ f(x\\plus{}2005) \\equal{} x \\plus{} f(2005)$ so I thought that the answer might be all functions of the form $ f(x) \\equal{} x \\plus{} k$ where k is some real, but how do I know that I have found all possible functions? Is there a general way to solve functional equations??? Like, what methods should I know?\n[/hide]\r\nThanks.", "Solution_1": "Let $ g(x)\\equal{}f(x)\\minus{}x$ then from $ f(x\\plus{}2005)\\equal{}x\\plus{}f(2005)$ we have $ g(x\\plus{}2005)\\equal{}g(2005)$, it mean $ g(x)\\equal{}c$, in which $ c\\equal{}g(2005)$: const!\r\nThen OK!", "Solution_2": "[quote=\"Nawahd Werdna\"]Find all functions from reals to reals such that:\n$ f(x \\plus{} f(y)) \\equal{} x \\plus{} f(f(y))$ and $ f(2004) \\equal{} 2005$\n[hide=\"my approach\"]\nThe way I solved this was that I showed that if $ y \\equal{} 2004$ then you see that $ f(x \\plus{} 2005) \\equal{} x \\plus{} f(2005)$ so I thought that the answer might be all functions of the form $ f(x) \\equal{} x \\plus{} k$ where k is some real, but how do I know that I have found all possible functions? Is there a general way to solve functional equations??? Like, what methods should I know?\n[/hide]\nThanks.[/quote]\r\nTake $ y\\equal{}0$ easy to check :$ f(x)\\equal{}x\\plus{}a$\r\nFrom $ f(2004)\\equal{}2005$ then $ a\\equal{}1$\r\n$ f(x)\\equal{}x\\plus{}1$", "Solution_3": "let $ y\\equal{}2004$,we get that:\r\n\r\n$ f(x\\plus{}2005)\\equal{}x\\plus{}f(2005)$ (*)\r\n\r\nnow in (*) set $ x\\equal{}\\minus{}1$,we get that:\r\n\r\n$ f(2004)\\equal{}\\minus{}1\\plus{}f(2005)\\Rightarrow f(2005)\\equal{}2006$\r\n\r\n$ (*)\\Rightarrow f(\\underbrace{x\\plus{}2005}_{t})\\equal{}x\\plus{}2006$\r\n\r\n$ \\Rightarrow\\boxed{f(t)\\equal{}t\\plus{}1}$\r\n\r\n------------------------------------------------\r\nanother solution is the following one:\r\n\r\nlet $ x\\equal{}\\minus{}f(f(y))$ we get that:\r\n\r\n$ f\\left(\\underbrace{f(y)\\minus{}f(f(y))}_{t}\\right)\\equal{}0$\r\n\r\n$ \\Rightarrow f(t)\\equal{}0$\r\n\r\nnow let $ y\\equal{}t$ we get that:\r\n\r\n$ f(x)\\equal{}x\\plus{}f(0)$\r\n\r\nnow let $ x\\equal{}2004$ we get that:\r\n\r\n$ f(2004)\\equal{}2004\\plus{}f(0)\\Rightarrow f(0)\\equal{}1$\r\n\r\n$ \\Rightarrow \\boxed{f(x)\\equal{}x\\plus{}1}$" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let $f\\in\\mathbb{R}[x],\\deg{f}=2003$ and $f(k)=\\frac{k^{2}}{k+1}(k=1,2,...,2004)$. Find value $f(2005)$.", "Solution_1": "[quote=\"N.T.TUAN\"]Let $f\\in\\mathbb{R}[x],\\deg{f}=2003$ and $f(k)=\\frac{k^{2}}{k+1}(k=1,2,...,2004)$. Find value $f(2005)$.[/quote]\r\n\r\nDefine the polynomial $h$ by: \\[h(x)=(x+1)f(x)-x^{2}\\], note that $\\deg{h}=2004$ and $0=h(1)=h(2)=\\ldots=h(2004)$, so $h$ has the form: \\[h(x)=K(x-1)(x-2)\\ldots(x-2004)\\] Also, since $h(-1)=0.f(-1)-1=-1$ we get the value of $K=\\frac{-1}{2005!}$. So \\[h(x)=\\frac{-1}{2005!}(x-1)(x-2)\\ldots(x-2004)\\] And $h(2005)=2006.f(2005)-2005^{2}=\\frac{-1}{2005!}.2004.2003\\ldots1=\\frac{-1}{2005}$. Finally $f(2005)=\\frac{2005^{3}-1}{2005.2006}$.\r\n\r\n$Tipe$", "Solution_2": "Me too! But I think this problem also sloved by using \r\nLagrange Interpolating Polynomial \r\nhttp://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html\r\nBecause $\\deg{f}=2003$ and we know $f(1),f(2),...,f(2004)$." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Show that there doesn't exist natural numbers $ k$ and $ m$ that satisfy\r\n\\[ k! \\plus{} 48 \\equal{} 48(k \\plus{} 1)^m\r\n\\]", "Solution_1": "The question $ \\Leftrightarrow 15 \\times \\frac {k!}{6!} \\plus{} 1 \\equal{} (k \\plus{} 1)^m$ , so $ k \\plus{} 1$ coprime with any of the $ n!$, where $ 7 \\le k \\plus{} 1 \\le k$. \r\nCase 1:$ k \\plus{} 1$ is not prime. It's easy to see that for $ k \\ge 10$ it is impossible, and it's impossible for $ 6 \\le k \\le 9$ too after checking. \r\nCase 2:$ k \\plus{} 1$ is a prime. Then we have $ k! \\plus{} 48 \\equal{} \\minus{} 1 \\plus{} 48 \\equal{} 47 \\equal{} 0(mod \\;\\; k \\plus{} 1)$ by Wilson, so $ k \\plus{} 1 \\equal{} 47$ , $ 46! \\plus{} 48 \\equal{} 48(47)^m$, which is impossible too.\r\n\r\nDone.", "Solution_2": "Please word your solution more clearly; I think you confused some variables.\r\n[hide=\"solution\"]\n\\[ (k\\minus{}1)!\\equal{}48((k\\plus{}1)^{m\\minus{}1}\\plus{}...\\plus{}1)\\]\nIf $ k\\plus{}1$ is prime then by Wilson's, $ 1\\equiv 48 \\mod{k\\plus{}1}$ so $ k\\equal{}46$; now, 46!+48=48(47)^m. But 47 is not a wilson prime.\nIf $ k\\plus{}1$ is not prime then $ 0\\equiv 48\\mod{k}$, a contradiction.\n[/hide]" } { "Tag": [ "geometry", "circumcircle" ], "Problem": "The number of diagonals of a regular polygon of 2n sides, not through the center of the circle to the polygon is given by:\r\n\r\na) 2n(n-3)\r\nb)2n(n-2)\r\nc) 2n(n-1)\r\nd) [n(n-5)]/2", "Solution_1": "[hide=\"Solution\"]The total number of diagonals a regular 2n sided polygon has is $ \\frac {2n(2n \\minus{} 3)}{2} \\equal{} n(2n \\minus{} 3) \\equal{} 2n^2 \\minus{} 3n$. The diagonals that cross through the center of the circle to the polygon (I assume circumcenter?) are the ones that have endpoints of verticies that are diametrically opposite each other. Each vertex (in a regular polygon with even side lengths) has another vertex that is diametrically opposite it. So it seems like we have $ 2n$ diagonals that cross the circumcenter. However, we over counted by a factor of $ 2$. Thus, we want to subtract out (from $ 2n^2\\minus{}3n$) $ \\frac {2n}{2} \\equal{} n$. Thus, our answer is $ 2n^2 \\minus{} 3n \\minus{} n \\equal{} 2n^2 \\minus{} 4n \\equal{} 2n(n \\minus{} 2)\\implies \\boxed{\\textbf{b)}}$[/hide]", "Solution_2": "$ \\underbrace{\\left(\\binom{2n}{2}\\minus{}2n\\right)}_{\\text{diagonals of a regular polygon of} \\ 2n \\ \\text{sides}}\\minus{}\\underbrace{n}_{\\text{not through the center of the circle to the polygon}}$\r\n$ \\equal{} \\boxed{2n(n\\minus{}2)}$", "Solution_3": "Ah ok, good explanation. Thanks ;)" } { "Tag": [ "linear algebra", "matrix", "college contests" ], "Problem": "Let A = matrix{[1, 0, 1];[0, 1, 2]; [0, 0, -1])\r\n\r\nFind all matirx B of size 3x3 such that AB+BA = 0.\r\n\r\nSource: VUMC 2001\r\n\r\nNamdung", "Solution_1": "well if u put \r\nB=\r\na b c\r\nd e f\r\ng h i \r\n\r\nand solve the sistem you get B.\r\na lot of computation, but I think it works!", "Solution_2": "Here the system like Lagrangia suggest us\r\n\r\n2a+g=0\r\n2b+h=0\r\ni+a+2b=0\r\n2d+2g=0\r\n2e+2h=0\r\n2i+d+2e=0\r\n-2i+g+2h=0\r\n\r\nNow your turn solve it", "Solution_3": "There is a non-computational sln:\r\n\r\nNote that A^2 = 1. If AB+BA=0 then B = - ABA, so B = X - AXA, where X = (1/2)B. Conversely, if B = X - AXA then AB + BA = AX - XA + XA - AX = 0.\r\n\r\nThus, we can conclude all B we have to find have form X - AXA where X - arbitrary matrix order 3x3.\r\n\r\nNamdung", "Solution_4": "[quote=\"Namdung\"]\n.....\nNote that A^2 = 1. If AB+BA=0 then B = - ABA, so B = X - AXA, where X = (1/2)B..... \n\nNamdung[/quote]\r\n\r\nWith your idea B is not only of the form X-AXA\r\n \r\nFrom B=-ABA => \r\nB = B/q - (q-1)ABA/q q any integer > 1\r\n => B=X-(q-1)AXA with X=B/q \r\n\r\nConversely if B=X-(q-1)AXA \r\nAB+BA = AX-(q-1)XA+XA-(q-1)AX=2XA+2AX-qXA-qAX=\r\n2/qBA+2/qAB-AB-BA=(2/q-1)(AB+BA)\r\n\r\nwe get (2/q-2)(AB+BA)=0 => AB+BA=0\r\n\r\n\r\n\r\nI solved the system here the solution\r\n\r\n(a b c)\r\n((2a) (2b) f)\r\n((-2a) (-2b) (-a-2b)) where a,b,c,f are reals" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Find all natural numbers $n$, for which the inequality $a(a+b)^{n}+b(b+c)^{n}+c(c+a)^{n}\\ge 0$ holds for all real numbers $a,b,c$", "Solution_1": "[quote=\"Beat\"]Find all natural numbers $n$, for which the inequality $a(a+b)^{n}+b(b+c)^{n}+c(c+a)^{n}\\ge 0$ holds for all real numbers $a,b,c$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=55723", "Solution_2": "how to prove it for n=3, and that there are no other n different from 1,3,5 that this inequality is true\r\n?", "Solution_3": "[quote=\"Beat\"]how to prove it for n=3...\n?[/quote]\nThe same idea helps:\nLet's $a+b=z,$ $a+c=y$ and $b+c=x.$\nHence, $\\sum_{cyc}a(a+b)^{3}\\geq0\\Leftrightarrow\\sum_{cyc}(x+z-y)x^{3}\\geq0\\Leftrightarrow\\sum_{cyc}(x^{4}-x^{3}y+xy^{3})\\geq0\\Leftrightarrow$\n$\\Leftrightarrow\\sum_{cyc}\\left((2x+y)^{2}(4x^{2}-12xy+11y^{2})+5x^{2}\\right)\\geq0.$\n[quote=\"Beat\"]... and that there are no other n different from 1,3,5 that this inequality is true?[/quote]\r\nFor $n\\geq7,$ $n\\in2\\mathbb N+1$ try $a=1,$ $b=-2n$ and $c=n.$ :wink:" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $PB$, $PC$ are tangents from $P$ to circle $(O)$ ($B,C$ in $(O)$).\r\nSuppose $PO$ meets $(O)$ at $A$ and $J$ ($J$ lies on the segment $PA$). $CO$ meets $(O)$ at $E$, the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively. $AE$ meets $PQ$ at $X$. \r\nProve that $XA=XD$.\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "[quote=\"Herodatviet\"] bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ \n[/quote]\r\n :?:", "Solution_2": "[quote=\"Herodatviet\"]Let $PB$, $PC$ are tangents from $P$ to circle $(O)$ ($B,C$ in $(O)$). Suppose $PO$ meets $(O)$ at $A$ and $J$ ($J$ lies on the segment $PA$). $CO$ meets $(O)$ at $E$, the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively, $AE$ meets $PQ$ at $X$. \nProve that $XA=XD$.[/quote]\r\n\r\nAre you sure you don't have some typos here? Anyway, I could imagine that you wanted to say \"the bisector of angle $ACO$ meets $OP$ and $(O)$ at $D$ and $Q$ respectively\" instead of \"the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively\"; this would make the problem, at least, look right on an imprecise sketch, but zooming in shows that the problem is still wrong in this case.\r\n\r\n darij", "Solution_3": "Im sorry it is :'bisector of $ABO$ meets $AC$,$(O)$ $Q,D$ '", "Solution_4": "[quote=\"Herodatviet\"]Im sorry it is :'bisector of $ABO$ meets $AC$,$(O)$ $Q,D$ '[/quote]\r\n\r\nStill wrong (the segments XA and XD are [i]almost[/i] equal). :(\r\n\r\n darij" } { "Tag": [ "algebra", "system of equations" ], "Problem": "For what value of the constant $ a$ does the system of equations below have infinitely many solutions? \\begin{align*} 3x + 2y &= 8,\\\\ 6x &= 2a - 7 - 4y \\end{align*}", "Solution_1": "[quote=\"GameBot\"]For what value of the constant $ a$ does the system of equations below have infinitely many solutions?\n\\begin{align*} 3x + 2y & = 8, \\\\\n6x & = 2a - 7 - 4y \\end{align*}\n[/quote]\r\n\r\nInfinitely many solutions simplifies to $ 0 = 0$.\r\nBring all variables to the left:\r\n$ 3x+2y=8$\r\n$ 6x-2a+4y=-7$\r\n\r\nMultiply first eq by 2 as you would to solve for a\r\n$ 6x+4y=16$\r\nSubtract the second eq from first, you get\r\n$ 2a=23$\r\n$ a=\\boxed{11.5}$" } { "Tag": [ "modular arithmetic", "number theory", "prime numbers" ], "Problem": "Hello. I recently looked at CMO #4 and tried to solve it. I was wondering if you guys could give me tips for writing (also I am kind of happy because even though it is sad that this is one of the hardest problems that I have solved, I solved it :) ). I hope that my solution is correct, however I have more faith in the AOPS community than anything else to help me check it over. I will conceal my solution for those who want to try the problem.\r\n\r\n[b]Problem[/b]\r\n\r\nProve that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^{n}\\minus{}n$.\r\n\r\n[hide] Firstly, I looked for a solution in terms of $ p$ with the knowledge that $ p>1$. Using Euler's Totient Theorem, we get the following.\n\n$ 2^{\\varphi(p)} \\equiv 1 \\pmod{p}$\n\nSince there exist $ p\\minus{}1$ numbers $ k$ such that $ 01$, $ (p\\minus{}1)^{2k\\minus{}1}>0$.\n\n$ 2^{(p\\minus{}1)^{2k}} \\equiv 1^{(p\\minus{}1)^{2k\\minus{}1}} \\pmod{p}$\n\n$ 2^{(p\\minus{}1)^{2k}} \\equiv 0\\plus{}1 \\pmod{p}$\n\nNow lets examine $ (p\\minus{}1)^{2k}\\minus{}1$ in $ \\pmod{p}$.\n\n$ (p\\minus{}1)^{2k}\\minus{}1 \\equiv (\\minus{}1)^{2k}\\minus{}1 \\equiv 1\\minus{}1 \\equiv 0 \\pmod{p}$\n\nNow we can substitute, yielding that,\n\n$ 2^{(p\\minus{}1)^{2k}} \\equiv (p\\minus{}1)^{2k}\\minus{}1\\plus{}1 \\equiv (p\\minus{}1)^{2k} \\pmod{p}$\n\nRearranging yields the following, that\n\n$ 2^{(p\\minus{}1)^{2k}}\\minus{}(p\\minus{}1)^{2k} \\equiv 0 \\pmod{p}$\n\nTherefore for all prime numbers $ p$ and positive integers $ k$, there will exist infinite solutions for $ n$ such that $ p|(2^{n}\\minus{}n)$ that take the form $ n\\equal{}(p\\minus{}1)^{2k}$. [/hide]", "Solution_1": "Would anyone mind giving feedback on my solution, or providing a more legitimate solution.", "Solution_2": "I'd be glad to check it for you, but I don't know some of the concepts involved in your solution (like Euler's totient). I think posting it in Pre-Olympiad might help." } { "Tag": [], "Problem": "A jar has 8 red, 9 blue and 10 green marbles. What is the least\nnumber of marbles you can remove from the jar to guarantee\nthat you have three of the same color?", "Solution_1": "[quote=\"GameBot\"]A jar has 8 red, 9 blue and 10 green marbles. What is the least\nnumber of marbles you can remove from the jar to guarantee\nthat you have three of the same color?[/quote]\r\n\r\nThe worst case scenario is 2 red,blue,green. We need to pick 1 more, so our answer is 7." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Sequence $ \\{ f_n(a) \\}$ satisfies $ \\displaystyle f_{n\\plus{}1}(a) \\equal{} 2 \\minus{} \\frac{a}{f_n(a)}$, $ f_1(a) \\equal{} 2$, $ n\\equal{}1,2, \\cdots$. If there exists a natural number $ n$, such that $ f_{n\\plus{}k}(a) \\equal{} f_{k}(a), k\\equal{}1,2, \\cdots$, then we call the non-zero real $ a$ a $ \\textbf{periodic point}$ of $ f_n(a)$.\r\n\r\nProve that the sufficient and necessary condition for $ a$ being a $ \\textbf{periodic point}$ of $ f_n(a)$ is $ p_n(a\\minus{}1)\\equal{}0$, where $ \\displaystyle p_n(x)\\equal{}\\sum_{k\\equal{}0}^{\\left[ \\frac{n\\minus{}1}{2} \\right]} (\\minus{}1)^k C_n^{2k\\plus{}1}x^k$, here we define $ \\displaystyle \\frac{a}{0}\\equal{} \\infty$ and $ \\displaystyle \\frac{a}{\\infty} \\equal{} 0$.", "Solution_1": "I think the narration must be wrong! You mean that Fn+1=2,then Fn=$\\infty$. What's that?", "Solution_2": "I don't think that def is needed... some other one!" } { "Tag": [ "conics", "parabola", "analytic geometry", "function" ], "Problem": "The equation y = (x - c)^2 defines a family of parabolas, one parabola for each value of c. On one set of coordinate axes, graph the memebers of the family for c = 0, c = 3, c = -2.", "Solution_1": "[hide=\"Hint\"] $f(x-c)$ is $f(x)$ shifted $c$ units right. $f(x+c)$ is $f(x)$ shifted $c$ units left. [/hide]", "Solution_2": "Thanks for the hint but can you show me how to do this type of question?\r\n\r\nYour hint does not lead me on the right track to the answer.", "Solution_3": "to shift the parabola $y=x^{2}$ to the right, say 5 units, we use the function $y=(x-5)^{2}$. This is because if by shifting the graph to the right 5 units, putting in the value of x+5 in the shifted graph would be the same as putting in the value of x in the original. We see this works because $y=(x+5-5)^{2}=x^{2}$.\r\n\r\nThus shifting the parabola to the right $c$ units, we get the function $y=(x-c)^{2}$", "Solution_4": "tjhance...thanks for the tips and extra reply." } { "Tag": [], "Problem": "A man realizes that the product of his two sons' ages is his age and that the product of his daughters' ages is his wife's age. He also realizes that the sum of his kids' ages will surpass the sum of him and his wife's ages in $25$ years. He also realizes that $n$ years ago, the product of his kids' ages was less than his own age. Compute the smallest possible $n$.", "Solution_1": "how old does the mom need to be to have children?...if its any age I get..[b]2[/b]..sounds weird.. :maybe: \r\n\r\n\r\nWhen the children are ages (right now) 28,2,6,6 and the dad is 56 and mom is 36?\r\n\r\nIm probalby wrong but I got this equaition... (a-1)(b-1)+(c-1)(d-1)=52", "Solution_2": "[hide=\"My original setup was\"]\nMan: 40\nWoman: 35\nSons: 4, 10\nDaughters: 5, 7\n[/hide]\r\n\r\nNote that I said the sum will SURPASS, not EQUAL. They are very different.", "Solution_3": "Actually, it doesnt make that big of a difference\r\n\r\nMy equation would just be changed to (a-1)(b-1)+(c-1)(d-1)<52..in your case you set (a-1)(b-1)+(c-1)(d-1)=51..\r\n\r\nwhich would your answer faulty because I can easily find another case:\r\n\r\nfor example- \r\ndad-48\r\nmom-38\r\nboys-4,12\r\ngirls-2,19\r\n\r\nthat contradicts your answer..and still get [b]2[/b]\r\n\r\nA better question would be to ask the largest possible value of n", "Solution_4": "Okay, then find the largeset possible $n$." } { "Tag": [ "function", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "f(x) is measurable on (0,1). And if we know that \r\nf(x)-f(y) is integrable on [0,]x[0,1]. \r\n\r\nProve: f(x) is integrable (that is f in L[0,1]).\r\n\r\n(I am supposed to use Fubini?)", "Solution_1": "In essence, yes. What does it mean for a function $g(x,y)$ to be integrable on $[0,1]^{2}$? By Fubini, the iterated integral $\\int_{o}^{1}\\int_{0}^{1}g(x,y)\\,dx\\,dy$ exists (and is equal to the integral of $g$). In particular, $\\int_{0}^{1}g(x,y)\\,dx$ exists and is finite for almost all $y$, since it is an integrable function of $y$.\r\n\r\nApplying the above to $g(x,y)=f(x)-f(y)$, $\\int_{0}^{1}f(x)-f(y)\\,dx=\\int_{0}^{1}f(x)\\,dx-f(y)$ exists and is finite for almost all $y$. For this to be true, $f$ must be integrable.", "Solution_2": "[quote=\"jmerry\"]$\\int_{0}^{1}f(x)-f(y)\\,dx=\\int_{0}^{1}f(x)\\,dx-f(y)$ exists and is finite for almost all $y$. [/quote]\r\nI finally work out a not straitforward pf:\r\nActually we don't know:\r\n$\\int_{0}^{1}f(x)-f(y)\\,dx=\\int_{0}^{1}f(x)\\,dx-f(y)$\r\nlinearity true for integrable functions, but we haven't prove f is integrable yet.\r\nCouter example: f=1/x, g=1/x,\r\n$\\int_{0}^{1}f(x)-g(x)\\,dx=\\int_{0}^{1}f(x)\\,dx-\\int_{0}^{1}g(x)\\,dx$\r\nis wrong.\r\n\r\nBut when C is a finite contant, and $\\int_{0}^{1}f(x)\\,dx$ exists, \r\n$\\int_{0}^{1}f(x)-C\\,dx=\\int_{0}^{1}f(x)\\,dx-C$ is true.\r\nSo need some extra work...", "Solution_3": "You've given the argument already: $f(y)$ is a constant." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc \\equal{} 3.$ Prove that:\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [11]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\r\n\r\n2)Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \\equal{} 1.$ Prove that:\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [8]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]", "Solution_1": "[quote=\"arqady\"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc \\equal{} 3.$ Prove that:\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [11]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\n[/quote]\r\nIt's very easy with Am-Gm, arqady. I'll post my proof later :)", "Solution_2": "The second is obviously trues because: $ \\sum\\ ab(a^2\\plus{}b^2) \\plus{} 8abc \\leq\\ 14$\r\nWith $ a\\plus{}b\\plus{}c\\equal{}3$ :maybe:", "Solution_3": "[quote=\"nguoivn\"]The second is obviously trues because: $ \\sum\\ ab(a^2 \\plus{} b^2) \\plus{} 8abc \\leq\\ 14$\nWith $ a \\plus{} b \\plus{} c \\equal{} 3$ :maybe:[/quote]\r\nVery nice proof! :lol:", "Solution_4": "[quote=\"arqady\"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc \\equal{} 3.$ Prove that:\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [11]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\n2)Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \\equal{} 1.$ Prove that:\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [8]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\n[/quote]\r\nFor the first,we can use this way.\r\n$ a^3(b \\plus{} c) \\plus{} b^3(c \\plus{} a) \\plus{} c^3(a \\plus{} b) \\equal{} 3p^2 \\minus{} 18 \\minus{} pr \\ge 3p^2 \\minus{} 18 \\minus{} \\frac {p^2(12 \\minus{} p^2)}{9}$\r\nFinally,we need to prove that:\r\n$ f(x) \\equal{} \\frac {x^{11}}{3^{11}} \\minus{} 3p^2 \\minus{} 18 \\minus{} \\frac {p^2(12 \\minus{} p^2)}{9} \\ge 0$\r\nWe have f\u2019\u2019\u2019(x) >0\r\nAnd f\u2019(x) $ \\ge f(3) \\equal{} 0$\r\nAnd$ f(x) \\ge f(3) \\equal{} 0$\r\nOur proof are completed.Equality occur if and only a=b=c.", "Solution_5": "[quote=\"arqady\"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \\plus{} ac \\plus{} bc \\equal{} 3.$ Prove that:\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [11]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\n[/quote]\r\nProof of weaker Inequality\r\n\r\nNotice the identity, then use the Am-Gm Inequality and $ 2\\sum a^4 \\geq \\sum a^3b \\plus{} \\sum a^3c$: \r\n\r\n$ (a \\plus{} b \\plus{} c)^4 \\equal{} 4(a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b) \\plus{} a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 6(ab \\plus{} bc \\plus{} ca)^2$\r\n\r\n$ \\equal{} 24\\left(\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{ 6}\\right) \\plus{} 3\\left(\\frac {a^4 \\plus{} b^4 \\plus{} c^4}{3}\\right) \\plus{} 54$\r\n\r\n$ \\geq 81 \\sqrt [81]{\\frac {(a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b)^{24}(a^4 \\plus{} b^4 \\plus{} c^4)^3}{6^{24} \\cdot 3^3}}$\r\n\r\n$ \\geq 81 \\sqrt [81]{\\frac {(a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b)^{27}}{6^{27}}}$\r\n\r\n$ \\equal{} 81\\sqrt [3]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}$\r\n\r\nSo, we have proved that:\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [12]{\\frac {a^3b \\plus{} a^3c \\plus{} b^3a \\plus{} b^3c \\plus{} c^3a \\plus{} c^3b}{6}}\\]\r\n:)", "Solution_6": "By the way, we can prove the following stronger inequality too: (with the same condition: $ ab \\plus{} bc \\plus{} ca \\equal{} 3$)\r\n\\[ \\frac {a \\plus{} b \\plus{} c}{3}\\geq\\sqrt [10.8]{\\frac {a^{3}b \\plus{} a^{3}c \\plus{} b^{3}a \\plus{} b^{3}c \\plus{} c^{3}a \\plus{} c^{3}b}{6}}\\]\r\nby using the same method above, but instead of using $ 2\\sum a^4 \\geq \\sum a^3b \\plus{} \\sum a^3c$, we will use\r\n\\[ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} \\frac{(ab \\plus{} bc \\plus{} ca)^2}{3} \\geq a^{3}b \\plus{} a^{3}c \\plus{} b^{3}a \\plus{} b^{3}c \\plus{} c^{3}a \\plus{} c^{3}\\]\r\nwhich directly follows from schur's inequality for degree 4. :)" } { "Tag": [ "projective geometry", "geometry unsolved", "geometry" ], "Problem": "Let $S_1$ and $S_2$ be two circumferences, with centers $O_1$ and $O_2$ respectively, and secants on $M$ and $N$. The line\n$t$ is the common tangent to $S_1$ and $S_2$ closer to $M$. The points $A$ and $B$ are the intersection points of $t$ with $S_1$ and $S_2$, $C$ is the point such that $BC$ is a diameter of $S_2$, and $D$ the intersection point of the line $O_1O_2$ with the perpendicular line to $AM$ through $B$. Show that $M$, $D$ and $C$ are collinear.", "Solution_1": "The problem is not clear to me. Should not the secant MN pass through the intersection H of the tangent t with the center line $O_1O_2$ and should not the points N, D, C be collinear? :?\r\n\r\nAssuming this is so, let the secant MN meet the circle $S_2$ again at P and let the normal to AM through B meet the circle $S_2$ again at Q. Because of similarity of the circles $S_1, S_2$ with the similarity center H, $BP \\parallel AM$, hence, $BP \\perp BQ$, so that PQ is another diameter of the circle $S_2$. The diameters BC, PQ of $S_2$ meet at its center $O_2$. Consider the cyclic pentagon BPCQN as a degenerate cyclic hexagon B'BPCQN with $B' \\equiv B$, so that $BB' \\equiv t$. By Pascal's theorem, the intersections $H = BB' \\cap NP \\equiv t \\cap MN,$ $O_2 \\equiv BC \\cap PQ,$ $D' \\equiv NC \\cap B'Q \\equiv NC \\cap BQ$ are collinear, i.e., $D' \\in HO_2$. But since $HO_2 \\equiv O_1O_2,$ we have $D' \\equiv O_1O_2 \\cap BQ \\equiv D,$ the points $D \\equiv D'$ are identical and N, D, C are collinear.", "Solution_2": "I read the original problem and it states that $M$ and $N$ are the intersection points of $S_{1}$ and $S_{2}$.", "Solution_3": "Assuming M, N are intersections of the circles $\\mathcal S_{1},\\ \\mathcal S_{2},$ let $H \\equiv AB \\cap O_{1}O_{2}$ be their external similarity center. Let a parallel to AM through B meet $\\mathcal S_{2}$ again at P, PM passes through H. Let the perpendicular from B to AM meet $\\mathcal S_{2}$ again at Q. $\\angle PBQ$ is right, PQ is a diameter of $\\mathcal S_{2}.$ By Pascal's theorem for the degenerate hexagon BBPCQM, the intersections $H \\equiv BB \\cap PM,$ $O_{2}\\equiv BC \\cap PQ,$ $D \\equiv BQ \\cap CM$ are collinear, $D \\in HO_{2}\\equiv O_{1}O_{2}.$" } { "Tag": [ "geometry", "trigonometry", "trig identities", "Law of Cosines", "complex numbers" ], "Problem": "Im terrible at geometry in math contests. Does anyone have any recommendations for some good geometry books that will help for math contests? \r\n\r\nThanks.", "Solution_1": "[u]Challenging Problems in Geometry[/u], by Alfred S. Posamentier and Charles T. Salkind. Published by Dover. Not all useful for competitions, but much of it is.", "Solution_2": "I had that book for 4 years now but i've never got thru the first 10 questions in chapter 1. in fact, i distinctly remember i stopped at #11 which i didnt know how to do", "Solution_3": "does that mean you didn't like it? I read the reviews on amazon and it seemed that they thought it was good.\r\n\r\nI bought geometry revisited today, too. Did any of you go through it? What did you think?", "Solution_4": "[quote=\"cats...\"]does that mean you didn't like it? I read the reviews on amazon and it seemed that they thought it was good.\n\nI bought geometry revisited today, too. Did any of you go through it? What did you think?[/quote]\r\n\r\ni got that a few days ago and so far it seems pretty good, havent been able to read past section 1.2.", "Solution_5": "[quote=\"cats...\"]I bought geometry revisited today, too. Did any of you go through it? What did you think?[/quote]\r\n\r\nyeah, i went through the whole thing. i don't know how much i'll remember, or how much it'll help on competitions, but i liked it.", "Solution_6": "well, thats fine too, so long as it's interesting.", "Solution_7": "Yes, Geometry Revisited has an excellent reputation.", "Solution_8": "[i]Geometry Revisited[/i] is a terrific book, but it's rather hard. I don't know of a book that's good but less challenging unfortunately.", "Solution_9": "Good, thanks for the help everyone.", "Solution_10": "well revisited is not really \"hard\" persay, but it is somewhat interesting... the volume of questions is hardly enough to be used in a competition. it mainly has proofs for classic theorems... and a section on projective/inversive geometry. challenging problems in geo is better for contest preparation... and i like the book, just didnt get around in working on the problems ;)\r\nif you want to choose between the two, i'd definately go with challenging", "Solution_11": "Euclid's Elements? lol", "Solution_12": "If someone can learn how to win math contests (or at least the geometry parts) by reading [i]Elements[/i], I will be extremely impressed.", "Solution_13": "the \"problems\" book, does it contain contest problems? or is it mostly problems by the authors?", "Solution_14": "By the authors. Maybe half of it would be competition-type stuff -- for example, it covers law of cosines, stewarts theorem, etc. It also has other cool stuff in it too, like Simpson lines and things of interest. It's all by the authors. It's largely proof-based, but there are plenty of \"compute\" questions.", "Solution_15": "[quote=\"cats...\"]Im terrible at geometry in math contests. Does anyone have any recommendations for some good geometry books that will help for math contests?[/quote]\r\n\r\nI agree with the other books already mentioned (and have them at home). The other book I would recommend is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883855100/learninfreed]Complex Numbers and Geometry[/url] by Liang-shin Hahn. That's in one of the useful MAA book series, at a very approachable reading level, full of interesting problems. \r\n\r\nHope this helps!" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Consider the equation $ x^2 \\minus{} 2kxy^2 \\plus{} k(y^3 \\minus{} 1) \\equal{} 0$,\r\nwhere $ k$ is some integer. \r\nProve that the equation has integer solutions $ (x, y)$ such that $ x > 0, y > 0$ if and only if $ k$ is a perfect square.", "Solution_1": "uhh... $ (x,y)\\equal{}(2k,1)$", "Solution_2": "[quote=\"indybar\"]Consider the equation $ x^2 \\minus{} 2kxy^2 \\plus{} k(y^3 \\minus{} 1) \\equal{} 0$,\nwhere $ k$ is some integer. \nProve that the equation has integer solutions $ (x, y)$ such that $ x > 0, y > 0$ if and only if $ k$ is a perfect square.[/quote]\r\nLook here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=262#p262" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove that the numbers $\\binom{2^n-1}{ i},i=0,1,...,2^{n-1}-1$,have pairwise different residues modulo $2^n$.", "Solution_1": "I don't really have time to get into the problem, but maybe this hint would help (it's a common idea for this sort of problems)\r\n\r\nLet's preassume that the hypothesis of the problem is wrong: that there exist two values of i (i.e. k, l) for which the residues are the same. Then, you will probably get a contradiction (k=l, or something like that)... I'm not sure this will help, but try it!", "Solution_2": "This problem is from the mathlinks contests,but I haven't seen the solution yet. :(", "Solution_3": "This problem is from the mathlinks contests,but I haven't seen the solution yet. :(", "Solution_4": "It is obviosly \r\n\\[ C_{2^n-1}^i=\\frac{(2^n-1)(2^n-2)...(2^n-i)}{i!}=\\prod_{k=1}^i \\frac{2^n-k}{k}=(-1)^i\\prod_{k=1}^i (1-\\frac{2^n}{k}). \\]", "Solution_5": "But it does not help.", "Solution_6": "\\[ C_{2^n-1}^i=\\frac{(2^n-1)(2^n-2)...(2^n-i)}{i!}=\\prod_{k=1}^i \\frac{2^n-k}{k}=(-1)^i\\prod_{k=1}^i (1-\\frac{2^n}{k})(mod \\ 2^n). \\]\r\nIt give $C_{2^n-1}^{k}=-C_{2^n-1}^{k-1}(1-\\frac{2^n}{k})\\not =C_{2^n-1}^{k-1}(mod \\ 2^n)$, because $\\frac{2^n}{k}-1\\not =1(mod \\ 2^n)$ when $k<2^{n-1}$.", "Solution_7": "And how does that solve the problem :huh:" } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "hello , can you prove:\r\n\r\n$ \\displaystyle\\lim_{x\\to 0\\plus{}}\\frac{\\sin x \\minus{} x}{x \\sin x}\\equal{}0$", "Solution_1": "Does not belong in Pre-Olympiad.\r\n\r\n[hide=\"Solution\"]Note that both the numerator and the denominator of this limit tends to 0, so we can use l'Hospital's Rule.\n\n$ \\lim_{x\\to 0^\\plus{}}\\frac{\\sin x\\minus{}x}{x\\sin x}\\equal{}\\lim_{x\\to 0^\\plus{}}\\frac{\\cos x\\minus{}1}{\\sin x\\plus{}x\\cos x}$.\n\nBoth the numerator and the denominator of the second limit tends to 0; let us use l'Hospital's Rule again.\n\n$ \\lim_{x\\to 0^\\plus{}}\\frac{\\cos x\\minus{}1}{\\sin x\\plus{}x\\cos x}\\equal{}\\lim_{x\\to 0^\\plus{}}\\frac{\\minus{}\\sin x}{\\cos x\\plus{}\\cos x\\minus{}x\\sin x}\\equal{}0$, as desired. $ \\blacksquare$[/hide]", "Solution_2": "[hide=\"Alternate\"]\nAs $ x \\to 0$ , $ \\sin{x}\\equal{}x$\n\nThus, $ \\frac{1}{\\sin{x}}\\equal{}\\frac{1}{x}$\n\n$ \\frac{1}{x}\\minus{}\\frac{1}{\\sin{x}}\\equal{}0$\n\n$ \\frac{\\sin{x}\\minus{}x}{x\\sin{x}}\\equal{}0$ as desired.[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "how many common elements do the follwing sets have ?\r\n \r\n$ A= ${ ${z\\in C , |z-5|=5} $}\r\nand\r\n$B= ${${ z\\in C , |z-12i|=8} $}", "Solution_1": "if there were a $ z $ common to the two sets then \r\n\r\n$ 13 = |5-12i| \\leq |5-z|+|z-12i| = 5+5 = 10 $, \r\n\r\na contradiction", "Solution_2": "sorry it was 8 and not 5 for the second set\r\ntry to solve it now... they say the answerar would be 4", "Solution_3": "lol, the answer is 1 though", "Solution_4": "please tell me how did you managed to solve it because that's what I want\r\nthank you", "Solution_5": "[quote=\"spx2\"]how many common elements do the follwing sets have ?\n \n$ A= ${ ${z\\in C , |z-5|=5} $}\nand\n$B= ${${ z\\in C , |z-12i|=8} $}[/quote]\r\n\r\n\r\n\r\n A:cup:B\r\nit would be more interesting if u do it:)))", "Solution_6": "8+5 = 13, the hypotenuse of the 5-12-13 right triangle that's formed..." } { "Tag": [ "inequalities" ], "Problem": "Fie $a,b,c \\geq 0$ cu $a+b+c=1$. Sa se demonstreze inegalitatea:\r\n\r\n$\\frac{a^{2}+b}{b+c}+\\frac{b^{2}+c}{c+a}+\\frac{c^{2}+a}{a+b}\\geq 2$", "Solution_1": "[hide=\"un fel de solutie\"]scriu asa\n$\\sum \\frac{a^{2}+b(a+b+c)}{(b+c)(a+b+c)}\\geq 2$\nse mai grupeaza umpic, se mai reduce umpic\nsi apoi aduc la acelasi numitor si iese cu metoda cu care v`am obisnuit deja :) insa nu mai am rabdarea sa scriu toate calculele \nsper sa ma credeti pe cuvant :)[/hide]", "Solution_2": "eu am inceput altfel rezolvarea cu cauchy , daka e poti sa vezi aici ca am postat-o si la preolimpiad[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114446[/url]. se mai pot vedea si alte 2 solutii frumoase la aceasta ineg", "Solution_3": "imi retrag cuvintele, inegalitatea nu este own, in momentu de fatza nu imi gaseam culegerea cu new and old inequalities, deci cer scuze fatza de omul caruia i-am furat-o :)\r\n\r\neu vazusem o generalizare urata tot aici pe ml si am particularizat pt 3 variabile, sry iar.", "Solution_4": "[quote=pohoatza]Fie $a,b,c \\geq 0$ cu $a+b+c=1$. Sa se demonstreze inegalitatea:\n\n$\\frac{a^{2}+b}{b+c}+\\frac{b^{2}+c}{c+a}+\\frac{c^{2}+a}{a+b}\\geq 2$[/quote]\n[url]https://artofproblemsolving.com/community/c6h1965107p13604675[/url]\n[url]https://artofproblemsolving.com/community/c6h1581044p9756632[/url]\n[url]https://artofproblemsolving.com/community/c6h1177493p5688614[/url]" } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "Suppose I have a list with $2007$ positive integers, including the number $2006$. When I take out the $2006$, the average of the numbers in the list becomes $2006$. What would be the average of the numbers in the list if I added a $2006$ to the [i]original[/i] list?", "Solution_1": "[hide]$\\boxed{2006}$ :D The average of the new list is 2006, when you add back the 2006 you took out, it remains 2006, and if you add another 2006, it's still 2006.[/hide]", "Solution_2": "[hide] $(x-2006)/2006 = 2006$ . \n\\[ x = 4026042 \\] where X is the sum of all 2007 original integers.\n$(4026042+2006)/2008 = 2006$[/hide]", "Solution_3": "[hide]the average would still be $\\boxed{2006}$[/hide]", "Solution_4": "[hide]When there are 2006 integers and the average of them 2006, that means all the integers add up to $2006^2$, or 4024036. Add 2*2006 and divide by 2008 to get [b]2006[/b][/hide]", "Solution_5": "[hide]Is this supposed to trick us?\n\nI get 2006, by setting some equations.\n\n$2006$[/hide]", "Solution_6": "i think it is supposed to be a trick.\r\n\r\ni got [hide]2006. but i didn't do anything fancy. it kind of popped into my head as the obvious answer. so i don't know how to put how i got it, except that when i put it in the problem, it works. :huh: [/hide]", "Solution_7": "[hide]So, basically if you add n to an set of numbers with an average of n, the average is still n. Wow.[/hide]", "Solution_8": "[quote=\"mathgeniuse^ln(x)\"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n. Wow.[/hide][/quote] Isn't it amazing? :D", "Solution_9": "[quote=\"lotrgreengrapes7926\"][quote=\"mathgeniuse^ln(x)\"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n. Wow.[/hide][/quote] Isn't it amazing? :D[/quote]\r\n\r\nI just love your sarcasm.\r\n[color=white][size=59]It is sarcasm, isn't it?[/size][/color]", "Solution_10": "[quote=\"SplashD\"][quote=\"lotrgreengrapes7926\"][quote=\"mathgeniuse^ln(x)\"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n. Wow.[/hide][/quote] Isn't it amazing? :D[/quote]\n\nI just love your sarcasm.\n[color=white][size=59]It is sarcasm, isn't it?[/size][/color][/quote]\r\nYes, it is.\r\n[color=white][size=59]That wasn't sarcastic.[/size][/color]", "Solution_11": "[quote=\"SplashD\"]\n[color=white][size=59]It is sarcasm, isn't it?[/size][/color][/quote]\r\n\r\nI thought you told me to stop writing in white text SplashD.\r\n\r\n[color=yellow][size=200]HYPOCRITE.[/size][/color]\r\n\r\nBesides, I still think that drunner2007 had a good question, and if it was Countdown, people better get the answer quickly.", "Solution_12": "[hide]\n\nDoesn't adding the average into a set keep the same average?\n\n[/hide]", "Solution_13": "[quote=\"mathmeister22\"][hide]\n\nDoesn't adding the average into a set keep the same average?\n\n[/hide][/quote]\n\n[quote=\"mathgeniuse^ln(x)\"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n. Wow.[/hide][/quote]\r\n\r\n...and I don't even have to say a single word.", "Solution_14": "[quote=\"mathmeister22\"][hide]\n\nDoesn't adding the average into a set keep the same average?\n\n[/hide][/quote]\r\n\r\nYes!\r\n\r\nIt should in the real world.", "Solution_15": "[quote=\"mathgeniuse^ln(x)\"][quote=\"mathmeister22\"][hide]\n\nDoesn't adding the average into a set keep the same average?\n\n[/hide][/quote]\n\nYes!\n\nIt should in the real world.[/quote]\r\nAnd imaginary and complex, too. :D", "Solution_16": "[quote=\"lotrgreengrapes7926\"][quote=\"mathgeniuse^ln(x)\"][quote=\"mathmeister22\"][hide]\n\nDoesn't adding the average into a set keep the same average?\n\n[/hide][/quote]\n\nYes!\n\nIt should in the real world.[/quote]\nAnd imaginary and complex, too. :D[/quote]\r\n\r\nActually, complex would just sum everything up. It includes real and imaginary components. So, you really just needed to say imaginary world. As in me doing well in a Math Competition.", "Solution_17": "Here is proof for a set of $N|N\\geq2$ numbers, with average of the $N-1$ numbers $n$.\r\n\r\n[hide]Call what is subtracted $x$. We know that\n$n-x\\equiv0\\pmod{x}$\n$n\\equiv x\\pmod{x}$\n$n\\equiv 0\\pmod{x}$\n$n+x\\equiv x\\pmod{x}$\n$\\therefore n+x\\equiv0\\pmod{x}$\nQED[/hide]", "Solution_18": "[quote=\"b-flat\"]Here is proof for a set of $N|N\\geq2$ numbers, with average of the $N-1$ numbers $n$.\n\n[hide]Call what is subtracted $x$. We know that\n$n-x\\equiv0\\pmod{x}$\n$n\\equiv x\\pmod{x}$\n$n\\equiv 0\\pmod{x}$\n$n+x\\equiv x\\pmod{x}$\n$\\therefore n+x\\equiv0\\pmod{x}$\nQED[/hide][/quote]\r\n\r\nNice proof.\r\n\r\nBut, why did we need to use modular arithmetic for this one though? We could use algebra without mods right?", "Solution_19": "[hide]The sum is then 2006*2007 with 2007 numbers, so the average is still 2006.[/hide]", "Solution_20": "[quote=\"mathgeniuse^ln(x)\"][quote=\"b-flat\"]Here is proof for a set of $N|N\\geq2$ numbers, with average of the $N-1$ numbers $n$.\n\n[hide]Call what is subtracted $x$. We know that\n$n-x\\equiv0\\pmod{x}$\n$n\\equiv x\\pmod{x}$\n$n\\equiv 0\\pmod{x}$\n$n+x\\equiv x\\pmod{x}$\n$\\therefore n+x\\equiv0\\pmod{x}$\nQED[/hide][/quote]\n\nNice proof.\n\nBut, why did we need to use modular arithmetic for this one though? We could use algebra without mods right?[/quote]\r\n\r\nThanks :lol: :lol: :) :) \r\n\r\nAs for proving it using algebra, it is going to be messier, and using mods is incredibly simple. So, yes, I am pretty sure you could prove it using Algebra." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let$d_1, d_2, ....... , d_k$be the positive divisors of $n = 1900!$. Prove that:\r\n$\\frac{d_1}{\\sqrt{n}}+...........+\\frac{d_k}{\\sqrt{n}}=\\frac{\\sqrt{n}}{d_1}+...........+\\frac{\\sqrt{n}}{d_k}$", "Solution_1": "If d divisor n, then n/d divisor. It give $\\sum_{d|n} d=\\sum_{d|n} \\frac nd .$", "Solution_2": "please remember you are posting in the advanced section." } { "Tag": [ "ratio" ], "Problem": "[url=http://img138.imageshack.us/my.php?image=8104ei9.png][img]http://img138.imageshack.us/img138/8026/8104ei9.th.png[/img][/url]\r\n\r\nIs there any relationships between theta1 and theta2?\r\nIf there is any, could anyone tell me the ratio or the equation to get one another please?\r\n\r\nThanks", "Solution_1": "ok, I will re write the problem like this: We have a right angled triangle $ABC$ ($\\angle B = 90^{\\circ}$) and $M$ is the midpoint of side $BC$. $\\angle AMB = \\theta_{1}, \\angle ACM = \\theta_{2}$, the relation between them:\r\n$tan \\theta_{1}= \\frac{AB}{BM}, tan \\theta_{2}= \\frac{AB}{BC}$ As a consequence :\r\n$\\frac{tan \\theta_{1}}{tan \\theta_{2}}= 2$. Is this what you wanted?", "Solution_2": "No. What he wants is the ratio of the angles, not of their tan's", "Solution_3": "is that possible because i think you need more information?", "Solution_4": "Hey Dude, I don't think there's any ratio for the angles then..." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "with a,b,c is three sidelenghts of triangle\r\n\r\nand $p=\\frac{a+b+c}{2}$\r\n\r\nprove that:\r\n\r\n$\\sum\\frac{a(b+c)}{p-a}\\geq8p$\r\ngoodluck :D", "Solution_1": "[hide]$a=x+y$,$b=x+z$,$c=z+y$\n$\\sum {(x+y)(2z+x+y) \\over z} \\geq 8(x+y+z)$\n$\\sum {(x+y)^2 \\over z}\\geq 4(x+y+z)$\nbut ${x^2 \\over y}+{y^2\\over x}\\geq x+y$\nand ${xy \\over z}+{xz \\over y}=x({y \\over z}+{z \\over y}) \\geq 2x$\n[/hide]", "Solution_2": "direct computation reduces this too\r\n$ \\frac{a}{b+c-a}+\\frac{b}{a+c-b}+\\frac{c}{a+b-c} \\ge 3$\r\nBy cauchy we hv to prove:\r\n$\\frac{(a+b+c)^2}{16(2ab+2bc+2ca-a^2-b^2-c^2)} \\ge 3$\r\nThis is true because $\\sqrt{a}, \\sqrt{b}, \\sqrt{c}$ are sidelengths of a triangle and the famous ineq (can anybody tell me the name of it?) $a^2+b^2+c^2 \\ge 4\\sqrt{3} S$", "Solution_3": "Sorry, siuhochung, but I don't understand how to deduce\r\n\r\n[quote=\"siuhochung\"]direct computation reduces this too\n$ \\frac{a}{b+c-a}+\\frac{b}{a+c-b}+\\frac{c}{a+b-c} \\ge 3$\n[/quote]\n\nand why this is Cauchy\n\n[quote=\"siuhochung\"]\nBy cauchy we hv to prove:\n$\\frac{(a+b+c)^2}{16(2ab+2bc+2ca-a^2-b^2-c^2)} \\ge 3$\n[/quote]\r\n\r\nYour used inequality is named Weitzenbock's inequality and you can find it with many proofs in Theorem and Formulas section.\r\n\r\nThank you.", "Solution_4": "manlio,\r\n\\[\r\n\\begin{array}{l}\r\n \\sum {\\frac{{a(2p - a)}}{{p - a}}} \\ge 8p \\\\ \r\n \\Leftrightarrow \\sum {a + } \\frac{{ap}}{{p - a}} \\ge 8p \\\\ \r\n \\Leftrightarrow \\sum {a + } \\sum {\\frac{{ap}}{{p - a}}} \\ge 8p \\\\ \r\n \\Leftrightarrow \\sum {\\frac{{ap}}{{p - a}}} \\ge 6p \\\\ \r\n \\Leftrightarrow \\sum {\\frac{a}{{p - a}}} \\ge 6 \\\\ \r\n \\Leftrightarrow \\sum {\\frac{a}{{b + c - a}}} \\ge 3 \\\\ \r\n \\end{array}\r\n\\]\r\nfor the second question,\r\n\\[\r\n\\left( {\\sum {\\frac{a}{{b + c - a}}} } \\right)\\left( {\\sum {a(b + c - a)} } \\right) \\ge \\left( {\\sum a } \\right)^2 \r\n\\]", "Solution_5": "Thank you very much.\r\n\r\nNice proof :)", "Solution_6": "$ \\frac{a}{b+c-a}+\\frac{b}{a+c-b}+\\frac{c}{a+b-c} \\geq 3 $ can be proved by putting $x=a+b-c,y=a-b+c,z=-a+b+c$ which are positive and writing as:\r\n$\\frac{x+y}{2z}+\\frac{y+z}{2x}+\\frac{z+x}{2y}\\geq\\frac{\\sqrt{xy}}{z}+\\frac{\\sqrt{yz}}{x}+\\frac{\\sqrt{zx}}{y}\\geq 3$\r\nUsing AM-GM two times." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": ":D find p is positive integer so that\r\n x 2 +y 2 -3xy-p=0", "Solution_1": "In other words,\r\n\\[ (2x\\minus{}3y)^2 \\minus{} 5y^2 \\equal{} 4p.\\]\r\nSo $ p$ must be a product of $ 4^k$ and primes of the form $ 5t\\pm 1$." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "email", "Support", "algebra" ], "Problem": "Folks, the AIME is a [b]copyrighted[/b] contest. We cannot have PDF versions of the AIME circulating around. Sorry.", "Solution_1": "What about the problems in the Resources section?", "Solution_2": "[quote=\"DPatrick\"]Folks, the AIME is a [b]copyrighted[/b] contest. We cannot have PDF versions of the AIME circulating around. Sorry.[/quote]\r\n\r\n\r\nI'm very sorry to hear that. \r\n\r\nWell what if you copyright the PDFs as property of AIME. \r\n\r\n4everwise spent all that time TO GET HIS TIME=MONEY work DELETED. TIME AWASTED. :(", "Solution_3": "He could use it himself to practice with it :wink:\r\n\r\nIs not the format copyrighted? Its in a PDF with just problems, not scanned versions.\r\n\r\nAlso, what about the old PDF generator? When will that work? Thanks.", "Solution_4": "True the AIME is copyrighted, but what about the AMCs, AIMEs, USAMOs, and IMOs in the Contests section?. \r\n\r\n4everwise's AIME PDF can just been seen as if he had just copied all the AIMEs from the Contests section and put them all in one easy file instead of 20+.", "Solution_5": "[quote=\"1/(ln x)\"]True the AIME is copyrighted, but what about the AMCs, AIMEs, USAMOs, and IMOs in the Contests section?. \n\n4everwise's AIME PDF can just been seen as if he had just copied all the AIMEs from the Contests section and put them all in one easy file instead of 20+.[/quote]\r\nuh, 4everwise didn't make one AIME pdf, he made one per year.", "Solution_6": "Um...aren't they still downloadable?\r\n\r\nhttp://www.artofproblemsolving.com/Forum/files/aime_1983.pdf\r\n\r\nstill works. :maybe:", "Solution_7": "[quote=\"anirudh\"]Um...aren't they still downloadable?\n\nhttp://www.artofproblemsolving.com/Forum/files/aime_1983.pdf\n\nstill works. :maybe:[/quote]The old ones are still downloadable. (only 4 or so) The new versions are not. (1983-2006, and the PDF with all the tests in a single file).", "Solution_8": "Unfortunately :(", "Solution_9": "[quote=\"anirudh\"]Unfortunately :([/quote]\r\nWhen I saw that every single AIME had been compiled in one source, I almost cried. And now I regret not having saved that PDF to my computer. \r\n\r\nMaybe 4everwise could PM or email the comprehensive PDF to various people who wanted it (cough, cough). :police:", "Solution_10": "That would be not right either :wink: \r\n\r\n[quote=\"DPatrick\"]Folks, the AIME is a [b]copyrighted contest[/b]. We cannot have PDF versions of the AIME circulating around. Sorry.[/quote]\r\n\r\nSo, he can't do that.", "Solution_11": "Yeah I don't know why they waited so long to take these off. It made people waste their time. Well, not really. I know A Lot of people already downloaded them. :lol:", "Solution_12": "Okay. People. To repeat what DPatrick said, the AIMES ARE COPYRIGHTED!!!\r\n\r\nAoPS cannot have PDF's floating around, in PM's or public posts. \r\n\r\nIf you want to be scared, maybe read Title 17. \r\n\r\nPlease, please, do not violate what DPatrick has said.\r\n\r\nThanks.[/hide]", "Solution_13": "Anirudh I think we get the idea. :lol: \r\n\r\n\r\nWho wants to be scared? I bet you've downloaded them already and then got scared. :rotfl:", "Solution_14": "It still seems rather silly to copyright the AIME, preventing interested students from practicing a test that was designed to inspire highschool students in mathematics.", "Solution_15": "[quote=\"DPatrick\"]\n2. I didn't say that the problems were copyrighted, I said that the [b]contests[/b] were copyrighted. Using an individual problem is \"fair use\" under copyright law, similar to excerpting a passage from a book. \n[/quote]\r\n\r\nOh, my bad. So what does it mean for a contest to be copyrighted, then? What are the details of what I can and cannot do?", "Solution_16": "[quote=\"purple plume\"][quote=\"anirudh\"]You have to buy rights? :maybe: \n\nThen couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]\n\n\nnothing's free in the world today, little grasshopper[/quote]\r\n\r\nuh huh...You can get rights from a certain publisher for free :wink: I got rights from History Channel once for a documentary to use a small clip that they had. :wink:", "Solution_17": "[quote=\"anirudh\"][quote=\"purple plume\"][quote=\"anirudh\"]You have to buy rights? :maybe: \n\nThen couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]\n\n\nnothing's free in the world today, little grasshopper[/quote]\n\nuh huh...You can get rights from a certain publisher for free :o I got rights from History Channel once for a documentary to use a small clip that they had. :wink:[/quote]\r\n\r\nwith connections maybe. unless they are ridiculously nice...", "Solution_18": "[quote=\"now a ranger\"][/quote][quote=\"anirudh\"][quote=\"purple plume\"][quote=\"anirudh\"]You have to buy rights? :maybe: \n\nThen couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]\n\n\nnothing's free in the world today, little grasshopper[/quote]\n\nuh huh...You can get rights from a certain publisher for free :o I got rights from History Channel once for a documentary to use a small clip that they had. :wink:[/quote][quote=\"now a ranger\"]\n\nwith connections maybe. unless they are ridiculously nice...[/quote]\r\nAre you kidding me? HC gave me permission to use a clip and I didn't have a connection", "Solution_19": "little grasshopper you have lots to learn in life. there's more to it than you think. you haven't been around long enough", "Solution_20": "[quote=\"PenguinIntegral\"][quote=\"DPatrick\"]\n2. I didn't say that the problems were copyrighted, I said that the [b]contests[/b] were copyrighted. Using an individual problem is \"fair use\" under copyright law, similar to excerpting a passage from a book. \n[/quote]\n\nOh, my bad. So what does it mean for a contest to be copyrighted, then? What are the details of what I can and cannot do?[/quote]\r\n\r\nYes, could an admin elaborate on what things are allowed and what things aren't allowed for copyrighted contests? Does the AoPS community have permission to discuss all copyrighted contests, just as long as those contests aren't reproduced for easy mass distribution? Are pdf versions of marathons legal, like the MATHCOUNTS marathons in the MC forum?", "Solution_21": "they do...\r\njorian", "Solution_22": "Discussing individual problems is fine, on the forum or in the wiki. Discussing contests in general is fine (like the post-contest discussion of the problems that occurs here). We also have permission from the AMC to use their problems in our classes and textbooks.\r\n\r\nProducing reproducible/transferable version of entire contests is where it crosses the line, as this interferes in a significant way with the AMC selling their product. \r\n\r\nThere is a good Wikipedia article on what is or is not \"fair use\" under copyright law: [url]http://en.wikipedia.org/wiki/Fair_use[/url]", "Solution_23": "well PDFs are easier to distribute.", "Solution_24": "hmmm On this year's(2007) AIME booklet it says, you can make copies(for non profit aid), So can we have the PDF back ? :maybe:", "Solution_25": "Again, it says you can make copies of INDIVIDUAL PROBLEMS. The best you can get is here:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2007", "Solution_26": "And the pdf's work now.", "Solution_27": "[quote=\"4everwise\"][quote=\"anirudh\"]Um...aren't they still downloadable?\n\nhttp://www.artofproblemsolving.com/Forum/files/aime_1983.pdf\n\nstill works. :maybe:[/quote]The old ones are still downloadable. (only 4 or so) The new versions are not. (1983-2006, and the PDF with all the tests in a single file).[/quote]\r\n\r\nReally? I got all of them a few weeks ago and it seems like they are still up. \r\n(The 1983+ ones, when I was preparing for the test.)\r\n\r\nWell, the 2007 works now so no comment from me.", "Solution_28": "Are there pdf's of solutions?", "Solution_29": "Nah, don't think so. Not all of the questions even have solutions posted; some just have a discussion thread." } { "Tag": [ "function", "induction", "algebra unsolved", "algebra" ], "Problem": "Find all strictly increasing functions $f: \\mathbb N \\to \\mathbb N$ such that\n\\[ f(n\\plus{}f(n))\\equal{}2f(n), \\quad \\forall n\\in \\mathbb {N}.\\]", "Solution_1": "[hide]$ f(n\\plus{}f(n))\\equal{}2f(n)$\n\n$ n\\plus{}f(n)\\plus{}f(n\\plus{}f(n))\\equal{}n\\plus{}3f(n)$\n\n$ f(n\\plus{}f(n)\\plus{}f(n\\plus{}f(n))\\equal{}f(n\\plus{}3f(n))$\n\n$ 2f(n\\plus{}f(n))\\equal{}f(n\\plus{}3f(n))$\n\n$ 4f(n)\\equal{}f(n\\plus{}3f(n))\\ge f(n\\plus{}3f(n)\\minus{}1)\\plus{}1\\ge f(n\\plus{}3f(n)\\minus{}2)\\plus{}2\\ge\\ldots\\ge f(n\\plus{}1)\\plus{}3f(n)\\minus{}1\\ge f(n)\\plus{}3f(n)\\equal{}4f(n)$\n\nEquality holds, so $ f(n\\plus{}1)\\equal{}f(n)\\plus{}1$. Thus $ f(n)\\equal{}n\\plus{}a$ for all $ n$. Easy to check the this function satisfies the given conditions for any $ a\\ge0$.[/hide]", "Solution_2": "we have $ f(n\\plus{}1) \\ge f(n)\\plus{}1$ by induction we attain $ f(m\\plus{}n) \\ge f(n)\\plus{}m$\r\nwe choose m=f(n) we have $ f(n\\plus{}f(n)) \\ge 2f(n)$ so f(n+m)=f(m)+f(n) with every natural m,n....", "Solution_3": "[quote=\"mathVNpro\"]Find all strictly increase function $ f$ from $ \\mathbb {N}$ to $ \\mathbb {N}$ such that:\n\\[ f(n \\plus{} f(n)) \\equal{} 2f(n), \\forall n\\in \\mathbb {N}\\]\n[/quote]\r\n\r\nIf $ \\exists u$ such that $ f(u\\plus{}1)>f(u)\\plus{}1$ then $ f(u\\plus{}f(u))>f(u)\\plus{}f(u)$, impossible. So $ f(u\\plus{}1)\\equal{}f(u)\\plus{}1$ $ \\forall u$. hence the unique solution $ f(x)\\equal{}x\\plus{}a$", "Solution_4": "f(n)>f(n-1)\r\n f(n)\u2265f(n-1)+1\r\n f(n)-n\u2265f(n-1) -(n-1)\u2265f(n-2)-(n-2)\u2265.............\u2265f(1)-1\r\n thus f(n+f(n))-(n+f(n))\u2265f(n)-n \r\n f(n+f(n))\u22652f(n)\r\n but we have f(n+f(n))=2f(n)\r\n equality holds f(n)-n=f(n-1)-(n-1)=.........=f(1)-1\r\n f(n)-n=f(1)-1 f(n)=n+c proven :roll:" } { "Tag": [ "inequalities" ], "Problem": "Let $0\\leq a \\leq b \\leq c$ be real numbers.\r\nProve that $(a+3b)(b+4c)(c+2a) \\geq 60abc$", "Solution_1": "EDIT: Oops, never mind :oops:\r\n\r\nEDIT #2: I have an extremely messy, but correct, solution:\r\n[hide]\nLet b = a+x and c = a + x + y with x,y >= 0. Expand everything and bring everything to one side to get\n5a^2 x + 20ax^2 + 15x^3 + 8a^2 y + 27axy + 27x^2 y + 16ay^2 + 12xy^2, where very term is non-negative, so it is >= 0\n[/hide]", "Solution_2": "it was posted in the inequalities forum recently.", "Solution_3": "hmm can anyone do this w/ AM-GM?\r\n\r\ni got $(a+3b)(b+4c)(c+2a) \\ge 16\\sqrt 6 (abc)$ by AM-GM and then multiplying them but i'm not sure on another more clever/clean way of doing this", "Solution_4": "Try and prove that\r\n\r\n$(b+4c)(c+2a)\\geq \\frac {15}{16}(b+3c)(c+3a)$", "Solution_5": "err after expanding everything out i got\r\n\r\n$bc + 19c^2 \\ge 7ac + 12ab$\r\n\r\nbut i dunno wat to do from there\r\n\r\nhaha sorry im still trying to learn this proof stuff :blush:", "Solution_6": "[quote=\"gotztahbeazn\"]err after expanding everything out i got\n\n$bc + 19c^2 \\ge 7ac + 12ab$\n\n[/quote]\r\n\r\nshould be $bc + 19c^2 \\ge 7ac + 13ab$\r\n\r\nProve it using $a\\leq b\\leq c$", "Solution_7": "Looks easy\r\n\r\n$(a+3b) \\geq 4a^{1/4}b^{3/4} $ by weighted amgm\r\n\r\nsimilarly with other two product terms on lhs, now divide by 60 and take away the denominators in the exponents:\r\n\r\nwe have now:\r\n\r\n$a^{55}b^{57}c^{68} \\geq a^{60}b^{60}c^{60}$\r\n\r\nwhich is obviously true", "Solution_8": "[quote=\"Singular\"]Looks easy\n\n$(a+3b) \\geq 4a^{1/4}b^{3/4} $ by weighted amgm\n\nsimilarly with other two product terms on lhs, now divide by 60 and take away the denominators in the exponents:\n\nwe have now:\n\n$a^{55}b^{57}c^{68} \\geq a^{60}b^{60}c^{60}$\n\nwhich is obviously true[/quote]\r\n\r\nYep, that's the nicer sol :D", "Solution_9": "How is that weighted AM-GM?", "Solution_10": "[quote=\"DPopov\"]How is that weighted AM-GM?[/quote]\r\n\r\nHow is it not? (a,b,b,b)", "Solution_11": "What is weighted AM-GM?", "Solution_12": "Isn't that just plain AM-GM \r\n\r\n\\[\\frac{a+b+b+b}{4} \\geq \\sqrt[4]{ab^3}\\]", "Solution_13": "[quote=\"DPopov\"]Isn't that just plain AM-GM \n\n\\[\\frac{a+b+b+b}{4} \\geq \\sqrt[4]{ab^3}\\][/quote]\r\n\r\nIndeed. But it's still called \"weighted\".\r\n\r\nhttp://www.artofproblemsolving.com/Forum/topic-32451.html" } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "How can I show in general that u=f(x-vt) and u=f(x+vt) satisfies the wave equation, where f is any function with second derivatives?.\r\n\r\nI figured this out easy enough for sin(x-vt) but in general it seems elusive. Any ideas?.\r\n\r\nThank you,\r\nCody", "Solution_1": "This is simply the chain rule. Here's one way of looking at it: Consider $f : \\mathbb{R}\\mapsto \\mathbb{R}$ and $g : \\mathbb{R}^2\\mapsto \\mathbb{R}$ where $w=g(x,t)=x-vt.$ We are looking at $u(x,t)=f(w)=f(g(x,t)).$ Then\r\n\r\n$\\frac{\\partial f}{\\partial x}=\\frac{df}{dw}\\frac{\\partial w}{\\partial x}=f'(w)\\cdot 1=f'(x-vt).$\r\n\r\nThen, $\\frac{\\partial^2 f}{\\partial x^2}=\\frac{d(f')}{dw}\\frac{\\partial w}{\\partial x}=f''(w)\\cdot 1=f''(x-vt).$\r\n\r\nAlso, $\\frac{\\partial f}{\\partial t}=\\frac{df}{dw}\\frac{\\partial w}{\\partial t}=f'(w)\\cdot (-v)=f'(x-vt)(-v).$\r\n\r\nFollowed by $\\frac{\\partial^2 f}{\\partial t^2}=(-v)\\frac{d(f')}{dw}\\frac{\\partial w}{\\partial t}=(-v)f''(w)\\cdot (-v)=v^2f''(x-vt).$\r\n\r\nHence if $u=f(x-vt),\\,\\frac{\\partial^2u}{\\partial t^2}=v^2\\frac{\\partial^2u}{\\partial x^2}.$\r\n\r\nIn your question, you assumed that $f$ was twice differentiable. Dealing with less than that as a hypothesis and the notion of \"weak derivative\" would be for another topic.", "Solution_2": "Thank you very much. I see now.\r\n\r\nCody" } { "Tag": [ "number theory", "prime numbers", "number theory proposed" ], "Problem": "Let $ a,n$ be positive integers such that $ a\\ge(n\\minus{}1)!$. Prove that there exist $ n$ [i]distinct[/i] prime numbers $ p_1,\\ldots,p_n$ so that $ p_i|a\\plus{}i$, for all $ i\\equal{}\\overline{1,\\ldots,n}$.", "Solution_1": "Maybe using Marriage Theorem??? NO?? :huh:", "Solution_2": "Yes, a solution using Hall's theorem is possible. (we only need to consider the numbers which have as prime divisors only primes 0) (1-cos(x)*cos(2x)*..*cos(nx))/x^2 for any n natural.\r\n\r\nnow.. there is a generalization for these kind of problems:\r\n\r\nprove that:\r\nIf lim(x-->0) (1-f(x))/x^p=L, L finite, then we have that:\r\nlim(x-->0) (1-f(x)f(2x)*..*f(nx))/x^p=L* (sum(k=1,n) k^(p+1)).\r\n\r\ncheers! :D :D", "Solution_1": "Think again about that p+1.\r\n\r\nWe have lim{x->00}f(x)=1 and\r\nThe limit is lim (1-f(x)+f(x)-f(x)(f2x)+...-f(x)f(2x)...f(nx))/x^p.\r\nCompute this and get what you said but with p instead of p+1.", "Solution_2": "If you don't belive me, take f(x)=e x.", "Solution_3": "yes.. u r right .. the problem is like so:\r\n\r\nIf lim(x-->0) (1-f(x))/x^p=L, L finite, then we have that: \r\nlim(x-->0) (1-f(x)f(2x)*..*f(nx))/x^p=L* (sum(k=1,n) k^p).\r\n\r\nsorry!", "Solution_4": "From the hypothesis we get \r\n\r\nf(x)=1-Lx^p+x^p.e(x) where lim_(x->0)e(x)=0 \r\n\r\n1-f(x)f(2x)...f(nx)=\r\n1-(1-Lx^p+x^p.e(x))(1-L2^px^p+2^px^p.e(x))...(1-Ln^px^p+n^px^p.e(x)) = \r\nLx^p \\sum_(k=1 to n ) k^p + term with x^(greater than p)+...\r\n\r\ndivide by x^p we get \r\n\r\nL \\sum_(k=1 to n ) k^p" } { "Tag": [ "function", "analytic geometry", "logarithms", "graphing lines", "slope", "geometry", "perpendicular bisector" ], "Problem": "Some points are chosen on the graph of lnx function on coordinate system. Here we have several questions:\r\n1) Is it possible to draw a isosceles triangle?\r\n2) Is it possible to find a piece of regular n-gon?\r\n3) Is it possible to draw a triangle whose biggest angle is 120? \r\n\r\n(Note: Above, lnx denotes the logarithm function base e?(also known as natural logarithm.)", "Solution_1": "[b](1)[/b] Obviously, the graph is continuous and is not a straight line. So choose any point of the graph as a centre and draw a circle, then the point with the two intersections form an isoscele triangle.\r\n\r\n[b](2)[/b] Draw the graph and then it will be very clear.\r\nAssume that there is such a n-gon.Then we have the perpendicular bisector of any sides of the n-gon intesect the graph.\r\n\r\nBut Slope of any of these side is of the form $ \\frac{x\\minus{}y}{In x\\minus{}In y}$ is positive,$ \\Rightarrow$ the perpendicular is of negative slope, $ \\Rightarrow$ the intesection of the perpendicular and the graph \r\n, say $ P(z,In z)$, satisfies $ \\begin{cases}z>x \\\\ In zIn x \\end{cases}$, impossible. Done.\r\n\r\n[b](3)[/b] It is possible. Indeed it's possible not only for $ 120^{\\circ}$, but also for any angle $ x^{\\circ}$ saisfying $ 90^{\\circ} x > 0\\] Thus, the solution $ (4y \\minus{} 1 \\minus{} x, y)$ works to the equation. Notice that the equation is symmetric in $ x$ and $ y$, so the solution $ (y, 4y \\minus{} 1 \\minus{} x)$ works. Yet, $ y \\plus{} 4y \\minus{} 1 \\minus{} x > y \\plus{} x$, so if $ a \\equal{} y$ and $ b \\equal{} 4y \\minus{} 1 \\minus{} x$, we have found the desired solution $ (a,b)$. [/hide]" } { "Tag": [ "vector", "calculus", "integration" ], "Problem": "A simple pendulum 1 m long has a bob of 10kg.\r\na)How much work will be required to move the pendulum from its vertical position to a horizontal position(it's easy)\r\nb)If the pendulum swings from a horizontal position, what will be the velocity and kinetic energy of the bob at the instant it passes through the lowest position of its path?", "Solution_1": "a) The gravitational field is a conservative vector field so the line integrals are independent of the path chosen. Therefore, work is simply the product of force and distance. The force is simply that to oppose gravity ($ F=mg$) and the distance is 1 m so $ W=98.0\\mbox{ N\\cdot\\text{m}}$\r\nb) It will travel a total distance of 1 m. I'm assuming the pendulum starts from rest so $ v_{i}=0$. Then\r\n$ v_{f}^{2}=0^{2}+2g$ (I substituted in the values)\r\nand so $ v_{f}=\\sqrt{2g}$\r\nThen $ KE=\\frac{1}{2}mv^{2}=10g=98.0\\text{ J}$\r\nwhich is what we hoped it would be." } { "Tag": [], "Problem": "is there anyone familiar with good software for capturing and burning web streams onto cds. Supposedly quick time pro is not very good, whats good for windows media player....is there a software program that is good for all of the players such as real player, itunes and windows media player that has gotten good reviews? this would be good for recording science lectures off the net, so it relates to this subject and others on this site", "Solution_1": "I hope these two links will help you in a way :):\r\nhttp://all-streaming-media.com/streaming-media-faq/faq-record-stream-windows-media-player.htm\r\nhttp://all-streaming-media.com/streaming-media-faq/faq-record-stream-real-player-realone.htm" } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "1) Let A\\in M_4(C) prove that A can be written as a sum at most of 4 matrices of rank 1\r\n2)Prove that I_4 cannot be written as a sum of less than four matrices\r\nof rank 1\r\nManuela Prajea, Ion Savu, Judet 2003\r\n---------------------------------------------------------------------------\r\n1) rank(A)=r, A is equivalent to \r\ndiag(1,0,0,0)=E_1 or\r\ndiag(1,1,0,0)= E_1+E_2 or\r\ndiag(1,1,1,0)=E_1+E_2+E_3 or\r\ndiag(1,1,1,1)=E_1+E_2+E_3+E_4 \r\nwhere E(i,j) is the matrix with 1 at the i-row, j-column when i=j, \r\nI wrote E(i,i)=E_i\r\nthere exist P,Q invertible in M_4(C) such that \r\nA=PE_1Q or\r\n=PE_1Q +PE_2Q or\r\n=PE_1Q +PE_2Q +PE_3Q or\r\n=PE_1Q +PE_2Q +PE_3Q +PE_4Q since P,Q are invertible \r\nrank(PE_iQ)=rank(E_i)=1\r\n\r\n2) We study each case \r\nIf I_4 = C with rankC=1 contradiction since rank(I_4)=4\r\n\r\nIf I_4=C+D with rank(C)=rank(D)=1\r\nI_4-C=D it is known that \r\n|rank(I_4)-rank(C)| \\leq |rank(I_4-C)=rank(D)| so 3 \\leq 1 contradiction\r\n\r\nIf I_4= C+D+F with rank(C)=rank(D)=rank(F)=1\r\n|rank(I_4)-rank(C)| \\leq |rank(I_4-C)=rank(D+F)| \\leq rank(D)+rank(F)\r\n4-1 \\leq 1+1 contradiction", "Solution_1": "For (a): If A=/=0_4 we can write it as sum of matrices which are all 0 except for a certain line (which is the same as in A). If A=0_4 we can write A=[1111] [1111]\r\n [0000] + (-1)*[0000]\r\n [0000] [0000]\r\n [0000] [0000]" } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "The function $ f: [\\minus{}1,1] \\rightarrow \\mathbb{R}$ is even and integrable. \r\n\r\n$ a)$ Prove that $ I\\equal{}\\displaystyle\\int^1_{\\minus{}1} f(x) \\arccos xdx\\equal{}\\frac {\\pi}{2} \\displaystyle\\int^1_{\\minus{}1} f(x) dx$\r\n\r\n$ b)$ Compute $ J\\equal{}\\displaystyle\\int^1_{\\minus{}1} \\frac {\\arccos x}{1\\plus{}x^2} dx$", "Solution_1": "Just take into account that $ \\arccos(\\minus{}x) \\equal{} \\pi \\minus{} \\arccos x$, then make the change of variable $ x \\rightarrow \\minus{}x$ and you're done.\r\nAt b) make $ f(x)\\equal{}\\frac{1}{1\\plus{}x^2}$ and use what you've learnt at a). :wink:", "Solution_2": "Divide [-1\uff0c1] into n subinteral denoted by [x(i),x(i+1)]\uff0ci=1,2,...,n.\u8bb0M(i)=sup{f(x),x belong to [x(i),x(i+1)] }and m(i)=inf{f(x),x belong to [x(i),x(i+1)]}.\r\n\\displaystyle\\int^1_{-1} f(x) \\arccos xdx<=\\sum_{i=1}^\\{i=n} {M(i)\\over n \\int_i\\over n^i+1\\over n arccos x\\,dx}=\\frac {\\pi}{2}\\sum_{i=1}^\\{i=n}{M(i)\\over n}\r\nto be samely we have \\displaystyle\\int^1_{-1} f(x) \\arccos xdx>=\\frac {\\pi}{2}\\sum_{i=1}^\\{i=n}{m(i)\\over n}\r\nsince f(x) is itegrable ,let n tends to infinite .wo draw that \\displaystyle\\int^1_{-1} f(x) \\arccos xdx=\\frac {\\pi}{2} \\displaystyle\\int^1_{-1} f(x) dx" } { "Tag": [ "inequalities", "geometric inequality", "inequalities proposed" ], "Problem": "[color=purple]That is a long time I don't post in this forum. Here is a nice geometric inequality I hope you will like it \n $ \\frac{r}{R}+\\frac{(a^{2}-b^{2})^{2}+(b^{2}-c^{2})^{2}+c^{2}-a^{2})^{2}}{16S^{2}}\\le \\frac{1}{2}$\n :D :D :P [/color]", "Solution_1": "[quote=\"evarist\"]$ \\frac{r}{R}+\\frac{(a^{2}-b^{2})^{2}+(b^{2}-c^{2})^{2}+c^{2}-a^{2})^{2}}{16S^{2}}\\le \\frac{1}{2}$\n [/quote]\r\nMaybe $ \\frac{r}{R}+\\frac{(a^{2}-b^{2})^{2}+(b^{2}-c^{2})^{2}+(c^{2}-a^{2})^{2}}{16S^{2}}\\le \\frac{1}{2}$ $ ?$\r\nIf so, try $ a=17,$ $ b=10$ and $ c=9.$ :wink:", "Solution_2": "[color=purple]Sorry it must be \n $ \\frac{r}{R}+\\frac{(a^{2}-b^{2})^{2}+(b^{2}-c^{2})^{2}+(c^{2}-a^{2})^{2}}{16S^{2}}\\ge \\frac{1}{2}$\n Thank arqady :D [/color]", "Solution_3": "[color=purple] What do you think about this problem gemath ? :D [/color]", "Solution_4": "Nothing for me to think about it! :D it is only easy algebraic problem by Ravi subtitude however I will help you to rewrite it :D and others will kill it for you, and note that it has a stronger version\r\n\\[ \\frac{r}{R}+\\frac{(a-b)^{2}c^{2}+(b-c)^{2}a^{2}+(c-a)^{2}b^{2}}{16S^{2}}\\ge \\frac{1}{2}\\]\r\nand algebraic version is\r\n\\[ \\frac{(x+y)(y+z)(z+x)}{4xyz}+\\frac{(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}+(x^{2}-y^{2})^{2}}{16(x+y+z)xyz}\\ge \\frac{1}{2}\\]\r\nwhich is trivial with mathlinkers in inequalities box :)" } { "Tag": [ "function", "LaTeX", "analytic geometry", "trigonometry", "calculus", "calculus computations" ], "Problem": "The function $ f: \\mathbb{R}^{2} \\minus{} \\{0\\}\\rightarrow\\mathbb{R}^{2}$ is given by $ f(x,y) \\equal{} \\left(\\frac {x}{x^{2} \\plus{} y^{2}},\\frac { \\minus{} y}{x^{2} \\plus{} y^{2}}\\right)$. \r\n\r\nShow that $ f$ is bijective. Find the inverse of $ f$. What are the images of the circles in their central position.", "Solution_1": "$ \\text{\\LaTeX}$ tip: the parentheses you want are \\left( and \\right). Don't use \\( and \\) - that does things you don't want to happen. (I edited for you.) And to indicate what is in set $ A$ but not set $ B$ use A \\setiminus B, not A - B. (I didn't edit that.)\r\n\r\nNow as for the problem: the bijection is from $ \\mathbb{R}^2\\setminus\\{(0,0)\\}$ to $ \\mathbb{R}^2\\setminus\\{(0,0)\\}.$ The function is actually its own inverse. This is one instance in which the algebra is much easier to see if you use complex arithmetic and think of $ f$ as mapping $ \\mathbb{C}\\setminus\\{0\\}$ to $ \\mathbb{C}\\setminus\\{0\\}.$ Thought of that way, it has a very nice algebraic form - can you see it?\r\n\r\nOne side effect: even the images of non-central circles are nice.", "Solution_2": "The denominator hints at the use of polar coordinates, so that $ r\\equal{}x^2\\plus{}y^2$", "Solution_3": "[quote=\"tmrfea\"]The function $ f: \\mathbb{R}^{2} - \\{0\\}\\rightarrow\\mathbb{R}^{2}$ is given by $ f(x,y) = \\left(\\frac {x}{x^{2} + y^{2}},\\frac { - y}{x^{2} + y^{2}}\\right)$. \n\nShow that $ f$ is bijective. Find the inverse of $ f$. What are the images of the circles in their central position.[/quote]\r\nSet ${{ \\alpha = \\frac {x^2}{(x^2 + y^2)^2}} + \\frac {y^2}{(x^2 + y^2)^2}} = \\frac {1}{x^2 + y^2}$. Then:\r\n$ f(f(x,y)) = f \\left(\\frac {x}{x^{2} + y^{2}},\\frac { - y}{x^{2} + y^{2}}\\right) = \\left( \\frac {\\frac {x}{x^2 + y^2}}{\\alpha}, \\frac {\\frac {y}{x^2 + y^2}}{\\alpha}\\right ) = (x,y)$. So $ f$ is it's own inverse.\r\nIf we consider $ z = x + yi$.\r\nThen $ g(z) = \\frac{x}{x^2+y^2} + i \\cdot \\frac{-y}{x^2+y^2} = \\frac{x-iy}{(x+iy)(x-iy)} = \\frac{1}{x+iy} = \\frac{1}{z}$ ;)", "Solution_4": "Yes, that complex form makes it pretty obvious why this function is its own inverse. And (like all linear fractional transformations) it maps [b]all[/b] circles or lines (not just circles centered at the origin) into circles or lines.", "Solution_5": "So we would use the same approach, just that $ z$ would be in polar coordinates for the following task: \r\n\r\n$ f: \\mathbb{R}^{2} \\rightarrow\\mathbb{R}^{2}$ and $ f(x,y) = (e^{x}\\cos{y},e^{x} \\sin{y})$\r\n\r\nLet $ z$ be $ z=|x|(\\cos{y}+i\\sin{y})$\r\n\r\nThen \r\n\\begin{eqnarray*}\r\nf(x,y)&=&e^{x}\\cos{y}+ie^{x}\\sin{y}=\\\\\r\n &=&e^{x}(\\cos{y}+i\\sin{y})=\\\\\r\n &=&\r\n\\{\r\n\\begin{array}{cc}\r\n e^{|z|}\\frac{z}{|z|} & \\mathrm{if} \\quad x \\geq 0\\\\\ne^{-|z|}\\frac{z}{|z|} & \\mathrm{if} \\quad x < 0\r\n\r\n\\end{array}\r\n\\end{eqnarray*}", "Solution_6": "That one is very simply $ z\\mapsto e^z.$ It is periodic of period $ 2\\pi i.$ (Assuming $ z\\equal{}x\\plus{}iy.$)", "Solution_7": "I see $ \\left( z = x + iy \\rightarrow f(x,y) = e^{x}(\\cos{y} + i\\sin{y}) = e^{x}e^{iy} = e^{x + iy} = e^{z}\\right)$.\r\n\r\nAnother interesting task for the of complex arithmetic would be:\r\n\r\nSimplify the following sums:\r\n\\begin{eqnarray*} A_{n} = 1 + r\\cos{\\alpha} + r^{2}\\cos{2\\alpha} + \\ldots + r^{n - 1}\\cos{(n - 1)\\alpha} & = & \\sum_{k = 0}^{n - 1}r^{k}\\cos{k\\alpha} \\\\\r\nB_{n} = r\\sin{\\alpha} + r^{2}\\sin{2\\alpha} + \\ldots + r^{n - 1}\\sin{(n - 1)\\alpha} & = & \\sum_{k = 0}^{n - 1}r^{k}\\sin{k\\alpha} \\end{eqnarray*}" } { "Tag": [], "Problem": "I thought about starting a problem of the day here, so I make up a problem and people answer it.\r\n\r\nProblem. Four students each earn one dollar and fifty cents a day. Each apple at the local grocery store costs two dollars. At the end of a standard three-hundred sixty-five day year, they all go and spend their savings on apples. Once each person can buy no more, they combine their funds and buy more apples, and if they can't share them equally, they cut them.\r\n(a) How many apples do they need to cut?\r\n(b) How many apples total do they buy?\r\n(c) Do (a) and (b) again, with the exception that they now go through a leap year in savings(366 days).", "Solution_1": "[hide]Ok, four students at $ \\$$1.50 a day. Multiply that so, $ \\$$1.50 times 365 equals $ \\$$547.50. But they earn that each, so $ \\$$547.50 times 4 equals $ \\$$2,190. Divide that by the cost of apples ($ \\$$2) which equals 1,095 apples. Divide that by 4, and that equals 273.75 apples. However, they won't get a fair share, so you have to cut up an apple into four slices. The total amount of apples is now 272 apples, and 7 apple slices. I'm kind of stuck at this point...Doing the math was just fun. Can someone also explain how to do the rest of it? (if I got the above information correct)[/hide]", "Solution_2": "[hide]Each student earns $ \\$$547.50 in the year, which is 273 whole apples. Pooling the remainder of their money, there is a total of $ \\$$6, or 3 apples. To divide these up, I think they would have to cut all 3 of them...maybe not. Is the answer to part a [b]3[/b]?\n(b)So, in total they bought (273*4)+3 apples = 1095 apples.\n(c) don't really want to do it right now\n[/hide]", "Solution_3": "@mathcountsfreak: Your answers are correct.\r\n\r\nCan anyone do (c)?", "Solution_4": "C is the same as A and B only with 366 instead of 365.\r\n[hide=\"Solution\"]\nEach person earns $ \\$1.50*366$, or $ \\$549$. This is 274 whole apples each. They end up with 4 more dollars, so that is 2 more apples. There is a total of $ 274*4\\plus{}2$ apples bought, so they must $ \\boxed {\\text{cut } 2 \\text{ apples and buy } 1,098 \\text{ apples}}$ [/hide]" } { "Tag": [], "Problem": "non asian gets -3; asian gets +2\r\n\r\nstarting off....\r\n\r\n\r\nnon-asian\r\n\r\n-3", "Solution_1": "-6\r\n\r\n(I'm not Asian.)", "Solution_2": "-9\r\n\r\n(European)", "Solution_3": "-7\r\n\r\nme=asian.", "Solution_4": "n.\r\ne.\r\ng.\r\na.\r\nt.\r\ni.\r\nv.\r\ne.\r\n\r\nt.\r\ne.\r\nn.", "Solution_5": "What are we supposed to do?", "Solution_6": "-13 \r\n\r\n(Indian)", "Solution_7": "-16\r\n\r\n@BOGTRO: Add 2 to the number in the previous post if you are Asian, and subtract 3 from the number if you are not Asian.\r\n@#H34N1: So, Indian counts as non-Asian for the purposes of this game?", "Solution_8": "[b]14-14-14[/b]", "Solution_9": "-12\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_10": "$ \\minus{}15$", "Solution_11": "-13\r\n\r\npshaw..", "Solution_12": "-11\r\n\r\n\r\n[sarcasm]yay.[/sarcasm]", "Solution_13": "-14\r\n\r\npshaw.." } { "Tag": [], "Problem": "Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?\r\n\r\n$\\text{(A) }29 \\qquad \\text{(B) }42 \\qquad \\text{(C) }45 \\qquad \\text{(D) }47 \\qquad \\text{(E) }50$", "Solution_1": "Ohh! I remember this from last year!! I put \"47\" but I never found out from other people if it was right. Can someone explain how to do this because I think I got this wrong!!", "Solution_2": "Solution:\r\n\r\n[hide]\nConverting their speeds to \"revolutions/minute\" we see that they have the same revolutional speed of about .796 rev/min, which means that they take 1.256 minutes per lap.\n\nThis means that they will cross when they come back to the start line and at the point directly across the start line.\n\n30/1.256 x 2 = 47.76...\n\nWe knock off the decimals because that just means that they were [i]about[/i] to cross again.\n\nAnswer is 47, D[/hide]" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Solve the equation:\r\n\r\n$ (x\\minus{}2xy\\minus{}y^2)y' \\plus{} y^2 \\equal{} 0$.\r\n\r\nI have the solution in the book, but I'm stuck at the following thing:\r\n\r\nI'll give the start also.\r\n\r\n$ y\\equal{}0$ is a solution. For $ y \\neq 0$ we have $ y' \\neq 0$ we will look at the equation having as unknown $ x(y)$. Hence the equation becomes: $ y^2 x' \\plus{} x(1\\minus{}2y)\\minus{}y^2 \\equal{} 0$. \r\n\r\nHere I'm stuck. Why does the equation become $ y^2 x' \\plus{} x(1\\minus{}2y)\\minus{}y^2 \\equal{} 0$ if we consider it into $ x(y)$? I can't see how they come to this. Can someone do it step by step or explain a little bit in detail?\r\n\r\nPlease help. Thank you very much! :blush:", "Solution_1": "It's a little algebra plus the fact that $ y' \\equal{} \\frac{1}{x'}$. (This is just the chain rule applied to $ \\frac{dy}{dx}$ and $ \\frac{dx}{dy}$.)" } { "Tag": [ "probability", "expected value" ], "Problem": "I had a question in the \"hard\" section questions of my textbook. Here's a variation of the problem:\r\n\r\nThe probability of Bob's team winning when Bob is tired is $\\frac{12}{25}$. The probability of Bob's team winning when Bob has lots of energy is $\\frac{14}{25}$. The probability of Bob's team winning when Bob isn't there is $\\frac{13}{25}$.\r\n\r\nOn any day, the probability of Bob being tired is $\\frac{1}{2}$ (guess his teachers torture him). The probability of Bob having lots of energy is $\\frac{1}{3}$. The probability of Bob not coming is $\\frac{1}{6}$.\r\n\r\n(1) On January 26, what is the probability that Bob's team won?\r\n(2) If Bob's team won, what is the probability that he:\r\n(a) was tired\r\n(b) had lots of energy\r\n(c) didn't come\r\n\r\n[hide=\"Weird thing to notice about problem (if you are bored)\"]If Bob doesn't come, the probability of them winning is greater than $50\\%$![/hide]", "Solution_1": "Here is my solution, but I am not sure if it is right:\r\n[hide] \n1. The probability that Bob is tired is 1/2 and that the team will win when he is tired is 12/25. This means that the chance that his team wins while he is tired is (1/2)(12/25)=6/25. You can do a similar thing for the rest and add everything up. So, for Bob having a lot of energy is (1/3)(14/25)=14/75. Also, for not showing up is (1/6)(13/25)=13/150. This means that when you add everything up, you get $\\frac{13}{150}+\\frac{14}{75}+\\frac{6}{25}=\\frac{36+28+13}{150}=\\frac{77}{150}.$ \n2's. Take 150 games played. The expected number of wins is 77 (since the chance of winning is 77/150). Out of these, the expected number of games that they would win with Bob tired is 36 (since 6/25=36/150). Similarly we can calculate that they probably won 28 games while Bob was energetic and 13 games when he was not there. Since the team won 77 games, and 36 of which were when he was tired, the probability that he is tired when they won would be $\\frac{36}{77}$, when they he was energetic and they would win is $\\frac{28}{77}=\\frac{4}{11}$, the probability that he wasn't there when they would won is $\\frac{13}{77}$. [/hide]", "Solution_2": "that's a cool way to solve #2. i think the answers are correct." } { "Tag": [ "geometry", "geometric transformation", "geometry solved" ], "Problem": "I would politely ask grobber (or anyone else), to please prove this \"easy-to-prove lemma\":\r\n\r\nGiven a center T and pairs of points (U,U'), (V,V') s.t. there is a similarity of center T which takes U to U' and V to V', then if UU' $\\cap$ VV' = P, the circles (UVP),(U'V'P) intersect again at T.", "Solution_1": "It is just an angle chase, man. The angle between $UU'$ and $VV'$ is equal to the angle between $TU$ and $TV$, because $TVV'$ is obtained from $TUU'$ through a spiral similarity centered at $T$. From here you get that $T,U,V,P$ are concyclic. Same for $T,U',V',P$." } { "Tag": [ "inequalities", "induction" ], "Problem": "Prove that if all $ x_i$ are positve, but less than 1, and $ \\displaystyle\\sum_{i \\equal{} 1}^{n}x_i \\equal{} \\frac {1}{y}$, then $ \\displaystyle\\prod_{i \\equal{} 1}^{n}\\frac {1 \\minus{} x_i}{1 \\plus{} x_i}\\ge \\frac {y\\minus{}1}{1 \\plus{} y}$.\r\n\r\nEDIT: Sorry, Jiang, you are correct. My post has been edited.", "Solution_1": "By induction on $ n$,\r\nthe inequality is trivial in case $ n\\equal{}1$.\r\nAssume $ \\prod^{n}_{i\\equal{}1}\\frac{1\\minus{}x_i}{1\\plus{}x_i}\\geq\\frac{y\\minus{}1}{y\\plus{}1}$\r\nFirst, $ \\sum^{n\\plus{}1}_{i\\equal{}1}x_i\\equal{}\\sum^{n}_{n\\equal{}1}x_i\\plus{}x_{n\\plus{}1}\\equal{}\\frac{1}{y}\\plus{}x_{n\\plus{}1}\\equal{}\\frac{1}{y/(1\\plus{}x_{n\\plus{}1}y)}\\equal{}\\frac{1}{y'}$.\r\nThen, $ \\prod^{n\\plus{}1}_{i\\equal{}1}\\frac{1\\minus{}x_i}{1\\plus{}x_i}\\equal{}\\prod^{n}_{n\\equal{}1}\\frac{1\\minus{}x_i}{1\\plus{}x_i}\\times\\frac{1\\minus{}x_{n\\plus{}1}}{1\\plus{}x_{n\\plus{}1}}\\geq\\frac{y\\minus{}1}{y\\plus{}1}\\times\\frac{1\\minus{}x_{n\\plus{}1}}{1\\plus{}x_{n\\plus{}1}}\\equal{}\\frac{y\\minus{}1\\minus{}x_{n\\plus{}1}y\\plus{}x_{n\\plus{}1}}{y\\plus{}1\\plus{}x_{n\\plus{}1}y\\plus{}x_{n\\plus{}1}}$\r\nNote that $ y\\minus{}1\\minus{}x_{n\\plus{}1}y0\nhence we can say that $f(h) = 1$ $sinh = h$\nhence we can write equation as $f(x)-f(x+h) = sinx.h$\nwe rearrange the equation as $\\frac{f(x+h)-f(x)}{h}=-sinx$\nnow L.H.S is $f`(x)$. now we got the differential equation \n$f`(x) =-sinx$. Integrating both the sides we get $f(x) = cos(x)+c$ now we use that $f(0) = 1$ we get $f(x) = cos(x)$[/hide]", "Solution_3": "Nice!\r\nBut you assume that $f$ is continuous at $0$,\r\n$\\lim_{h\\to 0}f(h)=f(0)=1$.\r\nMaybe it can be proved.", "Solution_4": "lordWings has proved that $f(-h) = f(h)$ hence we can say it is continious at x =0 .", "Solution_5": "Not every even function is continuous at $0$!\r\n\r\nBy the way, I didn't mean $f(x)=f(-x)$.\r\nI only meant to say that if $f(x)$ is a solution, then the other function $g(x)=f(-x)$ is a solution too (proved substituting $x$ by $-x'$ and $y$ by $-y'$ in the functional equation).\r\n\r\nI'm not sure if it's easy to prove $f(x)=f(-x)$.\r\n\r\nIt would be useful, because, taking $y=-x$,\r\n$f(x)f(-x)-1=-\\sin^{2}(x)$,\r\n$f(x)f(-x)=\\cos^{2}(x)$.", "Solution_6": "I'm very sory about my mistake. This problem have continuous condition. Thank you for your solution", "Solution_7": "Thank you. In this case, the solution is complete.\r\nashwinrk_jain proved that the only solution that is continuous at $x=0$ is $f(x)=\\cos x$.", "Solution_8": "ok then my solution is correct.", "Solution_9": "[quote=\"ashwinrk_jain\"] \n...\nwe get $f(x)f(h)-f(x+h) = sinx.sinh$\nnow we apply the limit h->0\nhence we can say that $f(h) = 1$ $sinh = h$\nhence we can write equation as $f(x)-f(x+h) = sinx.h$\nwe rearrange the equation as $\\frac{f(x+h)-f(x)}{h}=-sinx$\n...\n[/quote] \r\n\r\nI think this is almost a complete solution, but still not quite complete, \r\nbecause it assumes differentiability of $f$ in $0$:\r\n\r\nIf the above argument is written down carefully, it reads as follows: \r\n\r\n$\\forall h \\neq 0: \\frac{1}{h}[f(x)f(h)-f(x+h)] = \\sin x \\frac{\\sin h}{h}$\r\n$\\Rightarrow \\lim_{h \\to 0}\\frac{1}{h}[f(x)f(h)-f(x+h)] = \\sin x \\lim_{h \\to 0}\\frac{\\sin h}{h}= \\sin x$ \r\n$\\Rightarrow \\sin x = \\lim_{h \\to 0}\\Big[ \\frac{1}{h}[f(x)-f(x+h)]+\\frac{1}{h}[f(x)f(h)-f(x)] \\Big]$\r\n $= \\lim_{h \\to 0}\\Big[ \\frac{f(x)-f(x+h)}{h}+f(x) \\frac{f(h)-1}{h}\\Big]$\r\n\r\nIn order to continue we now have to distribute the limit to both summands. \r\nThis however is possibly only if at least one of the limits exists. \r\n\r\nThe existence of the first limit means that $f$ is differentiable in $x$, \r\nthe existence of the second limit means that $f$ is differentiable in $0$. \r\nSo mere continuity of $f$ does not help us here to continue the solution, \r\nand we either have to prove that the given equation implies the differentiability \r\nof $f$ in $0$, or we have to find a completely different solution... :wink:", "Solution_10": "[quote=\"solyaris\"] ...\n$\\sin x= \\lim_{h \\to 0}\\Big[ \\frac{f(x)-f(x+h)}{h}+f(x) \\frac{f(h)-1}{h}\\Big]$[/quote]\r\nBut [b]if[/b] there's continuity at $x=0$,\r\n$\\lim_{h \\to 0}f(h)=f(0)=1$ and then you have\r\n$-\\sin x= \\lim_{h \\to 0}\\frac{f(x+h)-f(x)}{h}$,\r\nso the function is differentiable because the limit exists.", "Solution_11": "[quote=\"lordWings\"][quote=\"solyaris\"] ...\n$\\sin x= \\lim_{h \\to 0}\\Big[ \\frac{f(x)-f(x+h)}{h}+f(x) \\frac{f(h)-1}{h}\\Big]$[/quote]\nBut [b]if[/b] there's continuity at $x=0$,\n$\\lim_{h \\to 0}f(h)=f(0)=1$ and then you have\n$-\\sin x= \\lim_{h \\to 0}\\frac{f(x+h)-f(x)}{h}$,\nso the function is differentiable because the limit exists.[/quote]\r\n\r\nI agree that $\\lim_{h \\to 0}f(h)=1$ by continuity, \r\nbut this does not imply $\\lim_{h \\to 0}\\frac{f(h)-1}{h}=0$.\r\nThe numerator and the denominator both go to zero, \r\nso the limit may be anything, and may even not exist.", "Solution_12": "You're right. :blush: \r\nEven if we have differentiability at $x=0$,\r\n$\\sin x= \\lim_{h \\to 0}\\Big[ \\frac{f(x)-f(x+h)}{h}+f(x) \\frac{f(h)-1}{h}\\Big]$,\r\n$\\Rightarrow\\lim_{h \\to 0}\\frac{f(x+h)-f(x)}{h}=-\\sin x+f(x)f'(0)$,\r\nthe function is differentiable for all $x\\neq 0$ too because the limit exists, but then we have\r\n$f'(x)=-\\sin x+f(x)f'(0)$.\r\n\r\nSo, we still have some work to do.", "Solution_13": "I think we had better tkinking of continuity of the function. And the function can solve easyly by change this one by another function so reduce to Cauchy function equation", "Solution_14": "Here's another thought.\r\nThe only odd function that is a solution from the functional equation is $f(x)=\\cos x$.\r\nNotice that the proof doesn't involve continuity.\r\n[hide=\"Proof\"](a) $y=0$ $\\to$ $f(x)(f(0)-1)=0$ $\\to$ $f(0)=1$,\nbecause $f(x)=0$ is not a solution to the functional equation.\n\n(b) $y=-x$ $\\to$ $f(x)f(-x)-f(0)=-\\sin^{2}x$ $\\to$ $f(x)f(-x)=\\cos^{2}x$,\nbecause $f(0)=1$ from (a).\n\n(c) $y=x$ $\\to$ $f^{2}(x)-f(2x)=\\sin^{2}x$.\n\n(d) = (b) - (c) $\\to$ $f^{2}(x)-f(2x)-f(x)f(-x)=\\cos^{2}x-\\sin^{2}x$\n$\\to$ $f(x)(f(-x)-f(x))+f(2x)=\\cos (2x)$.\n\nSo, if $f$ is odd, that is $f(x)=f(-x)$,\nfrom (b) we already had $|f(x)|=|\\cos x|$,\nfrom (d) we have $f(2x)=\\cos (2x)$.\n[/hide]Can we prove that $f$ must be an odd function?", "Solution_15": "[quote=\"lordWings\"]But at least it made me notice there are at least the two solutions:\n$f(x)=\\cos (x)$ and $f(x)=\\cos (-x)$.too.[/quote]\r\n\r\nAre these functions different?", "Solution_16": "No, of course not. :P\r\nThis is what I meant:\r\nI was looking for $g(x)$ such that $f(x)=\\cos (g(x))$ was a solution to the functional equation.\r\n(It's a restriction because it assumes $|f(x)|\\le 1$.)\r\nAnd there are at least two different solutions for $g(x)$:\r\n$g(x)=x$ and $g(x)=-x$.\r\n(I got them with another restriction: assuming $g(x)$ is a solution to Cauchy's functional equation.)\r\nOf course, both $g(x)$ lead to the same $f(x)$.", "Solution_17": "[Corrected]\r\nHello. I think I found short and nice solution without differentiability (I hope such word exists :lol: ). OK the key idea is to show that exist $x$ such that $f(x)=0$.\r\nFirst step is to show that $f(0)=1$. It can be achieved as you did before.\r\nNext putting $y=-x$ we get $f(x)\\cdot f(-x)-f(0)=\\sin x\\cdot\\sin(-x)$ or $f(x)\\cdot f(-x)=\\cos^{2}x$. Then pluging $x=\\pi/2$ we get $f(\\pi/2)\\cdot f(-\\pi/2)=0$. So indeed $f(\\pi/2)=0$ or $f(-\\pi/2)=0$. Let us say $f(\\pi/2)=0$. Assuming $y=\\pi/2$ into main equation we get $f(x+\\pi/2)=\\sin x$ and after argument shifting $x \\to x-\\pi/2$ $f(x)=\\cos x$ is achieved. Dealing with $f(-\\pi/2)=0$ returns the same results.\r\n\r\n[Thanks to hien for spotting out the mistyping]", "Solution_18": "Yes, very beatiful. I has checked carfully but didn'd find any error. Note that there is a mistake of typing.\r\nThen pluging $x=\\frac{\\pi}{2}$ we get\r\n $f(\\frac{\\pi}{2}).f(\\frac{-\\pi}{2})=0$." } { "Tag": [ "algebra", "polynomial", "trigonometry", "algebra solved" ], "Problem": "Prove that the polynomial $x^n + 4$ can be written as the product of two polynomials (each with degree less than $n$) with integer coefficients if and only if $n$ is a multiple of 4.", "Solution_1": "One implication is trivial: $x^{4k}+4=(x^{2k}-2x^k+2)(x^{2k}+2x^k+2)$.", "Solution_2": "Assume now that $x^n+4=f(x)g(x)$. All roots of $x^n+4$ (and thus of $f,g$) have modulus $\\sqrt [n]4$. This means that the free terms of $f,g$ must either be both $-2$ or $2$ (none of them can be $\\pm 1$). \r\n\r\nAssume $f,g$ have degrees $m,p$ respectively. Then $\\sqrt[n]4^m=\\sqrt[n]4^p=2\\Rightarrow 2^{\\frac {2m}n}=2^{\\frac {2p}n}=2\\Rightarrow \\frac {2m}n=\\frac {2p}n=1$. This means that $n$ is even and $m=p=\\frac n2$. Now assume $m=p$ is odd. The roots of $x^n+4$ can be partitioned in two sets with $m=p=\\frac n2$ elements each s.t. their products are equal (both $\\pm 2$). This can easily be shown to be impossible considering that the roots are $\\sqrt[n]4(\\cos\\frac {(2k+1)\\pi}n+i\\sin\\frac{(2k+1)\\pi}n),\\ k\\in\\overline{0,n-1}$ and assuming $n=2(2t+1)$. Even more can be done, we canshow that the product of any $\\frac n2$ roots (if \\frac n2$ is odd) isn't real." } { "Tag": [ "inequalities" ], "Problem": "Let $ a,b,c > 0$. Prove that:\r\n$ (\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a})^2\\geq\\frac32(\\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {a \\plus{} b}{c})$", "Solution_1": "[quote=\"tdl\"]Let $ a,b,c > 0$. Prove that:\n$ (\\frac {a}{b} \\plus{} \\frac {b}{c} \\plus{} \\frac {c}{a})^2\\geq\\frac32(\\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {a \\plus{} b}{c})$[/quote]\r\n\r\n$ \\Leftrightarrow \\sum_{cyc} \\frac{a^2}{b^2} \\plus{}2\\sum_{cyc} \\frac{a}{c} \\ge \\frac32\\left((\\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {a \\plus{} b}{c}\\right)$\r\n\r\n$ \\Leftrightarrow 2\\sum_{cyc} \\frac{a^2}{b^2} \\plus{}\\sum_{cyc} \\frac{a}{c} \\ge 3\\sum_{cyc} \\frac{a}{b}$\r\n\r\nBy AM - GM inequality:\r\n\r\n$ \\frac{a^2}{b^2} \\plus{}\\frac{a^2}{b^2} \\plus{}\\frac{b}{a} \\ge 3\\frac{a}{b}$\r\n\r\n$ \\frac{b^2}{c^2} \\plus{}\\frac{b^2}{c^2} \\plus{}\\frac{c}{b} \\ge 3\\frac{b}{c}$\r\n\r\n$ \\frac{c^2}{a^2} \\plus{}\\frac{c^2}{a^2} \\plus{}\\frac{a}{c} \\ge 3\\frac{c}{a}$\r\n\r\nAdding up then we are done!" } { "Tag": [], "Problem": "Let $ x,y,z$ natural numbers so that $ xyz \\equal{} 78$ and $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 206.$ What is $ x\\plus{}y\\plus{}z$?\r\n\r\nA. 18\r\nB. 20\r\nC. 30\r\nD. 42\r\nE. None of these", "Solution_1": "[hide]I tried doing manipulation, but it didn't work. So I tried factoring 78.\n\n$ xyz\\equal{}78\\equal{}(3)(2)(13)$\n\nThis limits the possibilities significantly. We see that one of the three numbers must be 13 because if it were 26, 39, or 78, the sum of squares would be too big. Thus $ z\\equal{}13$ and $ x^2\\plus{}y^2 \\equal{} 206\\minus{}13^2 \\equal{} 37$. Just looking at the numbers, it becomes evident that one of $ x,y$ is 6 and the other is 1.\n\n$ x\\plus{}y\\plus{}z\\equal{}13\\plus{}6\\plus{}1\\equal{}20\\equal{}>B$[/hide]", "Solution_2": "[hide=\"Why Manipulation Is a Bad Idea\"]\n\nThe problem tells us that $ x,y,z$ are all natural numbers. So, let's think simple and break 78 down. Very easily, we see that 2*3*13 = 78. So, simple trial-and-error or two tells us that $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 1 \\plus{} 36 \\plus{} 169$, not necessarily in that order but it doesn't matter. So, $ x\\plus{}y\\plus{}z \\equal{} 1\\plus{}6\\plus{}13 \\equal{} 20$. B [/hide]" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Can anybody please guide me to a set of good problems on the Chinese Remainder Theorem?", "Solution_1": "what do you mean with \"guide\"? give you problems? links? a summary?\r\ntry to search for this keyword here...\r\n\r\nNaphthalin", "Solution_2": ":blush: Good sites would do :maybe: ." } { "Tag": [ "function" ], "Problem": "Find $x,y>0$ such that \r\n$\\frac{1}{2}(x+y)=\\sqrt{3y}-\\frac{1}{\\sqrt{x}}$\r\n\r\n :)", "Solution_1": "write as $(\\sqrt{x}+1/x)^2+(\\sqrt{y}-\\sqrt{3})^2=3+1/x^2$, see that $(1,3)$ is a solution and then argue that it's the only possible value for $x$ based on how the function increases/decreases, like $(\\sqrt{x}+1/x)^2>3+1/x^2$ for positive $x$ except for $x=1$.", "Solution_2": "Yes,that is!! :) \r\nit can be solved either using your method or bringing in mind inequalities......\r\n\r\n$\\frac{2}{\\sqrt{x}}+x=\\frac{1}{\\sqrt{x}}+\\frac{1}{\\sqrt{x}}+x\\geq 3$\r\n\r\nwhile $2\\sqrt{3y}-y\\leq3$..... :)" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "ratio", "AMC", "AIME" ], "Problem": "Create a formula for the volume of a tetrahedron. :D \r\nthis should be a lot of fun...", "Solution_1": "i know \r\n$\\frac{\\sqrt2}{12}*s^{3}$\r\n\r\nwhat do you mean create?\r\nmake a new one?", "Solution_2": "if this question isn't challenging enough:\r\n\r\nsay there is a tetrahedron. another tetrahedron can be formed by connecting the centers of each of the faces of the tetrahedron. The ratio of the volumes of the larger tetrahedron to the smaller tetrahedron can be expressed as m/n, in which m and n are relatively prime. Find m+n", "Solution_3": "2003 AIME problem?", "Solution_4": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=713207[/url]\r\nHere.", "Solution_5": "[quote=\"8parks11\"]i know \n$\\frac{\\sqrt2}{12}*s^{3}$\n\nwhat do you mean create?\nmake a new one?[/quote]\r\nHe never said regular." } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "Let \\[p(x) = x^{n}+a_{n-1}x^{n-1}+...+a_{0}\\] be complex polynom. Prove that all his zeroes are inside the circle \\[|x| = 1+max|a_{i}|\\]", "Solution_1": "Any solutions?", "Solution_2": "But it is just the triangular inequality! If $M$ is $1+\\max(...)$ then if $|z|>1$ you have $|z|^{n}\\leq (M-1)(1+|z|+...+|z|^{n-1})<(M-1)\\frac{|z|^{n}}{|z|-1}$ and you are done." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Suppose that $ n\\geq 2$ be integer and $ a_1,\\ a_2,\\ a_3,\\ a_4$ satisfy the following condition:\r\n\r\ni) $ \\ n$ and $ a_i\\ (i \\equal{} 1,\\ 2,\\ 3,\\ 4)$ are relatively prime.\r\n\r\nii) $ \\ (ka_1)_n \\plus{} (ka_2)_n \\plus{} (ka_3)_n \\plus{} (ka_4)_n \\equal{} 2n$ for $ k \\equal{} 1,\\ 2,\\ \\cdots ,\\ n \\minus{} 1$.\r\n\r\nNote that $ (a)_n$ expresses the divisor when $ a$ is divided by $ n$.\r\n\r\nProve that $ (a_1)_n,\\ (a_2)_n,\\ (a_3)_n,\\ (a_4)_n$ can be divided into two pair with sum $ n$.", "Solution_1": "Sorry, I made a typo. I correct the problem as follows.\r\n\r\nSuppose that $ n\\geq 2$ be integer and $ a_1,\\ a_2,\\ a_3,\\ a_4$ satisfy the following condition:\r\n\r\ni) $ \\ n$ and $ a_i\\ (i \\equal{} 1,\\ 2,\\ 3,\\ 4)$ are relatively prime.\r\n\r\nii) $ \\ (ka_1)_n \\plus{} (ka_2)_n \\plus{} (ka_3)_n \\plus{} (ka_4)_n \\equal{} 2n$ for $ k \\equal{} 1,\\ 2,\\ \\cdots ,\\ n \\minus{} 1$.\r\n\r\nNote that $ (a)_n$ expresses the [color=red]remainder[/color] when $ a$ is divided by $ n$.\r\n\r\nProve that $ (a_1)_n,\\ (a_2)_n,\\ (a_3)_n,\\ (a_4)_n$ can be divided into two pair with sum $ n$.", "Solution_2": "hey, nobody answer this. I tried the problem and the only thing I have to prove for solve the problem is:\r\nIf we have 2 residues $ a$ and $ b$ such that:\r\n$ (ka)_n\\ge k \\Longleftrightarrow (kb)_n\\ge k$ for all $ 1\\le k \\le n\\minus{}1$\r\nThen $ a\\plus{}b\\equal{}n\\plus{}1$ or $ a\\equal{}b$.\r\nPlease [color=blue][i]kunny[/i][/color], if you have a solution, post it.", "Solution_3": "This problem is quite similar to (and more general than) this problem\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340038&sid=502505c7edf25f7b8c81bb871cdd3609#p340038[/url]\nIt's very hard..." } { "Tag": [ "geometry" ], "Problem": "Este foro en espec\u00edfico es para Discusiones Generales. Si se trata de un problema de nivel de secundaria, tratelo en el foro [url=http://www.mathlinks.ro/Forum/index.php?f=288]Matem\u00e1ticas de Preparatoria[/url], si se trata de un problema perteneciente a una competencia (o de nivel semejante a las mismas), tratelo en el foro [url=http://www.mathlinks.ro/Forum/index.php?f=287]Olimpiadas de Matem\u00e1tica[/url]. B\u00e1sicamente cualquier otro tema deberia ser tratado en este foro.\r\n\r\nModere el tono de la conversaci\u00f3n. Al igual que el resto de los foros en este sitio, este foro es visitado por miembros de potencialmente todas las edades, incluyendo miembros de edades inferiores a los 13 a\u00f1os, por lo tanto algunos temas deben ser tratados con mucho tacto, y otros no deber\u00edan ser tratados del todo. En caso de aparecer un mensaje inapropiado, [b]ser\u00e1 editado o borrado a discreci\u00f3n del moderador[/b] o de darse el caso, alguno de los administradores.\r\n\r\n[b]Trate tambi\u00e9n de moderar el lenguaje[/b]. Palabras que suelen considerarse ofensivas o inapropiadas en una conversaci\u00f3n semiformal, lo ser\u00e1n tambi\u00e9n aqu\u00ed. Del mismo modo, trate de ser [b]tolerante[/b]. Recuerde que nuestro idioma es hablado por una poblaci\u00f3n bastante numerosa y en una regi\u00f3n bastante extensa, por lo tanto, algunos modismos son distintos, algunas palabras pueden ser consideradas ofensivas en una regi\u00f3n y significar algo apropiado y distinto en otra. De todas maneras, en caso de aparecer un mensaje inapropiado, [b]ser\u00e1 editado o borrado a discreci\u00f3n del moderador[/b].\r\n\r\nCualquier consulta, puede enviarme un mensaje personal.", "Solution_1": "Hola,\r\n\r\nA ra\u00edz de algunos post recientes, uno de ellos en los foros en Espa\u00f1ol, se ha tomado una decisi\u00f3n, avalada por los administradores: \r\n\r\nSe considerar\u00e1 spam cualquier post en el cual se este procurando publicitar y vender un producto nuevo a varias personas, y en dicho caso, el mensaje ser\u00e1 borrado sin ning\u00fan previo aviso. \r\n\r\nEn el caso de una venta \u00fanica, de un articulo no nuevo, relacionado con la actividad del foro (ejemplo: Alg\u00fan libro usado de AoPS), se recomienda el uso de portales directamente dedicados a la venta de este tipo de productos (ejemplo: [url=http://www.ebay.com/]eBay[/url]), y en el caso de ser publicado en el foro, el thread ser\u00e1 cerrado o borrado a discreci\u00f3n del moderador.", "Solution_2": "Hola Comunidad Latina:\r\n\r\nComo ya sabran soy el nuevo moderador\r\n\r\nMe alegra esta posicion y espero poder contribuir mas a la comunidad\r\n\r\nContinuan las mismas reglas: En Espa\u00f1ol y MUCHA MATEMATICA!\r\n\r\nEspero cuelguen sus problemas aca:\r\n\r\n[b]En Discusiones Generales [/b]se discuten temas del folclor Matematico: Noticias, Olimpiadas, Eventos, Informacion, Fechas, Curiosidades, esperamos sean de Matematicas, no?\r\n\r\n[b]En Olimpiadas de Matematicas[/b] se ponen problemas de olimpiadas y se resuelven, favor poner la fuente del problema, el area y una descripcion para posiblemente organizarlos en un futuro. (y evitar el span!)\r\n\r\n[b]En Matematica de Preparatoria[/b] se ponen temas relacionados con las matematicas a nivel escolar, a modo de introduccion o nivel mas basico que el de olimpiadas.\r\n\r\nOtras actividades que tenemos son las de [b]EL reto de la semana[/b], simplemetne se trata de enviar soluciones a los problemas que semanalmente estamos proponiendo a mi mail (pascual2k1@hotmail.com) o al de Manuel u otro de los miembros del grupo creador.\r\n\r\nTambien pueden colgar articulos en espa\u00f1ol y aceptamos cualquier tipo de sugerencia para hacer de esta comunidad algo bien grande.[/b]", "Solution_3": "Pascual, gracias por administrar el foro.. esperamos aprender y disfrutar mucho de la buena matematica latina....", "Solution_4": "[b]Saludos a todos...[/b]\r\n\r\n Bueno, como ya lo habran notado... \r\n[b]Valentin V.[/b] nos ha designado a Pascual y a mi como moderadores de la comunidad ...\r\n\r\nYa Pascual lo ha dicho ...\r\nesperamos de parte de todos los miembros de la comunidad su participacion y colaboracion \r\npara mejorar este forum ...\r\n\r\n[b]Opinion..[/b]\r\nLa formacion de estas comunidades se dio como manera de evitar y/o reducir la barrera de los idiomas \r\nque aun mucha gente tiene y fomentar la colaboracion entre sus miembros ... eso es lo que esperamos\r\ntambien de esta comunidad, como podran notar, [b]ya somos la 2da Comunidad[/b] (numero de posts) en \r\nel Area Europa y estamos creciendo rapidamente ... esto amerita mayor organizacion y planificacion de\r\n nuestra parte ..\r\n\r\n[b]Finalidad...[/b]\r\nEstoy convencido que este forum generara un desarrollo en el nivel academico de la comunidad iberoamericana...\r\ny todo esto en pos de mejorar la participacion de las delegaciones latinoamericanas en las proximas olimpiadas (IMO, APMO, etc..)\r\n \r\n[b]Pedido..[/b]\r\n- Hago un pedido a todo miembro de esta comunidad a compartir la informacion con la que cuentan y asi difundirlo \r\naqui entre sus miembros, dentro de poco se habilitara una seccion donde podremos cargar articulos que escribamos, articulos traducidos, ebooks encontrados, etc ..\r\n- Pido difundir este forum entre los circulos de estudio en cada pais a los que tengamos acceso, pues mientras seamos m\u00e1s, MathLinks tendra mas consideraciones y apoyo a esta comunidad..\r\n\r\n[b]Finalmente..[/b]\r\nEstamos abiertos a sugerencias y nuevas propuestas en pos de mejorar esta comunidad, favoreciendo directamente a sus miembros y como no ... buscando que esta comunidad aporte al forum MathLinks en su conjunto, de inciativas de mejora.. \r\n\r\nCarlos Bravo :)\r\nLima -PERU", "Solution_5": "Just for the sake of it, it's MathLinks (one word, together) not \"Math Links\" which means a different thing. MathLinks = MathInterConnects, not a collection of links ;) :D", "Solution_6": "Creo que ya no hay actividad en el foro... parece que el moderador es Luis Gonzalez. Sr. Gonzalez, como cree que podemos generar mas actividad en el foro? Me falta practica de usar la computadora en espanol." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ x_0,x_1\\in\\mathbb{R}$ and\r\n\\[ x_{n}\\equal{}\\frac{(n\\minus{}1)g}{1\\plus{}(n\\minus{}1)g}x_{n\\minus{}1}\\plus{}\\frac{1}{1\\plus{}(n\\minus{}1)g}x_{n\\minus{}2}\\]\r\nfor some constant $ g>0.$\r\nFind $ \\lim_{n\\to\\infty}{x_n}.$", "Solution_1": "Let $ p_n\\equal{}\\frac{1}{1\\plus{}(n\\minus{}1)g}$, we have $ x_n\\minus{}x_{n\\minus{}1}\\equal{}(\\minus{}1)^{n\\minus{}1}p_2p_3\\cdots p_n(x_1\\minus{}x_0)$." } { "Tag": [ "limit", "calculus", "integration", "function", "complex analysis", "complex analysis unsolved" ], "Problem": "I wish to document a related problem in here as a separate item and hope that's alright:\r\n\r\nSuppose $ f(z)$ has an infinite set of isolated poles which accumulate to the non-isolated singularity $ z_0$ all of which are bounded by the contour $ \\Gamma$, that is:\r\n\r\n$ \\{z_i\\}$ is a sequence of poles of $ f(z)$such that:\r\n\r\n$ \\mathop\\lim\\limits_{n\\to\\infty} z_i\\to z_0$\r\n\r\nWhen is the following true:\r\n\r\n$ \\mathop\\oint\\limits_{|z|=\\Gamma} f(z)dz=2\\pi i\\sum_{n=1}^{\\infty}\\mathop\\text{Res}\\limits_{z=z_n} f(z)$\r\n\r\nJust seems to me this will always be true. Can someone give a counter-example in which it is not?\r\n\r\nThanks.", "Solution_1": "Take any $ f$ for which it is true and add $ \\frac 1{z\\minus{}z_0}$ to it. Then it'll become false.", "Solution_2": "Ok thanks. I'll work through it so that I can understand why.", "Solution_3": "Hello Fedja, I think I got it. I wonder if that construct is the only way? My justification of why I think that might be the case is:\r\n\r\nSuppose $ f(z)$ has a non-isolated singularity at $ z_0$ with the accumulation of isolated poles as described above, then (perhaps) the sum above is valid unless $ f(z)$ can be expressed as:\r\n\r\n$ f(z)\\equal{}g(z)\\plus{}h(z)$\r\n\r\nwith $ g(z)$ having the non-isolated singularity at $ z_0$ and $ h(z)$ having a pole at $ z_0$ with a non-zero residue. Take for example:\r\n\r\n$ f(z)\\equal{}\\frac{1}{Sin[1/z]}\\plus{}1/z$\r\n\r\nThe $ Sin[1/z]$ terms prevents $ f(z)$ from having a pole at the origin. However:\r\n\r\n$ \\oint f(z)dz\\equal{}\\oint \\frac{1}{Sin[1/z]}dz\\plus{}\\oint 1/z dz$ and thus the integral includes the residue of $ 1/z$. From this perspective, $ h(z)$ could have poles distributed elsewhere throughout $ \\Gamma$ and the sum I think would still apply. It's only when $ h(z)$ includes a pole at the non-isolated singularity does the idea break down. Or no?\r\n\r\nNot trying to prove myself right or nothing, just trying to better explore a concept I find interesting.", "Solution_4": "That is, indeed, essentially the only way just because if the formula is not true for $ f$ and the series of residues converges, then you can subtract $ \\frac c{z\\minus{}z_0}$ with appropriately chosen $ c\\in\\mathbb C$ from $ f$ and the formula will become true. On the other hand, it may happen that you encounter a function $ f$ for which the series of residues diverges and then you are in trouble from the very beginning. Try to construct such a function yourself :)." } { "Tag": [ "modular arithmetic" ], "Problem": "Find the last three digits of $2007^{2006^{2005^{\\cdots^{1}}}}$.", "Solution_1": "I'm not sure how useful most of that exponentiation is.\r\n\r\n[hide]We take it $\\pmod{8}$ and $\\pmod{125}$.\nSince $2007 \\equiv-1 \\pmod{8}$, and $2006^{\\cdots}\\equiv 0 \\pmod{2}$, $2007^{\\cdots}\\equiv 1\\pmod{8}$.\n\nNext $2007^{\\cdots}\\equiv 7^{\\cdots}\\pmod{125}$.\nThen, by Euler's totient theorem,\n$7^{2006^{\\cdots}}\\equiv 7^{6^{\\cdots}}\\pmod{125}$\n(Since $\\phi{125}= 100$, and $2006 \\equiv 6 \\pmod{100}$.)\nApply again to get $7^{6^{2005^{\\cdots}}}\\equiv 7^{6^{5^{\\cdots}}}\\pmod{125}$... and one last time - this time using Carmichael's theorem to cut the work - \n\n$7^{6^{5^{2004^{\\cdots}}}}\\equiv 7^{6^{5^{0}}}\\equiv 7^{6}\\equiv 24 \\pmod{125}$\n\nSo using CRT, we find the last three digits to be $649$. [/hide]", "Solution_2": "[quote=\"white_horse_king88\"]\n$7^{2006^{\\cdots}}\\equiv 7^{6^{\\cdots}}\\pmod{125}$\n(Since $\\phi{125}= 100$, and $2006 \\equiv 6 \\pmod{100}$.)\nApply again to get $7^{6^{2005^{\\cdots}}}\\equiv 7^{6^{5^{\\cdots}}}\\pmod{125}$[/quote]\r\n\r\nI don't believe this is true, as you can't apply that trick to powers, as they don't necessarily repeat in cycles of the totient function...", "Solution_3": "Oops... sorry. Actually, both of us were wrong, you being to the lesser extent:\r\n\r\n[hide]\nWe have:\n\n$x = 2006^{2005^{\\cdots}}$\n\n$2007^{x}\\equiv 7^{x}\\pmod{125}$\n\nNow this is where it should be changed.\n\nLooking at it $\\pmod{100}$, we need to use CRT since 6 and 100 aren't coprime.\n\nSo we need to find:\n\n1. $2006^{2005^{\\cdots}}\\bmod{25}$\n\nand\n\n2. $2006^{2005^{\\cdots}}\\bmod{4}$\n\n1.\n$2006^{2005^{\\cdots}}\\equiv 6^{2005^{\\cdots}}\\pmod{25}$\nUsing Carmichael,\n$2005^{2004^{\\cdots}}\\equiv 5^{2004^{\\cdots}}\\equiv 5 \\pmod{20}$, since $\\lambda(25) = 20$.\nSo $6^{5}\\equiv 1 \\pmod{25}$\n\n2.\n$2006^{2005^{\\cdots}}\\equiv 2^{2005^{\\cdots}}\\equiv 0 \\pmod{4}$\n\n\nUsing CRT on $x \\equiv 1 \\pmod{25}$ and $x \\equiv 0 \\pmod{4}$, we find $x \\equiv 76 \\pmod{100}$.\n\nNow we plug this back in: $2007^{x}\\equiv 7^{76}\\equiv 101 \\pmod{125}$.\n\nNow we use CRT on $2007^{\\cdots}\\equiv 101 \\pmod{125}$ and $2007^{\\cdots}\\equiv 1 \\pmod{8}$, and we see that:\n\n$2007^{\\cdots}\\equiv 601 \\pmod{1000}$\n\n[/hide]", "Solution_4": "Oh, wow. I don't know why I thought $(6,100) = 1$. :blush:" } { "Tag": [ "geometry", "geometric transformation", "rotation", "search", "combinatorics proposed", "combinatorics" ], "Problem": "Some $ m$ squares on the chessboard are marked. If among four squares at the intersection of some two rows and two columns three squares are marked then it is allowed to mark the fourth square. Find the smallest $ m$ for which it is possible to mark all squares after several such operations.", "Solution_1": "I guess $ m\\equal{}15$. Consider the marking where all the squares on the leftmost column and the topmost row are marked. I don't know how to prove.\r\n\r\nA different question:If we consider that board to be same after rotation, how many different markings are there it is possible to mark all squares after several such operations (of course with minimum number of markings) ?", "Solution_2": "here I assume the initial marking is already given and can't be changed by us.\nif it is an $ \\ n^2 $ checkboard then$ \\ m = (n-1)^{2}+1 $ \nsee $ \\ m >(n-1)^{2} $ \nbecause set all the row black apart from the last one.then we can't mark the whole board.\nnow to see why m=(n-1)^{2}+1 \nwork consider any white point.call it a. as there is atmost (n-2) unmarked left there is atleast one marked in that column call it b. now conside all other (n-1) column and search for c,d such that c, d are in same rows corr to a,b.then atleast\none such c,d is there s.t. both c,d are marked. so consider the square abcd.so now we can mark a.\nas a is arbitrary e are done\nso\n$ \\ m = (n-1)^{2}+1 $ is minimum no to guarranty the colouring of all the squares.", "Solution_3": "Suppose that we mark all cell.if it has a rows and b column,we prove m min=a+b-1.we will clean many cell to max such that it can recovery to mark all. Hence the first row we can clean max:b-1 cell.we prove that min cell of 3 rows consecutive=b+1.with 2 conditions we use fist condition for a-3 rows then use second conditions.( because don\u2019t have 4 cell with some shape and around white", "Solution_4": "there is only grade 8" } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "Find the smallest integer of the form $|f(m,n)|$ with \r\n\r\n$(a) f(m,n)=36^m-5^n.$ $(b) f(m,n)=12^m-5^n.$ \r\n[color=indigo][Please what does that mean : $f(m,n)$ is the problem asking for the minimum value of the function?][/color]\r\n\r\nDavron", "Solution_1": "$m=n=f(m,n)=0$ :D", "Solution_2": "I think it is for $m, n > 0$, otherwise, see Rust's post.\r\nIn this case, the answers are :\r\n(a) $(m, n) = (1, 2) \\leadsto |f(m, n)| = 11$\r\nand\r\n(b) $(m, n) = (1, 1) \\leadsto |f(m, n)| = 7$.\r\n\r\nIndeed, the problem asks you to find the minimal value $|f(m, n)|$ can be. So, now that you know what you're asked to do, try to solve it. You have the answers if you want to check. If you cannot solve it, just ask, and I'll help you. (But maybe someone else will post a proof soon ;).)" } { "Tag": [ "LaTeX" ], "Problem": "If $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{6}{7}$, where a, b, and c are positive integers, what is the smallest value of a+b+c?", "Solution_1": "[hide]Let one of them be $\\frac12$. That leaves $\\frac5{14}$.\n\nAfter guess-and-check, I get 1/3 and 1/42.\n\nSo, I get $47$.\n\n[/hide]", "Solution_2": "[hide]well i do guess and check, get 42, 3, and 2\ntherefore, 47[/hide]", "Solution_3": "[hide][i]Supposing a<=b<=c,[/i]\n1/a + 1/a + 1/a >= 6/7\n :arrow: 3/a >= 6/7; a <= 7/2, a = 1,2,3\n1/a < 6/7, a > 7/6 > 1\n[b]a = 2,3[/b]\n\n[b]a = 2[/b]\n1/b + 1/c = 5/14\n1/b + 1/b >= 5/14\n[b]b <= 28/5 :arrow: b = 3,4,5[/b]\n\na=2, b=3, c=42\na=2, b=4, c=28/3 \na=2, b=5, c=20/11\n\n[b]a = 3[/b]\n1/b + 1/c = 11/21\n1/b + 1/b >= 11/21\n[b] b <= 42/11 :arrow: b=1,2,3[/b]\nBut b connot equal 1 or 2 because it's has to be greater or equal to than a.\nIt also connot equal 3.\n\nIn the end, a=2, b=3, c=42 :arrow: so a+b+c=47. [/hide]", "Solution_4": "[quote=\"redsmartie\"][hide][i]Supposing a<=b<=c,[/i]\n1/a + 1/a + 1/a >= 6/7\n :arrow: 3/a >= 6/7; a <= 7/2, a = 1,2,3\n1/a < 6/7, a > 7/6 > 1\n[b]a = 2,3[/b]\n\n[b]a = 2[/b]\n1/b + 1/c = 5/14\n1/b + 1/b >= 5/14\n[b]b <= 28/5 :arrow: b = 3,4,5[/b]\n\na=2, b=3, c=42\na=2, b=4, c=28/3 \na=2, b=5, c=20/11\n\n[b]a = 3[/b]\n1/b + 1/c = 11/21\n1/b + 1/b >= 11/21\n[b] b <= 42/11 :arrow: b=1,2,3[/b]\nBut b connot equal 1 or 2 because it's has to be greater or equal to than a.\nIt also connot equal 3.\n\nIn the end, a=2, b=3, c=42 :arrow: so a+b+c=47. [/hide][/quote]\n\nHere is your answer written in $\\LaTeX$ so it is easier to read (I hope I didn't make any mistakes :D ):\n\n[hide][i]Supposing $a\\leq b\\leq c$,[/i]\n$\\frac{1}{a} + \\frac{1}{a} + \\frac{1}{a}\\geq \\frac67$\n :arrow: $\\frac{3}{a} \\geq \\frac67$; $a \\leq \\frac72, a = 1,2,3$\n$\\frac{1}{a} < \\frac67$, $a > \\frac76$ $> 1$\n[b]a = 2,3[/b]\n\n[b]a = 2[/b]\n$\\frac1{b} + \\frac1{c} = \\frac5{14}$\n$\\frac1{b} + \\frac1{b} \\geq \\frac5{14}$\n[b]$4b \\leq \\frac{28}{5}$ :arrow: b = 3,4,5[/b]\n\na=2, b=3, c=42\na=2, b=4, $c=\\frac{28}{3}$ \na=2, b=5, $c=\\frac{20}{11}$\n\n[b]a = 3[/b]\n$\\frac1{b} + \\frac1{c} = \\frac{11}{21}$\n$\\frac1{b} + \\frac1{b} \\geq \\frac{11}{21}$\n[b] $b \\leq \\frac{42}{11}$ :arrow: b=1,2,3[/b]\nBut b connot equal 1 or 2 because it's has to be greater or equal to than a.\nIt also connot equal 3.\n\nIn the end, a=2, b=3, c=42 :arrow: so a+b+c=47. [/hide]", "Solution_5": "[hide]47 I used guess and check (sad huh)[/hide]", "Solution_6": "[quote=\"moogra\"][hide]47 I used guess and check (sad huh)[/hide][/quote]\r\n\r\nIt's actually not that sad, because this problem actually requires some luck. A systematic way of doing this problem would take a while and is probably too advanced for a chapter level problem.", "Solution_7": "[hide]\nI think that guess and check is the easiest way to solve this problem:\nIf you have $\\frac{1}{2}$, you have $\\frac{5}{14}$ left. After guess and check, $\\frac{1}{3}$ and $\\frac{1}{42}$\n$\\boxed{47}$[/hide]" } { "Tag": [ "function", "ceiling function" ], "Problem": "Find a formula for the n[i]th[/i] term of the sequence\r\n\r\n$ 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...$\r\n\r\nWhere the integer $ m$ appears precisely $ m$ times.", "Solution_1": "[hide]\nwhen n is a triangle number (1, 3, 6, 10, 15...), this problem becomes trivial\n\njust reverse solve\n\n$ n = \\frac{a(a+1)}{2}$ for a\n\n$ 2n = a^{2}+a$\n$ a^{2}+a-2n = 0$\n\n$ a = \\frac{-1+\\sqrt{1+8n}}{2}$\n\nnow since each triangle number term is the final term with that given number, using the ceiling function will fill in all the in betweeners.\n\nFor example. solving for the 5th term will give you something between 2 and 3, which the ceiling function will resolve to 3\n\nthus:\n$ a_{n}= \\left\\lceil{\\frac{-1+\\sqrt{1+8n}}{2}}\\right\\rceil$\n[/hide]" } { "Tag": [], "Problem": "Let's say i have two forms: Form1 and Form2. On Form1 i have \r\na label1 and on Form2 i have a button. \r\nNow i want ,when i press the button from Form2 to write in label1 \r\nfrom Form1 the text \"HELLO\". \r\n\r\nHow do i make this??? Please give complet answer and source code. \r\nThank you!", "Solution_1": "You need some way to access the controls of the first form from the second form, right? One way to do this is to make the forms \"global variables\" or more precisely, static members of your main class:\r\n\r\n[code]class myapp {\n\tpublic static form1 f1;\n\tpublic static form2 f2;\n\tstatic void Main () {\n\t\tf1 = new form1 ();\n\t\tf2 = new form2 ();\n\t\tf1.Show ();\n\t\tf2.Show ();\n\t\tApplication.Run (f1);\n\t}\n}[/code]\n\nform1 and form2 are the classnames of Form1 and Form2. Now you can freely access the members of the two forms from anywhere in your application (provided that the members are public):\n\n[code]class form1 : Form {\n\tpublic Label label1;\n\t...\n}\n\nclass form2 : Form {\n\tvoid buttonclick (object sender, EventArgs e) {\n\t\tmyapp.f1.label1.Text = \"HELLO\";\n\t}\n\t...\n}\n[/code]\r\n\r\nI attach a complete sourcecode (I had to change the extension to .tex because of the restrictions....). Feel free to ask any questions you have about my program.", "Solution_2": "Well, i couldn't compile it because i'm not familliar with classes. Look my forms are:\r\n\r\n[code]\n/* form1.cs */\nusing System;\nusing System.Collections.Generic;\nusing System.ComponentModel;\nusing System.Data;\nusing System.Drawing;\nusing System.Text;\nusing System.Windows.Forms;\n\nnamespace forms\n{\n public partial class Form1 : Form\n {\n public Form1()\n {\n InitializeComponent();\n }\n\n private void button1_Click(object sender, EventArgs e)\n {\n Form2 f2 = new Form2();\n f2.Visible = true;\n }\n }\n}\n[/code]\n\nand\n\n[code]\n/* form2.cs */\nusing System;\nusing System.Collections.Generic;\nusing System.ComponentModel;\nusing System.Data;\nusing System.Drawing;\nusing System.Text;\nusing System.Windows.Forms;\n\nnamespace forms\n{\n public partial class Form2 : Form\n {\n public Form2()\n {\n InitializeComponent();\n }\n\n private void button1_Click(object sender, EventArgs e)\n {\n\n }\n }\n}\n[/code]\r\n\r\nNow : i have a label and a button in form1 and a button in form2.\r\nButon from form1 is for opening form2 and button from form2 is for put label1 = \"HELLO\".\r\nNow please help me to do this in my forms. I attached u my entire project. So please modify it to see the changes.\r\nThank you!" } { "Tag": [ "function", "calculus", "geometry" ], "Problem": "hi guys\r\nneed ur help\r\nif u guys know about site having some exceptionally good c++ program.\r\nPlz don't post ordinary program.\r\nI have to give it for my externals in just one week.\r\nPlzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz tell me some really nice programs.", "Solution_1": "not a single suggestion.\r\nI TEACHER will hang me if I will not submit it next Monday\r\n.", "Solution_2": "what kind of programs?", "Solution_3": "Any type of application based C++ program used in the field of technology , mathematics ,medical etc.", "Solution_4": "u can post any c++ program used to make thing quite easy.\r\nLike I have seen one program on how to update Criminal Datas?\r\nBelieve me \r\nIt was the worst program I have seen in my life .\r\nI m finding a program related to mathematics something dealing with graphs,functions ,calculas or even geometry but not does one dealing with how u can draw a 2-d graph or find the area with given sides blah..blah\r\n\r\n\r\nplzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz post it if u have something." } { "Tag": [], "Problem": "On a certain math exam, $ 10 \\%$ of the students got 70 points, $ 25 \\%$ got 80 points, $ 20 \\%$ got 85 points, $ 15 \\%$ got 90 points, and the rest got 95 points. What is the difference between the mean and the median score on this exam?\r\n\r\n$ \\textbf{(A)}\\ 0\\qquad\r\n\\textbf{(B)}\\ 1\\qquad\r\n\\textbf{(C)}\\ 2\\qquad\r\n\\textbf{(D)}\\ 4\\qquad\r\n\\textbf{(E)}\\ 5$", "Solution_1": "[hide=\"Answer\"]Simplest case, there are $20$ students. $2$ scored $70$, $5$ scored $80$, $4$ scored $85$, $3$ scored $90$, and $6$ scored $95$. Multiplying those out, the average score is $\\frac{2(70)+5(80)+4(85)+3(90)+6(95)}{20}=86$, while the median is $85$, so the answer is $86-85=1\\Rightarrow \\boxed{B}$.[/hide]" } { "Tag": [], "Problem": "The graph of $ f(x) \\equal{} x^4 \\plus{} 4x^3 \\minus{} 16x^2 \\plus{} 6x \\minus{} 5$ has a common tangent at $ x \\equal{} p$ and $ x \\equal{} q$. Compute the product $ pq$.", "Solution_1": "$ f(x)\\minus{}(46x\\minus{}105)\\equal{}(x^2\\plus{}2x\\minus{}10)^2\\equal{}(x\\minus{}p)^2(x\\minus{}q)^2\\equal{}\\{(x\\minus{}p)(x\\minus{}q)\\}^2,$ yielding $ pq\\equal{}\\minus{}10.$", "Solution_2": "How do you know that the common tangent is $ 46x \\minus{} 105$?", "Solution_3": "$ (x^2\\plus{}2x\\plus{}a)^2\\equal{}....$", "Solution_4": "Oh ok.. thanks :)" } { "Tag": [ "counting", "distinguishability" ], "Problem": "Three girls A, B and C and nine boys are to be lined up in a row. In how many ways can this be done if B must be between A and C and A,B must be separated by exactly 4 boys?", "Solution_1": "[hide=\"Boys are Indistinguishable\"]\nCase 1\nA_ _ _ _B_ _ _ _ _\n5 possible spots for C\n$ 5 \\cdot 9=45$\n\nCase 2\n_A_ _ _ _B_ _ _ _\n4 Possible spots for C\n$ 4 \\cdot 9 = 36$\n\nCase 3\n_ _A_ _ _ _B_ _ _\n$ 3 \\cdot 9= 27$\n\nCase 4\n_ _ _A_ _ _ _B_ _\n$ 2 \\cdot 9 = 18$\n\nCase 5\n_ _ _ _A_ _ _ _B_\n$ 9$\n\nAdd them $ \\boxed{9 (5+4+3+2+1) = 135}$.\n[/hide]\n[hide=\"Boys are Distinguishable\"]\nCase 1\nA_ _ _ _B_ _ _ _ _\n5 possible spots for C\n$ 5 \\cdot 9!$\n\nCase 2\n_A_ _ _ _B_ _ _ _\n4 Possible spots for C\n$ 4 \\cdot 9!$\n\nCase 3\n_ _A_ _ _ _B_ _ _\n$ 3 \\cdot 9!$\n\nCase 4\n_ _ _A_ _ _ _B_ _\n$ 2 \\cdot 9!$\n\nCase 5\n_ _ _ _A_ _ _ _B_\n$ 9!$\n\nAdd them $ \\boxed{15 \\cdot 9!=5443200}$.\n[/hide]" } { "Tag": [ "function", "topology", "real analysis", "real analysis theorems" ], "Problem": "A function is f : X -> Y is continuous iff f(Closure(A)) is contained in Closure(f(A)) for every subset of A of X. How can we prove this? Then \r\nf : X -> Y is cont. iff f-inv(Int(A)) is contained in Int(f-inv(A)), right?", "Solution_1": "$(\\Rightarrow)$ Assume $f$ to be continuous. Take $\\overline{x}\\in \\overline{A}$, and $V$ an open neighborhood of $f(\\overline{x})$. We want to show that $V \\cap f(A) \\neq \\emptyset$. By continuity, there is a neighborhood $U$ of $\\overline{x}$ such that $f(U) \\subseteq V$. But we also have $U \\cap A \\neq \\emptyset$, whence $f(U \\cap A) \\cap V \\neq \\emptyset$, and $f(A) \\cap V \\neq \\emptyset$.\r\n\r\n$(\\Leftarrow)$ If there is a point $x$ at which $f$ is not continuous, then there is an open neighborhood $V$ of $f(x)$ such that for every open neighborhood $U$ of $x$, $f(U) \\cap (Y-V) \\neq \\emptyset$. Assign a point $y_{i}$ in this intersection for every such $U_{i}$, and write $y_{i}=f(x_{i})$, $x_{i}\\in U_{i}$. Clearly enough $f(\\overline{\\{x_{i}\\}})$ contains $f(x)$ but $f(x)$ does not belong to $\\overline{f(\\{x_{i}\\})}\\subseteq Y-V$. Contradiction.", "Solution_2": "Thanks for the solution. It's simple and pretty good.", "Solution_3": "Well, one could actually wonder whether this problem can be solved without using the axiom of choice. I remember having this problem for my general topology exam a few years ago, but I don't remember solving it this way ... there must be something else without the axiom of choice." } { "Tag": [ "inequalities", "induction", "inequalities unsolved" ], "Problem": "The real numbers $a_0, a_1, ... , a_{n+1}$ satisfy $a_0 = a_{n+1} = 0$ and $| a_{k-1} - 2a_k + a_{k+1}| \\le 1$ for $k = 1, 2, ... , n$. Show that $|a_k| \\le \\frac{k(n + 1 - k)}{2}$ for all $k$.", "Solution_1": "[quote=\"James Potter\"]The real numbers $a_0, a_1, ... , a_{n+1}$ satisfy $a_0 = a_{n+1} = 0$ and $| a_{k-1} - 2a_k + a_{k+1}| \\le 1$ for $k = 1, 2, ... , n$. Show that $|a_k| \\le \\frac{k(n + 1 - k)}{2}$ for all $k$.[/quote]\r\nIt's easy using induction! The inequality is trivial for $n=0$ as well as for $k=0$ and $k=n+1$. If $0 \\frac{(k+1)(n-k)}{2}$$\nfor some $k$. It easy to see that\n$$(d_k+k)+(d_k+k-1)+...+(d_k+1)+d_k \\ge d_0+d_1+...+d_k > \\frac{(k+1)(n-k)}{2} \\Leftrightarrow 2d_k+2k > n. $$\nHence,\n$$0=d_0+d_1+...+d_n \\ge \\frac{(k+1)(n-k)}{2} + (d_k-1)+(d_k-2)+...+(d_k-n+k) = \\frac{(n-k)(2d_k+2k-n)}{2} > 0,$$\nwhich is a contraction. In like manner, we can show that\n$$d_0+d_1+...+d_k < -\\frac{(k+1)(n-k)}{2}$$\nleads to a contradiction. Therefore, we are done.", "Solution_3": "[quote=quangminh1173]\n$$(d_k+k)+(d_k+k-1)+...+(d_k+1)+d_k \\ge d_0+d_1+...+d_k > \\frac{(k+1)(n-k)}{2} \\Leftrightarrow 2d_k+2k > n. $$\n[/quote]\nIn 6th line you used inequality $d_0\\le d_k+k$. I wonder how do you know it's true since $d_0$ is included only in equation $d_0+d_1+...+d_n=0$, apart from inequalities $|d_{k+1}-d_{k}| \\le 1$.", "Solution_4": "[quote=WolfusA]\nIn 6th line you used inequality $d_0\\le d_k+k$. I wonder how do you know it's true since $d_0$ is included only in equation $d_0+d_1+...+d_n=0$, apart from inequalities $|d_{k+1}-d_{k}| \\le 1$.[/quote]\nedited\n\n" } { "Tag": [ "function", "trigonometry", "irrational number", "real analysis", "real analysis unsolved" ], "Problem": "I am learning signals and systems from oppenheim book.\r\nIn one page has a table about signals $ e^{(j\\omega _0 t)}$ and $ e^{(j\\omega _0 n)}$ for continuous and discrete time respectively\r\nand is says that for continious time the exponential is periodic for any choice of $ \\omega _0$ and for discrete time that is periodic ONLY if $ \\omega _0 \\equal{} m \\frac {2\\pi}{N}$\r\nwith $ m$ and $ N$ integers and $ N > 0$\r\n\r\ni believe that\r\nin continuous time we have also $ \\omega _0 \\equal{} k \\frac {2\\pi}{T_0}$ is this right??\r\n\r\nwhere is the difference?\r\ni can't understand and i am really confused\r\nThanks!", "Solution_1": "In discrete time, the exponential is periodic with periodicity $ P$ iff\r\n\r\n\\[ \\exp\\left(j\\omega_0n\\right)\\equal{}\\exp\\left(j\\omega_0(n\\plus{}P)\\right)\\]\r\nwhich implies that\r\n\\[ \\exp\\left(j\\omega_0P\\right)\\equal{}1\\]\r\n\r\nwhich in turn implies that $ \\omega_0P$ must be a multiple of $ 2\\pi$.\r\nBut in discrete time, unlike in continuous time, $ P$ is restricted to be an integer. Hence, $ \\omega$ must be a rational number $ m/N$ times $ 2\\pi$.\r\nIf $ \\omega_0$ is an irrational number times $ 2\\pi$, there is no way that $ P\\omega_0$ can ever be a multiple of $ 2\\pi$.", "Solution_2": "[quote=\"DK\"]\n Hence, $ \\omega$ must be a rational number $ m/N$ times $ 2\\pi$.\n[/quote]\r\n\r\nDid you mean \"Hence, $ \\omega$ must be a rational number $ m/P$ times $ 2\\pi$\" ?\r\n\r\nGenerally i think that\r\nwe define $ \\omega \\equal{} \\frac {2\\pi}{T}$ and here $ \\frac {2\\pi}{P}$\r\n\r\n$ \\exp\\left(j\\omega_{0}P\\right) \\equal{} 1$ means tha we want $ \\cos\\left(\\omega_{0}P\\right) \\equal{} 1$\r\nso we have $ cos(\\omega P) \\equal{} cos(\\frac {2\\pi}{P} P) \\equal{} cos(2\\pi) \\equal{} 1$\r\nwhy $ \\omega \\equal{} \\frac {2\\pi}{P} P$ isn't allowed?" } { "Tag": [ "calculus", "integration", "trigonometry", "function", "calculus computations" ], "Problem": "Find \r\n\r\n$\\int{\\frac{x^{2}}{\\sqrt{2-x^{2}}}}\\,dx$.\r\n\r\nUse the [b]result[/b], rather than the integrand, to show that $-\\sqrt{2}\\le{x}\\le\\sqrt{2}$ for the integral to be defined.", "Solution_1": "Make the substitution $x^{2}= 2\\sin^{2}u \\ du$, then we have\r\n\r\n${\\int \\frac{x^{2}}{\\sqrt{2-x^{2}}}\\ dx = \\int\\frac{2\\sin^{2}u \\sqrt{2}|\\cos u|}{\\sqrt{2-2\\sin^{2}u}}\\ du = 2\\int\\frac{\\sin^{2}u \\sqrt{2}|\\cos u|}{\\sqrt{2}|\\cos u|}}\\ du = 2\\int\\sin^{2}u \\ du$$= u-\\frac{1}{2}\\sin 2u$\r\n\r\n$=\\sin^{-1}\\frac{x}{\\sqrt{2}}-\\frac{x\\sqrt{2-x^{2}}}{2}$\r\n\r\nFrom this, we see that $\\sqrt{2-x^{2}}$ and $\\sin^{-1}\\frac{x}{\\sqrt{2}}$ is defined for $x\\in[-\\sqrt{2},\\sqrt{2}]$.", "Solution_2": "Perfect. \r\n\r\nAs an aside on notation, would you not tend to write $\\frac{1}{\\sqrt{2}}$ in rationalised form ($\\frac{\\sqrt{2}}{2}$) if you did this by hand? I ask because I've heard a few people swear by the rule \"never leave a denominator irrational\". :lol:", "Solution_3": "I don't see how the result in any way implies anything about where the integrand is defined. You must realize that finding an integrand is nonsense when it isn't defined...", "Solution_4": "[quote=\"coffeym\"]Perfect. \n\nAs an aside on notation, would you not tend to write $\\frac{1}{\\sqrt{2}}$ in rationalised form ($\\frac{\\sqrt{2}}{2}$) if you did this by hand? I ask because I've heard a few people swear by the rule \"never leave a denominator irrational\". :lol:[/quote]\r\n\r\nActually, it was just the opposite; when I did it by hand I wrote $\\frac{1}{\\sqrt{2}}$, then when I checked the answer on my calculator it was $\\frac{\\sqrt{2}}{2}$. Personally, I don't really have a preference about rationalizing the denominator.", "Solution_5": "[quote=\"coffeym\"]Use the [b]result[/b], rather than the integrand, to show that $-\\sqrt{2}\\le{x}\\le\\sqrt{2}$ for the integral to be defined.[/quote]\r\n\r\nThere could be many ways to write the result, and some of those may have meaningful extensions to values of $x$ outside the stated interval. When that happens it is often a feature, not a bug; for example, when showing that some function unexpectedly has an analytic continuation.", "Solution_6": "[quote=\"coffeym\"]Perfect. \n\nAs an aside on notation, would you not tend to write $\\frac{1}{\\sqrt{2}}$ in rationalised form ($\\frac{\\sqrt{2}}{2}$) if you did this by hand? I ask because I've heard a few people swear by the rule \"never leave a denominator irrational\". :lol:[/quote]\r\n\r\nThey only cared about that when I was in the 8th grade. So I sometimes do it by habit." } { "Tag": [ "vector", "Ring Theory", "superior algebra", "superior algebra theorems" ], "Problem": "$K$ a field, $A$ a commutative $K$-algebra of finite dimension. Prove that there is only finitely many maximal ideal.", "Solution_1": "It's always exciting when I get to use tough theorems, such as the [url=http://planetmath.org/encyclopedia/WedderburnArtinTheorem.html]Artin-Wedderburn Theorem[/url], in this case :).\r\n\r\nLet $\\mathcal J$ be the Jacobson radical of $A$. Now, $\\tilde A=A/\\mathcal J$ is a [url=http://planetmath.org/?op=getobj&from=objects&name=SemiprimitiveRing]semiprimitive[/url] ring, which is also a finite-dimensional algebra over $K$. Moreover, the maximal ideals of $\\tilde A$ are in one-to-one correspondence with those of $A$ via the canonical projection $A\\to\\tilde A$. It thus sufficces to prove that $\\tilde A$ has finitely many maximal ideals. \r\n\r\nNotice that $\\tilde A$ is also [url=http://planetmath.org/encyclopedia/Artinian.html]Artinian[/url]. This is because ideals of $\\tilde A$ are subspaces of the vector space $\\tilde A$ over $\\mathbb K$, and there are no infinite descending chains of subspaces of a finite-dimensional vector space. \r\n\r\nSince $\\tilde A$ is a semiprimitive and Artinian ring, we can apply the above-mentioned theorem to conclude that it must be a direct sum of rings of matrices over division rings that contain $K$. We also know that $\\tilde A$ is commutative, so it's in fact a fintie product of extension fields of $K$. Now it all comes down to proving that a finite direct product of fields , $\\prod_{i=1}^nF_i$, has finitely many maximal ideals. If an ideal of such a ring contains an element $(x_i)_{i=1}^n$ with $x_j\\ne 0$, then it contains all such elements, so the ideals of our ring are the products of several of the fields $F_i$, which means that the maximal ideals are precisely $M_j=\\prod_{i\\ne j}F_i$. \r\n\r\nI hope it's Ok." } { "Tag": [], "Problem": "A master is chasing after his dog. For every three steps the dog takes, the master\r\ntakes two. However one of the master\u2019s steps is equal to two of the dog\u2019s in length. The\r\ndog runs 10 steps before the master begins chasing. When the master finally catches up\r\nwith the dog, how many steps has the dog taken?", "Solution_1": "The man catches up one step on the dog for every 3 steps the dog takes. Thus the man catches up with the dog after 20 of his steps. In this time, the dog has traveled $ \\boxed{40}$ steps." } { "Tag": [ "abstract algebra", "algebra unsolved", "algebra" ], "Problem": "From the line segment $[-1; 1]$ $k\\geq 2$ different points have been choosen. For each point $i$ define $P_i$ as the product of distances to the other $k-1$ points. Define $S= \\sum \\frac{1}{P_i}$. Prove that $S \\geq 2^{k-2}$.\r\n\r\nActually, I have a solution to this problem but I didn't solved it by myself so I'm posting it here...", "Solution_1": "Given n points $x_1,x_2,...,x_n$\r\nConsider the Trebusep'polynomial T:degT$=n-1$\r\nThe coefficient of $x^{n-1}$is $2^{n-2}$\r\nBy using the Lagrange's theorem:\r\n$T_{n-1}(x)=\\sum_{i=1}^{n}\\frac{\\prod_{j\\neq i}(x-x_j)}{\\prod(x_i-x_j)}.T_{n-1}(x_{i})$\r\nThe coefficient of $x^{n-1}$ is $\\sum_{i=1}^{n}\\frac{T_{n-1}(x)}{\\prod(x_i-x_j)}$\r\n$\\Rightarrow 2^{n-2}=|\\sum\\frac{T_{n-1}(x)}{\\prod(x_i-x_j)}|=\\sum\\frac{|T_{n-1}(x)|}{\\prod|x_i-x_j|}$\r\n$\\Rightarrow 2^{n-2}\\leq \\sum\\frac{1}{\\prod|x_i-x_j|}=S_n$(q.e.d)", "Solution_2": "Yes, it's also a official solution :). Good work." } { "Tag": [ "graph theory" ], "Problem": "Find the greatest $ n$ for which there are points $ P_1$, $ P_2$, ... , $ P_n$ in the plane such that each triangle whose verticies are among $ P_1$, $ P_2$, ... , $ P_n$ has a side less than $ 1$ and a side greater than $ 1$.", "Solution_1": "I felt like this was not a good problem, but I'm probably wrong, as I have no right to be contradicting all these super-smart people. :oops: \r\n\r\nThere is no limit to the value $ n$. For any $ n$, no matter how large, we may arrange points $ P_1$ to $ P_{n \\minus{} 1}$ so that they are collinear and the greatest distance between them is less than 1 (i.e., they are all on, but not on the endpoints of, a segment of unit length). Then let $ P_n$ be far far away, at a distance greater than $ 1$ from any of the points.\r\n\r\nThus any triangle formed must include $ P_n$ and two of the points from $ P_1$ to $ P_{n \\minus{} 1}$, because if it did not include $ P_n$, the \"triangle\" would degenerate into a line. Now we placed $ P_n$ so that any line to it from another point is greater than $ 1$, and we also placed the other points so that any line between them is less than $ 1$.\r\n\r\nSo there is no greatest possible $ n$.", "Solution_2": "By triangles, I think they also included degenerate triangles (eg, any three points). \r\n\r\n[hide=\"Then\"]It's a well-known graph theory problem; consider a graph with $ \\ge 6$ points. We pick an arbitrary point $ P$, then of the remaining five points at least three belong to either $ S_1 \\equal{} \\{\\mathrm{points\\ less\\ than\\ 1\\ from\\ P}\\}, S_2 \\equal{} \\{\\mathrm{points\\ greater\\ than\\ 1\\ from\\ P}\\}$. Without loss of generality, let the first set have size $ |S_1| \\ge 3$; if any two points in $ S_1$ are less than $ 1$ apart those two points and $ P$ form a triangle with all sides less than $ 1$, contradiction. But otherwise at least three points in $ S_1$ are greater than $ 1$ apart, and the triangle with those three points generate a contradiction. Then just show $ n \\equal{} 5$ works.[/hide]\r\n\r\nI really should stop posting here .. :maybe:\r\n\r\nEdit: Wait, why not? Edit: Just read sticky ..", "Solution_3": "@ leoxnlin: When you are doing a problem, you assume that it isn't trivial...Your assumption that degenerate triangles are not triangles trivializes the problem. I suppose that it was not worded the best, but generally, you should attempt to solve a non-trivial problem if you see something on a contest or whatever. \r\n\r\n@azjps: what is greater or less than 1? Why can you say without loss of generality that it is the greater set...?", "Solution_4": "Does this question ask for the $ n$ for which such an arrangement is [b]possible[/b]?", "Solution_5": "The largest $ n$ such an arrangement of points exists..." } { "Tag": [ "email" ], "Problem": "Results are up: http://www.mathmeets.com/2008_state_meet/index.php.\r\n\r\nCongratulations to [b]Acton-Boxborough Regional High School [/b]for claiming first place in the Large School division by 10 points, with a score of 123.\r\nCongratulations to [b]Bedford High School [/b]for claiming first place in the Medium School division by 8 points, with a score of 71.\r\nCongratulations to [b]Worcester Academy[/b] for claiming first place in the Small School division by 11 points, with a score of 62.\r\n\r\nCongratulations to the following NEAML qualifiers from MA:\r\n[b]Large[/b]: Acton-Boxborough, Lexington, Belmont, Concord Carlisle, Algonquin, and Newton South.\r\n[b]Medium[/b]: Bedford, Winchester, Weston, Westborough, and Deerfield.\r\n[b]Small[/b]: Worcester, Maimonides, McDuffie, Dover-Sherborn, and Joseph Case.\r\n\r\nHow did everybody do? I got an 18 :lol: But only because I did Rounds 1, 5, and 6.\r\n\r\nCompared to last year's meet, this year's was much easier. (Note: ten 18's this year compared to a highest score of 13 last year.)\r\n\r\nAlso, this year, unlike last year, Bedford [b]got[/b] first place in the Medium Division without any ties!\r\n\r\n[hide=\"Stats\"]\n[b]Note:[/b] This was not on the website; I came up with these stats using Excel.\n\nAverage scores:\n[b]Round 1[/b] - 2.90625\n[b]Round 2[/b] - 1.42519685\n[b]Round 3[/b] - 1.03125\n[b]Round 4[/b] - 1.307086614\n[b]Round 5[/b] - 3.645669291\n[b]Round 6[/b] - 2.25\n\nSo the rounds, in order of difficulty, are: 5, 1, 6, 2, 4, 3.\n\nThe average total score of all participants was: [b]6.282352941[/b].\n\nScore distributions:\n\n[b]18[/b]: 10\n[b]17[/b]: 0\n[b]16[/b]: 4\n[b]15[/b]: 6\n[b]14[/b]: 4\n[b]13[/b]: 8\n[b]12[/b]: 7\n[b]11[/b]: 14\n[b]10[/b]: 8\n[b]9[/b]: 8\n[b]8[/b]: 14\n[b]7[/b]: 20\n[b]6[/b]: 26\n[b]5[/b]: 18\n[b]4[/b]: 21\n[b]3[/b]: 26\n[b]2[/b]: 19\n[b]1[/b]: 20\n[b]0[/b]: 22\n[/hide]", "Solution_1": "which additional teams qualified for NEAML?", "Solution_2": "There are definitely more NEAML qualifiers, seeing that the top 6 in large qualify.", "Solution_3": "Oh really? I never knew that. :blush: OK, I'll edit it.", "Solution_4": "I am pretty sure there are more teams from medium and small, which is what I'm wondering, but I don't know for a fact. Would you guys happen to know?", "Solution_5": "I believe the NEAML qualifiers are\r\n[b][u]LARGE[/u][/b]\r\n1. Acton-Boxborough\r\n2. Lexington\r\n3. Belmont\r\n4. Concord-Carlisle\r\n5. Algonquin\r\n6. Newton South\r\n[b][u]MEDIUM[/u][/b]\r\n1. Bedford\r\n2. Winchester\r\n3. Sharon\r\n4. Weston\r\n5. Westborough\r\n7. Deerfield [as WMML champion]\r\n[b][u]SMALL[/u][/b]\r\n1. Worcester\r\n2. MacDuffie\r\n3. Maimonides\r\n4. Dover-Sherborn\r\n7. Case [as SMCML champion]", "Solution_6": "Ooh, thanks! I thought someone told me only the top 3 schools get to NEAML... blah whatever. Editing again!", "Solution_7": "[quote=\"Yongyi781\"]I thought someone told me only the top 3 schools get to NEAML... [/quote] Quoting the rules: \"The selection of the teams for the New England Math Team Competition will be based upon the performance at the State Tournament. At least four Large schools, three Medium schools, and two Small schools will be selected to represent Massachusetts in the New England Tournament. The overall league champion from every league is automatically invited to attend the New Englands.\"", "Solution_8": "Ah, so they said \"at least.\"\r\n\r\nAnd additionally, I found out that Newton North only sent 7 team members instead of 8, and the MAML data listed that team member in the Individual results. So in fact, only 255 people went to the MAML, not 256.", "Solution_9": "Lexington didn't get first place, gasp!", "Solution_10": "loool\r\nhttp://www.wocomal.org/statistics/maml/alltime/team/championshipsbyyear.html\r\n\r\nhopefully lexington won't pummel us during NEAML for breaking their 9 year championship reign at MAML...", "Solution_11": "[quote=\"Pineappleperson\"]hopefully lexington won't pummel us during NEAML for breaking their 9 year championship reign at MAML...[/quote]\r\n\r\nhopefully", "Solution_12": "Here is the official list of qualifiers, which I received in email today. I had missed two (shown in boldface).\r\n\r\n[quote]Qualified teams:\n\n[b]Bromfield School[/b]\nJoseph Case High School\nDover Sherborn High School\nThe MacDuffie School\nMaimonides School\nWorcester Academy\n\nBedford High School\nDeerfield Academy\nSharon High School\nWestborough High School\nWeston High School\nWinchester High\n\nActon Boxborough\nAlgonquin\nBelmont High School\n[b]Canton High School[/b]\nConcord Carlisle High School\nLexington High School\nNewton South\n[/quote]", "Solution_13": "In Connecticut's state tournament, the school that won came from the Small Schools division.. (even though the Large Schools had 4 people per round, the top Large School still had less points than both the top Small and Medium school)\r\n\r\nSo yeah, don't expect much competition at NEAML from CT's Large Schools. Small Schools on the other hand though..", "Solution_14": "We got MAML (or maybe it was SAML) level 2 awards at the State Meet but are we ever going to find out our scores? I'm curious to know how many points were taken off for the oversight on the last problem, and I'm sure other people want to know their score as well.", "Solution_15": "I don't think scores are released.", "Solution_16": "Does anyone have a list of the top 20 on the MAML Level 2 for those that weren't at the state meet? (I actually know what I got but would like to see the whole list)", "Solution_17": "[quote=\"CatalystOfNostalgia\"][quote=\"Pineappleperson\"]hopefully lexington won't pummel us during NEAML for breaking their 9 year championship reign at MAML...[/quote]\n\nhopefully[/quote]\r\n\r\nHope springs eternal." } { "Tag": [ "email", "\\/closed" ], "Problem": "Art of Problem Solving now has a [url=http://www.facebook.com/pages/Art-of-Problem-Solving/322689570276?ref=ts]page on Facebook[/url] and a [url=http://twitter.com/AoPSNews]feed on Twitter[/url]. We'll be using these to make announcements about AoPS, and to spread the word about programs of interest to AoPSers. Those of you who host such programs should email us at aops@artofproblemsolving.com when you have news that AoPSers might want to know.", "Solution_1": "Thank you! This will be useful and exciting.", "Solution_2": "Will there be information posted on Facebook/Twitter that we couldn't find here, on AoPS?", "Solution_3": "I would highly doubt that. First of all, AoPS is the main place where the main announcements are posted and the activities are held. Also, some people don't have Facebook or Twitter where they would find information.\r\n\r\nI saw the AoPS page on Facebook! It's so cool that AoPS finally has one. :lol:", "Solution_4": "[quote=\"AIME15USAMO\"]Will there be information posted on Facebook/Twitter that we couldn't find here, on AoPS?[/quote]\r\n\r\nMy expectation is to use it as a news feed of sorts -- people will send me information about this program or that one, and I'll post it there.", "Solution_5": "Where could we find the similar announcements on AoPS?", "Solution_6": "Maybe [hide=\"here\"][url]http://www.artofproblemsolving.com/Forum/index.php?f=144[/url][/hide]?\r\nI guess the stuff posted on Facebook/Twitter isn't really news, more just like updates..if that makes sense :P \r\nAlso, can I add you as a friend of Facebook Mr Rusczyk? :maybe:", "Solution_7": "[quote=\"PowerOfPi\"]Where could we find the similar announcements on AoPS?[/quote]\r\n\r\nRight now, nowhere. AoPS specific announcements are made on the forum and our mailing lists. I hope to use FB/Twitter for more general announcements.", "Solution_8": "[quote=\"AIME15USAMO\"]\nAlso, can I add you as a friend of Facebook Mr Rusczyk? :maybe:[/quote]\r\n\r\nSure, but include your username in the Friend request, so I know who you are ;)", "Solution_9": "Are we allowed to add our instructors on Facebook? :D", "Solution_10": "Well I know Mr. Rusczyk friended a lot of us, but I don't know about the other instructors. But I don't see any reason why they won't either.\r\n(Or maybe they usually don't friend much users, but Mr. Rusczyk is very sociable. :wink: )" } { "Tag": [ "inequalities", "geometric inequality", "geometry", "cyclic quadrilateral" ], "Problem": "HaHaHaHa INMO is over\r\nNow i can rest in peace not pieces", "Solution_1": "How many didyou and others do ? What about Class XII students??? How many did they do usually - the top ones???", "Solution_2": "Hello,\r\n How was the INMO? How many did you do, Rushil, Vihang? I am Sanjith from Chennai. I managed only 2 and a half. In Chennai, one boy manged to do four. The rest of the boys,here, did around 2-3 like me. Let us start discussing about the problems. I got the second problem. I also did the one on combinatorics(Problem 4), I think. And the geometric inequality on cyclic quadrilateral, I got the first part.\r\nSanjith", "Solution_3": "Hello,\r\n How was the INMO?How many problems did you solve, Rushil,Vihang? Here, in Chennai, one boy managed four problems. The rest, it was 2-3. My brother Sanjith maged about 2 and a half. He got the second one, the fourth one and first part of the fifth problem. How did students do at Pune?\r\nSathej", "Solution_4": "hi,\r\ni did 3 and a half problems.the 2nd the 5th and the 6th and 1st part of the first one.does anybody hear know about the usual cut off for inmo. hope u all did well :)", "Solution_5": "I got 4 and a half (1,2,3,5 and 6(a)) :D \r\nSahil Mhaskar (Class XII) solved 5 and a half though. :o \r\nI have posted the problems in the respective sub forums and you may find answers there.", "Solution_6": "hi,\r\n i got three and a half problems right 1(a),2,5,6 and gave some uncooked answers for the rest\r\nhow did u all do??\r\n\r\nudbhav", "Solution_7": "[quote=\"Vihang\"]\nI have posted the problems in the respective sub forums and you may find answers there.[/quote]\r\nJust curious, where are the problems ? Thanks\r\n\r\nAdded later, Okay I see it. It is [url=http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=46&year=2006]Here[/url]", "Solution_8": "Well...\r\n\r\nI got 3 and a half : 1 , 3 , 5 , 6(a)\r\n\r\nI almost got Q 4 without the last step and I just missed the last step - dunno if they do partial marking but I'm not hopeful of getting much marks....\r\n\r\n6(b) - couldnt do but when came home and calculated on calculator, the required $n$ is same as in part (a)\r\n\r\nQ2 : Well, I have the following to say ---- the trick can only be found by those who have done extensive practice or those who have seen this exact question before( even for the former, they too mioght have seen this....)\r\nI am amazed that the INMO people put a question that can only be solved if you have done it before - something which happened here... a few XII students at my centre had been told the EXACT problem in an IIT-JEE prep tutition - this goes to say that this must be a very OLD olympiad question ---- why dont they INMO people realise who much unfairness this puts into the equation --- even those who did it and are here, be frank, dont you think you could do it only coz you had done something similar before!! \r\nReally shocking!!\r\n\r\nAnyways,,, it went good!! How did other XII students in your centres etc. do?? Fingers crossed!!!!", "Solution_9": "Where's ashwath.rabindranath.... he;s on this forum... how did he do???? Somebody PM him!!", "Solution_10": "I solved something between three and a half and four\r\n I solved 1 3 5 6A and am extremely doubtful about my solution for 2!", "Solution_11": "Hi,\r\nI agree with Rushil on the second problem, the problem is tough if one has not seen it before, actually I had seen the series( in the solution) in a book written by Prof.G.H.Hardy 'A Course in Pure Mathematics' . If I had not seen that before, I would not have got that.\r\nSanjith.", "Solution_12": "Well the 1/2(a+b-1)(a+b-2) reminded me of triangular numbers.\r\n(Thanks to the Proofs without words column here :roll: )\r\nThis cavee rise to an extremely simple solution.", "Solution_13": "See, I told ya!!! \r\n\r\nAnyways, best of luck to all of us!! :D \r\n\r\nAny ideas about the general performance of Class XII students?? Please tell !", "Solution_14": "Hello,\r\n I think this year's INMO was just like last years,maybe a bit easier. I gather that last year the cut off was about 31-32 for IX,X and XI students and 45 for XII students. This year maybe slightly higher, say by about 5-10 marks.And in INMO, the way you solve a problem matters.They give quite an importance to writing down proofs. Proofs given have to be conclusive, taking care of all nuances of the problem and be reasonably rigorous.\r\nSathej", "Solution_15": "I agree with Sathej.\r\nI know this because I missed clearing the INMO by 3 marks last year. :( \r\nThe proofs need to be up to the point and exact.(I learnt it the tough way!)\r\nAnyways best of luck to all of you.", "Solution_16": "last year the cut off was less than 2 qns.\r\nbut it varies every year.\r\n3 + is a sure in.(hopefully)", "Solution_17": "hey guys,\r\nits really cool to answer inmo in the 9th.\r\nbut then, there have been guys who have cleared inmo in the 8th itself!!(they r unique)\r\ninfact 3 from india have been to the imo in the 9th, out of which 2 have been bronze medallists and 1 silver medallist.\r\nso....", "Solution_18": "rushil,\r\nlast time the cut off for 12thers was 4 problems (approx).\r\nits usually tough for 12thys, but then u seem to be quite good.\r\nhope to c u at the maths camp this yr.", "Solution_19": "Hi there,\r\n\r\nDo you when the INMO results shall be announced?? I got 3 and a half questions right, hopefully. :? \r\n\r\n I'm eagerly looking for the results to come up...... When do they usually arrive?? \r\nI thought this year's paper was easier than last years... what do you say :?: \r\n\r\nsigning off......\r\n Sudharsha (A math lunatic!) :lol: [/quote]", "Solution_20": "hi!\r\n the inmo results are out but i can't seem to find them on the net.do you ppl have any ideas where i can find them except on bhaskaracharya's sit because they don't have it.btw congarats to everyone who is selected for the camp.\r\n\r\n\r\nUdbhav", "Solution_21": "I doubt they'll put them online so soon like the Physics/Chem etc.... it seems Maths is toooo underfunded in India --- we'll have to wait for the letters!! I'm not aware of any up to date site!", "Solution_22": "Well the result have arrived at Bhaskaracharya :D \r\nI don't know the full list but will let you know about it soon", "Solution_23": "I'm in!!! I got 51 marks --- round about what I expected! In fact, the guy in XII who I told you abt that he had done one more question coz he had already done it, got just 53!! From Punjab, its just me and another guy in XII and just one in XI( he got around 35 --- so the cut off hasnt increased much!) \r\n\r\nSo, who others qualified??\r\n\r\nPlease post the full list when its out!!!", "Solution_24": "hey guys it seems that even maths is enjoying a special status in india. the whole list is online on the hbcse site.here's the link.i got in :lol: \r\n \r\nhttp://www.hbcse.tifr.res.in/olympiads/inbo06_results/inmo_results06\r\n\r\ncongrats to all others who qualify.\r\n\r\nUdbhav", "Solution_25": "That's true, venkata - the paper was easier --- but it seems the cut off is still low --- the XI guy from Punjab got in with 35 marks --- dont know abt other regions!", "Solution_26": "[color=darkblue][size=200]Please note: Kindly keep all INMO post-result discussion to the INMO 2006 result Sticky..... I shall lock this now --- reason: this was a 'pre-result' discussion!!! [/size][/color]" } { "Tag": [ "calculus", "integration", "trigonometry", "induction", "algebra", "polynomial", "real analysis" ], "Problem": "A nice integral in my textbook, but I cannot get it:\r\n${ F(x)=\\int^{2\\pi}_{0}e^{xcos\\alpha}cos(xsin\\alpha)}\\,d\\alpha$, prove that $ F(x)=2\\pi$ for all x in real.", "Solution_1": "Notice that $ e^{x \\cos \\alpha} \\cos (x \\sin \\alpha) \\equal{} \\text{Re} \\left[ \\exp(x e^{i \\alpha}) \\right]$.\r\n\r\nThen for $ x \\neq 0$,\r\n\r\n$ F(x) \\ \\equal{} \\ \\text{Re} \\left[ \\oint_{|z|\\equal{}|x|} \\frac{e^z}{iz} \\, dz \\right]$\r\n\r\nwith $ z \\equal{} x e^{i \\alpha}$. You may use residue thorem to evaluate this integral.", "Solution_2": "Seem a nice solution. I can't , however, get it, for I am just learning multivariable calculus therefore I don't understand more than an element solution...", "Solution_3": "[hide=\"My second trial\"]\nProposition) $ F^{(n)}(x) = \\int_{0}^{2\\pi} e^{x \\cos\\alpha}\\cos(n\\alpha+x\\sin\\alpha) \\, d\\alpha$.\n\nProof) For $ n = 0$, this equality obviously holds. Now assume this equality holds for $ n$. Then\n\n\\begin{eqnarray*}\nF^{(n+1)}(x)\n& = & \\frac{d}{dx}\\int_{0}^{2\\pi} e^{x \\cos\\alpha}\\cos(n\\alpha+x\\sin\\alpha) \\, d\\alpha\\\\\n& = & \\int_{0}^{2\\pi} \\frac{\\partial}{\\partial x} e^{x \\cos\\alpha}\\cos(n\\alpha+x\\sin\\alpha) \\, d\\alpha\\\\\n& = & \\int_{0}^{2\\pi} \\left[ e^{x \\cos\\alpha}\\cos\\alpha \\cos(n\\alpha+x\\sin\\alpha) - e^{x \\cos\\alpha}\\sin\\alpha \\sin(n\\alpha+x\\sin\\alpha) \\right] \\, d\\alpha\\\\\n& = & \\int_{0}^{2\\pi} e^{x \\cos\\alpha}\\cos((n+1)\\alpha+x\\sin\\alpha) \\, d\\alpha\n\\end{eqnarray*}\n\nand by induction, the proposition is proved.\n\n\nNow we have $ F(0) = 2\\pi$ and $ F^{(k)}(0) = 0$ for all positive integer $ k$.\n\nThus if $ F$ is analytic at $ x = 0$, then $ F(x) = 2\\pi$ for all $ x$.\n\nHmm... well, this method also seems to need somewhat advanced arguments, since we have to prove that $ F$ is indeed analytic.\n[/hide]\r\n\r\n\r\nSlightly modified version:\r\n\r\n$ e^{x \\cos \\alpha} \\cos (x \\sin \\alpha) = \\text{Re} \\left[ \\exp(x e^{i \\alpha}) \\right] = \\text{Re} \\left[ \\sum_{n=0}^{\\infty} \\frac{1}{n!}x^n e^{i n\\alpha} \\right] = \\sum_{n=0}^{\\infty} \\frac{1}{n!}x^n \\cos (n \\alpha)$\r\n\r\nFor each $ x$, this series converges uniformly for all $ \\alpha$. So we can integrate this series term-by-term, and we have\r\n\r\n$ F(x) = \\sum_{n=0}^{\\infty} \\int_{0}^{2\\pi} \\frac{1}{n!}x^n \\cos (n \\alpha) \\, d\\alpha = 2\\pi$", "Solution_4": "Thank you, sos440. I find that I can use the remains of Taylor polynomial to solve it with the $ F^{(n)}$. Actually it is not difflcult in that way. Thank you.", "Solution_5": "[quote=\"dingdongdog\"]A nice integral in my textbook, but I cannot get it:\n${ F(x) = \\int^{2\\pi}_{0}e^{x \\cos\\alpha}\\cos(x \\sin\\alpha)}\\,d\\alpha$, prove that $ F(x) = 2\\pi$ for all x in real.[/quote]Here's a simple solution, with credit to one of my colleagues: Clearly $ F(0) = 2 \\pi$. Then\r\n\\[ F'(x) = \\int_{0}^{2\\pi} e^{x \\cos \\alpha} \\left( \\cos \\alpha \\cos (x \\sin \\alpha) - \\sin \\alpha \\sin (x \\cos \\alpha) \\right) d \\alpha \\\\= \r\n\\int_0^{2\\pi} \\frac{d}{d \\alpha} \\frac{1}{x} e^{x \\cos \\alpha} \\sin ( x \\sin \\alpha) d \\alpha = \\frac{1}{x} \r\ne^{x \\cos \\alpha} \\sin ( x \\sin \\alpha) \\bigg|^{\\alpha = 2\\pi}_{\\alpha = 0} = 0\\]\r\nand hence $ F(x) = 2 \\pi$ for all $ x$. :lol:", "Solution_6": "[quote=\"hpe\"][quote=\"dingdongdog\"]A nice integral in my textbook, but I cannot get it:\n${ F(x) = \\int^{2\\pi}_{0}e^{x \\cos\\alpha}\\cos(x \\sin\\alpha)}\\,d\\alpha$, prove that $ F(x) = 2\\pi$ for all x in real.[/quote]Here's a simple solution, with credit to one of my colleagues: Clearly $ F(0) = 2 \\pi$. Then\n\\[ F'(x) = \\int_{0}^{2\\pi} e^{x \\cos \\alpha} \\left( \\cos \\alpha \\cos (x \\sin \\alpha) - \\sin \\alpha \\sin (x \\cos \\alpha) \\right) d \\alpha \\\\\n= \\int_0^{2\\pi} \\frac {d}{d \\alpha} \\frac {1}{x} e^{x \\cos \\alpha} \\sin ( x \\sin \\alpha) d \\alpha = \\frac {1}{x} e^{x \\cos \\alpha} \\sin ( x \\sin \\alpha) \\bigg|^{\\alpha = 2\\pi}_{\\alpha = 0} = 0\n\\]\nand hence $ F(x) = 2 \\pi$ for all $ x$. :lol:[/quote]\r\nMaybe what you mean is $ F'(x) = \\int_{0}^{2\\pi} e^{x \\cos \\alpha} \\left( \\cos \\alpha \\cos (x \\sin \\alpha) - \\sin \\alpha \\sin (x \\sin \\alpha) \\right) d \\alpha$ ..." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "How many distinct prime divisors does 900 have?", "Solution_1": "To answer this question, we would first need to do a prime factorization. \r\n\r\nThere are many ways to find the prime factorization of 900. Here is one method: \r\n$ 900$\r\n$ \\equal{}9\\times100$\r\n$ \\equal{}3\\times3\\times10\\times10$\r\n$ \\equal{}3\\times3\\times2\\times5\\times2\\times5$\r\n$ \\equal{}2^2\\times3^2\\times5^2$\r\n\r\nWe thus see that 900 is divisible by $ \\fbox{3}$ different primes: $ 2,3,$ and $ 5$." } { "Tag": [ "geometry", "probability" ], "Problem": "It\u2019s raining! Each raindrop can hit the ground anywhere within a fixed n x n region on a flat plane. The plane starts out completely dry. On impact, each raindrop leaves a wet circle of radius r, with the impact point as its center. After t raindrops have fallen, what is the area of the part of the plane that will end up wet, on average?", "Solution_1": "all of the plane, right?\r\n\r\nor am I missing something? :?", "Solution_2": "You're missing something -- try re-reading, perhaps?\r\n\r\nThis doesn't seem easy at all. All that seems obvious to me is that as $r \\to 0$, the solution approaches $\\pi \\cdot r^2 t$, but beyond that I really don't have any clue. Any approach based on putting points on a plane seems unlikely to result in anything nice at all.", "Solution_3": "Look at it from the point of view of a point on the plane.\r\n\r\n[hide=\"Solution\"]The probability that any particular raindrop makes the point wet is $\\frac{\\pi\\cdot r^2}{n^2}$. Since each raindrop is independent, the probability that this point is not wet after $t$ drops is $\\left(1-\\frac{\\pi\\cdot r^2}{n^2}\\right)^t$. This is the average fraction of the region still dry.[/hide]\r\nActually, that isn't quite enough; the above only applies when the point is more than $r$ away from the boundary. It could be worked out for the other points as well, but the answer would be hopelessly ugly, requiring numerical methods.", "Solution_4": "[quote=\"JBL\"]You're missing something -- try re-reading, perhaps?\n\nThis doesn't seem easy at all. All that seems obvious to me is that as $r \\to 0$, the solution approaches $\\pi \\cdot r^2 t$, but beyond that I really don't have any clue. Any approach based on putting points on a plane seems unlikely to result in anything nice at all.[/quote]\r\n\r\nsorry, I just thought of a rainstorm, and did not see the $t$ part. oops :blush:" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $a,b,c$ positive reals.\r\n\r\nProve that\r\n$(a+b)^{c}+(b+c)^{a}+(c+a)^{b}>2$", "Solution_1": "[quote=\"Till\"]Let $a,b,c$ positive reals.\n\nProve that\n$(a+b)^{c}+(b+c)^{a}+(c+a)^{b}>2$[/quote]\r\nThe hint see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=133925" } { "Tag": [ "videos", "Gauss" ], "Problem": "list em here!", "Solution_1": "Anything by hannah montana\r\n\r\nAnything by anyone like hannah montana", "Solution_2": "i know her songs scare me", "Solution_3": "Umm........\r\n\r\nThere was a total failure by Beatles in White Album\r\n\r\nCalled Revolution 9, you will creep out once you hear it or just feel weird.\r\n\r\nhttp://youtube.com/watch?v=_wCJ9WmX9Zw not interesting vid to watch\r\n\r\nhttp://youtube.com/watch?v=PG0wksBzKSc this one gives me the creeps", "Solution_4": "thats weird....", "Solution_5": "NOOOOOOOOOOOOOOOOOOOOOOOOO\r\nHANNAH MONTANA IS LIFE!!\r\n\r\nok i admit... shes not veryy er...professional???\r\n :rotfl:", "Solution_6": "She's only 15 years old. How professional can a 15 year old be? \r\n\r\nSummer Love by Justin Timberlack. Fail.", "Solution_7": "ahhhh that terrible songg\r\ni heard it on the radio last week... why would they bring that up again...", "Solution_8": "anything from disney channel, american idol, mtv, justin timberlake, soulja boys, and me.", "Solution_9": "[quote=\"7h3.D3m0n.117\"]She's only 15 years old. How professional can a 15 year old be? \n\nSummer Love by Justin Timberlack. Fail.[/quote]\r\nnot very professional", "Solution_10": "You're Beautiful by James Blunt.\r\n\r\nThis song is annoying.\r\n\r\nEverything else said on this thread I agree with so far", "Solution_11": "cant believe no one said jonas brothers....yet\r\n\r\n\r\nand yea the other ones are really annoying", "Solution_12": "[youtube]D-Kfcq5KM9o[/youtube]", "Solution_13": "In my opinion Hillary Duff's cover of The Who's My Generation is the worst.", "Solution_14": "WTF? Dream Theater is great.\r\n\r\nMost popular songs nowadays.", "Solution_15": "Any song that my sister listens to.", "Solution_16": "Russellicious.", "Solution_17": "anything by miley cyrus makes my ears bleed. :( \r\n\r\n@1=2: What? that songs pretty good.....", "Solution_18": "Yea, I agree DT is an awesome band, even if LaBrie's vocals get a little annoying at times.\r\n\r\nAnyway, to continue with the thread...This is not necessarily the worst, but I personally dislike most songs on Metallica's St. Anger album. I really hope they pick up the Master of Puppets, and Ride the Lightning feel with their next release (Death Magnetic).", "Solution_19": "Anything by Soulja Boy (duh)", "Solution_20": "[quote=\"MustItAlwaysBeMe\"]Anything by Soulja Boy (duh)[/quote]\r\n\r\nThe only song that was ever on air by Soulja Boy was Crank Dat.", "Solution_21": "rap of all sorts. yeah, i hate rap.", "Solution_22": "Crank Dat was OK for like 2 days. Then it got old...\r\nAn annoying song is Bleeding Love (Leona Lewis) or just Miley Cyrus (always bad)", "Solution_23": "Avril Lavigne is pretty annoying. And any song similar in style like hers...", "Solution_24": "Dora Dora Dora the explorer...", "Solution_25": "Barbie Girl", "Solution_26": "To Mewtoo555...you created your account on my birthday. \r\n\r\nMy least favorite song: Anything by anyone who thinks they can sing but really cant ie. miley cyrus.", "Solution_27": "[quote=\"RussianRocket\"]\nAn annoying song is Bleeding Love (Leona Lewis) [/quote]\r\n\r\n**gasp** how dare you?", "Solution_28": "[quote=\"basted\"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]\r\n\r\n\r\n :mad:", "Solution_29": "[quote=\"basted\"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]\nNOOOOOOO!!!!! Avril Lavigne rocks!\n\n[quote=\"mewto55555\"]Barbie Girl[/quote]\r\n :rotfl: agreed :rotfl:", "Solution_30": "[quote=\"RussianRocket\"]\nAn annoying song is Bleeding Love (Leona Lewis) [/quote]\r\n\r\ni agree. it keeps repeating the same line over and over.\r\n\r\n\r\nI also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.", "Solution_31": "[quote=\"cyberspace\"][quote=\"basted\"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]\nNOOOOOOO!!!!! Avril Lavigne rocks![/quote]\r\n\r\n\r\n\r\nTHank you cyberspace.", "Solution_32": "[quote=\"aanjohri\"]I also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.[/quote]\r\n\r\nWell then. I guess no one else but me likes these song...\r\n\r\nBut I also love every type of music (especially Bengali music my friend lets me listen to on her iPod...:D) except country. The one thing people here like. IN ALABAMA.", "Solution_33": "\"I've got my sights set on you, \r\nAnd I'm ready to aim, \r\nI have a heart that will, \r\nNever be tamed, \r\nI knew you were something special\r\nWhen you spoke my name, \r\nNow, I can't wait to see you again... \r\n\r\nI've got a way of knowing, \r\nWhen something is right, \r\nI feel like I must have known you, \r\nIn another life, \r\nCause I felt this deep connection, \r\nWhen you looked in my eyes, \r\nNow I can't wait to see you again, \r\n\r\nThe last time I freaked out, \r\nI just kept looking down, \r\nI st-st-stuttered when\r\nYou asked me what I'm thinkin' bout\r\nFelt like I couldn't breathe\r\nYou asked what's wrong with me\r\nMy best friend Lesley said\r\n\"Oh she's just being Miley!\"\r\n\r\nThe next time we hang out, \r\nI will redeem myself, \r\nMy heart it can rest till then, \r\nWhoa! Whoa! I, \r\nI can't wait to see you again, \r\n\r\nI got this crazy feeling\r\nDeep inside\r\nWhen you called and ask to see me\r\nTomorrow night, \r\nI'm not a mind reader, \r\nBut I'm reading the signs\r\nThat you can't wait to see me again\r\n\r\nThe last time I freaked out, \r\nI just kept looking down, \r\nI st-st-stuttered when\r\nYou asked me what I'm thinkin' bout\r\nFelt like I couldn't breath\r\nYou asked what's wrong with me\r\nMy best friend Lesley said\r\n\"Oh she's just being Miley!\" \r\n\r\nThe next time we hang out, \r\nI will redeem myself \r\nMy heart can rest till then, \r\nWhoa! Whoa! I, \r\nI can't wait to see you again, \r\n\r\nI've got my sight set on you, \r\nAnd I'm ready to aim, \r\n\r\nThe last time I freaked out, \r\nI just kept looking down, \r\nI st-st-stuttered when\r\nYou asked me what I'm thinkin' bout\r\nFelt like I couldn't breathe\r\nYou asked what's wrong with me\r\nMy best friend Lesley said, \r\n\"Oh she's just being Miley\"\r\n\r\nThe next time we hang out, \r\nI will redeem myself\r\nMy heart can rest till then, \r\nWhoa whoa I, \r\nI can't wait to see you again, \r\nWhoa whoa I, \r\nI can't wait to see you again.\"\r\n\r\n[size=200][u][i][b]HAD 2 SHARE THE PAIN... SORRY... COULDNT HELP IT[/b][/i][/u][/size] :furious: :noo: :cursing:", "Solution_34": "......what is that filth........", "Solution_35": "\"See You Again\"--miley cyrus\r\n\r\n\r\nTHe pain is too much russianrocket delete it!!", "Solution_36": "[quote=\"aanjohri\"][quote=\"RussianRocket\"]\nAn annoying song is Bleeding Love (Leona Lewis) [/quote]\n\ni agree. it keeps repeating the same line over and over.\n\n\nI also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.[/quote]\r\n\r\nChocolate Rain is probably the most repetitive song there is, also Say by John Mayer is pretty repetitive too.", "Solution_37": "[quote=\"SnipedYou\"][quote=\"aanjohri\"][quote=\"RussianRocket\"]\nAn annoying song is Bleeding Love (Leona Lewis) [/quote]\n\ni agree. it keeps repeating the same line over and over.\n\n\nI also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.[/quote]\n\nChocolate Rain is probably the most repetitive song there is, also Say by John Mayer is pretty repetitive too.[/quote]\r\nDude, Chocolate Rain is pretty awesome (*moves away from mic to breath*). DnB remix is pretty good, too. \r\nDude, it's scary, and that's an understatement; let me read my book marks bar which got screwed got by my sister:\r\nYahoo E-mail\r\nAoPS\r\nGmail\r\nYoutube - Hannah Montana rock star FULL + lyrics video (WTF?)\r\nYoutube - Hannah Montana True friend FULL + lyrics\r\nYoutube - Miley Cyrus Hannah Montana I Miss You FULL + lyrics\r\nYoutube -Let's Dance! (Lyrics) Hannah Montana/ Miley Cyrus\r\n\r\nAUGHAUGUAGAHG! And she sings this crap all the time.\r\n\r\n\r\n\r\nThat aside, Soulja boy is pretty dang crappy too. Actually, pretty much all the stereotype bands (Nickelback, ewww, Greenday, okayyy, Linkin Park, wtf?, etc.)", "Solution_38": "[quote=\"Nerd_of_the_Ages\"]stereotype bands (Nickelback, ewww, Greenday, okayyy, Linkin Park, wtf?, etc.)[/quote]\r\n\r\nOMG ARE YOU FRIKIN KIDDING ME LINKIN PARK IS AWESOME!\r\n[color=white][size=9](maybe thats cuz i like emo music....)[/size][/color]", "Solution_39": "[quote=\"myyellowducky82\"]i know her songs scare me[/quote]\r\nTHAT'S WHY SHE GOT OWNED BY KEVIN CHEN!!!!", "Solution_40": "OMG I WAS SO HAPPY MILEY GOT OWNED BY KEVIN CHEN!!!!", "Solution_41": "Time to bash.\r\n\r\nLinkin fart has about the same musical talent as a rock with testosterone.\r\n\r\nMiley Cyrus is Mainstream. Mainstream is just popular for being mainstream. Most of the musicians there have less musical talent than linkin fart.\r\n\r\nAnd wtf? gauss, you're like a fan girl. \r\n\r\nWith that being said. I hate tool.", "Solution_42": "umm...\r\n\r\ni get annoyed by miley cyrus.\r\na lot of high school musical songs.\r\n\r\nnow that i think about it, anything from disney channel.\r\n\r\nbut i do like the stuff thats on the radio. like Ne-o, chris brown's voice is just annoying, rihanna (except for umbrella. umbrella was the worst), and Cascada.", "Solution_43": "[quote=\"hunter34\"]Time to bash.\n\nLinkin fart has about the same musical talent as a rock with testosterone.\n\nMiley Cyrus is Mainstream. Mainstream is just popular for being mainstream. Most of the musicians there have less musical talent than linkin fart.\n\nAnd wtf? gauss, you're like a fan girl. \n\nWith that being said. I hate tool.[/quote]\r\nTotally...", "Solution_44": "[quote=\"hunter34\"]Time to bash.\n\nLinkin fart has about the same musical talent as a rock with testosterone.\n\nMiley Cyrus is Mainstream. Mainstream is just popular for being mainstream. Most of the musicians there have less musical talent than linkin fart.\n\nAnd wtf? gauss, you're like a fan girl. \n\nWith that being said. I hate tool.[/quote]\r\nI'm not a fan girl, I'm a tomboy. So i guess the whole AoPS community owns miley cyrus. and that's only because we're all smarter than her in math :D \r\n\r\nGO NERDS!!!!!!!!! :D" } { "Tag": [ "inequalities", "abstract algebra", "triangle inequality", "geometry proposed", "geometry" ], "Problem": "Let a acute-triangle $ ABC$. Find point $ M$ inside triangle so that $ AM.BM.AB \\plus{} BM.CM.BC \\plus{} CM.AM.CA$ min!", "Solution_1": "We have the following inequality for every point $ M$ in the triangle $ ABC$:\r\n\\[ a\\cdot MB \\cdot MC \\plus{} b\\cdot MC \\cdot MA \\plus{} c\\cdot MA \\cdot MB \\geq abc\r\n\\]\r\nI shall prove this inequality using complex numbers.\r\nLet $ M$ be the center of the complex plane where $ A(a)$,$ B(b)$,$ C(c)$.\r\nWe have the following identity:\r\n\\[ ab(a \\minus{} b) \\plus{} bc(b \\minus{} c) \\plus{} ca(c \\minus{} a) \\equal{} \\minus{} (a \\minus{} b)(b \\minus{} c)(c \\minus{} a)\r\n\\]\r\nAplying module and using the triangle inequality we obtain the desired inequality:\r\n\\[ \\sum {\\left|a\\right|\\left|b\\right|\\left|a \\minus{} b\\right|}\\geq \\left|\\sum{ab(a \\minus{} b)}\\right| \\equal{} \\left|a \\minus{} b\\right|\\left|b \\minus{} c\\right|\\left|c \\minus{} a\\right|\r\n\\]\r\nSo $ min\\{ a\\cdot MB \\cdot MC \\plus{} b\\cdot MC \\cdot MA \\plus{} c\\cdot MA \\cdot MB\\} \\equal{} abc$", "Solution_2": "[quote=\"Vrajitoarea Andrei\"]mininum for $ M\\in\\{A,B,C\\}$[/quote]\r\nBut $ M$ must be inside triangle!\r\nAnd min when $ M$ be orthocentre of triangle!", "Solution_3": "Sorry.My mistake. Indeed , we obtain the minimum for $ a\\plus{}b\\plus{}c\\equal{}0$ meaning that $ M\\equal{}H$", "Solution_4": "[hide=\"Another hint:\"]Let $ E,F$ are points so that: $ BCME,BCAF$ are parallelograms.\nHence $ AF\\equal{}EM\\equal{}BC,EF\\equal{}AM,EB\\equal{}CM,BF\\equal{}AC$.\nApply Ptoleme inequality for $ ABEF,AEBM$ we have solution! [/hide]", "Solution_5": "[quote=\"Vrajitoarea Andrei\"]Sorry.My mistake. Indeed , we obtain the minimum for $ a \\plus{} b \\plus{} c \\equal{} 0$ meaning that $ M \\equal{} H$[/quote]\r\n\r\n\r\nWhy for $ a\\plus{}b\\plus{}c\\equal{}0$ ?" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Could someone help me with this problem:\r\nIf f in L^1 is differentiable a.e., and f' in L^1, what is the Fourier transform of f' ?? Is is tiF(t), where F(t) is the Fourier transform of t?\r\nI think I can show this by using integration by parts, but I wanted to check that I am right.", "Solution_1": "Yes. (up to constant multipliers depending on which definition you're using)" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Coach Frost had fifteen students attend his first MATHCOUNTS practice. Coach Frost asked the students to shake hands with each other. Assume each student shook hands with every other student exactly once, and no student shook hands with Coach Frost. How many total handshakes were exchanged?", "Solution_1": "[hide=\"My Answer\"]15 choose 2 \n\n15*14/2=15*7=105\n\n14+13+12+11+10+9+8=7+6+5=4+3+2+1 also gives you the same result.[/hide]", "Solution_2": "[hide]\n${15 \\choose 2}=\\framebox{105}$\nOr by doing addition, 14+13+12+11+10+9+8+7+6+5+4+3+2+1=105[/hide]", "Solution_3": "this is soooo easy....................i can't believe this...\r\n[hide=\"easy way\"]\n15 choices for person one\n14 choices for person two\n\ndivide by two because you doublecount\n\n<_<\n>_>\n<_<\n>_>\n[/hide]", "Solution_4": "well...[hide=\"solution\"]this is a combination problem...so just do $\\frac{15*14}{2}$ to get $15*7$ which is [b]105[/b].\nu also can do (tho i dont suggest it) 14+13+12+11+10+9+8+7+6+5+4+3+2+1 which comes out being the same answer, 105.[/hide]", "Solution_5": "[hide] (14*15)/2=7*15=105 [/hide]", "Solution_6": "[hide][quote=\"flamesofdestiny\"]u also can do (tho i dont suggest it) 14+13+12+11+10+9+8+7+6+5+4+3+2+1 which comes out being the same answer, 105.[/quote]\n\nThat's because the formula for the sum of the first $n$ integers is $\\frac{n(n+1)}{2}=\\frac{14(15)}{2}$, which is actually where your first solution came from.[/hide]", "Solution_7": "[quote=\"Treething\"]this is soooo easy....................i can't believe this...\n[hide=\"easy way\"]\n15 choices for person one\n14 choices for person two\n\ndivide by two because you doublecount\n\n<_<\n>_>\n<_<\n>_>\n[/hide][/quote]\r\nFolks: If you think a problem is very easy, why don't you leave it for those who don't think it is so easy.\r\n\r\nWhile you didn't give a final answer, your solution seems to lead to an incorrect answer. [Maybe this wasn't as easy as you thought? :) ]", "Solution_8": "[hide]its 15 choose 2=105 :idea: [/hide]", "Solution_9": "[quote=\"rcv\"][quote=\"Treething\"]this is soooo easy....................i can't believe this...\n[hide=\"easy way\"]\n15 choices for person one\n14 choices for person two\n\ndivide by two because you doublecount\n\n<_<\n>_>\n<_<\n>_>\n[/hide][/quote]\nFolks: If you think a problem is very easy, why don't you leave it for those who don't think it is so easy.\n\nWhile you didn't give a final answer, your solution seems to lead to an incorrect answer. [Maybe this wasn't as easy as you thought? :) ][/quote]\r\n\r\nIt looks right to me :? $\\frac{15 \\cdot 14}{2}=15 \\cdot 7 = 105$...", "Solution_10": "[quote=\"chess64\"]It looks right to me :? $\\frac{15 \\cdot 14}{2}=15 \\cdot 7 = 105$...[/quote]\r\n\r\nI think he got lucky.... sure works for these types of problems.", "Solution_11": "[quote=\"236factorial\"]I think he got lucky.... sure works for these types of problems.[/quote]\r\nWhat do you mean? That's what I did... What's wrong with that method? It's what you're supposed to do! 15 choices for the 1st, 14 for the 2nd, divide by 2 to account for repeats. (A-B is the same as B-A)", "Solution_12": "[quote=\"chess64\"]What do you mean? That's what I did... What's wrong with that method? It's what you're supposed to do! 15 choices for the 1st, 14 for the 2nd, divide by 2 to account for repeats. (A-B is the same as B-A)[/quote]\r\n\r\nOh, I was just thinking that the first person had only 14 people, second had 13, third had 12, etc.... so we wouldn't worry about repeats. :oops:", "Solution_13": "[quote=\"chess64\"]It looks right to me :? $\\frac{15 \\cdot 14}{2}=15 \\cdot 7 = 105$...[/quote]\r\n\r\nTreething's solution is correct. My apologies to you, Treething. I misunderstood your brief solution." } { "Tag": [ "complex numbers", "algebra solved", "algebra" ], "Problem": "determine a,b if ax^17+bx^16-1 divides x^2-x-1", "Solution_1": "A solution using complex numbers : \r\n\r\n\r\np(x)=ax^17+bx^16-1 divides x^2-x-1 iff p(phi) = p(1/phi) = 0\r\n\r\nwith phi=(1+sqrt(5))/2\r\n\r\nfrom phi\u00b2=phi+1 we quickly get phi^17 =1597*phi+987 and phi^16=987*phi+610\r\n\r\nThen p(phi) = p(1/phi) = 0 gives (after some boring :( and unchecked :) algebra) :\r\n\r\na = -987, b = 1497.", "Solution_2": "Can anyone call that a problem?" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Prove that: $sin(3x)=4sina.sin(a+\\frac{\\pi}{3}).sin(a+\\frac{2\\pi}{3})$", "Solution_1": "${\\sin (3 x)=4 \\sin (a) \\sin \\left(a+\\frac{\\pi }{3}\\right) \\sin \\left(a+\\frac{2 \\pi }{3}\\right)}$\r\n\r\nFrom Trig Rules\r\n\r\n${\\sin (\\theta+\\phi )=\\cos (\\phi ) \\sin (\\theta )+\\cos (\\theta ) \\sin (\\phi )}$\r\n\r\nSimplifying\r\n\r\n${\\sin \\left(a+\\frac{\\pi }{3}\\right)=\\cos (a) \\sin \\left(\\frac{\\pi }{3}\\right)+\\cos \\left(\\frac{\\pi }{3}\\right) \\sin (a)}$\r\n${\\sin \\left(a+\\frac{\\pi }{3}\\right)=\\frac{1}{2}\\left(\\sqrt{3}\\cos (a)+\\sin (a)\\right)}$\r\n\r\n${\\sin \\left(a+\\frac{2 \\pi }{3}\\right)=\\cos (a) \\sin \\left(\\frac{2 \\pi }{3}\\right)+\\cos \\left(\\frac{2 \\pi }{3}\\right) \\sin (a)}$\r\n${\\sin \\left(a+\\frac{2 \\pi }{3}\\right)=\\frac{1}{2}\\left(\\sqrt{3}\\cos (a)-\\sin (a)\\right)}$\r\n\r\nSubstituing\r\n\r\n${4 \\sin (a) \\sin \\left(a+\\frac{\\pi }{3}\\right) \\sin \\left(a+\\frac{2 \\pi }{3}\\right)=\\sin (a) \\left(\\sqrt{3}\\cos (a)-\\sin (a)\\right) \\left(\\sqrt{3}\\cos (a)+\\sin (a)\\right)}$\r\n${4 \\sin (a) \\sin \\left(a+\\frac{\\pi }{3}\\right) \\sin \\left(a+\\frac{2 \\pi }{3}\\right)=3 \\cos^{2}(a) \\sin (a)-\\sin^{3}(a)}$\r\n\r\nBut we have...\r\n\r\n${\\sin (3 x)=3 \\cos^{2}(x) \\sin (x)-\\sin^{3}(x)}$\r\n\r\n${3 \\cos^{2}(a) \\sin (a)-\\sin^{3}(a)=3 \\cos^{2}(a) \\sin (a)-\\sin^{3}(a)}$\r\n\r\nSo it is correct only if\r\n\r\n${a=x}$\r\n\r\nAnd it is the proof" } { "Tag": [], "Problem": "Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna. In how many ways can we choose 6 starters if all three triplets are in the starting lineup?", "Solution_1": "If all three triplets are in the starting line, then there are 11 players left to choose 3 people from. Thus our answer is $ \\binom{11}{3} \\equal{} 165$.", "Solution_2": "how is this level 12\n\nin total, the volleyball team has $14$ players. the triplets are automatically starters, so exclude them which gives $11$ remaining players.\n\nwe want to pick $6$ starters but the triplets are automatically starters, so exclude them which gives $3$ remaining starters.\n\ntherefore the answer is $\\binom{11}{3}=\\boxed{165}$.", "Solution_3": "[quote=OlympusHero]how is this level 12\n\nin total, the volleyball team has $14$ players. the triplets are automatically starters, so exclude them which gives $11$ remaining players.\n\nwe want to pick $6$ starters but the triplets are automatically starters, so exclude them which gives $3$ remaining starters.\n\ntherefore the answer is $\\binom{11}{3}=\\boxed{165}$.[/quote]\n\nlvl 12 is supposed to be easy..." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "vector", "parameterization", "search", "linear algebra" ], "Problem": "I posted these in the Intermediate forums, but no one answered :( ...\r\n\r\n[quote=\"leoxnlin\"]Let $ S$ be a set of $ n$ ordered sextuples where the $ i$th ordered sextuple is $ (i_1,i_2,i_3,i_4,i_5,i_6)$. For all $ i_u$, $ i_u \\in \\mathbb{Z} , 0 \\le i_u \\le 9$. Find the maximum possible value of $ n$ if there do not exist any $ i_u, i_v, j_u, j_v$ such that $ i_u \\equal{} j_u$ and $ i_v \\equal{} j_v$.[/quote]\n\n[quote=\"leoxnlin\"]Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one [i]or more[/i] distinct names written on it, although different cards may share some names. We define a \"shuffle with respect to a name\" the process of moving every card that has that name written on it to the opposite stack. Prove that it is always possible have more cards in the right stack after finitely many shuffles.[/quote]\r\n\r\nThank you :) .", "Solution_1": "[hide=\"Second problem\"]Shuffle with respect to all names.[/hide]", "Solution_2": "Um, no MellowMelon, I believe there must be a misunderstanding,... counterexamples are easily constructed... perhaps you missed the \"one [i]or more[/i] distinct names\" part?", "Solution_3": "Some weakish bounds on the first problem:\r\n\r\n[hide]If we have more than 100 sextuples then two of them have the same first two elements, a contradiction. Therefore $ n\\le 100$.\n\nA construction for $ n \\equal{} 52$: For 49 of our sextuples, we only use the digits 0-6. Let the elements of a general sextuple be $ a,b,a \\plus{} b,a \\plus{} 2b,a \\plus{} 3b,a \\plus{} 4b$, reducing modulo 7. If two pairs of corresponding elements of different sextuples are the same, then some linear combination of their formulas above gives us $ a$ or a multiple of $ b$, and from there we can show that the $ a$ and $ b$ values of the two sextuples are the same. Therefore, making the $ a$ and $ b$ values of each sextuple different is sufficient to ensure that the 49 sextuples satisfy the requirements of the problem.\n\nWe can also throw in $ (7,7,7,7,7,7),(8,8,8,8,8,8)$, and $ (9,9,9,9,9,9)$ to give us a total of 52 sextuplets.\n\nTherefore $ 52\\le n\\le 100$.[/hide]\r\nWhat is the source of this problem? It's similar to an old and fairly easy USAMO problem, but this one doesn't seem like it has a nice solution.", "Solution_4": "Thanks matt276eagles :) That's about the only useful response I've ever received at all on these questions...", "Solution_5": "[hide=\"Second problem\"]\nLet $ N$ be the set of all names which appear on any of the cards. for every $ A \\subseteq N$ consider the effect of shuffling all the names in $ A$. If we do this for all possible $ A$'s, then each card remains on the lefts stack precisely half the time, and moves to the right half of the time. So, on average, the number of cards on the left and on the right must be equal. Since $ A \\equal{} \\emptyset$, which corresponds to doing nothing, yields a larger left stack, there must be at least one $ A$ yielding a larger right stack.\n[/hide]\r\n\r\nThis problem can be found in Peter Winkler's \"Mathematical Puzzles: A Connoisseur's Collection\", p. 3, \"Tipping the Scales\". It's on Google Books.", "Solution_6": "Can you rephrase your first problem? I can't make sense of it... sorry.", "Solution_7": "Essentially it is asking for the maximum number of six-digit integers (where we allow leading zeroes) such that no two numbers share more than one digit.", "Solution_8": "Not quite. The precise problem is this: find a large set $ A$ of vectors $ {\\bf x} \\equal{} (x_1,x_2,\\ldots, x_6)$ such that, for any two indices $ 1 \\leq i < j \\leq 6$ and any two vectors $ \\bf{x}$, $ \\bf{y}$ we have that $ (x_i,x_j) \\neq (y_i,y_j)$. This is part of a family of such problems determined by the parameters: alphabet size (10), length (6), and length of sub-vectors we care about (2). For other parameter sets this is easily solved via linear algebra, but the choice of parameters here is a bit nastier. When I get a chance I'll see if I can imporove the bounds.", "Solution_9": "I think (2) is actually a weaker version of http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1509886817&t=5976", "Solution_10": "i think the answer is 10.\r\nsince i_u is in the interval[0,9], there are only 10 alternatives for it,\r\none for each set." } { "Tag": [ "abstract algebra", "number theory proposed", "number theory" ], "Problem": "solve: 2$x^4+1=y^2$", "Solution_1": "It's really obvious.\r\nWe have the only solution $(x,y)=(0,1)$.\r\nAssume that $y>1$.\r\nIf $2x^4=(y-1)(y+1)$ then $y$ is odd (and $\\gcd(y-1,y+1)=2$).\r\nSo ($y-1=2\\alpha^4$ and $y+1=\\beta^4$) or ($y+1=2\\alpha^4$ and $y-1=\\beta^4$) \r\nThe first case is impossible because $2(1+\\alpha^4)=\\beta^4 \\Rightarrow \\alpha^4\\equiv-1[4]$.\r\nThus $\\beta^4=2(\\alpha^4-1)=2(\\alpha-1)(\\alpha^2+1)(\\alpha+1)$ and $\\alpha$ is odd.\r\nWe have $\\gcd(\\alpha-1,\\alpha+1)=\\gcd(\\alpha-1,\\alpha^2+1)=\\gcd(\\alpha+1,\\alpha^2+1)=2$, so:\r\n$\\alpha-1=2\\gamma^4$ and $\\alpha+1=2\\chi^4$ $\\Rightarrow 1+\\gamma^4=\\chi^4 \\Rightarrow$ impossible.", "Solution_2": "[quote=\"nguyenquockhanh\"]solve: 2$x^4+1=y^2$[/quote]\r\nEasy, only need to work module $5$ and then module 2, geting that $\\tilde X^4=2+5D$, so if $x\\ne 0$ the congruence is IMPOSSIBLE. Then the only solution is $(0,1)$ and $(0,-1)$.\r\n\r\nBest Regards.\r\n\r\nRX TCM", "Solution_3": "okie :D \r\nI have three solutions for this pro, one of them similar above solutions.\r\nAnd this morning, my teacher gave me below pro:\r\n[b]PRO-2[/b]: solve: $8\\cdot x^4+1=y^2$\r\nAnd i have 2 sol for this pro.\r\nI think very cleary if use THEOREM-FERMAT:\r\n[b]THEOREM-FERMAT[/b]: if $a^2-b^2=c^2$ and $a^2+b^2=d^2$ then $b=0$ you agree, aren't you!", "Solution_4": "[b]PRO[/b]: assume a,b possitive integer \r\nsolve : $(a^2+b^2)\\cdot x^4+1=y^2$", "Solution_5": "THEOREM-FERMAT: if $ a^2-b^2=c^2 $ and $ a^2+b^2=d^2 $ then $ b=0 $ \r\n Can you post the proof for this problem ?", "Solution_6": "[quote=\"romano\"]THEOREM-FERMAT: if $ a^2-b^2=c^2 $ and $ a^2+b^2=d^2 $ then $ b=0 $ \n Can you post the proof for this problem ?[/quote]\r\nokie: here is solution:assume $b>0$, $gcd(a,b)=1$ and $d$ min \r\n$2a^2=c^2+d^2$ so $2|c-d$, we obtain $a^2=(\\frac{c+d}{2})+(\\frac{d-c}{2})$ \r\nnow, applying PITAGO-EQUATION then $2b^2=d^2-c^2=2\\cdot{4mn(m^2-n^2)}$ so $b=2t$ and $t^2=mn(m^2-n^2)$ thus $m=x^2$, $n=y^2$ and $m^2-n^2=z^2$, but now, we have $x^2-y^2$ and $x^2+y^2$ are both squares, contradiction with the hypothesis $d$ min", "Solution_7": "[b]PITAGO-EQUATION[/b]:all solutions of the equation: $x^2+y^2=z^2$ are: $(k(m^2-n^2),2mn,(m^2+n^2))$ in here $m\\neq {n}(mod2)$, $gcd(m,n)=1$", "Solution_8": "[quote=\"nguyenquockhanh\"][b]PRO[/b]: assume a,b possitive integer \nsolve : $(a^2+b^2)\\cdot x^4+1=y^2$[/quote]\r\n\r\nNOTE: when $a=b=6$ then equation $ 72\\cdot x^4+1=y^2$ has no roots, okie?", "Solution_9": "By Lagrange:Solve the equation x 2 +y^4=2*z^4 in Z+", "Solution_10": "Here is my solution for proving x^4-2y^4=\u00b11 has no solution and i think it can be used in the general equation http://www.mathlinks.ro/Forum/viewtopic.php?t=21818&highlight=\r\nANd:[quote=\"nguyenquockhanh\"][b]PRO[/b]: assume a,b possitive integer \nsolve : $(a^2+b^2)\\cdot x^4+1=y^2$[/quote]\r\nCould you post your solution,nguyenquockhanh???" } { "Tag": [ "Support", "LaTeX", "\\/closed" ], "Problem": "I am not sure maybe the question was already asked. Is the limited version of the forum, no avatars and other stuff which is not necessarily required but causes a lot of traffic, already implemented on the new forum. I assumed the topic should be covered at: \r\n\r\nForum index \"user\" Options Board layout \r\n\r\nThanks.", "Solution_1": "I see no problem ... what is the problem? (remember this is the support forum not the question & suggestion one ;) ).", "Solution_2": "[quote=\"orl\"]stuff which is not necessarily required but causes a lot of traffic[/quote]\r\nOh, yes. Huge traffic is a big problem for me... :( \r\nAnd avatars and signatures are not main reason for this, but entire structure.", "Solution_3": "[quote=\"Valentin Vornicu\"]I see no problem ... what is the problem? (remember this is the support forum not the question & suggestion one ;) ).[/quote]\r\n\r\nFeel free to move the post. What I meant is what for instance was requested by Darij once. That you could choose a more text based version where some graphical components were not displayed. I supposed that it can be found here:\r\n\r\nForum index orl Options Board layout \r\n\r\nSo please tell me how I can change the forum view to this limited edition in my profile, options or anywhere else...", "Solution_4": "[quote=\"Myth\"][quote=\"orl\"]stuff which is not necessarily required but causes a lot of traffic[/quote]\nOh, yes. Huge traffic is a big problem for me... :( \nAnd avatars and signatures are not main reason for this, but entire structure.[/quote]\r\nMyth, the new structure is massively improved. It caches most of the stuff down into your hard disk (thus the massive speed improvement), and all the java scripts are downloaded as a temporary file. \r\n\r\nI tried it on Romanian dial-up and it was very quick :) \r\n\r\nBut again I fail to see what this topic is about :?", "Solution_5": "I am Opera user, so all graphical components are not important for me, because all images are cached (as stylesheets and other things). And only images I download are latex-images.\r\n\r\nSpeed were quite well all the time, even if I used dial-up. But I pay for traffic, and that is very annoying thing for me, because cost is quite high. :(", "Solution_6": "[quote=\"Myth\"]Speed were quite well all the time, even if I used dial-up. But I pay for traffic, and that is very annoying thing for me, because cost is quite high. :([/quote]Ok you so you want another text-based (or something like that) version right?", "Solution_7": "I think, it is impossible at the current stage. Just see source of this page. It is about 60KB, but useful information occupies at most 3-4KB. What are we talking about?\r\nAt the same time, I don't think we are able to compact large part of the source code without lost of site functionality.\r\n\r\nI don't see appropriate solution for reducing page weight at the moment.", "Solution_8": "[quote=\"Myth\"]I think, it is impossible at the current stage. Just see source of this page. It is about 60KB, but useful information occupies at most 3-4KB. What are we talking about?\nAt the same time, I don't think we are able to compact large part of the source code without lost of site functionality.\n\nI don't see appropriate solution for reducing page weight at the moment.[/quote]The size of the page also includes cached elements, so the download you make is actually much smaller than before.", "Solution_9": "[quote=\"Valentin Vornicu\"][quote=\"Myth\"]I think, it is impossible at the current stage. Just see source of this page. It is about 60KB, but useful information occupies at most 3-4KB. What are we talking about?\nAt the same time, I don't think we are able to compact large part of the source code without lost of site functionality.\n\nI don't see appropriate solution for reducing page weight at the moment.[/quote]The size of the page also includes cached elements, so the download you make is actually much smaller than before.[/quote]\r\n????\r\nPage is what browser gets for his request. Nothing else. But page source can refer to external files, which are cached.", "Solution_10": "I looked at this and it seems that the page sizes are indeed large :( ... I can scrap off some 20kB / page if I'll get the jumpbox off. \r\n\r\nOther stuff might affect the forum navigation ... but if you have any other suggestions I'd be glad to hear them.", "Solution_11": "Is it possible to make new style without jumpbox? You can add link to the page with the forum site map instead of it.", "Solution_12": "Maybe the main style with jumpbox and an alternative style without. I personally use the jumpbox very often.\r\n\r\n Darij", "Solution_13": "I said \"new\", not \"instead\".", "Solution_14": "[quote=\"Myth\"]Is it possible to make new style without jumpbox? You can add link to the page with the forum site map instead of it.[/quote]:what?: I won't make a new style for that ... I'll put an option in the board layout menu, so users can select it on/off ;)\r\n\r\nNew style means different colors and images, so I won't do that right now. We have two different styles, and that's enough worries for now :cool:", "Solution_15": "[quote=\"Valentin Vornicu\"][quote=\"Myth\"]Is it possible to make new style without jumpbox? You can add link to the page with the forum site map instead of it.[/quote]:what?: I won't make a new style for that ... I'll put an option in the board layout menu, so users can select it on/off ;)[/quote]\r\nIndeed, I mean such solution (in other words... :blush: )" } { "Tag": [], "Problem": "What is the greatest prime factor of 221?", "Solution_1": "$ 221 \\equal{} 13 * 17$ The greater prime factor is $ \\boxed{17}$." } { "Tag": [ "function", "integration" ], "Problem": "\u039a\u03ac\u03c0\u03bf\u03b9\u03b1 (elementary) \u03bb\u03cd\u03c3\u03b7? \u0388\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03c7\u03c1\u03ae\u03c3\u03b7 (\u03bc\u03b9\u03b1\u03c2 \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae\u03c2) Gronwall.\u039a\u03b1\u03bc\u03b9\u03ac \u03ac\u03bb\u03bb\u03b7 \u03b9\u03b4\u03ad\u03b1?", "Solution_1": "\u0397 Gr\u00f6nwall \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c3\u03c4\u03b9\u03c2 ODEs.\r\n\r\n\u039f\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ae \u03b7 elementary \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ac\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03ae \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03b7\u03c2 Gr\u00f6nwall (\u03b1\u03c0\u03bb\u03ac \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c2 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b5\u03c2 weight/decay functions) -- \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf elementary \u03b4\u03b5 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9, \u03ba\u03b1\u03bd\u03ac \u03b4\u03c5\u03bf \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9.", "Solution_2": "\u039f\u03ba,thx (\u03c1\u03c9\u03c4\u03bf\u03cd\u03c3\u03b1, \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03c4\u03b7 \u03b2\u03c1\u03ae\u03ba\u03b1 \u03b7 Gronwall \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd long shot :wink: )", "Solution_3": "[quote=\"Fischerman\"]\u039f\u03ba,thx (\u03c1\u03c9\u03c4\u03bf\u03cd\u03c3\u03b1, \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03c4\u03b7 \u03b2\u03c1\u03ae\u03ba\u03b1 \u03b7 Gronwall \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd long shot :wink: )[/quote]\r\n\r\n\u039d\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ac \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b1 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03c4\u03b7\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 more or less straightforward, \u03b1\u03ba\u03cc\u03bc\u03b1 \u03c0\u03c7. \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03bc\u03b1\u03b8\u03b7\u03c4\u03ad\u03c2 \u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5 (\u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03b5\u03bb\u03c0\u03af\u03b6\u03c9...).", "Solution_4": "\u0398\u03ad\u03c4\u03c9 $ g(x)\\equal{}\\int^x_0f(s)ds$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ g$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b7 $ e^{\\minus{}x}g(x)$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1. \u0386\u03c1\u03b1 $ g(x)e^{\\minus{}x} \\le g(0).e^0\\equal{}0$ \u03bf\u03c0\u03cc\u03c4\u03b5 $ g(x) \\le 0$ \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03ba\u03b1\u03b9 $ f(x) \\le 0$ . \u038c\u03bc\u03c9\u03c2 $ f(x) \\ge 0$ \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $ f(x)\\equal{}0$ \u03c3\u03c4\u03bf $ [0,1]$", "Solution_5": "[quote=\"r_boris\"]\u0398\u03ad\u03c4\u03c9 $ g(x) \\equal{} \\int^x_0f(s)ds$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ g$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b7 $ e^{ \\minus{} x}g(x)$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1. \u0386\u03c1\u03b1 $ g(x)e^{ \\minus{} x} \\le g(0).e^0 \\equal{} 0$ \u03bf\u03c0\u03cc\u03c4\u03b5 $ g(x) \\le 0$ \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03ba\u03b1\u03b9 $ f(x) \\le 0$ . \u038c\u03bc\u03c9\u03c2 $ f(x) \\ge 0$ \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $ f(x) \\equal{} 0$ \u03c3\u03c4\u03bf $ [0,1]$[/quote]\r\n\r\n\u039c\u03b5 \u03b4\u03b9\u03ad\u03c8\u03b5\u03c5\u03c3\u03b5 \u03bf r_boris: \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c7\u03c9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 :D", "Solution_6": "[quote=\"r_boris\"]\u0398\u03ad\u03c4\u03c9 $ g(x) \\equal{} \\int^x_0f(s)ds$ \u03c4\u03cc\u03c4\u03b5 \u03b7 $ g$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03b3\u03c9\u03b3\u03af\u03c3\u03b9\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b7 $ e^{ \\minus{} x}g(x)$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1. \u0386\u03c1\u03b1 $ g(x)e^{ \\minus{} x} \\le g(0).e^0 \\equal{} 0$ \u03bf\u03c0\u03cc\u03c4\u03b5 $ g(x) \\le 0$ \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03ba\u03b1\u03b9 $ f(x) \\le 0$ . \u038c\u03bc\u03c9\u03c2 $ f(x) \\ge 0$ \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $ f(x) \\equal{} 0$ \u03c3\u03c4\u03bf $ [0,1]$[/quote]\r\n\r\n\u039d\u03b1\u03b9." } { "Tag": [], "Problem": "A free living one celled organism, lacking a cell wall is collected from a stream containing mineral rich water. The organism is placed on a microscope slide in distilled water.\r\n1-What microscopic changes in the cell would the observer see?\r\n2-What are the causes of the changes observed?", "Solution_1": "Did you even read your textbook?\r\n[hide]\n1. The cell will expand.\n2. Osmosis. (Water will enter the cell, because the cell is hypotonic to the distilled water. Its natural environment is the stream, which is \"mineral rich\", so the concentration of water is higher outside the cell when it is placed in distilled water.)\n\n[/hide]" } { "Tag": [ "function", "induction" ], "Problem": "A function is defined by $ f(0) \\equal{} 1$ and $ f(n) \\equal{} f(n \\minus{} 1) \\plus{} n \\plus{} 1$. Find $ f(5)$.", "Solution_1": "$ f(0)\\equal{}1 \\implies f(1)\\equal{}1\\plus{}1\\plus{}1\\equal{}3\r\n\\\\ f(1) \\implies f(2)\\equal{}3\\plus{}2\\plus{}1\\equal{}6\r\n\\\\ f(2) \\implies f(3)\\equal{}6\\plus{}3\\plus{}1\\equal{}10\r\n\\\\ f(3) \\implies f(4)\\equal{}10\\plus{}4\\plus{}1\\equal{}15\r\n\\\\ f(4) \\implies f(5)\\equal{}15\\plus{}5\\plus{}1\\equal{}\\boxed{21}$\r\n\r\nWe can use induction to show that $ f(n)$ is the $ n\\plus{}1$th triangular number." } { "Tag": [ "inequalities" ], "Problem": "Find min :\r\n$ \\frac{a^8}{(a^2\\plus{}b^2)^2}\\plus{}\\frac{b^8}{(b^2\\plus{}c^2)^2}\\plus{}\\frac{c^8}{(c^2\\plus{}a^2)^2}$\r\nLet a,b,c>0 and ab+bc+ca=1", "Solution_1": "[hide]\n\nFirst note that if $ a\\equal{}b\\equal{}c$ then it evaluates to $ \\frac{1}{12}$.\nWe'll prove that this is indeed the minimum.\n\nBy $ \\sqrt{\\frac{x^2 \\plus{} y^2 \\plus{} z^2}{3}} \\ge \\frac{x\\plus{}y\\plus{}z}{3}$ we have $ x^2 \\plus{} y^2 \\plus{} z^2 \\ge \\frac{(x\\plus{}y\\plus{}z)^2}{3}$\nand applying that here we get:\n\n$ \\frac{a^8}{(a^2 \\plus{} b^2)^2} \\plus{} \\frac{b^8}{(b^2\\plus{}c^2)^2} \\plus{} \\frac{c^8}{(c^2 \\plus{} a^2)^2} \\ge \\frac{1}{3}\\left(\\frac{a^4}{a^2 \\plus{} b^2} \\plus{} \\frac{b^4}{b^2 \\plus{} c^2} \\plus{} \\frac{c^4}{c^2 \\plus{} a^2}\\right)^2$\n\nThus it's enough to prove that\n\n$ \\frac{1}{3}\\left(\\frac{a^4}{a^2 \\plus{} b^2} \\plus{} \\frac{b^4}{b^2 \\plus{} c^2} \\plus{} \\frac{c^4}{c^2 \\plus{} a^2}\\right)^2 \\ge \\frac{1}{12} \\equal{} \\frac{(ab \\plus{} bc \\plus{}ca)^2}{12}$\n\nThis is equivalent with\n\n$ 2\\frac{a^4}{a^2 \\plus{} b^2} \\plus{} 2\\frac{b^4}{b^2 \\plus{} c^2} \\plus{} 2\\frac{c^4}{c^2 \\plus{} a^2} \\ge ab \\plus{} bc \\plus{} ca$\n\nMoreover $ \\frac{a^4}{a^2 \\plus{} b^2} \\equal{} a^2 \\minus{} \\frac{a^2b^2}{a^2\\plus{}b^2}$ and similarly for the other three. Thus we get the equivalent inequality\n\n$ 2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\ge ab \\plus{} bc \\plus{} ca \\plus{} 2\\frac{a^2b^2}{a^2 \\plus{} b^2} \\plus{} 2 \\frac{b^2c^2}{b^2 \\plus{} c^2} \\plus{} 2\\frac{c^2a^2}{c^2 \\plus{} a^2}$\n\nBut now\n\n$ \\frac{a^2 \\plus{} b^2}{2} \\ge ab$ and\n\n$ \\frac{a^2 \\plus{} b^2}{2} \\ge 2\\frac{a^2b^2}{a^2 \\plus{} b^2} \\Leftrightarrow (a^2 \\plus{} b^2)^2 \\ge 4a^2b^2 \\Leftrightarrow (a^2 \\plus{} b^2 \\minus{} 2ab)(a^2 \\plus{} b^2 \\plus{} 2ab) \\ge 0$\n\nThus $ a^2 \\plus{} b^2 \\ge ab \\plus{} 2\\frac{a^2b^2}{a^2 \\plus{} b^2}$ and we get the desired inequality by doing the same for $ b,c$ and $ c,a$ and adding up.\n\n[/hide]" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "solve :lol: \r\n\r\n$ \\int_0^1\\;\\sec^{\\minus{}1}\\,(1\\plus{}\\sec\\,x)\\;\\textbf dx$", "Solution_1": "put $ {1\\plus{}secx\\equal{}secz}$ and solve it by integration by parts...", "Solution_2": "[OT]For misan: your avatar is very nice!!! :lol: :rotfl: [/OT]" } { "Tag": [ "HMMT", "geometry", "3D geometry", "pyramid" ], "Problem": "A man named Juan has 3 rectangular solids, each having volume 128. Two of the faces of one solid have areas 4 and 32. Two faces of another solid have areas 64 and 16. Finally, two faces of the last solid have areas 8 and 32. What is the minimum possible exposed surface area of the tallest tower Juan can construct by stacking his solids one on top of the other, face to face? (Assume that the base of the tower is not exposed.)", "Solution_1": "well if you know ab, and ac, you can find bc (xy represents a face area) by doing a^2b^2c^2/(ab*ac), so from that then you can find a,b,c for the three cubes by plugging abc/(ab) or /(bc) or /(ac) (its pretty straightforward)\r\n\r\nanyways, then you know the side lengths are (1,4,32) for cube 1 (2,8,8) for cube 2 and (2,4,16) from cube three.\r\n\r\nif you want tallest, then 32,8, and 16 have to be the heights.\r\n\r\nminimum surface area means that the largest base will go on bottom, second largest in middle, thrid largest at top (so like a pyramid shape when the three cubes are stacked)\r\n\r\nso put the 8 height cube on bottom with base area of 16, 16 height in middle with base area of 8, and 32 height on top with base area of 4.\r\n\r\nSo total surface area that is not exposed is 16+2(8)+2(4)=40, multiply by 2 for the others because its an area of 8 for first and second cube when stacked on top of eachother.\r\n\r\nThe total surface area of the 3 cubes is 728, so 728-40=688. I probalby made a mistake through all that though" } { "Tag": [ "FTW", "Support" ], "Problem": "obama ftw,\r\nbut i supported clinton in the primaries\r\n\r\nnote: i don't know if this has already been set up, if yes, then just move it to there\r\nnote number 2!!!!: give reasons, not just \"he's stupid\" or anything like that", "Solution_1": "i feel really bad for nader,\r\nbut he supported bush for the war\r\n\r\nand that's what made people hate him.\r\n\r\nit must be depression knowing you wasted 20 years of your life trying to run for president. :(", "Solution_2": "1) well he didn't JUST run for president in 20 years, cheer up\r\n\r\n2) it's Nader", "Solution_3": "Obama has an age older than his IQ. I think we will have the Republican win by a long shot.\r\nI wish Hillary was the democratic candidate...\r\nHave you heard people like Rush Limbaugh or Hillary? They are like \"SUPPORT OUR OWN PARTY!!!\" but that is not necessarily saying \"Vote for the person best suited for president.\" Also, people who were democrats/republicans went to the republican/democratic convention and voted for the person whom the democrats/republicans could beat easily, not who would lead this country best. Now their age is TWICE their IQ!", "Solution_4": "While I don't necessarily disagree with some of your conclusions, I don't think you have any idea what you're talking about.", "Solution_5": "actually no I don't; dad just tells me what he makes of it and forbids me to read the news. :(\r\nI think I should start to be a very naughty boy soon..." } { "Tag": [ "complex analysis", "complex analysis unsolved" ], "Problem": "$ K^{p,q} :$ faisceau des germes de formes diff\u00e9rentielles de type $ (p,q)$ \u00e0 coefficients distributions. Je ne sais pas ce concept? Aide-moi", "Solution_1": "Posts in the non-country-specific forums should be in English.", "Solution_2": "http://en.wikipedia.org/wiki/Dolbeault_cohomology\r\n\r\nthis might help? it looks like the general idea is simply a generalization of de Rham cohomology to the complex setting -- now one considers sheaves of holomorphic differential forms ?\r\n\r\nactually I don't know if you know english, so this might help more:\r\n\r\nhttp://fr.wikipedia.org/wiki/Cohomologie_de_Dolbeault \r\n\r\nje suis d\u00e9sol\u00e9 de ne pas \u00eatre capable de t'aider directement mais je ne sais presque rien de la g\u00e9om\u00e9trie complexe; peut-\u00eatre que tu pourrais poser la question encore dans << advanced topics >>. si je peux expliquer ce que je comprends, il semble que ce n'est qu'une g\u00e9n\u00e9ralisation de la cohomologie de de Rham si on consid\u00e8re les formes holomorphiques. \r\n\r\n(I'm not sure where the differential forms with coefficients as distributions comes from, though)", "Solution_3": "I read this concept in artile \"Un theor\u00e8me de dualit\u00e9\" of Jean - Pierre Serre. But I do not understand this concept. Help me, please.", "Solution_4": "Which concept is \"this concept\"? Sheaf? Germ? Differential form? Distribution? De-Rham cohomology? There's some pretty heavy machinery here (and in your other post on fibred spaces). Without a [b]lot[/b] more information about what you already know and what it is exactly that you don't understand, I doubt that anyone here (or elsewhere) will be able to help you.\r\n\r\n(I trust that you are thoroughly familiar with, at least, the concepts of sheaves, differential forms, and cohomology. If you do, it would help us if you explained what it is about $ K^{p,q}$ that you're not sure about. If you don't, you should not be attempting to read this paper by Serre.)", "Solution_5": "Serre signed the $ K^{p,q}$ is the germs of the differential forms of type (p, q) with the coefficient distribution. I really do not know what $ K^{p,q}$ is ? . You help me,please.", "Solution_6": "I think I'm giving up. Just in case:\r\n\r\n[url=http://en.wikipedia.org/wiki/Distribution_(mathematics)]Distribution[/url]\r\n\r\n[url=http://en.wikipedia.org/wiki/Complex_differential_form]Differential forms[/url] of type $ (p,q)$\r\n\r\n[url=http://en.wikipedia.org/wiki/Germ_(mathematics)]Germs[/url]" } { "Tag": [ "inequalities", "induction", "calculus", "derivative", "inequalities unsolved" ], "Problem": "Let $ a_1,a_2,...,a_n\\geq0$.Prove that:\r\n$ (a_1^t\\plus{}a_2^t\\plus{}...\\plus{}a_n^t)^{t\\plus{}1}\\geq(a_1^{t\\plus{}1}\\plus{}a_2^{t\\plus{}1}\\plus{}...\\plus{}a_n^{t\\plus{}1})^t$,\r\nwhere $ t,n$ are naturals.", "Solution_1": "Are you asking for a proof of the power mean inequality? :huh: \r\n\r\nYou could just look it up... it's Jensen on stuff that I don't quite remember.\r\n\r\nEDIT: wait, darn there's no $ n$ and it's reversed :(\r\n\r\nHowever, induction on $ n$ should work nicely... group $ a_{n}$ and $ a_{n \\plus{} 1}$ together\r\n\r\nYou know $ \\sqrt [t]{a_1^{t} \\plus{} a_2^{t}}$ is decreasing in $ t$ by derivative testing. Then, if\r\n\r\n$ \\sqrt [t]{a_1^{t} \\plus{} \\ldots \\plus{} a_n^{t}} \\ge \\sqrt [t \\plus{} 1]{a_1^{t \\plus{} 1} \\plus{} \\ldots \\plus{} a_n^{t \\plus{} 1}}$, we want to show\r\n\r\n$ \\sqrt [t]{a_1^{t} \\plus{} \\ldots \\plus{} a_{n \\plus{} 1}^{t}} \\ge \\sqrt [t \\plus{} 1]{a_1^{t \\plus{} 1} \\plus{} \\ldots \\plus{} a_{n \\plus{} 1}^{t \\plus{} 1}} (*)$\r\n\r\nBut as $ \\sqrt [t]{a_n^{t} \\plus{} a_{n \\plus{} 1}^{t}}\\ge \\sqrt [t \\plus{} 1]{a_n^{t \\plus{} 1} \\plus{} a_{n \\plus{} 1}^{t \\plus{} 1}}$, we have the LHS of $ (*)$ is at least \r\n\r\n$ \\sqrt [t]{a_1^{t} \\plus{} \\ldots \\plus{} a_{n \\minus{} 1}^{t} \\plus{} \\left((a_n^{t \\plus{} 1} \\plus{} a_{n \\plus{} 1}^{t \\plus{} 1})^{\\frac {1}{t \\plus{} 1}}\\right)^t}$\r\n\r\nwhich by the inductive hypothesis is at least\r\n\r\n$ \\sqrt [t \\plus{} 1]{a_1^{t \\plus{} 1} \\plus{} \\ldots \\plus{} a_{n \\plus{} 1}^{t \\plus{} 1}}$ as desired.\r\n\r\n(note n=1 is trivial)" } { "Tag": [ "geometry", "area of a triangle" ], "Problem": "I realized today that I had done the entirety of number 5 correctly, except I had used the wrong formula for the area of a triangle: I used ab sin C insetad of 0.5ab sin C. So my answer came out to exactly twice the correct answer (1004/671007), presumably without any other mistakes made. I would say that the error was simply a memory lapse, so I'm rather disappointed. Is there any chance that a grader might overlook that and still give full credit, if the entire solution was correct except for that one recollection of the formula?", "Solution_1": "They'll probably take off one point.", "Solution_2": "I figured. I just want to bash my head against the wall though! What a dumb thing to make a mistake on!" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "We define a function $f$ on the positive integers as follows:\\[f(1)=1,f(3)=3\\]\\[f(2n)=f(n)\\]\\[f(4n+1)=2f(2n+1)-f(n)\\]\\[f(4n+3)=3f(2n+1)-2f(n)\\]for all positive integers n\r\n(i) For $x=a_k.2^k+a_{k-1}.2^{k-1}+...+a_0$, where $a_j$ is either $0$ or $1$, evaluate $f(x)$\r\n(ii) Determine the number of positive integers $n$, less than or equal to $1994$, for which $f(n)=n$", "Solution_1": "We can easily proove that if Ks=1+2+ 2^2 + 2^3 + ....+ 2^s then f(Ks)=Ks and this can give the solution for the second question.\r\nFor the first question,I considred the fact that every natural number can be written as sums of terms like (2^a)*Ks.\r\nThis consideration leaded me to see that f behave as a sequence of the form Un=2Un-1 + b,but i really failed to reach the closed form for f(n).I have a real feeling that this way is to be considered .", "Solution_2": "He-he.\r\nWe are to notice that $f(x)=a_02^k+a_12^{k-1}+...+a_k$ (reversed representation of $x$).\r\n\r\nThat's all." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Is there a function $f: \\mathbb N \\rightarrow \\mathbb N$ such that for every $n \\in \\mathbb N$, \\[ f(f(n-1)) = f(n+1) - f(n) \\, ?\\]", "Solution_1": "[quote=\"X-man\"]Is there a function $ f: \\mathbb N \\rightarrow \\mathbb N$ such that for every $ n \\in \\mathbb N$ is\n\\[ f(f(n \\minus{} 1)) \\equal{} f(n \\plus{} 1) \\minus{} f(n)\n\\]\n[/quote]\r\n[hide=\"My solution\"]\n$ f(n)\\in\\mathbb N$ $ \\implies$ $ f(n)\\geq 1$ $ \\forall n$\nThen $ f(n\\plus{}1)\\equal{}f(n)\\plus{}f(f(n\\minus{}1))$ $ \\forall n>1$ $ \\implies$ $ f(n\\plus{}1)>f(n)$ $ \\forall n>1$\nSo $ f(n)\\geq n\\minus{}1$ $ \\forall n$\nSo $ f(f(n))\\geq n\\minus{}2$ $ \\forall n$\n\nSo $ f(n\\plus{}1)\\geq f(n)\\plus{}n\\minus{}3\\geq n\\minus{}1\\plus{}n\\minus{}3\\equal{}2n\\minus{}4$ $ \\forall n>1$ and so $ f(8)\\geq 10$ and so $ f(f(8))\\geq f(10)$\nBut $ f(f(8))\\equal{}f(10)\\minus{}f(9)$ and so $ f(f(8))\\infty}\\frac{X_n}{\\ln{n}}\\equal{}1)\\equal{}1$.", "Solution_1": "Use Borel-Cantelli:\r\n\\[ \\sum_n P(X_n\\geq (1 \\minus{} \\epsilon)\\ln(n)) \\equal{} \\sum_n n^{\\minus{}(1 \\minus{} \\epsilon)} \\equal{} \\infty\r\n\\]\r\n\r\n\\[ \\sum_n P(X_n\\geq (1 \\plus{} \\epsilon)\\ln(n))\\equal{}\\sum_n n^{\\minus{}(1 \\plus{} \\epsilon)} < \\infty\r\n\\]\r\nThis means exactly what you need: $ \\limsup \\frac {X_n}{\\ln(n)} \\equal{} 1 \\ a.e.$" } { "Tag": [ "vector", "linear algebra", "matrix" ], "Problem": "Lines L and M are given by the parametric equations\r\n\r\nL: x = 2 + s, y = -s, z = 2 - s\r\nM: x = -1 - 2t, y = t, z = 2 + 3t\r\n\r\n(i) Show that L and M do not intersect\r\n\r\n(ii) Calculate the shortest distance between L and M", "Solution_1": "[hide]Um, I've never worked with parametric equations before, but it doesn't look too intimidating... \nWe eliminate s from the first line by taking the first equation times two and adding it to the rest.\n$2x+y+z=6$\nWe eliminate t from the second line by taking the first equation times two and adding it to the rest.\n$2x+y+z=4$\n\nSo that's the answer for a).\n[/hide]", "Solution_2": "[quote=\"archimedes1\"][hide]Um, I've never worked with parametric equations before, but it doesn't look too intimidating... \nWe eliminate s from the first line by taking the first equation times two and adding it to the rest.\n$2x+y+z=6$\nWe eliminate t from the second line by taking the first equation times two and adding it to the rest.\n$2x+y+z=4$\n\nSo that's the answer for a).\n[/hide][/quote]\n\nwouldn't those be plains, not lines?\n\n[hide=\"a\"]\ny=2-x\nz=2+y\n\ny=-.5(x+1)\nz=3y+2\n\nif there were a point x,y,z that was on both lines, then\n2+y=z=2+3y -> y=0 \ny=-.5(x+1) means that x=-1\n0= 2-(-1)\n0=3\n\n \n[/hide]", "Solution_3": "[quote=\"0714446459923\"]\r\nwouldn't those be plains, not lines?\r\n[quote]\r\nI guess so. But does it matter for the first part? We're still relating x y and z to some number and 4 is never equal to 6.", "Solution_4": "By the definition of intersection:\r\n\r\n[hide]$(x_{L}, y_{L}, z_{L}) \\equiv (x_{M}, y_{m}, z_{m}) \\implies x_{L}=x_{m}, \\; y_{L}=y_{M}, \\; z_{L}=z_{M}$\n\n(I) $2+s=-1-2t$\n(II) $-s = t$ \n(III) $2-s=2+3t$\n\nBy (II), (I) becomes $2+s=-1+2s \\implies s=3 \\implies t=-3$\n\nHowever, by (II), (III) becomes $2+t=2+3t \\implies t=0 \\implies s=0$\n\nThis leads to a contradiction, and all three equations cannot hold true simultaneously.\n\nHence, the lines will never intersect.[/hide]", "Solution_5": "[hide=\"ii\"]\nSince the lines do not intersect, they are skew lines, and there must be two parallel planes such that each plane contains one of the lines. Written in vector form, the two lines are\n\\[<2, 0, 2>+s<1,-1, 1>, <-1, 0, 2>+t<-2, 1, 3>. \\]\nLet $S$ and $T$ denote the vectors parallel to the two lines. To find equations for the planes, we take their cross product\n\\[S\\times T = \\left|\\begin{matrix}i&j&k\\\\ 1&-1&-1\\\\-2&1&3 \\end{matrix}\\right| = <-2,-1,-1> \\]\nUsing the two points $(2, 0, 2)$ and $(1,-1,-1)$, we can write the two planes in point-norm form as\n\\[-2x-y-z+6 = 0,\\\\-2x-y-z = 0 \\]\nThe distance between the two is then\n\\[d = \\frac{|6-0|}{\\sqrt{(-2)^{2}+(-1)^{2}+(-1)^{2}}}= \\sqrt{6}\\]\n[/hide]", "Solution_6": "[quote=\"archimedes1\"][hide]Um, I've never worked with parametric equations before, but it doesn't look too intimidating... \nWe eliminate s from the first line by taking the first equation times two and adding it to the rest.\n$2x+y+z=6$\nWe eliminate t from the second line by taking the first equation times two and adding it to the rest.\n$2x+y+z=4$\n\nSo that's the answer for a).\n[/hide][/quote]\r\n\r\nThis is certainly not the way that I would solve the problem, but it is certainly correct. Let me clarify your argument:\r\n\r\nSince those planes do not intersect, and the lines are in those planes, the lines must not intersect. This is a nice argument; it does not merit a \"spam\" rating." } { "Tag": [ "algorithm", "function", "Harvard", "college", "ceiling function", "algebra", "floor function" ], "Problem": "I have posed this problem here before, but noone did post a solution. I then recall that I somehow managed to show it (I hope I managed), but now I'm stuck again when I try to solve it (I didn't write down the approach I did before), so then I suspect that I maybe didn't prove it correctly before. Anyway, here it is:\r\n\r\nLet A = {1,2,...,2002}\r\n\r\nLet B be a subset of A such that:\r\n1) There are no two consecutive elements in B\r\n2) There are no two elements in B such that one of them divides the other.\r\n\r\nWhat is the size (amount of elements) of the largest subset B with those properties above?\r\nI get the answer to be 834, by choosing the following subset: B = {2001,1999,...,669,666,...,334}, that is I choose first as many odd numbers as possible (starting with the largest one) and then switch over to even numbers and choose as many as possible. This algorithm seem to generate maximal subsets for smaller sets A (I have verified it for smaller sets A), but I find it hard to prove that the maximal subset must have that many elements. \r\n\r\nVery thankful for a solution or some new insights in how to prove it.", "Solution_1": "Anyone has insights into this problem?\r\n\r\nI haven't been able to solve it before, and neither has the new idea helped me:\r\nThe set I constructed has the maximal sum of all those subsets B with maximal size. \r\n\r\nThat didn't help me, but maybe I've missed out on something? Any thoughts would be welcome!", "Solution_2": "Hmm. I'm wondering if anyone has tried this problem? I see that many people have seen it...\r\n\r\nIt would be really interesting to see a solution for this one. And for the generalized version of it.\r\n\r\nI have no good ideas at the moment :S", "Solution_3": "your B is not optimal (834 members), for example B={2002,2000,1998,1996, 1993,1991,...,665, 662,660,...,332} have 835 members.", "Solution_4": "True :)\r\n\r\nDo you have any idea on how to find the correct answer? :S", "Solution_5": "it would be nice to study \r\n$f(n)=2n-[\\sqrt{n}]^2$ for the subsets $A \\subset N$\r\nso that $f(A) \\subseteq A$\r\n\r\nthere is some stuff that resembles to this problem", "Solution_6": "Hmm. What is so special with that function? No two values of it divide each other or what?", "Solution_7": "yes and none are consecutive...", "Solution_8": "Hm. I don't know what [x] is denoting. Ceiling function? Floor function? A combination of the two? \r\n\r\nIf it's any of those, then look at f(4) = 2*4 - 4 = 4 and f(16) = 2*16 - 16 = 16. f(4) divides f(16)...\r\n\r\nSo it probably means something else.", "Solution_9": "it says what u say it says", "Solution_10": "What do you mean? Does [ ] denote any of those I mentioned? Then the function isn't so good for our problem (at least as far as I can see).", "Solution_11": "oh...ur right..", "Solution_12": "I've been thinking about this problem a lot lately as well (I noticed you're Swedish - did you get it from fr\u00e5ga lund om matematik too?), but I haven't found the answer. I too found that first taking as many odd numbers as possible from the \"top\", and then as many even numbers as possible, gives a good result.\r\n\r\nI've written a program to brute force subsets of {1,...,n} for all n up to 104 (after that, it was too slow to continue). For these n, the above kind of subset gives the optimal number of elements except when n=52 and n=76. For these, the optimal subsets have one more element than the other ones. These were:\r\n\r\nn: 52\r\nexpected: 21 (taking odd numbers, then even numbers, as above)\r\noptimal subsets: 22 members\r\n8 10 12 14 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 52 \r\n6 8 10 14 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 52\r\n\r\nn: 76\r\nexpected: 31\r\noptimal subsets: 32 members\r\n12 14 16 18 20 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76 \r\n10 12 14 16 18 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76 \r\n8 12 14 18 20 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76 \r\n8 10 12 14 18 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76", "Solution_13": "maybe if you post your solution(the algo in the language of choice) we can make some optimization and make it\r\nprint the solution for the problem faster", "Solution_14": "I first saw the problem on a homepage of a Ph.D student in Harvard, Gabriel Carroll. He had a collection of math problems and this one was rated as hard. I haven't been able to find that homepage again (it seems that it's erased). So I asked this question to 'Fr\u00e5ga Lund' forum, but you saw the answer :(\r\n\r\nIt's interesting that all those sets your algorithm produced are generated quite simply. By substracting a 2 or a 3 from every number. Similar to the one I got before (by taking odd numbers first). But it seems to be very hard to prove that it's the optimal way. And it seems equally hard to just prove, without constructing an algorithm, that the maximal set must have so and so many numbers. I have given up on the problem for a while now...hopefully something will happen soon.\r\n\r\nAlso, you can send your algorithm if you want as spx2 wrote.", "Solution_15": "The program I mentioned was simply a DFS on all \"B\"-sets. I kept a list of \"candidate members\" of the set, initially the whole range {2,...,n}, and then in succession chose members from it, sieving out candidates that are multiples of it. As an optimization, I wouldn't generate sets with less members than the expected value. The code was ugly, so I'd rather not post it.\r\n\r\n[quote=\"Rust\"]your B is not optimal (834 members), for example B={2002,2000,1998,1996, 1993,1991,...,665, 662,660,...,332} have 835 members.[/quote]\r\nThis, by the way, is wrong. 1001 divides 2002, and both are in the set.\r\n\r\nSo... the 835-member set is still a possibility! :) The algorithm for constructing it wasn't valid for all n, but maybe it is for most n - maybe for all odd ones?", "Solution_16": "Eighty there is no expected size of a set in this problem so dont expect one in the algorithm.\r\n2nd thing please write the algorithm here dont matter if its ugly i want to see it", "Solution_17": "I used the word \"expected\" in the exact same sense as in my first post, ie as the size of the \"B-set\" generated by taking as many odd numbers \"from the top\" as possible, and after that, as many even numbers as possible.\r\n\r\nI'll clean up the code a little, and post it shortly. My code can be found on: [url=http://paste.axpr.net/?show=395]http://paste.axpr.net/?show=395[/url]. The possible bottlenecks are the excessive copying of arrays (there's probably nothing that can be done about that), and the extreme recursion." } { "Tag": [ "pigeonhole principle", "combinatorics unsolved", "combinatorics" ], "Problem": "Hello . \r\nGiven $ 2^{2n\\minus{}1} \\plus{}1$ odd numbers from $ {{2^{2n} \\plus{}1 , 2^{2n} \\plus{}2 , 2^{2n} \\plus{}3 , ... ,2^{3n}}}$ . Prove that there exist two number like $ a,b$ such that $ a^2$ not divisible by $ b$ and $ b^2$ not divisible by $ a$ .", "Solution_1": "I just know that `Pigeonhole principle` helps and no more !!! Plz help me . :blush:", "Solution_2": "Is there any idea ?" } { "Tag": [ "geometry", "trapezoid" ], "Problem": "How many incongruent trapezoids ABCD with side AB parallel to side CD have sides of lengths AB=16, BC=13, CD=10, and DA=6? \r\n\r\na) 0\r\nb) 1\r\nc) 2\r\nd) 3\r\nd) 4", "Solution_1": "[quote=\"236factorial\"]How many incongruent trapezoids ABCD with side AB parallel to side CD have sides of lengths AB=16, BC=13, CD=10, and DA=6? \n\na) 0\nb) 1\nc) 2\nd) 3\nd) 4[/quote]\r\n\r\n[hide]is it a) [size=200]0[/size][/hide]", "Solution_2": "[hide]I got $0$ too. Draw $2$ altitudes from $D$ and $C$ to side $AB$. Let the intersection of the altitude from $D$ to $AB$ be $E$ and the one from $C$ to $AB$ be $F$. Let the length of the altitude be $h$. Let $AE$ be $x$, which makes $BF$ $6-x$. We have $2$ right triangles now, $AED$ and $BFC$. These give the equations $x^2+h^2=6^2$ and $(6-x)^2+h^2=13^2$. Substitute the first equation into the second and we get $x=-11\\frac{1}{12}$, which cannot be possible, so the answer is $0$.[/hide]" } { "Tag": [ "function", "integration", "trigonometry", "Euler", "real analysis", "real analysis unsolved" ], "Problem": "Prove that $\\forall 0 0.\r\na: max f = + infinity.\r\nwe can see that f(n,n,n) = 3n/5 --> infinity as n --> infinity. Thus the maximum does not exist or if you want it is + infinity.\r\n\r\nsuggestion: if we modify the hypothesis and put: a + b + c = 1 and a , b , c > 0. what is then the maximum value of f ? i bet you can find an answer." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "In $ABC$, $2AC^2=AB^2+BC^2$.\r\nProve that $cot^2B \\geq cotAcotC$", "Solution_1": "[quote=\"indybar\"]In $ABC$, $2AC^2=AB^2+BC^2$.\nProve that $cot^2B \\geq cotAcotC$[/quote]\r\n$\\begin{eqnarray*}\r\n(*)\\ \\ cot^2 B\\ge cotA\\cdot cotC & \\Longleftrightarrow & accos^2 B\\ge b^2 cosAcosC\\\\\r\n & \\Longleftrightarrow & (a^2+c^2-b^2)^2\\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\\\\\r\n & \\Longleftrightarrow & b^4-2b^2(a^2+c^2)+(a^2+c^2)^2\\ge b^4-(a^2-c^2)^2\\\\\r\n & \\Longleftrightarrow & b^2(a^2+c^2)\\le a^4+c^4.\r\n\\end {eqnarray*}$\r\n\r\n\\begin{eqnarray*}\r\n(**)\\ \\ cot^2\\frac{B}{2}\\le cot\\frac{A}{2}\\cdot cot\\frac{C}{2} & \\Longleftrightarrow & (s-a)(s-c)\\le (s-b)^2\\\\\r\n& \\Longleftrightarrow & b^2-(a-c)^2\\le (a+c-b)^2\\\\\r\n& \\Longleftrightarrow & b(a+c)\\le (a^2+c^2).\r\n\\end{eqnarray*}\r\n\r\nIf $2b^2=a^2+c^2$, then:\r\n\r\n$(*)\\Longleftrightarrow (a^2+c^2)^2\\le 2(a^4+c^4)$, true.\r\n\r\n$(**)\\Longleftrightarrow b(a+c)\\le 2b^2\\Longleftrightarrow a+c\\le 2b\\Longleftrightarrow (a+c)^2\\le 2(a^2+c^2)$, true.\r\n\r\n[b]Remark.[/b] Originally I thought that the relation (*) is $cot^2\\frac{B}{2}\\ge cot\\frac{A}{2}\\cdot cot\\frac{C}{2}\\ !$. Thus appeared the relation (**).", "Solution_2": "[quote=\"levi\"]\n$\\begin{eqnarray*}\n(*)\\ \\ cot^2 B\\ge cotA\\cdot cotC & \\Longleftrightarrow & accos^2 B\\ge b^2 cosAcosC\\\\\n & \\Longleftrightarrow & (a^2+c^2-b^2)^2\\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\\\\\n\\end {eqnarray*}$\n[/quote]\r\n\r\nHow do you get this?", "Solution_3": "With the relations $\\frac {a}{sinA}=\\frac{b}{sinB}=\\frac{c}{sinC}=2R.$ and cosA=cotA.sinA a.s,o.", "Solution_4": "A further problem:\r\nAssuming the given condition on the sides of the triangle, show that CotACotC <= (CotB)^2 when B>=60\r\n\r\n CotACotC<=[(CotB)^2]/4 + 1/4 when B<=60", "Solution_5": "$4cotAcotC\\le 1+cot^2B\\Longleftrightarrow 4b^2cosAcosC\\le ac\\Longleftrightarrow (b^2+c^2-a^2)(b^2+a^2-c^2)\\le a^2c^2\\Longleftrightarrow b^4\\le a^2c^2+(a^2-c^2)^2.\\ \\ (*)$\r\n\r\n$B\\le 60\\Longleftrightarrow 2cosB\\ge 1\\Longleftrightarrow a^2+c^2-b^2\\ge ac\\Longleftrightarrow b^2\\le a^2+c^2-ac\\Longleftrightarrow$ \r\n$b^4\\le (a^2+c^2-ac)^2=(a^2+c^2)^2-2ac(a^2+c^2)+a^2c^2=a^2c^2+(a^2-c^2)^2+$\r\n$4a^2c^2-2ac(a^2+c^2)=(a^2-c^2)^2+a^2c^2-2ac(a-c)^2\\le a^2c^2+(a^2-c^2)^2,$ i.e. $b^4\\le a^2c^2+(a^2-c^2)^2.$ Thus, the relation (*) is true." } { "Tag": [ "geometry", "inequalities", "calculus", "integration", "logarithms", "number theory", "complex analysis" ], "Problem": "What is your favorite topic in math? (Of the following) Mine is close between number theorey and geometry, but I will pick number theorey.", "Solution_1": "You forgot [i]inequalities[/i]\r\n\r\n :P", "Solution_2": "Inequalties is algebra, is it not ;)", "Solution_3": "[quote=\"jhollenbeck\"]Inequalties is algebra, is it not ;)[/quote]\r\n\r\nYeah..\r\nHowever, in some books (or math sites ;) ) are considered a seperate topic....\r\n\r\n :)", "Solution_4": "calculus :D i even find myself using calculus on problems that have an obvious and somewhat easy non-calculus approach.", "Solution_5": "algebra and algebra 2\r\neasy classes", "Solution_6": "I used to like number theory but then now I just seem to have developed a serious liking towards calculus -- Especially after reading russian books on integration from first principles -- The concept is simply amazing !!", "Solution_7": "Surely Number Theory.\r\nBut what is Algebra II for you (especially when inequalities shall belong to Algebra)\u00bf\r\nBut Complex Analysis is missing :( (and no, there is a huge difference to normal calculus).\r\nGame theory and other themes of combinatorics are also missing...\r\n\r\nPS: I'm sure such poll allready occured ;)", "Solution_8": "I am currently in Algebra and I like it.\r\n\r\nI have looked into the geometry and algebra II book, and it lookd nice! :D", "Solution_9": "[quote=\"ZetaX\"]PS: I'm sure such poll allready occured ;)[/quote]\r\nYeah, there was at least one :roll: \r\n\r\nAnyway, I definitely like Geometry the most :D", "Solution_10": "[size=150]Geometry is :first: [/size]\r\n\r\nFunny, I used to like Algebra and stuff best and I hated Geometry...", "Solution_11": "I love algebra becuase I have MASTERED IT. I understand everything in Algebra. Geometry is difficult for me......good luck to those who like Geo the best. You are sure to get stumped on some really weird problem.", "Solution_12": "I like Geometry, I think most of people who love geometry used to hate it. It's weird\r\n\r\n :first: Geometry :first:", "Solution_13": "Algebra II, definitely. Yay for logs and complex numbers! :lol:", "Solution_14": "OMG Logs and Complex numbers make my head spin in tiny little circles....and I faint....they are so hard and annoying.", "Solution_15": "I have always liked number theory the best (starting... way back)\r\n\r\nRecently I have seen some of the beauty of geometry, and also some inequalities... why do they always want me to learn what I'm bad at? WHY?!", "Solution_16": "Calculus - I just enjoy the fact that it's much different and can do much more amazing things than everything before it. I mean, yeah Algebra can get pretty complex and interesting (Geometry too), but even the simplest idea in Calculus uses some idea of...infinity!\r\n\r\nAfter that, I love complex numbers and logarithms - logarithms are the only type of problem where I can easily do say AMC 12 #25. \r\nComplex numbers are fun too, hmm *wonders what wonders could be created if we combined complex numbers and calculus :) , hint hint complex analysis*", "Solution_17": "[quote=\"dogseatcheese\"]OMG Logs and Complex numbers make my head spin in tiny little circles....and I faint....they are so hard and annoying.[/quote]\r\n\r\nLogs and complex numbers together are rather straightforward :)\r\n\r\n$\\ln (z) = \\ln (re^{i \\theta}) = \\ln r + i \\theta + 2 \\pi k, k \\in \\mathbb{Z}$. :)\r\n\r\nPersonally, my vote is for complex analysis. Even if I don't understand it, it's very cool :)", "Solution_18": "Actually mine is Combinatorics...it rules!" } { "Tag": [ "function", "search", "limit", "calculus", "calculus computations" ], "Problem": "Does any one have interesting problems on limits?", "Solution_1": "any problems???", "Solution_2": "Which limits -- sequences, functions? You might also try to search forum for \"limit\".", "Solution_3": "Suppose $ f(x)$, $ f'(x)$ and $ f''(x)$ exist and are continuous in $ x \\in (0, \\infty)$, and that\r\n\\[ \\lim_{x \\rightarrow \\infty} (x^2 \\, f''(x) \\plus{} 4x\\, f'(x) \\plus{} 2f(x)) \\equal{} 1.\\]Find $ \\lim_{x \\rightarrow \\infty} f(x)$ and $ \\lim_{x \\rightarrow \\infty} x\\, f'(x)$.", "Solution_4": "any more problems", "Solution_5": "You've already solved this one?!", "Solution_6": "aidan, look through debank2's posts -- there's basically zero worthwhile content, mostly posts of the same useless form as in this thread. There's really not much point in responding to someone who comes to a forum with thousands of calculus problems (and a search function) and makes posts like this -- just ignore them.", "Solution_7": "hello, here you will find some problems\r\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html\r\nSonnhard.", "Solution_8": "[quote=\"JBL\"]aidan, look through debank2's posts -- there's basically zero worthwhile content, mostly posts of the same useless form as in this thread. There's really not much point in responding to someone who comes to a forum with thousands of calculus problems (and a search function) and makes posts like this -- just ignore them.[/quote]\r\n\r\nRight.. will do. Anyway I'd appreciate some help with that problem I posted above: I've repackaged it into\r\n\\[ \\lim_{x \\rightarrow \\infty} [(x^2 \\plus{} 2x)(f'(x) \\plus{} f(x))]' \\equal{} 1\\] and can't think of any way to proceed. Does anyone have a hint to point me in the right direction?", "Solution_9": "Let\r\n $ \\begin{array}{l} g\\left( x \\right) \\equal{} {x^2}f\\left( x \\right),x \\in \\left( {0, \\plus{} \\infty } \\right) \\Rightarrow \\\\\r\n\\Rightarrow g'\\left( x \\right) \\equal{} {x^2}f'\\left( x \\right) \\plus{} 2xf\\left( x \\right) \\Rightarrow \\\\\r\n\\Rightarrow g''\\left( x \\right) \\equal{} {x^2}f''\\left( x \\right) \\plus{} 4xf'\\left( x \\right) \\plus{} 2f\\left( x \\right),\\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } g''\\left( x \\right) \\equal{} 1 \\\\\r\n\\end{array}$\r\nAlso\r\n$ \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } f\\left( x \\right) \\equal{} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\frac {{g\\left( x \\right)}}{{{x^2}}}\\mathop \\equal{} \\limits^{DLH} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\frac {{g'\\left( x \\right)}}{{2x}}\\mathop \\equal{} \\limits^{DLH} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\frac {{g''\\left( x \\right)}}{2} \\equal{} \\frac {1}{2}$\r\n\r\nand\r\n$ \\begin{array}{l} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\frac {{g'\\left( x \\right)}}{{2x}} \\equal{} \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\frac {{{x^2}f'\\left( x \\right) \\plus{} 2xf\\left( x \\right)}}{{2x}} \\equal{} \\frac {1}{2} \\Rightarrow \\\\\r\n\\Rightarrow \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left[ {\\frac {{xf'\\left( x \\right)}}{2} \\plus{} f\\left( x \\right)} \\right] \\equal{} \\frac {1}{2} \\\\\r\n\\end{array}$\r\nso\r\n$ \\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } xf'\\left( x \\right) \\equal{} 2\\mathop {\\lim }\\limits_{x \\to \\plus{} \\infty } \\left[ {\\frac {{xf'\\left( x \\right)}}{2} \\plus{} f\\left( x \\right) \\minus{} f\\left( x \\right)} \\right] \\equal{} 2 \\cdot \\frac {1}{2} \\minus{} 2 \\cdot \\frac {1}{2} \\equal{} 0$", "Solution_10": "Thanks I think I get it now, but I have a question about this part:\r\n\r\n$ \\mathop{\\lim }\\limits_{x\\to\\plus{}\\infty }f\\left( x\\right) \\equal{}\\mathop{\\lim }\\limits_{x\\to\\plus{}\\infty }\\frac{{g\\left( x\\right)}}{{{x^{2}}}}\\mathop \\equal{}\\limits^{DLH}\\mathop{\\lim }\\limits_{x\\to\\plus{}\\infty }\\frac{{g'\\left( x\\right)}}{{2x}}\\mathop \\equal{}\\limits^{DLH}\\mathop{\\lim }\\limits_{x\\to\\plus{}\\infty }\\frac{{g''\\left( x\\right)}}{2}\\equal{}\\frac{1}{2}$\r\n\r\nHow do I justify using l'Hopital's rule here? Do we know that $ \\frac{g(x)}{x^2}$ or $ \\frac{g'(x)}{2x}$ is indeterminate as $ x \\rightarrow \\infty$?", "Solution_11": "Since $ g'' \\to 1$, there exists some $ N$ such that $ x > N$ implies $ g''(x) > \\frac{1}{2}$. Then $ g'(x) > \\frac{1}{2}(x \\minus{} N) \\plus{} g(N)$, a linear function of positive slope. Then do the same thing again.", "Solution_12": "How does $ g'(x) > \\frac{1}{2} (x \\minus{} N) \\plus{} g(N)$ allow l'Hopital's? :|", "Solution_13": "It implies $ g' \\to \\infty$ as $ x \\to \\infty$.", "Solution_14": "Oh right :blush: Thanks!" } { "Tag": [ "inequalities", "algebra", "polynomial", "function", "rational function", "superior algebra", "superior algebra open" ], "Problem": "I think almost all the inequalities I have ever seen were derived from the fact that $x^2\\geq 0$ for all real numbers $x$. Is every inequality a consequence of this fundamental inequality? One way to make the question precise is the following. \r\n\r\nIs there a polynomial $f$ with integer coefficient, which is positive indefinite in $\\mathbb{R}$, and an ordered ring $R$ in which $x^2\\geq 0$ such $f$ is [i]not[/i] positive indefinite in $R$?\r\n\r\nIf $R$ is an $\\mathbb{R}$-algebra, then since every positive indefinite polynomial in $\\mathbb{R}[X]$ is a sum of squares of polynomials, we see that there is no one-variable counterexample. It is not clear to me that there are no one-variable counterexamples if $R$ is not an $\\mathbb{R}$-algebra. Of course, it is possible that there are multivariable counterexamples even if $R$ is an $\\mathbb{R}$-algebra.\r\n\r\nIf the answer some of the above question is negative, then it is interesting to ask the same questions when $R$ is a field.\r\n\r\nBoris", "Solution_1": "i think it has been proven that every positive definite polynomial is the sum of squares of quotients of polynomials. this immediately proves your claim.", "Solution_2": "Can you refer me to a proof of that for multivariable polynomials?\r\n\r\nAlso, what if $R$ is not $\\mathbb{R}$-algebra? For example, $2x^2=(\\sqrt{2}x)^2$, but $R$ does not necessarily have a $\\sqrt{2]}$. Of course, in this case $2x^2=x^2+x^2$. However, the polynomial $3x^2-3x+1$ is positive definite, but there is no way of writing it as a sum of squares of polynomials with integer coefficients. Also, what if $R$ is not a field?\r\n\r\nBoris", "Solution_3": "I think there are several ways this can go wrong. I'm pretty sure there are multivariable counterexamples, and even $\\mathbb{R}[X,Y]$ allows them.\r\n\r\nWe can also cause trouble with a choice of field. Consider $\\mathbb{Q}(\\sqrt{2})$, with the usual order. It is impossible to determine algebraically which square root of 2 is positive because there is a field automorphism mapping it to a negative number, so $\\sqrt{2}$ cannot be written as a sum of squares.\r\n\r\nThis example could be made larger; consider the smallest field $F$ containing $\\mathbb{Q}$ such that for any $a,b\\in F$, $a^2+b^2=c^2$ for some $c\\in F$. Since it can be embedded in $\\mathbb{R}$, it can be ordered, but it has infinitely many automorphisms which scramble the order, so there must be many elements which cannot be written as sums of squares, and $\\sqrt{2}$ is still one of them.", "Solution_4": "i'm sorry - indeed we can get some problems. but it least it answers the question whether one can prove every algebraic inequality with the inequality $x^2\\geq 0$ in the real numbers.", "Solution_5": "Hilbert's 17th problem asked something very similar to what you're asking. He asked whether every rational function (i.e. quotient of polynomials) with real coefficients, which is everywhere greater or equal to 0, is the quotient of two polynomials, each of which is a sum of squares. This was answered affirmatively by Emil Artin in the 1920s. Note that the analogous questions for polynomials is false. There exist polynomials that are everywhere greater or equal to 0, but cannot be written as the sum of squares of polynomials. (It can however be written as a sum of squares of rational functions.) See the Wikipedia page on Hilbert's 17th Problem." } { "Tag": [ "inequalities", "search", "inequalities proposed" ], "Problem": "Let $ a,b,c\\geq 0$ such that:$ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$.Prove that$ : \\sum\\frac{a}{a^3\\plus{}bc}\\geq 3$", "Solution_1": "it is a russian inequality if my memory is good , a hint is to use the substitution : $ x\\equal{}\\frac{ab}{c} ...$ and there is another nice simple sol but hard to find ( at least for me) , search for it in the forum :wink:" } { "Tag": [ "MATHCOUNTS", "algebra", "system of equations", "AMC" ], "Problem": "If you want to get technical, this is actually based on a mathcounts national problem - but it's hard enough to count as an AMC-level problem. \r\n\r\nExpress the following as a difference of two square roots:\r\n:sqrt:(5 - 2:sqrt:6)\r\nOr in exponents, this is (5 - 2*6^1/2)^1/2.\r\n\r\nShow how you arrived at your answer.\r\n\r\nP.S. If you already have done this, don't say it. There's a little \"trick\" to finding it.", "Solution_1": "Is it sqrt(5-2*sqrt(6)) or sqrt(5-sqrt(6))? You wrote each one in some place. If it's the first, it's sqrt(3)-sqrt(2). I don't know how to do it nicely and properly. I just solved it ``by inspection.'' It's possible to consider the two cases a*sqrt(2)+b*sqrt(3) and a+b*sqrt(6) and solve a system of equations, but that takes a fairly large amount of time. A friend of mine from school has a set of Chinese training books that he allowed me to borrow, and there were a lot of problems having to do with nested radicals, so after going through a lot of their exercises, I can now do most simple ones in my head very quickly.", "Solution_2": "can someone please give the solution to this problem. i still dont know how to do it. thanks in advance", "Solution_3": "It has to be a+b*sqrt(6) or a*sqrt(2)+b*sqrt(3). Just square each one with the original number and then solve the systems of equations.", "Solution_4": "Sorry, I was on vacation. Cut off from computers for a whole week! Sadness!\r\n\r\nBut back to the point. ComplexZeta, it was the former. That is, sqrt(5-2*sqrt(6)) \r\n\r\nAnd the answer [i]is[/i] sqrt(3)-sqrt(2). Here's my solution:\r\n\r\nsqrt(5-2*sqrt(6)) \r\nIgnore the first square root for a sec. We have\r\n\r\n5-2*sqrt(6)\r\n= (3+2) - 2*sqrt(6)\r\n= 3 - 2*sqrt(6) + 2\r\n= (:rt3:):^2: - 2* :rt3: * :rt2: + (:rt2:) :^2: \r\nThis is of the form a^2 - 2ab + b^2 = (a-b)^2, with a = :rt3: , b = :rt2: \r\nSo 5-2*sqrt(6) = ( :rt3: - :rt2:) :^2: \r\n\r\nPut that back under the radical and you get:\r\nsqrt(( :rt3: - :rt2:) :^2: )\r\n= :rt3: - :rt2:\r\n\r\nIn other words, sqrt(3) - sqrt(2), as Simon said.", "Solution_5": "thanks, i finally understand. CELEBRATE!", "Solution_6": "Anytime. 'nother problem?", "Solution_7": "The more problems the better. I finished 2 math books this week, and I don't know what to do now other than IMO problems.", "Solution_8": "4 sheezy. i may not be able to solve it though. oh well. make it quick. i leave tomorrow", "Solution_9": "ComplexZeta - see my challenge #1 under the Challenge section. If you haven't seen it already or have any other reason for not wanting to try it.", "Solution_10": "how bout me. oh god. \"uu\"", "Solution_11": "Seems like it's just you and me on the forum right now. We should PM. I'll send you a message.", "Solution_12": "k, but you left, oh well" } { "Tag": [ "logarithms", "geometry", "3D geometry", "algebra", "polynomial", "parallelogram", "ratio" ], "Problem": "[color=blue][size=150]2005 mock AMC D[/size][/color]\r\nby amirhtlusa(Amir Talebi)\r\n\r\nyou guys know the rules, \r\ntake 75 mins to answer the questions, when done PM your answers to me.\r\nyou can use a SAT approved calculator and other aids.\r\nMANY THANKS TO CHESS64 FOR PDF FILE\r\ni'll try to be online Most of the time\r\n[hide]\n1.If $x=1+2^p$ and $y=1+2^{-p}$ then $y$ in terms of $x$ is:\n\nA)$\\frac{x+1}{x-1}$\nB)$\\frac{x+2}{x-1}$\nC)$\\frac{x}{x-1}$\nD)$2-x$\nE)$\\frac{x-1}{x}$\n\n2.The smallest value of $x^2+8x$ for all real values of $x$ is:\n\nA)-4\nB)4\nC)-8\nD)-16\nE)NOTA\n\n3.The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$\n\nA)$\\frac{-2}{3}$\nB)$\\frac{-1}{3}$\nC)$0$\nD)$\\frac{1}{4}$\nE)$\\frac{2}{3}$\n\n4. Define the operation $\\star$ for numbers $\\in\\mathbb{R^+}$ as $a\\star b=\\frac{ab}{a+b}$ then $4\\star(4\\star 4)=$\n\nA)$\\frac{3}{4}$\nB)$1$\nC)$\\frac{4}{3}$\nD)$2$\nE)$\\frac{16}{3}$\n\n5. $(\\sqrt{1+\\sqrt{1+\\sqrt{1}}})^4$ equals:\n\nA)$\\sqrt{2}+\\sqrt{3}$\nB)$\\frac{1}{2}(7+3\\sqrt{5})$\nC)$1+2\\sqrt{5}$\nD)$3$\nE)$3+2\\sqrt{2}$\n\n6. If $x=(\\log_{8}{2})^{(\\log_{2}{8})}$ then $\\log_{3}{x}$ equals:\n\nA)$-3$\nB)$-\\frac{1}{3}$\nC)$\\frac{1}{3}$\nD)$3$\nE)$9$\n\n7. If the sum of two numbers is $1$ and their product is $1$ then the sum of their cubes is:\nA)$2$\nB)$-2-\\frac{3\\sqrt{3}}{4}$\nC)$0$\nD)$-\\frac{3\\sqrt{3}}{4}$\nE)$-2$\n\n8.a sequence of natural numbers is made as following: the first term in this sequence is equal to $1$, then every term afterwards is equal to the sum of all the previous terms before it plus one. The $n$ th term of this sequence is\nA)$2n-1$\nB)$n$\nC)$2^n-1$\nD)$2^{n-1}$\nE)$n(n+1)(2n+1)$\n\n9. If $x,y$ are positive integers, then the number of ordered pairs $(x,y)$ which satisfy: $3x+5y=501$ is :\n\nA)$33$\nB)$34$\nC)$35$\nD)$100$\nE)NOTA\n\n\n\n\n10. we have an equilateral triangle. We construct a new equilateral triangle by connecting the midpoints of the sides of the bigger triangle. After performing this operation $n$ times we see that there is a TOTAL number of $45$ triangles in the picture. $n$ is :\n\nA)9\nB)10\nC)11\nD)12\nE)22\n\n11. if the polynomial $ax^3+bx+c$ is divisible by $x^2+tx+1$ then which of the following is correct?\nA)$a^2-2c\\geq b$\nB)$a+c>3$\nC)$a^2-c\\geq ab$\nD)$a^2+c^2=ab$\nE)$a^2-c^2=ab$\n\n12.How many distinct values of $n$ will make $1!+2!+3!+4!+\\cdots+n!$ a perfect square?\nA) Infinite\nB)2\nC)3\nD)4\nE)5\n\n13. in $\\triangle ABC$ point $M$ is on the side $BC$. From $M$ we draw lines parallel to $AB$ and $AC$. if the area of the created parallelogram is $\\frac{5}{18}$ of the area of $\\triangle ABC$; What ratio does $M$ divide the side $BC$ into?\nA)1 to 5\nB)1 to 6\nC)1 to 4\nD)2 to 9\nE)it can not be determined\n\n14. What is the sum of the coefficients in the expansion $(2x+y-3z)^{2006}$\nA)$2006$\nB)$2^{2006}$\nC)0\nD)1\nE) $3^{2006}$\n\n15. let $d(n)$ denote the number of (positive) divisors of the number $n$. find $d(24898)$\nA)4\nB)6\nC)8\nD)16\nE)10\n\n16. in the game of chess, the king can attack any of the 8 possible squares around itself. In how many ways can we place three kings in a $3\\times 3$ square such that none of the kings could attack each other?\nA)5\nB)7\nC)9\nD)8\nE)10\n\n17. what is the maximum number of terms that can be picked from the following set of numbers such that the mean of the picked numbers would be $15$?\n$\\{21,14,13,17,15,16,23,12,8,11,12,9,5,4,16,2,14,15,18,8,3,16\\}$\n\nA)13\nB)14\nC)15\nD)16\nE)17\n\n18. if $f(x)=\\log{(\\frac{1+x}{1-x})}$ for $-1_<\r\n\r\n[b]EDIT:[/b] Err, wait, now I'm confused. Is time up now? What did you mean when you said we had until 7:30 Eastern? :huh:", "Solution_30": "[quote=\"amirhtlusa\"][quote=\"chess64\"]hm i don't get one of the answer choices given for #18 :([/quote]\n\numm.. do you mean you dont see a choice?\nthat might be because in the PDF file the choice A is right infront of the question\n ;)[/quote]\r\n\r\ni know, but i didn't get one of those answers :( oh well", "Solution_31": "[quote=\"amirhtlusa\"][b]time is not UP, please send in your answers[/b][/quote]\r\n\r\n?????????????? :?", "Solution_32": "not probably a typo for now\r\n\r\nyea for 18 I also didn't get one of those answers, but I wasn't really trying that one, so I might have overlooked some clever factorization - didn't look too interesting, though.", "Solution_33": "if some of the powers on 18 were switched (the bottom would be a 3x^2 instead of 3x and the top would be an x^3 instead of an x^2 I think), it might work out better... but eh, maybe it works out this way", "Solution_34": "ooops. :P \r\ntypo for 18 the bottom was supposed to $1+3x^2$\r\nsorry about that", "Solution_35": "[quote=\"amirhtlusa\"]ooops. :P \ntypo for 18 the bottom was supposed to $1+3x^2$\nsorry about that[/quote]\r\nThis problem is familiar - could it be from the qualifying exam for the UCSB C of CC qualifying exam? :lol: I'm sad that I was too busy last weekend to take the test - I'll sit down eventually and actually do it under exam conditions, though I've sort of skimmed the problems and some of the solutions thread, so the value will be slightly diminished.", "Solution_36": "I saw it in the AoPS book too.", "Solution_37": "i think igot it from a old mao book, maybe around 1960's" } { "Tag": [ "inequalities" ], "Problem": "Prove that $\\frac{a^{n-1}+a^{n-2}+...+a+1}{a^n+a^{n-1}+...+a+1} > \\frac{b^{n-1}+b^{n-2}+...+b+1}{b^n+b^{n-1}+...+b+1}$ for $0\\frac{b^n-1}{b^{n+1}-1}$[/hide]", "Solution_2": "And to ruin it!\r\n\r\n\r\n$f(x)=\\displaystyle \\frac{x^n-1}{x^{n+1}-1}$\r\n\r\nNow how would we prove that this is always decreasing??? Hmmm, let me think....", "Solution_3": "Unless I made a mistake, the expression is equal to $\\frac{1}{1+\\frac{1}{\\frac{1}{x}+\\frac{1}{x^2}+....\\frac{1}{x^n}}}$, which clearly decreases as x increases" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive numbers such that:$ a\\plus{}b\\plus{}c\\equal{}3$.Prove that:\r\n\r\n$ 6(4\\minus{}\\sqrt{3})(\\sqrt{3a\\plus{}b}\\plus{}\\sqrt{3b\\plus{}c}\\plus{}\\sqrt{3c\\plus{}a}) \\ge 3(27\\minus{}5\\sqrt{3})\\plus{}7(3\\minus{}\\sqrt{3})(ab\\plus{}bc\\plus{}ca)$\r\n\r\nSince this inequality,we have an nice inequality of nguoivn:\r\n\r\n$ 10(\\sqrt{3a\\plus{}b}\\plus{}\\sqrt{3b\\plus{}c}\\plus{}\\sqrt{3c\\plus{}a}) \\ge 39\\plus{}7(ab\\plus{}bc\\plus{}ca)$\r\n :)", "Solution_1": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that:$ a \\plus{} b \\plus{} c \\equal{} 3$.Prove that:\n\n$ 6(4 \\minus{} \\sqrt {3})(\\sqrt {3a \\plus{} b} \\plus{} \\sqrt {3b \\plus{} c} \\plus{} \\sqrt {3c \\plus{} a}) \\ge 3(27 \\minus{} 5\\sqrt {3}) \\plus{} 7(3 \\minus{} \\sqrt {3})(ab \\plus{} bc \\plus{} ca)$\n\nSince this inequality,we have an nice inequality of nguoivn:\n\n$ 10(\\sqrt {3a \\plus{} b} \\plus{} \\sqrt {3b \\plus{} c} \\plus{} \\sqrt {3c \\plus{} a}) \\ge 39 \\plus{} 7(ab \\plus{} bc \\plus{} ca)$\n :)[/quote]\r\nWe can solve this inequality by Schur inequality :)", "Solution_2": "They are also results from this problem of can_hang:\r\nhttp://mathvn.org/forum/viewthread.php?thread_id=1180" } { "Tag": [], "Problem": "Four whole numbers, when added three at a time, give the sums 180,197,208 and 222. What is the largest of the four numbers?\r\n\r\n$\\text{(A)} \\ 77 \\qquad \\text{(B)} \\ 83 \\qquad \\text{(C)} \\ 89 \\qquad \\text{(D)} \\ 95 \\qquad \\text{(E)} \\ \\text{cannot be determined from the given information}$", "Solution_1": "[hide=\"How to solve\"]Add all four of the numbers together, divide by three, then subtract the smallest triple. \n\nEasy. :) [/hide]", "Solution_2": "[hide]\nLet the numbers be $a$, $b$, $c$, and $d$. Add all of the weights together and you get $3(a+b+c+d)=807$, so $a+b+c+d=269$. Now notice that if you subtract 269 from the weight, you get negative the non-added number. It's clear that the highest number you can find is $|180-269|=89$, which is $\\boxed{C}$.\n[/hide]", "Solution_3": "[hide=\"Answer\"]$(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)=3(a+b+c+d)$\n$a+b+c+d=269\\Rightarrow (a+b+c+d)-(a+b+c)=269-180=89\\Rightarrow \\boxed{C}$[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "hi\r\nquestion:assume that \"n\" is anteger ,n>1 and \"p\" is a prime number \r\nand we have n|p-1 and p|(n^3)-1\r\nprove that 4p-3 is a perfect squar.\r\ntnx", "Solution_1": "$ P \\minus{} 1\\geq(n)$ so $ (P,n \\minus{} 1) \\equal{} 1$. so $ P|n^2 \\plus{} n \\plus{} 1$ and so $ (nm \\plus{} 1)k \\equal{} n^2 \\plus{} n \\plus{} 1$ ($ p \\equal{} nm \\plus{} 1$) so we must have $ n(n \\plus{} 1 \\minus{} km) \\equal{} k \\minus{} 1$ so $ k\\geq(n \\plus{} 1)$ so $ n^2 \\plus{} n \\plus{} 1\\geq(nm \\plus{} 1)(n \\plus{} 1)$ so $ k \\minus{} 1 \\equal{} 0$ so $ nm \\plus{} 1 \\equal{} n^2 \\plus{} n \\plus{} 1 \\equal{} p$ so $ 4p \\minus{} 3 \\equal{} 4n^2 \\plus{} 4n \\plus{} 1$" } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "There are two tennis teams,each team has 1000 players.Knowing that any two competitors of the two teams copete with each other once only and there is no game ending in a tie \r\n Prove that :Existing 10 competitor of one team so that there is one player of the other losing at least one among that 10 competitors .", "Solution_1": "You've posted almost exactly the same thing before, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1344065672&t=158664]here[/url]. This time, the statement is trivial. Probably what you mean is,\r\n\r\n\"There are two tennis teams with 1000 players. Each player plays one game against every player on the other team, and no game ends in a tie. Show that we can find 10 players on one team such that every player on the other team lost to at least one of those players.\"\r\n\r\nAlso, give your posts better titles." } { "Tag": [ "number theory", "number theory proposed" ], "Problem": "Assume : $ a+b =p-1 $ , where $ a,b,p $ are natural numbers and p is a prime number . Prove that : $ a!b! +(-1)^a \\equiv 0 [p] $", "Solution_1": "By Wilson's Theorem, taking $\\mod p$, we have $(p-1)! = b! (b+1) (b+2) \\ldots (p-1) \\equiv b! (-a) (-a + 1) \\ldots (-1) \\equiv b! a! (-1)^a \\equiv -1$, hence result follows. :)", "Solution_2": "Err, this may be stupid...\r\nhow is $a!b!(-1)^a \\equiv -1 \\mod p$?", "Solution_3": "[quote=\"coolamit\"]Err, this may be stupid...\nhow is $a!b!(-1)^a \\equiv -1 \\mod p$?[/quote] \r\n I don't understand you ! :roll: This expression is exactly what we need to prove . ;)" } { "Tag": [ "algebra", "polynomial", "function" ], "Problem": "For each polynomial function below, find the complex zeros.\r\n\r\n[1] f(x) = x^4 - 1\r\n\r\n[2] f(x) = 2x^4 + x^3 - 35x^2 - 113x + 65", "Solution_1": "For #1, try factoring it.\r\n\r\nI just looked at these for about 5 seconds.", "Solution_2": "[hide=\"Hint for #1\"]\n$x^{4}-1=(x^{2}+1)(x+1)(x-1)$\n[/hide]", "Solution_3": "Can you elaborate more on the given hints?\r\n\r\nI am not sure what the questions are asking for, you see?", "Solution_4": "For each question, find all values of $x$ for which $f(x)=0$. So for the first one, it's to solve for $x$ when $x^{4}-1=0$.", "Solution_5": "Yes. And its really easy because you know that:\r\n\r\n[hide]\n$x^{4}-1=(x^{2}+1)(x+1)(x-1)=0$\n\nSo, just figure out values of $x$!\n\n$x^{2}=0$\n$x=0$\n\n$x+1=0$\n$x=-1$\n\n$x-1=0$\n$x=1$\n\n$x\\in\\{-1,0,1\\}$\n\nSo, that should be an even bigger (hint)\n[/hide]", "Solution_6": "[quote=\"#H34N1\"]Yes. And its really easy because you know that:\n\n[hide]\n$x^{4}-1=(x^{2}+1)(x+1)(x-1)=0$\n\nSo, just figure out values of $x$!\n\n$x^{2}=0$\n$x=0$\n\n$x+1=0$\n$x=-1$\n\n$x-1=0$\n$x=1$\n\n$x\\in\\{-1,0,1\\}$\n\nSo, that should be an even bigger (hint)\n[/hide][/quote]\r\nBy the way $x\\ne0$.\r\n\r\nI think you mean $x^{2}+1=0\\rightarrow x=\\pm i$", "Solution_7": "[quote=\"Interval\"]For each polynomial function below, find the complex zeros. \n\n[1] f(x) = x^4 - 1 \n\n[2] f(x) = 2x^4 + x^3 - 35x^2 - 113x + 65[/quote]\r\n[hide=\"Numero Uno (1)\"]\n$x^{4}-1=0$\n$(x^{2}+1)(x+1)(x-1)=0$\nOne of them must be zero so $x=1,-1$ or $x^{2}+1=0\\rightarrow x^{2}=-1$ so $x=\\pm i$[/hide][/hide]", "Solution_8": "[quote=\"Quevvy\"][quote=\"#H34N1\"]Yes. And its really easy because you know that:\n\n[hide]\n$x^{4}-1=(x^{2}+1)(x+1)(x-1)=0$\n\nSo, just figure out values of $x$!\n\n$x^{2}=0$\n$x=0$\n\n$x+1=0$\n$x=-1$\n\n$x-1=0$\n$x=1$\n\n$x\\in\\{-1,0,1\\}$\n\nSo, that should be an even bigger (hint)\n[/hide][/quote]\nBy the way $x\\ne0$.\n\nI think you mean $x^{2}+1=0\\rightarrow x=\\pm i$[/quote]\r\n\r\nYes; i meant that :blush:" } { "Tag": [ "algebra", "system of equations", "algebra proposed" ], "Problem": "Solve the system of equations:\r\n\r\n$ y^2\\equal{}(x\\plus{}8)(x^2\\plus{}2),$\r\n$ y^2\\minus{}(8\\plus{}4x)y\\plus{}(16\\plus{}16x\\minus{}5x^2)\\equal{}0.$", "Solution_1": "[quote=\"moldovan\"]Solve the system of equations:\n\n$ y^2 \\equal{} (x \\plus{} 8)(x^2 \\plus{} 2),$\n$ y^2 \\minus{} (8 \\plus{} 4x)y \\plus{} (16 \\plus{} 16x \\minus{} 5x^2) \\equal{} 0.$[/quote]\r\n\r\n$ y^2 \\minus{} (8 \\plus{} 4x)y \\plus{} (16 \\plus{} 16x \\minus{} 5x^2)\\equal{}[y\\minus{}(4\\plus{}5x)] \\cdot [y\\minus{}(4\\minus{}x)]\\equal{}0$\r\n\r\n[b]Case 1:[/b] $ y\\equal{}4\\plus{}5x$, then $ y^2\\equal{}(x\\plus{}8)(x^2\\plus{}2)\\equal{}x^3\\plus{}8x^2\\plus{}2x\\plus{}16$\r\n$ \\Leftrightarrow x(x\\minus{}19)(x\\plus{}2)\\equal{}0$\r\n$ \\Leftrightarrow (x,y)\\equal{}(0,4),(19,99),(\\minus{}2,\\minus{}6)$ as solutions.\r\n\r\n[b]Case 2:[/b] $ y\\equal{}4\\minus{}x$, then $ (4\\minus{}x)^2\\equal{}y^2\\equal{}x^3\\plus{}8x^2\\plus{}2x\\plus{}16$\r\n$ \\Leftrightarrow x(x\\plus{}2)(x\\plus{}5)\\equal{}0$\r\n$ \\Leftrightarrow (x,y)\\equal{}(0,4),(\\minus{}2,6),(\\minus{}5,9)$ as solutions.\r\n\r\n\r\nSo $ (x,y)\\equal{}(0,4),(19,99),(\\minus{}2,\\minus{}6),(\\minus{}2,6),(\\minus{}5,9)$ are the only solutions for the system of equations.", "Solution_2": "hello,eliminating $ y$ we get $ y\\equal{}\\minus{}\\frac{(x\\plus{}2)(x^2\\plus{}x\\plus{}16)}{(\\minus{}8\\minus{}4x)}$ inserting this in our equationsystem\r\nand factoring we get $ x(x\\plus{}5)(x\\minus{}19)(x\\plus{}2)\\equal{}0$ from here we get all solutions as\r\n$ x\\equal{}0,y\\equal{}4$\r\n$ x\\equal{}\\minus{}5,y\\equal{}9$\r\n$ x\\equal{}\\minus{}2,y\\equal{}6$\r\n$ x\\equal{}\\minus{}2,y\\equal{}\\minus{}6$\r\n$ x\\equal{}19,y\\equal{}99$\r\nSonnhard." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given a mxn grid .We color its vertex with red,blue and green.Firstly,color the vertex of boundary with red.A cell is called properly if it has exactly 2 adjacent vertex which have same colour.Show that: the number of properly cell is always even.", "Solution_1": "To each side of an unit square, give the number $2$ if its endpoints have the same color, and $1$ otherwise.\r\nThen define the weight of an unit square as the sum of the numbers attached to its four sides.\r\nIt is easy to verify that an unit square is a properly cell iff its weight is odd.\r\nThus, the total sum $S$ of all cells has the same parity than the number of properly colored cells.\r\nBut $S$ is even since in this sum each side is counted twice except those which form the boundary of the whole grid, and these ones are attached to $2$.\r\n\r\nPierre.", "Solution_2": "Wasthis asked in the Indian Namtional Olympiad 2000?", "Solution_3": "[quote=\"thealchemist\"]Wasthis asked in the Indian Namtional Olympiad 2000?[/quote]\r\nNo, it was not. Next time just check the olympiad resources section. :wink:" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "$(X_i,d_i)$ are metric spaces, where the topology on each space is induced by each $d_i$, respectively. Prove that \r\n\\[ d(x,y)=\\sum_{i=1}^\\infty \\dfrac{1}{2^i}\\dfrac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)} \\]\r\ndefines a metric in $\\prod_{i=1}^\\infty X_i$, and the topology which it induces is exactly the product topology.\r\n\r\nI think it must be easy, but I don't know how to verify the topology that $d(x,y)$ induces.", "Solution_1": "To verify the topology, show that an epsilon-ball lies between two set of the basis for the product topology, which have the form (open in first k coordinates)X(unrestricted in all other coordinates). In addition, show that as the ball shrinks to 0, the other sets do as well.", "Solution_2": "[quote=\"jmerry\"]In addition, show that as the ball shrinks to 0, the other sets do as well.[/quote]\r\nWhat are 'the other sets'? And what does it verify the topology? It looks so different from the definition of an induced topology that I know.", "Solution_3": "That was badly phrased. Here's what I should have said the first time:\r\n\r\nGiven an open set (an $\\epsilon$-ball) $U$ in the metric topology and a point $x$ in that set, find $V$ an open set (in the basis) in the product topology with $x\\in V\\subset U$.\r\nGiven an open set (in the basis) $U$ in the product topology and a point $x$ in that set, find $V$ an open set (an $\\epsilon$-ball) in the metric topology with $x\\in V\\subset U$.\r\n\r\nThe first shows that every open set in the metric topology is open in the product topology, and the second shows that every open set in the product topology is open in the metric topology.\r\nWe have to work with the open sets, because we don't know in advance that the product topology is determined by sequences.", "Solution_4": "[quote=\"jmerry\"]Given an open set (an $\\epsilon$-ball) $U$ in the metric topology and a point $x$ in that set, find $V$ an open set (in the basis) in the product topology with $x\\in V\\subset U$.\nGiven an open set (in the basis) $U$ in the product topology and a point $x$ in that set, find $V$ an open set (an $\\epsilon$-ball) in the metric topology with $x\\in V\\subset U$.\n[/quote]\r\nI get it now, thank you." } { "Tag": [ "function", "number theory unsolved", "number theory" ], "Problem": "$ 2002^{2002} = 3^{a_1} + 3^{a_2} + ... + 3^{a_n} $ , \r\n $ a_i \\in N $\r\n Find min{n} .", "Solution_1": "Are you sre about the $3^{a_i}$? In that case, I think the problem follows easily from the greedy algorithm.\r\nBut according to your other recent post, I think the problem deals with $a_i^3$, and in that case, it is another one from ISL 2002 :\r\nhttp://www.kalva.demon.co.uk/short/soln/sh02n1.html\r\n\r\nPierre.", "Solution_2": "I had written something for this one but it didnt posted!\r\n\r\nIf it is the problem Pierre is talking about by locking mod9 one finds that \r\na^3=0,-1,1 so we get the minimal number is 4 because 2002^2002=4mod9, and we can achieve this with 1000*2002^2001+1000*2002^2001+2002^2001+2002^2001.\r\n\r\nBut the problem as it is is not easy. Obviously the best representation is the base 3 one. but how can we count the number of powers used on it?\r\n\r\nI thought about a function f(1)=1 f(2)=2 f(3)=1 and f(3n)=f(n) f(3n+1)=f(n)+1 and f(3n+2)=f(n)+2 and our number is f(2002^2002) but we should find propertys to make easy the calculations." } { "Tag": [ "AMC", "AIME", "number theory", "AMC 10", "AMC 8", "\\/closed" ], "Problem": "I'm currently debating weather to take Algebra 3 or Intermediate C+P. I think my Algebra is OK but...it's still just Algebra 2.3 level. On the other hand, counting made me FAIL in the AIME. I don't even want to talk about it.\r\n\r\nThe only class I've taken is Intermediate Number theory seminar.\r\n\r\n138 AMC 10, 24 AMC 8, never mind the AIME, or the MC, but I think I improved a lot since then.\r\n\r\nThey are both pretty long classes I think, and I can probably only take 1.", "Solution_1": "Let the post-test be your guide. Do the Do You Need This test for Algebra 3. If you can't get more than half of them, then you should take Algebra 3. If you can get 1/2 - 3/4 of them without too much difficulty, then you can probably feel OK getting the book, focusing on the parts you don't know (and the Challenge Problems in the parts you do), and heading into C&P. If you're 3/4+, then skip the class and go to C&P.", "Solution_2": "So I'm assuming C+P is harder?\r\n\r\nOK guess I'll take Algebra III.", "Solution_3": "Yes, C&P is harder than Algebra 3." } { "Tag": [ "geometry", "perimeter" ], "Problem": "On the 5 by 5 square grid below, each dot is 1 cm from its\nnearest horizontal and vertical neighbors. What is the product of\nthe value of the area of square $ ABCD$ (in\ncm$ ^2$) and the value of the perimeter of\nsquare $ ABCD$ (in cm)? Express your\nanswer in simplest radical form.\n\n[asy]unitsize(1inch);\ndot((0,0));\ndot((0,1));\ndot((0,2));\ndot((0,3));\ndot((0,4));\ndot((1,0));\ndot((1,1));\ndot((1,2));\ndot((1,3));\ndot((1,4));\ndot((2,0));\ndot((2,1));\ndot((2,2));\ndot((2,3));\ndot((2,4));\ndot((3,0));\ndot((3,1));\ndot((3,2));\ndot((3,3));\ndot((3,4));\ndot((4,0));\ndot((4,1));\ndot((4,2));\ndot((4,3));\ndot((4,4));\ndraw((0,3)--(3,4)--(4,1)--(1,0)--cycle);\nlabel(\"$A$\",(3,4),N);\nlabel(\"$B$\",(4,1),E);\nlabel(\"$C$\",(1,0),S);\nlabel(\"$D$\",(0,3),W);[/asy]", "Solution_1": "the length of one side of the square is $ \\sqrt{10}$.\r\n\r\nso the area of the square is 10, and the perimeter of the square is $ 4\\sqrt{10}$.\r\n\r\nanswer : $ 40\\sqrt10$", "Solution_2": "[hide = Solution]The length of the square's side is $\\sqrt{10}$ because it is the hypotenuse of a right triangle with legs $3$ and $1:$ $\\sqrt{3^2 + 1^2} = \\sqrt{10}.$ So the area of square $ABCD$ is $(\\sqrt{10})^2 = 10.$ The perimeter of the square is $4\\sqrt{10}.$ The question asks for the product of the area and perimeter: $10 \\cdot 4\\sqrt{10} = \\boxed{40\\sqrt{10}}.$[/hide]\n\n[hide = Alternative Way to Find Area]We could have found the area of square $ABCD$ a different way, though it is best to do it the first way, since the side length of $ABCD$ needs to be found anyway for perimeter. The area of the entire 4 by 4 square is $16,$ and the area of one 1 by 3 triangle is $\\frac{3}{2}.$ There are four of these, giving the left-out portions a total area of $6.$ $16 - 6 = 10,$ square $ABCD$'s area.[/hide]" } { "Tag": [], "Problem": "$ 15$", "Solution_1": "[b]14[/b]", "Solution_2": "$ 13$", "Solution_3": "[b]12[/b]", "Solution_4": "$ 14$", "Solution_5": "[b]13[/b]", "Solution_6": "spaceguy, I'll pretend you posted 13.\r\n\r\n$ 12$", "Solution_7": "[b]11[/b]", "Solution_8": "$ 12$ \r\n\r\nHelp boys...", "Solution_9": "[b]11[/b]", "Solution_10": "$ 12$", "Solution_11": "[b]11[/b]", "Solution_12": "$ 10$", "Solution_13": "$ 11$\r\n\r\nBackup please...", "Solution_14": "[b]10[/b]", "Solution_15": "[b]47[/b]\r\nboyz pwn\r\nboyz rule :D \r\n\r\ngirls are lazy\r\ngirls are dumb", "Solution_16": "[b]48[/b]woohoo!", "Solution_17": "49 mmmm :D :D :D", "Solution_18": "[b]50[/b]", "Solution_19": "WE WIN!!! 3-3 now!", "Solution_20": "boyz win!!! 3-3 tie now!", "Solution_21": "Hammered :D", "Solution_22": "lol the girlz all eating lunch :rotfl: :rotfl: \r\n :D", "Solution_23": "all the boyz , rejoice\r\ncoz we won the game\r\nand put the girls to shame\r\nshowed them who the man is", "Solution_24": "Come on, stop insulting girls... And end this topic", "Solution_25": "ooh\r\n\r\nI never meant to insult them ...\r\njust serves them right though...\r\n\r\nand next time you should vote for girls :rotfl:", "Solution_26": "aww.. i leave for four hours and this is wut hapenns?? sigh", "Solution_27": "I go to eat lunch, and.... :(", "Solution_28": "Sigh... good job boys... @westiepaw: go to round seven!!!", "Solution_29": "Wow, don't check the forum for 24 hours and... :( FF changes really fast! These Boy v. Girl threads are getting kinda annoying. Shouldn't we just start a Boy vs. Girl thread that resets after it reaches 0 or 50?" } { "Tag": [ "analytic geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "How many squares are on chessboard n * n :)", "Solution_1": "Do you mean squares formed by the unit squares of the board or squares whose vertices are vertices of the unit squares of the board? In the second case, are you allowing only squares whose sides are parallel to the sides of the board or not?\r\n\r\nPierre.", "Solution_2": "I mean all squares, not the small one. All squares that are formed by lines parallel lines to the sides of board.", "Solution_3": "Thus, the answer is $\\frac {n(n1)(2n+1)} 6.$\r\nLet $U_n$ be the desired number.\r\nClearly $U_1 = 1.$\r\n\r\nNow, draw the $n \\times n$ board, and consider it as the grid of integer points $(i,j)$ where $0 \\leq i,j \\leq n$. These integer points are the vertices of the uit squares forming the board, and they are the vertices of the squares we are looking for.\r\nThere are $U_{n-1}$ squares whose vertices are not points with at least one coordinate equal to $n$ (that is whose vertices belong to the $(n-1) \\times (n-1)$ board).\r\n\r\nLet $A$ be the point with coordinates $(n,n).$\r\nThere are $C_n^2$ squares with two vertices having their $x$-coordinate equal to $n$ but not being $A$ (just have to choose two points among these $n$ points, which determine uniquely the whole square).\r\n\r\nThe same for the $y$-coordinate, and all these squares are pairwise distinct, because we did not choose $A$.\r\n\r\nThen, there are $n$ squares with one vertex being $A$ (just choose one point among the $n$ with $x$-coordinate equals to $n$, which determine uniquely the square).\r\n\r\nTherefore, $U_n = U_{n-1} + 2C_n^2 + n = U_n + n^2.$\r\nIt follows that $U_n = 1^2 + 2^2 +\\cdots + n^2 = \\frac {n(n+1)(2n+1)}6$, as claimed.\r\n\r\nPierre." } { "Tag": [], "Problem": "Find all primes $p > 0$ and all integers $q \\geq 0$ such that $p^{2}\\geq q \\geq p$ and $\\binom{p^{2}}{q}-\\binom{q}{p}= 1$.\r\n\r\n[i]Costel Chite\u015f, Adrian Stoica[/i]\r\n\r\n\r\n[i]Note.[/i] I slightly modified the problem. The original version didn't make sense.", "Solution_1": "[hide]I know by inspection that $(p,q)=(2,3)$ works.[/hide]", "Solution_2": "Does anyone have a solution to this nice problem? :dry:" } { "Tag": [], "Problem": "I'm confused with how to solve this problem, so help would be great! \r\n\r\nThe volume of any spherical balloon can be found by using the formula V= 4/3 pi r^3. Write an equation for r in terms of V and pi", "Solution_1": "[hide]\nIt is just asking you to solve the equation $V=\\frac{4}{3}\\pi r^{3}$ for $r$.\n\n$V=\\frac{4}{3}\\pi r^{3}\\implies\\frac{3}{4}\\cdot V=\\pi r^{3}\\implies r^{3}=\\frac{3V}{4\\pi}\\implies r=\\sqrt[3]{\\frac{3V}{4\\pi}}$.[/hide]", "Solution_2": "thanks i kept multiplying it by 3 and wondered why i was wrong" } { "Tag": [], "Problem": "A man parked his car in a carpark at 0830 and retrieved it at 1545 on the same day.\r\na) How long did he park his car in the carpark?\r\nb) If the parking charges are $1.50 for the first hour and 80 cents for each subsequent half hour or part thereof, how much must he pay on parking his car there?", "Solution_1": "[quote=\"risinglegend\"]A man parked his car in a carpark at 0830 and retrieved it at 1545 on the same day.\na) How long did he park his car in the carpark?\nb) If the parking charges are 1.50 for the first hour and 80 cents for each subsequent half hour or part thereof, how much must he pay on parking his car there?[/quote]\r\n\r\n[hide]If he started at 8:30 AM and ended at 3:45 PM, then he stayed for 7 hours 15 minutes.\n\nFor the first hour, he pays 1.50 on the first hour, leaving 6 hours and 15 minutes for the 80 cent charge.\n\nThat charge would be $80*(6*2+1)=80*13=1040$ cents, or 10.40\n\nSo the total would be $1.5+10.4=11.9$ so it would be 11.90[/hide]" } { "Tag": [ "topology" ], "Problem": "Let $ d: X\\times X \\rightarrow \\mathbb{R}$ determine $ \\displaystyle{(x,y)\\equal{}\\sum_{k\\equal{}1}^{\\plus{}\\infty}\\frac{\\min(|x_k\\minus{}y_k|,1)}{k^2}}$\r\nwith $ x\\equal{}\\{x_n\\}_n,y\\equal{}\\{y_n\\}_n\\subset X$. Prove $ (X,d)$ is a metric\r\nspace", "Solution_1": "you might want to make your notation more precise...", "Solution_2": "[quote=\"sondhtn\"]Let $ d: X\\times X \\rightarrow \\mathbb{R}$ determine $ \\displaystyle{(x,y) \\equal{} \\sum_{k \\equal{} 1}^{ \\plus{} \\infty}\\frac {\\min(|x_k \\minus{} y_k|,1)}{k^2}}$\nwith $ x \\equal{} \\{x_n\\}_n,y \\equal{} \\{y_n\\}_n\\subset X$. Prove $ (X,d)$ is a metric\nspace[/quote]\r\nSorry. Where $ \\displaystyle{d(x,y) \\equal{} \\sum_{k \\equal{} 1}^{ \\plus{} \\infty}\\frac {\\min(|x_k \\minus{} y_k|,1)}{k^2}}$", "Solution_3": "Proving that $ d(x,x) \\equal{} 0\\quad \\forall x$ and $ d(x,y) \\equal{} d(y,x)\\quad \\forall x,y \\in X$ is trivial. Further,\r\n\\[ \\min \\left( |x_i \\minus{} y_i|,1\\right) \\plus{} \\min \\left( |y_i \\minus{} z_i|,1\\right) \\ge \\min \\left( |x_i \\minus{} z_i|,1\\right)\\]\r\nis true for every $ i$.\r\n\r\nSince, the distance is bounded ($ d(x,y) < k \\plus{} \\frac {\\pi^2}{6}$ with some appropriate constant), it follows that its a metric space." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that for any positive numbers $ x$ , $ y$ , $ z$ exists the relation \r\n\r\n$ 3\\cdot\\sum_{\\mathrm{sym}}\\frac {y\\plus{}z}{x(2x\\plus{}y\\plus{}z)}\\ge4\\cdot \\frac {(x\\plus{}y\\plus{}z)^2}{3(y\\plus{}z)(z\\plus{}x)(x\\plus{}y)}\\ .$", "Solution_1": "[quote=\"Virgil Nicula\"]Prove that for any positive numbers $ x$ , $ y$ , $ z$ exists the relation \n\n$ \\cdot\\sum_{\\mathrm{sym}}\\frac {y \\plus{} z}{x(2x \\plus{} y \\plus{} z)}\\ge4\\cdot \\frac {(x \\plus{} y \\plus{} z)^2}{3(y \\plus{} z)(z \\plus{} x)(x \\plus{} y)}\\ .$[/quote]\r\n\r\nWe have\r\n\\[ \\sum {\\frac {{y \\plus{} z}}{{x\\left( {2x \\plus{} y \\plus{} z} \\right)}}} \\equal{} \\sum {\\frac {1}{x} \\cdot \\left( {1 \\minus{} \\frac {{2x}}{{2x \\plus{} y \\plus{} z}}} \\right)} \\equal{} \\frac {{xy \\plus{} yz \\plus{} zx}}{{xyz}} \\minus{} 2\\sum {\\frac {x}{{2x \\plus{} y \\plus{} z}}} \\ge \\frac {{xy \\plus{} yz \\plus{} zx}}{{xyz}} \\minus{} \\sum {\\frac {1}{{x \\plus{} y}}}\r\n\\]\r\nIt's enough to show that\r\n\\[ \\frac {{xy \\plus{} yz \\plus{} zx}}{{xyz}} \\minus{} \\frac {1}{{x \\plus{} y}} \\minus{} \\frac {1}{{y \\plus{} z}} \\minus{} \\frac {1}{{z \\plus{} x}} \\ge \\frac {{4\\left( {x \\plus{} y \\plus{} z} \\right)^2 }}{{3\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow x^2 \\plus{} y^2 \\plus{} z^2 \\ge xy \\plus{} yz \\plus{} zx\r\n\\]\r\nDone.", "Solution_2": "Using Cauchy Schwarts, we have:\r\nLHS $ \\geq\\ \\frac {4(x \\plus{} y\\plus{}z)^2}{3(x \\plus{} y)(y \\plus{} z)(z \\plus{} x)} \\equal{}$ RHS\r\n :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "prove that if $n | (p-1)^{n}+1$ then $p|n$", "Solution_1": "Can be n=1. Exactly if n natural >1 and $n|(p-1)^{n}+1$, then $p|n.$\r\nLet $q$ is minimal prime divisor n. If q=2, then p=2, n=2. If q>2 n is odd p>2 and $(q,p-1)=1$. Let $2k=ord_{q}(p-1)$, then $k|n$ and $k n$ with\r\n\\[m\\{|f_{n,k(n)}-f_{n}| > \\frac{1}{n}\\}< 2^{-n}\\]\r\nThis is possible because the measure of the domain is finite.\r\n\r\nBy [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=143796]strong convergence in measure implies pointwise convergence[/url], we have $|f_{n,k(n)}-f_{n}| \\to 0$ a.e. Therefore $f_{n,k(n)}$ and $f_{n}$ have the same limit a.e." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ ABC$ be a triangle. A circle $ P$ is internally tangent to the circumcircle of triangle $ ABC$ at $ A$ and tangent to $ BC$ at $ D$. Let $ AD$ meets the circumcircle of $ ABC$ agin at $ Q$. Let $ O$ be the circumcenter of triangle $ ABC$. If the line $ AO$ bisects $ \\angle DAC$, prove that the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent.", "Solution_1": "[quote=\"Raja Oktovin\"]Let $ ABC$ be a triangle. A circle $ P$ is internally tangent to the circumcircle of triangle $ ABC$ at $ A$ and tangent to $ BC$ at $ D$. Let $ AD$ meets the circumcircle of $ ABC$ agin at $ Q$. Let $ O$ be the circumcenter of triangle $ ABC$. If the line $ AO$ bisects $ \\angle DAC$, prove that the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent. [/quote]\r\nLet E be the intersection of (Q) and AC, O is the circumcenter of triangle ABC\r\nWe have $ \\frac{AP}{AO}\\equal{}\\frac{AD}{AQ}$ then $ PD//OQ$ or $ OQ\\perp BC$\r\nSo Q is the midpoint of arc BC\r\nSince PO bisects $ \\angle DAC$, it's easy to see triangle AQC and ADE are isosceles then DECQ is a isosceles trapezium.\r\nWe get $ \\angle QED\\equal{}\\angle EDC\\equal{}\\angle DAE$\r\nSo QE is a tangent of (P)\r\nThus $ QE^2\\equal{}QD.QA\\equal{}QB^2$\r\nWe obtain $ QB\\equal{}QE$ or E lies on (Q)\r\nMoreover, $ \\angle DQE\\equal{}\\angle DCE\\equal{}\\angle AQB$ then $ BQE$ is an isosceles triangle\r\nTherefore $ AD\\perp BE$\r\nHence the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent at E" } { "Tag": [ "inequalities", "calculus", "integration", "search", "inequalities proposed" ], "Problem": "(1) Suppose that real numbers $ a_1,\\ a_2,\\ x_1,\\ x_2,\\ y_1,\\ y_2$ satisfy : \r\n\r\n$ 00 \\Rightarrow x_{1} \\leq y_{1}$\r\n\r\n$ a_{1}x_{1}\\plus{}a_{2}x_{2} \\leq a_{1}y_{1}\\plus{}a_{2}y_{2}$\r\n$ a_{1}(x_{1}\\minus{}y_{1}) \\leq a_{2}(y_{2}\\minus{}x_{2})$\r\n\r\nWe have 2 different cases:\r\n1. $ x_{1}\\minus{}y_{1}\\equal{}0$\r\nAnd we get $ 0 \\leq a_{2}(y_{2}\\minus{}x_{2}) \\Rightarrow y_{2}\\minus{}x_{2}\\geq 0 \\Rightarrow y_{2}\\geq x_{2}$ \r\nSince $ y_{1} \\geq x_{1}$ and $ y_{2}\\geq x_{2}$ we can conclude $ x_{1}\\plus{}x_{2}\\leq y_{1}\\plus{}y_{2}$\r\n\r\n2. $ x_{1}\\minus{}y_{1}\\neq 0$\r\n$ a_{1}(x_{1}\\minus{}y_{1}) \\leq a_{2}(y_{2}\\minus{}x_{2})$\r\nMultiply this by $ \\minus{}1$\r\nAnd we get $ a_{1}(y_{1}\\minus{}x_{1}) \\geq a_{2}(x_{2}\\minus{}y_{2})$\r\nDivide by $ a_{1}$ to get $ (y_{1}\\minus{}x_{1}) \\geq \\frac{a_{2}}{a_{1}}(x_{2}\\minus{}y_{2})$\r\nSince $ a_{2}\\geq a_{1} \\Rightarrow \\frac {a_{2}}{a_{1}}\\geq 1$\r\nTherefore $ (y_{1}\\minus{}x_{1}) \\geq (x_{2}\\minus{}y_{2})$\r\nHence $ x_{1}\\plus{}x_{2}\\leq y_{1}\\plus{}y_{2}$", "Solution_2": "Sorry for that, I forgot to write the last part.\r\n \r\n(2) Let $ n\\geq 2$ be integers.\r\n\r\nIf $ 3n's$ real numbers $ a_1,\\ a_2,\\ \\cdots ,\\ a_n\\ ,\\ x_1,\\ x_2,\\ \\cdots ,\\ x_n,\\ y_1,\\ y_2,\\ \\cdots ,\\ y_n$ satisfy $ 0 < a_1,\\ a_2\\leq \\cdots \\leq a_n$ and $ n's$ inequalities $ \\sum_{i \\equal{} 1}^j a_ix_i\\leq \\sum_{i \\equal{} 1}^j a_iy_i\\ (j \\equal{} 1,\\ 2,\\ \\cdots n)$,\r\n\r\n prove : $ \\sum_{i\\equal{}1}^n x_i\\leq \\sum_{i\\equal{}1}^n y_i$.\r\n\r\n@Moderator, please edit the context of question (2). Thanks in advance.\r\n\r\nkunny", "Solution_3": "Hi,\r\n\r\nThe theme of this problem is Abel Summation, or Integration by Parts in discrete form:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=135850088&t=192043\r\n\r\nwe have\r\n$ \\sum_{i\\equal{}1}^n x_i\\equal{}\\sum_{i\\equal{}1}^n (\\frac {1}{a_i})(a_ix_i) \\equal{} \\sum_{i\\equal{}1}^m (\\sum_{j\\equal{}1}^i a_jx_j)(\\frac {1}{a_i}\\minus{}\\frac {1}{a_i\\plus{}1}) \\plus{} \\frac {1}{a_n} (\\sum_{i\\equal{}1}^n a_ix_i) \\leq \\sum_{i\\equal{}1}^m (\\sum_{j\\equal{}1}^i a_jy_j)(\\frac {1}{a_i}\\minus{}\\frac {1}{a_i\\plus{}1}) \\plus{} \\frac {1}{a_n} (\\sum_{i\\equal{}1}^n a_iy_i)\\equal{} \\sum_{i\\equal{}1}^n (\\frac {1}{a_i})(a_iy_i) \\equal{} \\sum_{i\\equal{}1}^n y_i$\r\nwhere $ m \\equal{} n\\minus{}1$ and $ \\frac {1}{a_i} \\minus{} \\frac {1}{a_i\\plus{}1} \\geq 0$ for $ i \\equal{} 1,2 \\cdots m$\r\n\r\n\r\n\r\nRegards,\r\nLei Lei", "Solution_4": "Thank you.\r\n\r\nkunny" } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "$ p(x)$ is a polynomial in $ \\mathbb{Z}[x]$ such that for each $ m,n\\in \\mathbb{N}$ there is an integer $ a$ such that $ n\\mid p(a^m)$. Prove that $0$ or $1$ is a root of $ p(x)$.", "Solution_1": "[hide=\"Hint\"]For some prime $ q$, let $ n \\equal{} q$, $ m \\equal{} q \\minus{} 1$.[/hide]", "Solution_2": "Let $P(x)=\\sum_{k=0}^n a_k x^k$. Note that if $a_0=0$, the condition trivially holds. Now let $a_0\\neq 0$. Fix a prime $p$, such that $p>|a_0|$ and $p>\\sum_{k=0}^n |a_k|$. Observe that, for $m=p-1$ and $n=p$, we have $p\\mid P(a^{p-1})$. If $p\\mid a$, then we earn a contradiction since $p>|a_0|$. Thus, $p\\nmid a$, hence, $a\\not\\equiv 0\\pmod{p}$, and by Fermat, $a^{p-1}\\equiv 1\\pmod{p}$. This yields, $P(a^{p-1})\\equiv P(1)\\pmod{p}$. Since $p>\\sum_{k=0}^n |a_k|\\geqslant |P(1)|$ (by triangle inequality), it follows that $P(1)=0$." } { "Tag": [ "geometry", "perimeter", "FTW" ], "Problem": "The area of square $ A$ is $ \\frac14$ of the area of square $ B$. The diagonal of square $ A$ is $ n$. Find the perimeter of square $ B$ in terms of $ n$. Express your answer in simplest radical form.", "Solution_1": "Let x be the side length of A, and 2x be the side length of B. Thus $ n \\equal{} x\\sqrt {2}$. We know that the diagonal of B is $ 2x\\sqrt{2}$, which is $ 2n$. Thus each side length is (in terms of n) $ \\frac{2n}{\\sqrt{2}}\\equal{}\\frac{2n\\sqrt{2}}{2}\\equal{}n\\sqrt{2}$. We want 4 times this, or $ 4n\\sqrt{2}$.", "Solution_2": "We can just do everything with $n$. Knowing that the diagonal of square $A$ is $n$, we know each side of square $A$ must be $\\frac{n}{\\sqrt{2}}$. Knowing that the area of square $B$ is four times larger, we can write the area of square $B$ as $\\frac{4n^{2}}{2}$. Taking the square root of that and multiplying it by four gives us the the perimeter of square $B$. Hence $\\boxed{4n\\sqrt{2}}$.", "Solution_3": "Okay, that wasn't fair. In the FTW game I put 4sqrt2n, but then the answer is 4nsqrt2.", "Solution_4": "Well that is because $4\\sqrt{2}n$ can be interpreted as $4\\sqrt{2n}$, which is wrong." } { "Tag": [ "induction", "limit", "integration", "function", "real analysis", "algebra proposed", "algebra" ], "Problem": "Suppse that $k$ be a positive integer. Prove that (without induction) \\[ \\lim_{n \\rightarrow \\infty}\\frac{1^k+2^k+...+n^k}{n^{k+1}}=\\frac{1}{k+1} \\]", "Solution_1": "Just use that for large enough $n$: $\\sum_{i=1} ^n i^k$ tends to $\\int_0 ^n x^k dx = \\frac{n^{k+1}}{k+1}$", "Solution_2": "Use Stolz-Cesaro Theorem (see http://planetmath.org/encyclopedia/StolzCesaroTheorem.html ),\r\nthen binomial expansion of $(n+1)^k$ and $(n+1)^{k+1}$.", "Solution_3": "Another solution: \r\n$\\frac{1^k+2^k+...+n^k}{n^{k+1}}$ is a Riemann sum associated to function $f: [0,1]\\rightarrow \\mathbb {R}, f(x)=x^k$ (and k may be an arbitrary positive number), so the limit in the statement is equal to $\\int_0^1x^kdx=\\frac 1{k+1}$.", "Solution_4": "Thanx ;) but who can solve it by the Sqeezing Lemma ????" } { "Tag": [ "geometry", "incenter", "circumcircle", "trigonometry", "geometry proposed" ], "Problem": "Let $ I$ be incenter of triangle $ ABC$, $ M$ be midpoint of side $ BC$, and $ T$ be the intersection point of $ IM$ with incircle, in such a way that $ I$ is between $ M$ and $ T$. Prove that $ \\angle BIM\\minus{}\\angle CIM\\equal{}\\frac{3}2(\\angle B\\minus{}\\angle C)$, if and only if $ AT\\perp BC$.", "Solution_1": "Beautiful problem. [hide=\"A hint\"] when $ AT\\perp BC$ then $ T$ is on the nine-point circle.[/hide]\n[hide=\"Solution\"] Assume $ AT\\perp BC$ and incircle meets $ BC$ at $ E$. Now note that $ \\angle BIM\\minus{}\\angle BIC\\equal{} 2\\angle EIM\\minus{}\\frac{1}{2}(\\angle C\\minus{}\\angle B)$ So we should prove $ \\angle EIM\\equal{}\\angle B\\minus{}\\angle C$.\nNow note that if we invert to $ M$ with respect to the incircle and since the invertion preserves the angles, nine-point circle changes to the tangent line that is perpendicular to $ IM$ at it's other intersection with incircle beside of $ T$ which means that $ T$ was initially on the \nnine-point circle.\nWe have $ \\angle EIM\\equal{}\\angle H_{a}TM$ where $ H_{a}$ is the feet of the altitude from $ A$. So it comes down to showing that arc $ H_{a}M$ is twice the $ \\angle B\\minus{}\\angle C$. Let $ N$ be the mid point of $ AC$ then note that $ \\angle H_{a}NC\\equal{}180\\minus{}2\\angle C$ since $ H_{a}N$ is median in the right triangle $ AH_{a}C$ and $ \\angle MNC\\equal{}A$. hence $ \\angle H_{a}NM\\equal{}\\angle B\\minus{}\\angle C$. Since $ N$ is on the nine-point circle $ \\angle H_{a}NM\\equal{}\\angle H_{a}TM$ which finishes the proof. [/hide]", "Solution_2": "[color=darkred][b][u]Prpblem.[/u][/b] $ w = C(I,r)$ is the incircle of $ \\triangle ABC$. Denote the points $ \\{\\begin{array}{c}M\\in (BC)\\ ,\\ MB = MC\\\\ \\ D\\in BC\\ ,\\ AD\\perp BC\\\\ \\ T\\in MI\\cap w\\end{array}$.\nProve that $ T\\in AD\\Longleftrightarrow$ $ m(\\widehat{BIM}-m(\\widehat{CIM}) =\\frac{3}{2}\\cdot (B-C)$.[/color]\r\n\r\n[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $ b > c$. Calculate easily $ ME =\\frac{b-c}{2}$ and $ MD =\\frac{b^{2}-c^{2}}{2a}$. Denote the points $ \\{\\begin{array}{c}E\\in BC\\cap w\\\\ \\ S\\in MI\\cap AD\\end{array}$ and $ AS = x$.\n\nObserve that $ IE\\parallel AD$ in the $ D$- right triangle $ MDA$. Thus, $ IE\\parallel AD\\Longleftrightarrow$ $ \\frac{IE}{SD}=\\frac{ME}{MD}$ $ \\Longleftrightarrow$ $ \\frac{r}{h_{a}-x}=\\frac{a}{b+c}$ $ \\Longleftrightarrow$\n\n$ ax = ah_{a}-(b+c)r$ $ \\Longleftrightarrow$ $ ax = 2pr-(b+c)r$ $ \\Longleftrightarrow$ $ x = r$, i.e. $ \\boxed{\\ AS = r\\ }$. Therefore, $ \\boxed{\\ T\\in AD\\Longleftrightarrow T\\equiv S\\ }$.\n\n==============================================================================================================\n\nSince $ \\triangle ATI$ is $ A$- isosceles and $ m(\\widehat{TAI}) =\\frac{B-C}{2}$, obtain $ m(\\widehat{DTM}) = B-C$. Since $ IE\\parallel TD$, obtain $ T\\in AD$ $ \\Longleftrightarrow$ $ \\phi\\equiv\\boxed{m(\\widehat{EIM}) = B-C}$.\n\nSince $ \\{\\begin{array}{c}m(\\widehat{BIM}) =(90^{\\circ}-\\frac{B}{2})+\\phi\\\\ \\ m(\\widehat{CIM}) =(90^{\\circ}-\\frac{C}{2})-\\phi\\end{array}\\|$ $ \\implies$ $ m(\\widehat{BIM})-m(\\widehat{CIM}) =$ $ 2\\phi-\\frac{B-C}{2}=\\frac{3}{2}\\cdot (B-C)$, obtain $ m(\\widehat{BIM})-m(\\widehat{CIM}) =\\frac{3}{2}\\cdot (B-C)$. \n\n[b]Remark.[/b] $ T\\in AD\\Longleftrightarrow$ the circumcenter $ O$ of $ \\triangle ABC$ belongs to the $ A$- Nagel's line, i.e. $ O\\in AF$, where $ M\\in (EF)$ and $ MF = ME$ $ \\Longleftrightarrow$ \n\n$ OM = r$ $ \\Longleftrightarrow$ $ r = R\\cos A$ (the our triangle is acute !) $ \\Longleftrightarrow$ $ AT = OM$ $ \\Longleftrightarrow$ the point $ T$ is middle of the segment $ AH$,\n\nwhere $ H$ is the orthocenter of $ \\triangle ABC$ $ \\Longleftrightarrow$ the point $ T$ belongs to the Euler's circle of $ \\triangle ABC$ $ \\Longleftrightarrow$ $ \\cos B+\\cos C = 1$ $ \\Longleftrightarrow$ $ \\frac{r}{R}+\\frac{a}{b+c}= 1$.\n\n[b]Commentary.[/b] The remarkable relation $ \\boxed{\\ AS = r\\ }\\ \\ (*)$ is well-known. From this moment the our nice problem became more easily !\n\nBy the way, I don't like the right hand of the equivalence from the conclusion, is even nastily. In the my opinion, $ m(\\widehat{EIM}) = B-C$ is more nicely.\n\nIn fact, this problem is the property $ (*)$ somehow \"painted\".\n[/color]", "Solution_3": "I think this is closely related to the solution given.\r\n\r\nDenote by D the intersection of AI and the circumcircle; it is not difficult to see that D is the center of the circle passing through B, I, C, and the A-excenter. In particular, we will use that DM is perpendicular to BC. The angle condition is easily seen to be equivalent to DIM = IDM, so we wish to show that AT is perpendicular to BC if and only if MID is isosceles. This follows easily once we note that IT / IA = DM / DI, so both conditions are in fact equivalent to the similarity of the triangles AIT and DIM.", "Solution_4": "Let $D_{1}$ be the touch point if the $A$-excircle,let $AD_{1}\\cup \\odot I={Z}$,$\\odot I$ touches $BC$ in $D$.\n\n$IM$ is the $I$ midline in $\\triangle DZD_{1}$ and hence $IM||D_{1}Z$.$ID\\perp BC$,$AT\\perp BC$ and hence $AT||IZ$ $\\implies$\n$AZIT$ is a parallelogram but as $IT=IZ$ its a rhombus and hence $AD_{1}$ is the isogonal to $AT$ $\\implies$ $AD_{1}$ passes thru\nthe center of $\\odot ABC$.\n$$\\angle CIM=\\angle \\frac{\\pi}{2}-\\frac{\\gamma}{2}+ \\angle DIM=\\angle \\frac{\\pi}{2}-\\frac{\\gamma}{2}+\\angle HAO$$\n$$=\\angle \\frac{\\pi}{2}-\\frac{\\gamma}{2}+\\frac{\\pi}{2}-\\beta-\\frac{\\pi}{2}+\\gamma=\\frac{\\pi}{2}+\\frac{\\gamma}{2}-\\beta$$\nAnalogously we have $\\angle BIM=\\frac{\\pi}{2}-\\gamma+\\frac{\\beta}{2}$ so we are done.$\\clubsuit$" } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "Let $ z(x,y)$ be a two-variable $ C^1$ class function on $ \\{(x,y): x,\\ y > 0\\}$. Show that $ z(x,y)$ can be expressed in the form\r\n\\[ z \\equal{} yf\\left(\\frac {y}{x}\\right),\\]\r\nwhere $ f(s)$ is a one-variable $ C^1$ class function on $ (0,\\infty)$, if and only if\r\n\\[ x\\frac {\\partial z}{\\partial x} \\plus{} y\\frac {\\partial z}{\\partial y} \\equal{} z.\\]", "Solution_1": "$ \\frac{\\partial z}{x} \\equal{} \\minus{}\\frac{y^2}{x^2} f '\\left(\\frac{y}{x}\\right)$\r\n$ \\frac{\\partial z}{y} \\equal{} f\\left(\\frac{y}{z}\\right) \\plus{} \\frac{y}{x} f '\\left(\\frac{y}{x}\\right)$\r\n\r\nThus, $ x\\frac{\\partial z}{x}\\plus{}y\\frac{\\partial z}{y} \\equal{} \\minus{}\\frac{y^2}{x} f '\\left(\\frac{y}{x}\\right) \\plus{} y f\\left(\\frac{y}{z}\\right) \\plus{} \\frac{y^2}{x} f '\\left(\\frac{y}{x}\\right) \\equal{} y f\\left(\\frac{y}{z}\\right) \\equal{} z$", "Solution_2": "Indeed that's true, but observe it's a double implication. I certainly had no problem showing that if the function satisfies $ z\\equal{}yf(y/x)$ then the relation of the partial derivatives is satisfied, but I find the converse somewhat difficult to prove in a formal way. Can you help me finish the proof please?", "Solution_3": "Define a new function $ Z(t,y)\\equal{}z(ty,y)$; we want to show that $ Z(t,y)\\equal{}yg(t)$, where $ g(t)\\equal{}f(\\frac1t)$. This is effectively a substitution.\r\n\r\nDifferentiating, $ \\frac{\\partial Z}{\\partial t}\\equal{}y\\frac{\\partial z}{\\partial x}$ and $ \\frac{\\partial Z}{\\partial y}\\equal{}t\\frac{\\partial z}{\\partial x}\\plus{}\\frac{\\partial z}{\\partial y}$. Multiply the latter equality by $ y$ to get $ y\\frac{\\partial Z}{\\partial y}\\equal{}ty\\frac{\\partial z}{\\partial x}\\plus{}y\\frac{\\partial z}{\\partial y}\\equal{}z(ty,y)\\equal{}Z(t,y)$.\r\n\r\nThat's a simple differential equation $ h'(y)\\equal{}\\frac{h(y)}{y}$ for the one-variable function $ h(y)\\equal{}Z(t,y)$, and its solutions are $ h(y)\\equal{}cy$ for constants $ c$, which extend to the whole region $ y>0$. This $ c$ is independent of $ y$, but may depend on $ t$; call it $ g(t)$, and we have written $ Z$ in the desired form.", "Solution_4": "For all such problem,You can solver like this:\r\nLet \r\n$ u\\equal{} z\\minus{}yf(\\frac{y}{x})$\r\n$ \\frac{\\partial u}{\\partial x}\\equal{} \\frac{y}{x^2} f'(\\frac{y}{x})$\r\n$ \\frac{\\partial u}{\\partial y}\\equal{} \\minus{}f(\\frac{y}{x} )\\minus{}\\frac{y}{x}f'(\\frac{y}{x})$\r\n$ \\frac{\\partial u}{\\partial z}\\equal{} 1$\r\nSo\r\n$ \\frac{\\partial z}{\\partial x}\\equal{} \\frac{y}{x^2} f'(\\frac{y}{x})$\r\n$ \\frac{\\partial z}{\\partial y}\\equal{} \\minus{}f(\\frac{y}{x} )\\minus{}\\frac{y}{x}f'(\\frac{y}{x})$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $ g: [1;\\plus{}\\infty)\\to[1;\\plus{}\\infty)$ a funtion with the property $ g(g(x))\\equal{}x^2\\minus{}2x\\plus{}2$ . Prove that $ g$ is injective.", "Solution_1": "Isn't it quite easy?\r\nIf $ g(a) \\equal{} g(b)$ then $ a^2\\minus{}2a\\plus{}2 \\equal{} b^2\\minus{}2b\\plus{}2$\r\nso that $ (a\\minus{}b)(a\\plus{}b\\minus{}2)\\equal{}0$.\r\nSince $ a,b \\geq 1$ the conclusion holds easily.... :wink: \r\n\r\nPierre." } { "Tag": [ "Asymptote" ], "Problem": "Hi there,\r\n\r\nI am a new user to Asymptote.\r\n\r\nand I need some help in drawing the following:\r\n\r\n[asy]size(200);\npair A=(0,5), B=(-5,0), C=(5,0);\ndraw(circle(A,1));\ndraw(circle(B,1));\ndraw(circle(C,1));\nlabel(\"$1$\",A);\nlabel(\"$4$\",B);\nlabel(\"$5$\",C);\n\npair D=(-5,-5), F=(5,-5);\ndraw(circle(D,1));\ndraw(circle(F,1));\nlabel(\"$2$\",D);\nlabel(\"$3$\",F);[/asy]\r\n\r\nThat's what I can do so far.\r\n\r\nThe original picture is in the PDF file.\r\n\r\nThank you for help :lol:", "Solution_1": "Do you know how to make lines?" } { "Tag": [ "Combinations" ], "Problem": "A set of 1990 persons is divided into non-intersecting subsets in such a way that\r\n\r\n1. No one in a subset knows all the others in the subset,\r\n\r\n2. Among any three persons in a subset, there are always at least two who do not know each other, and\r\n\r\n3. For any two persons in a subset who do not know each other, there is exactly one person in the same subset knowing both of them.\r\n\r\n(a) Prove that within each subset, every person has the same number of acquaintances.\r\n\r\n(b) Determine the maximum possible number of subsets.\r\n \r\nNote: It is understood that if a person $A$ knows person $B$, then person $B$ will know person $A$; an acquaintance is someone who is known. Every person is assumed to know one's self.", "Solution_1": "Let $A,B$ be two people who do not know each other. They have exactly one common acquaintance $C$, and for each acquaintance $D_a\\ne C$ of $A$ there is exactly one common acquaintance $D_b\\ne C$ of $D_a,B$. This way, we can pair up the acquaintances of $A$ and those of $B$ which are different from $C$, and the conclusion is that every two people who do not know each other have the same number of acquaintances. \r\n\r\nIn the graph $G$ which we associate to this situation in an obvious manner, two vertices can have different degrees only if they lie in different connected components of the complementary graph $\\overline G$. Now, if $\\overline G$ has at least two components each of which has $\\ge 2$ vertices, there will be $4$ vertices $a_1,a_2,b_1,b_2$ such that $a_i,b_j$ are connected for $i,j=1,2$, which contradicts the hypotheses, and if it has at least two components, one of which has exactly one vertex, then that vertex is connected to all the others, so we get a contradiction again. This means that $\\overline G$ is connected, and it follows from this and the first sentence in the paragraph that all vertices of $G$ have the same degree.\r\n\r\nIt's easy to see that there are no graphs $G$ with the required properties and with fewer that $5$ vertices, and that a $5$-cycle satisfies all the conditions; it follows from this that the largest possible number of subsets is achieved when we divide the set $1990$ vertices (well, people :)) into $398=\\frac{1990}5$ subsets.", "Solution_2": "[quote=\"grobber\"]Now, if $ \\overline G$ has at least two components each of which has $ \\ge 2$ vertices, there will be $ 4$ vertices $ a_1,a_2,b_1,b_2$ such that $ a_i,b_j$ are connected for $ i,j \\equal{} 1,2$, which contradicts the hypotheses,[/quote]Sorry to revive this old thread. I don't understand this part of the solution. Can you explain why is it a contradiction?", "Solution_3": "A square in this graph either requires an additional edge (so a triangle) or has two disconnected vertices with two common neighbors (contradicting the \"exactly 1\" in condition 3).", "Solution_4": "I'm fairly sure such a graph must be a 5 cycle or 2 attached 5 cycles. if any vertex has degree less than 2, since it can't be zero it has degree one but then the one it is attached to is itself attached to all others since that first one is not attached to any =><=\r\n\r\nNow since each vertex has degree 2 we have a cycle. We clearly do not have a 4 or 3 cycle but if the length is 6 or greater take 2 verticies 2 apart if they are connected we have a smaller cycle if not then they are connected to another which if put together with the part of the cycle that vertex is not part of gives us a smaller cycle so we have a 5 cycle.\r\n\r\nEvery vertex not connected to the 5 cycle that we will call ABCDE is connected to at most of them. We consider pairs X, Y with X not in ABCDE and Y in ABCDE. And note that X,Y have a common neighbor outside ABCDE if and only if they are not connected X is not connected to a neighbor of Y in ABCDE. If any vertex is not attached to any of ABCDE it contributes 5 pairs while those attached to one contribute 2 since it is not possible for a vertex to be common neighbors for three such pairs ( if two are connected to one vertex then those 2 have two common neighbors one outside ABCDE and one of ABCDE they are both connected to) thus we only consider the case where each vertex outside is attached to exactly one of ABCDE and each is a common neighbor of exactly 2 pairs X,Y. So each vertex outside ABCDE has degree 3. Now if a vertex is connected to A then we need another vertex to connect it to C going around like this we find there is a vertex connected to each of ABCDE. Now consider a neighbor of A and one of C if they are not connected then their neighbor T is needs to be connected to a vertex of ABCDE which is not a neighbor of A or C which is impossible thus each neighbor of A is connected to each neighbor of C. Thus there is at most 1 neighbor of C so C has degree 2+1=3 so we are done.\r\n\r\nIf the 5 cycle is the entire graph we are also done. :starwars:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let ABC be a right triangle with hypotenuse c.\r\n\r\nProve the inequality\r\n\r\n$a+b+2(c+\\sqrt{2}) \\geq 4\\sqrt{a+b+c}$", "Solution_1": "But the inequality is not homogenous... \r\n\r\nSo in fact it should be true even if the $\\sqrt{2}$ is not there... :huh:", "Solution_2": "If it's not homogenous and there's no $\\sqrt{2}$, we can easily derive a contradiction.\r\n\r\nThat's why there's $\\sqrt{2}$.\r\n\r\nBecause it's not homogenous.\r\n\r\nThe trick is pure evil.", "Solution_3": "Hm, I see, I sort of misread the question. I'll think about it :)", "Solution_4": "I don't know what your evil trick is, but here's my solution.\r\nIf $c \\ge 4-2\\sqrt{2}$, \r\n$a+b+c+c+2\\sqrt{2} \\ge 2\\sqrt{(a+b+c)(c+2\\sqrt{2})} \\ge 4 \\sqrt{a+b+c}$ because $c \\ge 4-2\\sqrt{2}$ implies $\\sqrt{c+2\\sqrt{2}} \\ge 2$\r\nIf $c \\le 4-2\\sqrt{2}$,\r\n$8(\\sqrt{2}-1)^2 \\ge c^2=a^2+b^2 \\ge 2(a+b)^2$ which gives $2\\sqrt{2} - 2 \\ge a+b$, then\r\n$a+b+2c+2\\sqrt{2} \\ge 2(a+b+c+1) \\ge 4\\sqrt{a+b+c}$", "Solution_5": "You beat me there, Soarer. It's not straightforward, but only a little tricky :)", "Solution_6": "[quote=\"Soarer\"]\nIf $c \\le 4-2\\sqrt{2}$,\n$8(\\sqrt{2}-1)^2 \\ge c^2=a^2+b^2 \\ge 2(a+b)^2$ which gives $2\\sqrt{2} - 2 \\ge a+b$, then\n$a+b+2c+2\\sqrt{2} \\ge 2(a+b+c+1) \\ge 4\\sqrt{a+b+c}$[/quote]\r\n\r\nWrong!!! since when do we have $a^2+b^2 \\ge 2(a+b)^2$???\r\n\r\nIf it's that easy, i won't post it here.", "Solution_7": "No one?\r\nYeah the trick is evil...\r\nbut you can manage to find it...", "Solution_8": "Still no one?\r\n\r\nOkay it seems too hard for you guys...\r\n\r\nI'll give you a little hint. (For those who want to solve it on their own, don't look.)\r\n\r\n[hide]\n\nHere's the hint\n\nobserve that\n\n$a+b+c= \\sqrt{a^2+b^2+c^2+2ab+2ac+2bc}$\n\nthat's it.\n\nNow you are much closer to the solution.[/hide]", "Solution_9": "I think I have found a really nice and simple solution to this:\r\n\r\n$a+b+2c+2\\sqrt{2} \\ge \\sqrt{2}(a+b+c) + 2\\sqrt2 \\geq 4\\sqrt{a+b+c}$\r\n\r\nThe first step follow from AM-QM in the form $(2-\\sqrt{2})\\sqrt{a^2+b^2}\\ge(\\sqrt{2}-1)(a+b)$, the second step is just AM-GM in the classical way.", "Solution_10": "Detective, what was your solution? It seems you had a much bigger one in mind? (seeing your comments and hint in the end)", "Solution_11": "ah okay i admit that your solution is far better than mine\r\n\r\nhere's mine by AM-GM four variables\r\n\r\n$a+b+2(c+\\sqrt{2}) = (a+c) + (b+c) + \\sqrt{2} + \\sqrt{2} \\geq 4\\sqrt{\\sqrt{2(a+c)(b+c)}}$\r\n\r\nNotice that by pythagoras theorem we have\r\n\r\n$2(a+c)(b+c) = 2ab + 2ac + 2bc + 2c^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$\r\n\r\nSince\r\n\r\n$\\sqrt{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc} = a+b+c$\r\n\r\nwe have the desired result.\r\n\r\nit's pretty tricky right?" } { "Tag": [ "quadratics", "algebra" ], "Problem": "Simplify: $\\frac{a^{3}}{(a-b)(a-c)}+\\frac{b^{3}}{(b-a)(b-c)}+\\frac{c^{3}}{(c-a)(c-b)}$", "Solution_1": "[hide=\"Solution\"] The quadratic equation in $a$ given by\n\n$a^{3}-\\frac{ b^{3}(a-c)}{b-c}-\\frac{c^{3}(a-b)}{(c-b)}= (a+b+c)(a-b)(a-c)$\n\nHas roots $a = b, a = c, a = 0$ (yes, it is quadratic because the cubic terms cancel), hence it is an identity. Dividing by $(a-b)(a-c)$ we have the original expression is equal to $\\boxed{ a+b+c }$. [/hide]", "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Solution\"] The quadratic equation in $a$ given by\n\n$a^{3}-\\frac{ b^{3}(a-c)}{b-c}-\\frac{c^{3}(a-b)}{(c-b)}= (a+b+c)(a-b)(a-c)$\n\nHas roots $a = b, a = c, a = 0$ (yes, it is quadratic because the cubic terms cancel), hence it is an identity. Dividing by $(a-b)(a-c)$ we have the original expression is equal to $\\boxed{ a+b+c }$. [/hide][/quote] \r\n\r\nThanks for the solution....." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $ ABC$ be an acute triangle.Let $ S_{BC}$ has segment $ BC$ as its diameter.Circle $ W_A$ is to lines $ AB$ and $ AC$ and is tangent externally to $ S_{BC}$ at $ A_1$ .Points $ B_1,C_1$ are difined analogously.Prove that $ AA_1,BB_1,CC_1$ are concurent.", "Solution_1": "This solution belongs to a friend of mine \r\nDenote M=A1 N=B1 P=C1 W is the circumcenter of WA\r\nM1,M2 is the projection of M to AB,AC \r\nWM/MD=2r/a\r\nMM1=(r+2ra\u22c5a2\u22c5sinB)/(1+2ra)\uff0cMM2=(r+2ra\u22c5a2\u22c5sinC)/(1+2ra)\r\nSo MM1/MM2=(1+sinB)/(1+sinC) \r\nsin\u2220MAB/sin\u2220MAC=(1+sinB)/(1+sinC) \r\nthe same way sin\u2220NBC/sin\u2220NBA=(1+sinC)/(1+sinA) sin\u2220PCAsin\u2220PCB=(1+sinA)/(1+sinB)\r\ndone!" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "Is a group of order 39 always cyclic?", "Solution_1": "No.\r\n\r\nConstruct the nonabelian group with a semidirect product.", "Solution_2": "Yes it IS!\r\n\r\nIn fact we can say something stronger. Assume [G] = pq with p and q relatively prime. \r\n\r\nThen by sylows theorem there exists sylow subgroups say P and Q with [P] = p and [Q] = q. Now the number of p-sylow subgroups is congruent to 1modq. Moreover the number of p-sylow subgroups must divide [G]/p. So in our case it must divide q. This forces the number of p-sylow subgroups to be one. Since there is only one p-sylow this p-sylow must be normal in G.\r\n\r\nA similar argument shows there is only one q-sylow and it is therefore normal in G.\r\n\r\nSo P and Q are the unique p and q sylows, respectively. Since P and Q have prime order they must be cyclic. Suppose = P and = Q. \r\n\r\nWe will show [] = pq so that generates all of G making G a cyclic group. \r\n\r\nFirst we will show xy = yx. It suffices to show xyx~y~ = 1. [by x~ i mean x inverse]\r\n\r\nSince is normal in G xyx~ must also lie in . But [xyx~] = q. Thus we must have xyx~ = y. Hence, xyx~y~ = yy~ = 1. \r\n\r\nNow look at (xy)^(pq) = x^(pq)*y^(pq) = 1*1 = 1. Hence [xy] = pq so G = .", "Solution_3": "No it ISN'T!\r\nYou have $ 13 \\equiv 1 \\mod 3$ and $ 13$ divides $ \\frac {39}{3}$. Thus there can be $ 13$ $ 3$-Sylows.", "Solution_4": "oops my mistake :blush: i was thinking of when p1$ then $f(x)$ is an [b]increasing function[/b].\nIf $x>1$, then $3x-3>0$. \nHence $4x - x+3 > 0$.\nWhich means $4^x - 4^{x+3} > 0$\nAnd since $2x-1>1$, so $9^{2x-1}-9>0$.\nHence the $3^{4x}-3^{x+3}+9^{2x-1}-9 > 0$. Impossible.\n\nThe proof of $x<1$ is the same. \n\nHence we see $x=1$ is the only solution so $k=1$.\n\nP.S. We can set $x \\neq 1$ and the proof can become shorter.[/hide]", "Solution_2": "Well i did it this way: Let $3^x=a>0$\r\n\r\n\\begin{eqnarray*}\r\n3^{x+3}+9&=&\\sqrt{81^{2x-1}}+3^{4x}\\\\\r\n27\\cdot 3^x+9&=&3^{2(2x-1)}+3^{4x}\\\\\r\n27\\cdot 3^x+9&=&3^{4x}\\cdot\\frac{1}{9}+3^{4x}\\\\\r\n27a+9&=&a^4\\cdot\\frac{1}{9}+a^{4}\\\\\r\n10a^4-243a-81&=&0 \\end{eqnarray*}\r\n\r\nThe solution of the last equation is not so obviously ($a=3$). Can anybody post a method of solving this ?:)", "Solution_3": "I believe there is formula for the roots of 4 degree equations.", "Solution_4": "[quote=\"Slizzel\"]$27a+9=a^4\\cdot\\frac{1}{9}+a^{4}$\n$10a^4-234a-81=0$[/quote] you must have mistyped or something it's not 234 thou\r\n$10a^4-243a-81=0$\r\n\r\nshobber: the quartic formula is WAY too ugly and i dont recommend it unless that is absolutely the last option to go for", "Solution_5": "Sorry , fixed. I must have hurried :). I think there is a way to find the roots. Using [i][b]Horner's Method[/b][/i] we obtain the root $a=3$. Then we know that $10a^4-243a-81$ is divisible by $a-3$ so we just divie it. After that we get: $(a-3)(10a^3+30a^2+90a+27)=0$. But we also know that $a>0$ which means that $(10a^3+30a^2+90a+27)$ is always different by $0$. So the only root is $a=3\\Rightarrow x=1$ :)", "Solution_6": "Whats Horner Method ? :? :? \r\n\r\nAnyway I make a conclusion after seeing those solution ....\r\n\r\nAfter simplying , we can get \r\n\r\n1+3^(x+1) = 3^(4x-4) . 10 \r\n\r\nIf x is rational , then notice that LHS doesnt have factor of 3 , so the only possible rational solution for x is when 3^(4x-4) = 1 which is x=1 \r\n\r\nIf x is irrational , just subsitute 3^x = a like what Slizzel had done and you will see there is no solution ." } { "Tag": [ "linear algebra", "matrix", "vector", "function", "linear algebra solved" ], "Problem": "$A=(a_{ij}) \\in M_n(R)$ where $a_{ij}$ is the number of common divisors of $i$ and $j$\r\n\r\nProve that $A$ is invertible", "Solution_1": "I don't know if this would help much, but I think this comes more or less from the same applications than the Hilbert Matrix. Basically the idea is to look for a combinatorical explanation of the entries of the matrix, and then, somehow it becomes self-evident that the matrix is invertible.\r\n\r\nSorry, but I can't find much about the Hilbert Matrix right now in the web, but if you need a solution, probably find something about it may help in your problem.", "Solution_2": "Following djimenez's idea, we show that $a_{ij}=\\langle v_i,v_j\\rangle$ for some linearly independent vectors $v_i$ in an inner product space. Let the space be the set of functions from $\\{1,2,\\dots,n\\}$ to $\\mathbb{R}$ with the standard inner product $\\langle a,b\\rangle=\\sum_k a(k)b(k).$ Let $v_i(k)=1$ if $k|i$ and $v_i(k)=0$ otherwise. This clearly gives the appropriate result (interpreting the question as asking for positive divisors). These vectors are clearly linearly independent; the change of basis matrix from the standard basis is triangular, with each diagonal entry equal to 1.\r\nThis shows that $A$ is positive definite, hence invertible.", "Solution_3": "[quote=\"jmerry\"]Following djimenez's idea, we show that $a_{ij}=\\langle v_i,v_j\\rangle$ for some linearly independent vectors $v_i$ in an inner product space. Let the space be the set of functions from $\\{1,2,\\dots,n\\}$ to $\\mathbb{R}$ with the standard inner product $\\langle a,b\\rangle=\\sum_k a(k)b(k).$ Let $v_i(k)=1$ if $k|i$ and $v_i(k)=0$ otherwise. This clearly gives the appropriate result (interpreting the question as asking for positive divisors). These vectors are clearly linearly independent; the change of basis matrix from the standard basis is triangular, with each diagonal entry equal to 1.\nThis shows that $A$ is positive definite, hence invertible.[/quote]\r\n\r\njmerry, brilliant, just a detail that may make easier to understand the problem to the reader. The fact that $a_{ij}=\\langle v_i,v_j\\rangle$ means that, if you consider $V=[v_1,\\dots,v_n]$ (looking the $v_i$'s as colum vectors), then $A=VV^T$. The fact that $V$ is triangular with diagonal entries $1$ makes it invertible (what makes $A$ automatically invertible), but also, $x^TAx=x^TVV^Tx=\\langle V^Tx,V^Tx\\rangle=\\|V^Tx\\|^2>0$ if $x\\neq 0$, therefore $A$ is possitive definite.\r\n\r\nBest," } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "1) x^2 - 3 (x ^2 + 3)^1\\2 =< 1\r\n\r\n\r\n\r\n\r\n2) x ^2 + (5 x \\(x - 5))^2 =11\r\n\r\n\r\n\r\n3) ( x^3 +2 x)^1\\5 =( x^5 - 2x )^1\\3", "Solution_1": "I think you write that juste for fun :rotfl: ?" } { "Tag": [ "algebra", "polynomial", "LaTeX", "number theory unsolved", "number theory" ], "Problem": "Given a polynomial $P(x)$ with integer coefficients. For each natural number $n$ the number $P(n)$ is greater than $n$. Consider the sequence $x_1=1 , x_2=P(x_1), ... ,x_n=P(x_{n-1}),....$ .Given that for each netural number $N$ there exists a member of the sequence which is divisible by $N$. Prove that $P(x)=x+1$ ;)", "Solution_1": "nice problem.\r\n\r\nWe first show that $x_2 = 2$, which implies that $P(1)=2$.\r\nSuppose $x_2 = P(1) = \\sum _{i=0}^{m} a_i = k$, where $k$ is a positive integer. (Note that $P(x)$ has only integer coefficients and $P(x) \\geq x+1$ for all positive integers $x$.) Now $x_3 = P(2) = \\sum _{i=0}^m a_i * k^i = k (mod (k-1))$. Similarly, $x_n = 1 (mod (k-1))$ for all positive integers $n$. But if $k>2, k-1$ has a prime divisor, which will divide no term in the sequence ${x_n}$. Contradiction. Hence $P(1)=1$, as desired.\r\nBy similar reasoning (start from $x_3$), we get $P(2)=3$, and $P(n)=n+1$ for all natural $n$. QED :)\r\n\r\n[color=red]Edited by Megus: al.M.V - it isn't hard to use LaTeX - you can learn it to make your posts more readable [/color]", "Solution_2": "It's been posted before at least twice :).", "Solution_3": "Check http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6448 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10285 (problem 6).", "Solution_4": "alright Megus, I will try to learn Latex... :P I'm just super lazy to bother most of the times... ;)" } { "Tag": [ "function" ], "Problem": "Let $f(x)$ be a continous function for all real values of $x$ such that $f(a) \\le f(b)$ whenever $a \\le b$. Prove that, for ever real number $r$, the equation $x+f(x)=r$ has exactly one solution.\r\n\r\nThanks", "Solution_1": "Both x and f(x) are strictly increasing, meaning x + f(x) is also strictly increasing, meaning that each distinct input will result in a distinct output.\r\n\r\nThis means that x + f(x) = r must have exactly one solution for each value of r.", "Solution_2": "Is it common knowledge that this is true because how do we know that no real value will be \"skipped\".", "Solution_3": "Then it wouldn't be continuous.", "Solution_4": "I meant how can we show that x+f(x) is continuous...oh...I see...thanks very much...I forgot that x is really a constant in this case at any point...\r\n\r\nThanks Treething..." } { "Tag": [], "Problem": "Crux is a good problem solving journal by Canadian Mathematical Society\r\nand some issue of this journal is public and you can use free", "Solution_1": "Can you plz give the link!", "Solution_2": "[url=http://camel.math.ca/CMS/Publication/crux.html]the link[/url].", "Solution_3": "Oh,I am sorry \r\nI have work and I could`nt replay \r\nThis is the Link :\r\n\r\n[url=http://www.journals.cms.math.ca/CRUX/]www.journals.cms.math.ca/CRUX/[/url]", "Solution_4": "Another thing\r\nonly 96...99 is public and you can download it", "Solution_5": "Sometimes ago, I can take volume 9 of the year 1991, v10 is unavaillable, but now it's out. What's the matter?", "Solution_6": "[quote=Ali_aomt]Crux is a good problem solving journal by Canadian Mathematical Society\nand some issue of this journal is public and you can use free[/quote]\n\nI have some articles I want to recommend to Crux. What should I do?", "Solution_7": "[quote=Pham_Quoc_Sang][quote=Ali_aomt]Crux is a good problem solving journal by Canadian Mathematical Society\nand some issue of this journal is public and you can use free[/quote]\n\nI have some articles I want to recommend to Crux. What should I do?[/quote]\n\nOne can submit their articles on the website freely although read [url=https://cms.math.ca/publications/crux/information-for-contributors/]this[/url] first." } { "Tag": [ "search" ], "Problem": "Can anyone explain balls and urns to me? i believe that's what it's called. I see AoPSers use very often.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1904611533&t=145047[/url]\r\n\r\nThis should help :wink: ." } { "Tag": [ "inequalities" ], "Problem": "Let be $ a,b,c > 0$ . Prove that :\r\n\r\n$ \\frac {2a^3}{a^6 \\plus{} bc} \\plus{} \\frac {2b^3}{b^6 \\plus{} ca} \\plus{} \\frac {2c^3}{c^6 \\plus{} ab}\\le \\frac {a}{bc} \\plus{} \\frac {b}{ca} \\plus{} \\frac {c}{ab}$", "Solution_1": "[quote=\"alex2008\"]Let be $ a,b,c > 0$ . Prove that :\n\n$ \\frac {2a^3}{a^6 \\plus{} bc} \\plus{} \\frac {2b^3}{b^6 \\plus{} ca} \\plus{} \\frac {2c^3}{c^6 \\plus{} ab}\\le \\frac {a}{bc} \\plus{} \\frac {b}{ca} \\plus{} \\frac {c}{ab}$[/quote]\r\n$ a^6\\plus{}bc \\ge 2\\sqrt{a^6bc}\\equal{}2a^3\\sqrt{bc}$\r\nSo we have to prove that:\r\n$ \\sum\\sqrt{\\frac{1}{bc}} \\le \\frac{a^2\\plus{}b^2\\plus{}c^2}{abc}$\r\nSetting $ \\sqrt{a}\\equal{}x,\\sqrt{b}\\equal{}y,\\sqrt{c}\\equal{}z$,\r\nAnd ginally the inequality is:\r\n$ a^4\\plus{}b^4\\plus{}c^4 \\ge a^2bc\\plus{}ab^2c\\plus{}abc^2$\r\nWhich is true by Muirhead for the triples \r\n$ (4,0,0) \\succ (2,1,1)$" } { "Tag": [ "trigonometry", "geometry", "angle bisector" ], "Problem": "given angle xOy , one arbitrary circle with center I inside angle xOy . circle (I) has constant radius R . M on the circle , MH drawn perpendicular to Ox at H and MK drawn perpendicular to Oy at K ,\r\n find M where MH+MK min", "Solution_1": "Isn't the circle with centre I touching the arms of $ \\angle XOY$??\r\nIf so, let $ M$ be the point such that $ MH\\plus{}MK$ is minimum. That is,\r\n$ (\\sin \\angle HOM\\plus{}\\sin \\angle MOK)OM$ is minimum.\r\n$ \\Leftrightarrow (2\\sin \\frac {XOY}{2} \\cos \\frac {(\\angle HOM \\minus{} \\angle MOK)}{2})OM$ is minimum. But $ 2\\sin \\frac {XOY}{2}$ is constant.So, we have to consider when $ \\cos \\frac {(\\angle HOM \\minus{} \\angle MOK)}{2}OM$ is minimum. \r\nLet $ OM_1M_2$ be the angle bisector of $ \\angle XOY$ with $ M_1$ and $ M_2$ on the circle such that $ OM_2>OM_1$.\r\n$ M$ lies in the $ arc AM_1B$ where $ A,B$ are the points of contact of the arms of $ \\angle XOY$.\r\n$ \\cos \\frac {(\\angle HOM \\minus{} \\angle MOK)}{2}\\equal{}\\cos \\frac {2\\angle MOM_1}{2}\\equal{}\\cos \\angle MOM_1$\r\nBut, $ (\\cos \\angle MOM_1)OM > OM_1\\cos 0$ since $ \\angle OM_1M > 90$.\r\nBut this is contradictory Since $ (\\cos \\angle MOM_1)OM$ is minimum.\r\nIt follows that point $ M$ coincides with point $ M_1$.", "Solution_2": "the circle I does not touch the arms of xOy\r\nhere you can download the figure of this problem\r\nhttp://www.mediafire.com/?d1bjdnj1j4g" } { "Tag": [ "abstract algebra", "linear algebra", "linear algebra unsolved" ], "Problem": "Suppose that $ f: V\\rightarrow V$ is a linear transformation and $ f\\circ f\\equal{}f$ (where the dimension of $ V$ is finite). Prove that\r\n\r\n\\[ V\\equal{}\\textrm{Image}\\left(f\\right)\\plus{}\\textrm{ker}\\left(f\\right).\\]\r\nClearly, the RHS is a subset of the LHS. Also, I've shown that the intersection of the image of $ f$ and the kernel of $ f$ is the zero vector. How do we show that if $ v\\in V$, then $ v\\in\\textrm{Image}\\left(f\\right)\\plus{}\\textrm{ker}\\left(f\\right)$?", "Solution_1": "[quote=\"amcavoy\"]How do we show that if $ v\\in V$, then $ v\\in\\textrm{Image}\\left(f\\right) \\plus{} \\textrm{ker}\\left(f\\right)$?[/quote]\r\nWhat about $ v\\equal{}f(v) \\plus{} v \\minus{} f(v)$", "Solution_2": "Alternatively compare the dimensions of LHS and RHS using the dimension formula applied to the endomorphism f." } { "Tag": [], "Problem": "I don't get the grammar of this problem no matter how many times I read it. Could some one explain it to me?\r\nThe length of a median to the hypotenuse of an isosceles, right triangle is 10 units. What is the length of a leg of the triangle , in units? Express your answer in simplest radical form.\r\nThanks!", "Solution_1": "isosceles right triangle: a right triangle, meaning it has one right angle. since it's isosceles, the angles are 45-45-90, and the two legs are congruent.\r\n\r\nmedian: the line segment that connects one of the vertices to the middle of the opposite side\r\n\r\nhope this helps", "Solution_2": "Could you draw it for me please? :) I'm still confused about the median part...", "Solution_3": "The median is $ 10$, so half of the hypotenuse is $ 10$ also. The whole hypotenuse is $ 20$. The length of a side is then $ \\sqrt{10 \\times 20} \\equal{} \\sqrt{200} \\equal{} \\boxed{10\\sqrt2}$" } { "Tag": [ "calculus", "derivative", "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f$ be a function which is differentiable on $ [a.b]$. Is it always true that $ f'$ is integrable on $ [a,b]$?", "Solution_1": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=136158]here[/url]" } { "Tag": [ "floor function", "geometry", "3D geometry" ], "Problem": "How many subsets of the set $[1,2,3...29,30]$ have the sum of its elements greater than $232$ ?\r\n[hide=\"ACoPS hint\"]\nWhat is special about the number 232?\n[/hide]", "Solution_1": "[quote=\"Leao\"]How many subsets of the set $[1,2,3...29,30]$ have the sum of its elements greater than $232$ ?\n[hide=\"ACoPS hint\"]\nWhat is special about the number 232?\n[/hide][/quote]\n[hide]\nThe max sum of the elements in a subset is 465. $\\lfloor 465/2\\rfloor=232$ :) \n\nfind the number of subsets and divide by two...[/hide]", "Solution_2": "How do you know it's half and half?", "Solution_3": "For each subset with sum greater than 232, there is a corresponding subset with sum less than 232: its complement.", "Solution_4": "[quote=\"Bictor717\"]For each subset with sum greater than 232, there is a corresponding subset with sum less than 232: its complement.[/quote]\r\nIt should be \"For each subset with sum greater than 232, there is a corresponding subset with sum less [b]or equal[/b] than 232: its complement\".\r\nAnyway, my full solution would be:\r\nThe cardinal of the set $S\\equiv\\{1,2,\\dots,30\\}$ is $|S|=30$, so it has $2^{30}$ possible subsets (each element can appear or not in a subset, so there are two choices for every element).\r\nLet's call $s(A)$ the sum of the elements of a set $A$.\r\n$s(S)=1+2+\\dots+30=30\\times 31/2=465$.\r\nIf $S'$ is a subset of $S$ and $S''$ is the complementary subset $S''\\equiv S\\setminus S'$ , then $s(S')+s(S'')=s(S)=465$.\r\nSo, for every subset $S'$ with $s(S')\\leq 232$ the complementary subset $S''$ has $s(S'')\\geq 233$ and vice versa.\r\nTherefore, there are $2^{29}$ subsets with sum greater than $232$.", "Solution_5": "Just to avoid starting a new thread, here is another problem that I didn't quite understand, it also came from ACoPS:\r\nTen children order ice-cream cones at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor? \r\nThanks in advance for your help!\r\n[hide=\"my attempt\"]\nThe book's solution involved looking at the complement, and I understood that solution, but why doesn't mine work:\nYou choose 2 of the ten children at the start to get the same flavor, and this can be done in ${10\\choose 2}*31$ ways. Then, the remaining 8 children can get their ice creams in $31^{8}$ ways, so the total is $45*31^{9}$ which is totally wrong...[/hide]", "Solution_6": "What's the solution?\r\n\r\n[hide]I think it is $31^{9}$, because you just need the 10th kid to have the same flavor as any of the other kids.[/hide]", "Solution_7": "[quote=\"Leao\"]Just to avoid starting a new thread, here is another problem that I didn't quite understand, it also came from ACoPS:\nTen children order ice-cream cones at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor? \nThanks in advance for your help!\n[hide=\"my attempt\"]\nThe book's solution involved looking at the complement, and I understood that solution, but why doesn't mine work:\nYou choose 2 of the ten children at the start to get the same flavor, and this can be done in ${10\\choose 2}*31$ ways. Then, the remaining 8 children can get their ice creams in $31^{8}$ ways, so the total is $45*31^{9}$ which is totally wrong...[/hide][/quote]\r\n\r\nYou overcount.\r\n\r\nSay you choose two children at the beginning to get vanilla, and then two other children get vanilla, and the others get some other flavors. But you could have chosen the two other children to get vanilla at the beginning, and then the original two could have been assigned vanilla, etc.", "Solution_8": "If you don't want to count the complement, you need to use PIE. First suppose two children get the same flavor. This can happen in $\\binom{10}{2}31^{9}$ ways. As blahblahblah explained, we overcounted for situations in which more than two children got the same flavor.\r\n\r\nSuppose three children get the same flavor. This can happen in $\\binom{10}{3}31^{8}$ ways. Now we need to know how many times we counted each one of these possibilities in the previous case. This is just $\\binom{3}{2}=3$, because we counted each possibility once for each pair of people. Therefore, we need to subtract each possibility for this case twice.\r\n\r\nYou can eventually get the answer this way after considering 4 people, 5 people, etc., but you have to do a lot of calculations. The method using the complement is much easier.", "Solution_9": "[quote=\"Leao\"]Just to avoid starting a new thread, here is another problem that I didn't quite understand, it also came from ACoPS:\nTen children order ice-cream cones at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor? \nThanks in advance for your help!\n[hide=\"my attempt\"]\nThe book's solution involved looking at the complement, and I understood that solution, but why doesn't mine work:\nYou choose 2 of the ten children at the start to get the same flavor, and this can be done in ${10\\choose 2}*31$ ways. Then, the remaining 8 children can get their ice creams in $31^{8}$ ways, so the total is $45*31^{9}$ which is totally wrong...[/hide][/quote]\r\n\r\nTo avoid overcount, you have to separate the problem into two scenarios:\r\n1. more than 2 children got the same flavor\r\n2. all the children get the DIFFERENT flavor\r\n\r\nso the number of ways for more than 2 children get the same flavor is equal the number of ways for 10 children to select icecream minus the number of ways that all the children get the different flavor.\r\n\r\nnumber of ways for 10 children to select icecream=${31}^{10}$\r\nnumber of ways all the children get different flavor=$_{31}P_{10}$ or $31*30*29*28*27*26*25*24*23*22$\r\n\r\nTherefore, the answer should be $31^{10}-_{31}P_{10}$\r\n\r\nwhat you think?", "Solution_10": "[quote=\"ch1n353ch3s54a1l\"][quote=\"Leao\"]How many subsets of the set $[1,2,3...29,30]$ have the sum of its elements greater than $232$ ?\n[hide=\"ACoPS hint\"]\nWhat is special about the number 232?\n[/hide][/quote]\n[hide]\nThe max sum of the elements in a subset is 465. $\\lfloor 465/2\\rfloor=232$ :) \n\nfind the number of subsets and divide by two...[/hide][/quote]\r\nyou are right, however, you need to show that every number under 465 can be construct with the elements 1...30.\r\nor formally, prove that every number under 1+2+3+...+n can be construct with the number 1...n, using every number only once. (it isn't difficult as it seems...)", "Solution_11": "Going along with what Allan Z said, but I think this solves it for if it is asking for unique orders, as in one kids takes all the orders in a list and goes to place the order, aka assignment of flavors to child does not matter.\r\n\r\nThen the number of unique orders is $\\binom{10+31-1}{10}$\r\n\r\nSo the total number is $\\binom{10+31-1}{10}-\\binom{31}{10}$", "Solution_12": "[quote=\"srulikbd\"]you are right, however, you need to show that every number under 465 can be construct with the elements 1...30.\nor formally, prove that every number under 1+2+3+...+n can be construct with the number 1...n, using every number only once. (it isn't difficult as it seems...)[/quote]\r\n\r\n[hide]Use induction. The subset $\\{1\\}$ has sum 1. If some subset $A$ has sum $n<465$, then a subset $B$ with sum $n+1$ can be constructed as follows:\n\nIf $A$ does not contain 1, then take $B=A\\cup\\{1\\}$. If the subset $A$ does contain 1, then consider the least element of $\\{1,2,\\cdots ,30\\}$ that $A$ doesn't contain, where $1\\le k\\le 30$. (If $A$ contained all elements, its sum would be 465, a contradiction.) We get $B$ by replacing $k-1$ with $k$.[/hide]", "Solution_13": "[quote=\"breez\"]What's the solution?\n\n[hide]I think it is $31^{9}$, because you just need the 10th kid to have the same flavor as any of the other kids.[/hide][/quote]\r\n\r\nSolution (from ACoPS): We assume the children are distinguishable. First, we count all possible orders without restrictions, which is just $31{}^{1}{}^{0}$. Now we count orders where there is no duplication of flavor; this is just $31*30*29...*22 = P(31,10)$. The answer is then the difference $31{}^{1}{}^{0}-P(31,10).$", "Solution_14": "[quote=\"Leao\"][quote=\"breez\"]What's the solution?\n\n[hide]I think it is $31^{9}$, because you just need the 10th kid to have the same flavor as any of the other kids.[/hide][/quote]\n\nSolution (from ACoPS): We assume the children are distinguishable. First, we count all possible orders without restrictions, which is just $31{}^{1}{}^{0}$. Now we count orders where there is no duplication of flavor; this is just $31*30*29...*22 = P(31,10)$. The answer is then the difference $31{}^{1}{}^{0}-P(31,10).$[/quote]\r\n\r\n\r\nOh, so my solution is correct :P" } { "Tag": [ "inequalities", "quadratics", "inequalities proposed" ], "Problem": "($ x,y \\in R$)\r\n\r\n$ 3 (x\\plus{}y\\plus{}1)^2 \\plus{}1 \\ge 3xy$", "Solution_1": "hello, it is aequivalent to\r\n$ \\left(x\\plus{}1\\plus{}\\frac{y}{2}\\right)^2\\plus{}\\frac{3}{4}\\left(y\\plus{}\\frac{2}{3}\\right) \\ge 0$.\r\nSonnhard.", "Solution_2": "Consider the inequality as quadratic trinomial,and then just prove that the discriminant $ D\\leq 0$. :wink:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x\\geq0,$ $y\\geq0,$ $z\\geq0.$ Prove that\r\n$(x+y+z)^{8}\\geq243(xy+xz+yz)^{2}(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}).$", "Solution_1": "Let $a=\\frac{x+y+z}{3}, b=(\\frac{xy+yz+zx}{3})^{\\frac{1}{2}}, c=(xyz)^{\\frac{1}{3}}$, then the inequality is equivalent to $a^{8}+2ab^{4}c^{3}\\geq 3b^{8}$.\r\n\r\nAs $3a^{3}+c^{3}\\geq 4ab^{2}\\iff \\sum_{cyc}x^{3}+3xyz \\geq \\sum_{sym}x^{2}y$, we get $a^{8}+2ab^{4}c^{3}\\geq a^{8}+2ab^{4}(4ab^{2}-3a^{3})$\r\n$=a^{8}-6a^{4}b^{4}+8a^{2}b^{6}$.\r\nSo it is sufficient to prove that $a^{8}-6a^{4}b^{4}+8a^{2}b^{6}\\geq b^{8}$. This is equivalent to $(a^{2}-b^{2})^{3}(a^{2}+3b^{2}) \\geq 0$, which is true by $a \\geq b \\iff (x+y+z)^{2}\\geq 3(xy+yz+zx)$.", "Solution_2": "Cool! :ninja: Thank you very much. :)", "Solution_3": "The following is true and beautiful: \\[(a+b+c)^{6}\\geq32(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}).\\]", "Solution_4": "[quote=\"pvthuan\"]The following is true and beautiful: \\[(a+b+c)^{6}\\geq32(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}).\\] [/quote]\r\nBut SOS kills it (for $a,b,c$ non-negative) enough easy. :wink:", "Solution_5": "What is SOS?", "Solution_6": "[quote=\"SpongeBob\"]What is SOS?[/quote]\r\nSum Of Squares. See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127", "Solution_7": "[quote=\"arqady\"]\nBut SOS kills it (for $a,b,c$ non-negative) enough easy. :wink:[/quote]\r\n\r\nYes, arqady. But I solved it in one line without any expansion. And an even harder is \r\n\r\n\r\nLet $x,y,z$ be real nonegative numbers such that $x+y+z=2$, prove that \\[(x^{4}+y^{4}+z^{4})(xy+yz+zx)\\leq2.\\]", "Solution_8": "[quote=\"pvthuan\"]The following is true and beautiful: \\[(a+b+c)^{6}\\geq32(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}).\\] [/quote]\r\nCheck out this solution:\r\n$(a+b+c)^{4}=[(a^{2}+b^{2}+c^{2})+(2ab+2bc+2ca)]^{2}\\geq 8(a^{2}+b^{2}+c^{2})(ab+bc+ca)$\r\nSo all we need to prove is :\r\n$(a+b+c)^{2}(ab+bc+ca)\\geq 4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$\r\nBut $(a+b+c)^{2}(ab+bc+ca)=2(ab+bc+ca)^{2}+\\sum_{sym}a^{3}b+abc(a+b+c)\\geq$\r\n$4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$", "Solution_9": "That is right, Eagle. And the following is also true and solved \\[(x+y+z)^{6}\\geq 32(x^{4}+y^{4}+z^{4})(xy+yz+zx).\\] Another question is to find the greatest possible number $a$ such that the following inequality holds for all nonegative real numbers $x,y,z$ \\[(x+y+z)^{8}\\geq a(x^{5}+y^{5}+z^{5})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2}).\\]", "Solution_10": "[quote=\"pvthuan\"] And the following is also true and solved \\[(x+y+z)^{6}\\geq 32(x^{4}+y^{4}+z^{4})(xy+yz+zx).\\] [/quote]\r\nIt's wrong: $x=3$ and $y=z=1.$ :wink:" } { "Tag": [ "trigonometry", "algebra", "polynomial", "quadratics", "logarithms", "symmetry", "induction" ], "Problem": "If $ x\\plus{}\\dfrac{1}{x}\\equal{}\\sqrt{2}$, what is $ x^{2004}\\plus{}\\dfrac{1}{x^{2004}}$?", "Solution_1": "[hide=\"Answer\"]$ x \\plus{} \\frac {1}{x} \\equal{} \\sqrt {2} \\implies x^2 \\minus{} \\sqrt {2} x \\plus{} 1 \\equal{} 0 \\implies \\left( x \\minus{} \\frac {1}{\\sqrt {2}} \\right)^2 \\equal{} \\minus{} \\frac {1}{2}$\n\n$ \\therefore x \\equal{} \\frac {1}{\\sqrt {2}} \\pm \\frac {i}{\\sqrt {2}} \\equal{} \\text{cis} \\left( \\pm \\frac {\\pi}{4} \\right)$\n\n$ \\therefore x^{2004} \\plus{} x^{ \\minus{} 2004} \\equal{} \\text{cis} (501 \\pi) \\plus{} \\text{cis} ( \\minus{} 501 \\pi) \\equal{} ( \\minus{} 1) \\plus{} ( \\minus{} 1) \\equal{} \\boxed{\\minus{} 2}$[/hide]", "Solution_2": "[hide]\nSquaring and rearranging, we know \n$ x^2\\equal{}\\minus{}\\frac{1}{x^2} \\implies x^{2004}\\equal{}\\frac{1}{x^{2004}}$\n\nSo we only need to find $ 2x^{2004}$\n\n$ x^2\\equal{}\\minus{}\\frac{1}{x^2} \\implies x^4\\equal{}\\minus{}1$ if $ x\\not\\equal{}0$\n$ x^{2004}\\equal{}x^{4(501)}\\equal{}\\minus{}1$\n\nSo the answer is $ \\boxed{\\minus{}2}$\n\n[/hide]", "Solution_3": "[hide]Let $ x \\equal{} \\cos \\theta \\plus{} i\\sin \\theta$. Then $ x \\plus{} \\frac {1}{x} \\equal{} 2\\cos \\theta \\equal{} \\sqrt {2}$, so say $ \\theta \\equal{} \\frac {\\pi}{4}$. By [url=http://mathworld.wolfram.com/deMoivresIdentity.html]de Moivre's identity[/url],\n\\[ x^{2004} \\plus{} \\frac {1}{x^{2004}} \\equal{} 2 \\cos 2004\\theta \\equal{} \\minus{} 2.\n\\]\n[/hide]", "Solution_4": "[hide=\"Wow, these problems have tons of solutions\"]\nDefine a sequence $ a_n = x^n + \\frac {1}{x^n}$. Obviously, $ a_1 = \\sqrt2$ and $ a_0 = 2$. To come up with a recurrence relation, we have:\n\\begin{eqnarray*} \\sqrt2a_n & = & a_1a_n \\\\\n& = & \\left(x + \\frac {1}{x}\\right)\\left(x^n + \\frac {1}{x^n}\\right) \\\\\n& = & a_{n + 1} + a_{n - 1} \\\\\na_{n + 1} & = & \\sqrt2a_n - a_{n - 1} \\end{eqnarray*}\n\n$ a_{n + 2} - \\sqrt2a_{n + 1} + a_n = 0$\n\nThis recurrence relation's characteristic polynomial is $ x^2 - \\sqrt2x + 1$, whose roots are $ \\cos\\frac {\\pi}{4} + i\\sin\\pm\\frac {\\pi}{4}$. Plugging in, we have that the two coefficients of the closed form are both $ 1$, so $ a_n = 2\\cos\\frac {\\pi}{4}n$ by de Moivre. Therefore, $ a_n = 2\\cos(501\\pi) = \\boxed{ - 2}$.\n[/hide]", "Solution_5": "[quote=\"chess64\"][hide]Let $ x \\equal{} \\cos \\theta \\plus{} i\\sin \\theta$. Then $ x \\plus{} \\frac {1}{x} \\equal{} 2\\cos \\theta \\equal{} \\sqrt {2}$, so say $ \\theta \\equal{} \\frac {\\pi}{4}$. By [url=http://mathworld.wolfram.com/deMoivresIdentity.html]de Moivre's identity[/url],\n\\[ x^{2004} \\plus{} \\frac {1}{x^{2004}} \\equal{} 2 \\cos 2004\\theta \\equal{} \\minus{} 2.\n\\]\n[/hide][/quote]You can apply de Moivre's identity to the whole expression $ x^{2004} \\plus{} \\frac {1}{x^{2004}}$?", "Solution_6": "chess64 is skipping a few steps. \r\n\r\n[b]Lemma:[/b] If $ x \\plus{} \\frac {1}{x} \\equal{} 2 \\cos \\theta$ then $ x^n \\plus{} \\frac {1}{x^n} \\equal{} 2 \\cos n \\theta$.\r\n\r\nProof: Let $ x \\equal{} e^{i \\theta}$ and use de Moivre's. Note that $ \\theta$ can be complex, although in this case it isn't.", "Solution_7": "[quote=\"t0rajir0u\"]chess64 is skipping a few steps. \n\n[b]Lemma:[/b] If $ x \\plus{} \\frac {1}{x} \\equal{} 2 \\cos \\theta$ then $ x^n \\plus{} \\frac {1}{x^n} \\equal{} 2 \\cos n \\theta$.\n\nProof: Let $ x \\equal{} e^{i \\theta}$ and use de Moivre's. Note that $ \\theta$ can be complex, although in this case it isn't.[/quote]\r\n\r\nIt isn't that big a point, but I think that as long as we are trying to be rigorous: we don't just \"Let $ x \\equal{} e^{i \\theta}$\"...we have to show this (by using the quadratic formula). I also believe that this is how you would go about trying to solve the problem if you didn't already know the result.", "Solution_8": "[quote=\"Altheman\"]we don't just \"Let $ x \\equal{} e^{i \\theta}$\"...we have to show this (by using the quadratic formula). [/quote]\r\n\r\nWe do if we define $ \\cos \\theta \\equal{} \\frac{ e^{i \\theta} \\plus{} e^{\\minus{}i \\theta} }{2}$. :) Like I said, $ \\theta$ can be complex and the identity is still valid, so in that context it makes more sense to adopt this definition of the cosine.", "Solution_9": "I know that any real number can be written in that form. But you cannot start with one relation between r and theta, and assume another. Of course you can say that $ x \\equal{} e^{i\\gamma}$, but how do you know $ \\theta \\equal{} \\gamma$ or for that matter that only one such $ \\theta$ exists (well thats not true anyway).", "Solution_10": "[quote=\"t0rajir0u\"]chess64 is skipping a few steps. \n\n[b]Lemma:[/b] If $ x \\plus{} \\frac {1}{x} \\equal{} 2 \\cos \\theta$ then $ x^n \\plus{} \\frac {1}{x^n} \\equal{} 2 \\cos n \\theta$.\n\nProof: Let $ x \\equal{} e^{i \\theta}$ and use de Moivre's. Note that $ \\theta$ can be complex, although in this case it isn't.[/quote]\r\n\r\nI have to double check on this but this looks like an old AHSME problem (back in long time ago...).", "Solution_11": "[quote=\"Altheman\"]Of course you can say that $ x \\equal{} e^{i\\gamma}$, but how do you know $ \\theta \\equal{} \\gamma$ or for that matter that only one such $ \\theta$ exists (well thats not true anyway).[/quote]\r\n\r\nTrue. The complex logarithm is multivalued. The quadratic method is fine as far as rigor, then.", "Solution_12": "Again, it isn't that big a point and it is probably spam to post all these posts.\r\n\r\nBut another thing to note is that there are two principal values of $ \\theta$. If we were say, trying to find the value of $ x^{10}$, we would not get a unique answer. But the symmetry of $ x^{10} \\plus{} x^{ \\minus{} 10}$ in this case makes it so there is only one answer.", "Solution_13": "Alternately, a fairly elaborate proof would be to show by induction that the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomials of the first kind[/url] describe the behavior of both functions, but that would only prove for positive integral $ n$.", "Solution_14": "[quote=\"t0rajir0u\"]chess64 is skipping a few steps. \n\n[b]Lemma:[/b] If $ x \\plus{} \\frac {1}{x} \\equal{} 2 \\cos \\theta$ then $ x^n \\plus{} \\frac {1}{x^n} \\equal{} 2 \\cos n \\theta$.\n\nProof: Let $ x \\equal{} e^{i \\theta}$ and use de Moivre's. Note that $ \\theta$ can be complex, although in this case it isn't.[/quote]Could you explain this thoroughly? Sorry I'm not understanding.", "Solution_15": "Recall the identity $ e^{i \\theta} \\equal{} \\cos \\theta \\plus{} i \\sin \\theta$. Negating $ \\theta$ gives $ e^{ \\minus{} i \\theta} \\equal{} \\cos \\theta \\minus{} i \\sin \\theta$. Adding the two yields $ e^{i \\theta} \\plus{} e^{ \\minus{} i \\theta} \\equal{} 2 \\cos \\theta$\r\n\r\nIf $ x \\plus{} \\frac {1}{x} \\equal{} 2 \\cos \\theta$ for some $ \\theta$, then $ x \\equal{} e^{\\pm i \\theta}$ by the identity above. Since this provides two solutions and the original equation was a quadratic, these are the only values of $ x$. Let's just use $ x \\equal{} e^{i \\theta}$, since the answer will be the same.\r\n\r\n$ x^n \\equal{} (e^{i \\theta})^n \\equal{} e^{i(n \\theta)}$, and similarly $ x^{ \\minus{} n} \\equal{} e^{ \\minus{} i (n \\theta)}$\r\n\r\nAdd to get $ x^n \\plus{} x^{ \\minus{} n} \\equal{} e^{i(n \\theta)} \\plus{} e^{ \\minus{} i (n \\theta)}$\r\n\r\nUsing the cosine formula we derived, $ x^n \\plus{} x^{ \\minus{} n} \\equal{} 2 \\cos n \\theta$", "Solution_16": "Thank you; I understand now.", "Solution_17": "[hide=\"very very sweet solution\"]\nlet $ a_n \\equal{} x^n\\plus{}\\frac 1{x^n}$\nnotice that $ a_2\\equal{}0$ and that $ a_2a_m\\equal{}a_{m\\plus{}2}\\plus{}a_{m\\minus{}2}$\nthis implies that $ a_n\\equal{}a_{n\\plus{}8}$, so $ a_{2004}\\equal{}a_4\\equal{}\\minus{}2$\n[/hide]", "Solution_18": "Yeah, you just used recurrence, but I like Zuton's solution the best. :D", "Solution_19": "I liked RisingStar's solution because I thought it was the most elementary" } { "Tag": [ "calculus", "integration", "induction", "calculus computations" ], "Problem": "Please excuse me of my naivety (if you will) as I am in only my first quarter of calculus. I would be so grateful if you could help me on all (any) of the following problems; for I am preparing for a final exam that will occur this comming Thursday. Anyway, the \"questions\" are...\r\n\r\n1.) evaluate the SUM {n=0->infinity} of 1/[n! (n+2)], I evaluated SUM {n=o->5} which indicates that the sum is ~1, but I'm not sure how to irrefutably demonstrate this via \"Power Series\" (or what have you).\r\n\r\n2.) Prove: for all n an element of positive integers, that the integral {o->infinity} of (x^n)*(e^-x)dx = n!\r\n\r\n3.) Please confirm that the \"Power Series Representation\" for xe^x is the SUM {n=0->infinity} of [x^(n+1)]/n!. If it isn't then please help me to discover it's true PSR (...over all real numbers).\r\n\r\nThnx,\r\nChris", "Solution_1": "#1: Write the term as $\\frac{n+1}{(n+2)!}$. Turn it into a telescoping series.\r\nThe sum is exactly 1.\r\n\r\n#2: Induction. The induction step is integration by parts.\r\n\r\n#3: Yes, that's it.", "Solution_2": "Thx for your hasty response jmerry, that helps a lot.", "Solution_3": "One more question; I don't necessarily want you to solve this for me (but at least give me a few hints):\r\n\r\nProblem:\r\n\r\n Let B be the set of all binary sequences. Prove or disprove: there is a sequence whose image (range) is B.\r\n\r\n------------------------------------------------------------------------------------------------------------\r\n\r\nnote: \u00ce denotes \u201can element of\u2026\u201d\r\n\r\n\r\n\r\nMy Attempt to Prove (disprove):\r\n\r\nAssume a sequence (call it \u201cg\u201d) such that the \u201cImage of g\u201d equals B: I g = B,\r\nThen for each n \u00ce Z+ (Positive Integers), g(n) \u00ce B.\r\n\r\nTherefore, for each k \u00ce Z+, g(k) = 0 or g(k) = 1\r\n\r\n\r\nLet N be a subset of Z+, then\r\n\r\nFor any N, let g(k) = 1, if k is in the subset \r\n g(k) = 0, if k is not in the subset\r\n\r\nTherefore g(k) has image B, and is an infinite \u201cbinary sequence\u201d.\r\n\r\n\r\nTherefore, B (set of all binary sequences) has one-to-one correspondence with the set of all subsets of the positive integers (N)\u2026 or natural numbers.\r\n\r\n------------------------------------------------------------------------------------------------------------\r\n\r\nI don't think that this satisfies the conditions of the proof. I'm pretty ignorant when it comes to sequences and proofs for that matter (if you haven't already deduced that); but I think I actually established (in this proof) that there's a binary sequence that has image B, where B = the set {0,1}, not the \"set of all binary sequences\" like it is supposed to be. Can someone plz help me out here? I'm in desperate need for help...", "Solution_4": "The 1-1 correspondence between infinite binary sequences and subsets of the natural numbers is easy- we map a sequence to the set of indices at which the sequence is 1.\r\n\r\nThis isn't the question. We want to know whether there is a surjective map from the natural numbers to the infinite binary sequences.\r\n\r\n[hide=\"Hint\"]The answer is no. Given a list of sequences, construct a sequence that isn't in the list. This is known as Cantor's diagonal argument.[/hide]", "Solution_5": "thnx yet again jmerry,\r\n\r\nnow you are positive about this right?\r\n\r\nDon't worry, I'm joking...no...wait, no I'm not. I don't understand how my teacher could expect me to use a method in which he never even took the time to mention during his lectures. Can this be interpreted another way (or is my teacher just evil)? :? \r\n\r\nMy intention isn't to mock you or anything like that, it's just that I want to be 100% sure that I'm being guided in the right direction. I cannot think of a worse scenario than putting this much time into a proof, having had the correct answer, then erasing it, only to replace it with a proof that I (quite frankly) have to decipher like a secret code and then try to put into the correct context (I'm about to go insane). \r\n\r\nAgain, I'm sry about my level ignorance relative to the average person that posts here.", "Solution_6": "It's a famous theorem and a famous proof. The proof is worth knowing. It doesn't belong in a calculus class.\r\n\r\nThe problem doesn't come out clearly in the language given, but I'm pretty sure it is what I said.\r\n\r\n[quote=\"Calc_quandary\"]Again, I'm sry about my level ignorance relative to the average person that posts here.[/quote]\r\nThere's nothing wrong with ignorance. You're trying to learn.\r\n\r\nMy responses in this thread have been short largely because I'd rather guide you toward the right idea than simply give the answers." } { "Tag": [ "modular arithmetic" ], "Problem": "Source: myself\r\n\r\n$f_n = f_{n-1} + f_{n+1}$\r\nIf $f_0 = 42$ and $f_{1109} = 858$, then compute $f_{2006}$.", "Solution_1": "[hide]\n$f_{n+1} = f_n - f_{n-1}$\n$f_{n+1} = (f_{n-1} - f_{n-2}) - f_{n-1} = -f_{n-2}$\n$f_n = - f_{n-3}$\n\n$1109 + 3(299) = 2006 \\implies f_{1109} = -f_{2006}$\n$f_{2006} = -858$\n[/hide]", "Solution_2": "That is correct.\r\n(after realizing I can't divide by 6)", "Solution_3": "[quote=\"LordoftheMorons\"]Source: myself\n\n$f_n = f_{n-1} + f_{n+1}$\nIf $f_0 = 42$ and $f_{1109} = 858$, then compute $f_{2006}$.[/quote]\r\n\r\n[hide]$f_{n+1} = f_{n} - f_{n-1}$ \n$f_{n+2} = f_{n+1} - f_{n}$ \n$f_{n+2} + f{n+1} = f_{n+1} - f_{n-1}$ \n$f_{n+2} = - f_{n-1}$, so\n$f_{n} = f_{n+6}$\n\nTherefore:\n\n$f_{1112} = -858$\n\nAnd since 1112 and 2006 are both 2 mod 6,\n\n$f_{2006} = \\fbox{-858}$.[/hide]", "Solution_4": "How about this one:\r\nStill using the equation $f_n = f_{n-1} + f_{n+1}$\r\nFind the sum of the first $10^n - 4$ terms in terms of $f_0, f_1, f_2,$ and $n$.\r\n(No terms defined)", "Solution_5": "Let $f_0 = a, f_1 = b, f_2 = c$. Then the series alternates $a, b, c, -a, -b, -c, \\cdots$ as shown before. Therefore, the sum of every next 6 is 0. So we merely have to find $10^n - 4 \\pmod 6$. \r\nWe have $4^n - 4 \\pmod 6$. Clearly this is $0 \\pmod 2$, and we see it is equal to $1^n - 1 \\equiv 0 \\pmod 3$, which implies that it is always $0 \\pmod 6$. Therefore, the sum is always $0$.", "Solution_6": "the sum is always $\\boxed{0}$" } { "Tag": [ "geometry", "circumcircle", "inequalities unsolved", "inequalities" ], "Problem": "Let $ ABC$ be a triangle such that $ \\frac{a}{c\\minus{}a}\\equal{}\\frac{c\\plus{}a}{b}$. Find $ \\angle A: \\angle C$.", "Solution_1": "[quote=\"johan.gunardi\"]Let $ ABC$ be a triangle such that $ \\frac {a}{c \\minus{} a} \\equal{} \\frac {c \\plus{} a}{b}$. Find $ \\angle A: \\angle C$.[/quote]\r\n\r\n$ ab\\equal{}c^{2}\\minus{}a^{2}$ --> $ 4R^{2}sinAsinB\\equal{}4R^{2}(sin^{2}C\\minus{}sin^{2}A)$ ----> $ sinAsinB\\equal{}\\frac{1\\minus{}cos2C}{2}\\minus{}\\frac{1\\minus{}cos2A}{2}\\equal{}\\frac{1}{2}(cos2C\\minus{}cos2A)\\equal{}\\frac{1}{2}*2sin(C\\plus{}A)sin(C\\minus{}A)\\equal{}sin(\\pi\\minus{}B)sin(C\\minus{}A)$ ---> $ sinAsinB\\equal{}sinBsin(C\\minus{}A)$---> $ sinB(sinA\\minus{}sin(C\\minus{}A))\\equal{}2sinBsin(\\frac{C}{2})sin(\\frac{2A\\minus{}C}{2})\\equal{}0$\r\n\r\n$ sinB\\equal{}0$ -----> $ B\\equal{}\\pi k$\r\n$ sin(\\frac{C}{2})\\equal{}0$ ---> $ \\frac{C}{2}\\equal{}\\pi n$ ---> $ C\\equal{}2\\pi n$\r\n$ sin(\\frac{2A\\minus{}C}{2})\\equal{}0$ ----> $ \\frac{2A\\minus{}C}{2}\\equal{}\\pi m$ ----> $ 2A\\minus{}C\\equal{}2\\pi m$ ---> $ 2A\\minus{}C\\equal{}0$ ----> $ \\angle A: \\angle C\\equal{}2$", "Solution_2": "The condition is equivalent to $ c^2 \\equal{} a(a \\plus{} b)$.\r\n\r\nDefine a point D on the ray $ \\vec{BC}$ such that CD = b, and B and D are on the opposite side to C.\r\n\r\nThen, from the condition we obtain that the circumcircle of $ \\triangle ACD$ is tangent to $ AB$.\r\nThis implies that $ \\angle A \\equal{} \\angle CBA \\equal{} \\frac {\\angle C}{2}$.", "Solution_3": "[quote=\"johan.gunardi\"]Let $ ABC$ be a triangle such that $ \\frac {a}{c \\minus{} a} \\equal{} \\frac {c \\plus{} a}{b}$. Find $ \\angle A: \\angle C$.[/quote]\r\n$ \\frac {a}{c \\minus{} a} \\equal{} \\frac {c \\plus{} a}{b}$ deduces $ \\frac {sinA}{sinC \\minus{} sinA} \\equal{} \\frac {sinC \\plus{} sinA}{sin(C\\plus{}A)}$. Thus: $ \\frac {sinA}{2cos\\frac{C\\plus{}A}{2}sin\\frac{C\\minus{}A}{2}} \\equal{} \\frac {2sin\\frac{C\\plus{}A}{2}cos\\frac{C\\minus{}A}{2}}{2sin\\frac{C\\plus{}A}{2}cos\\frac{C\\plus{}A}{2}}$\r\nTherefore: $ sinA \\equal{} 2sin\\frac{C\\minus{}A}{2}cos\\frac{C\\minus{}A}{2} \\equal{} sin(C\\minus{}A)$. Since $ 0 < C < 180$ we deduce $ {A \\equal{} C\\minus{}A}$ --> $ \\frac{C}{A} \\equal{} 2$" } { "Tag": [ "inequalities" ], "Problem": "If $x,y,z \\in R$ then:\r\n\r\n$ min ( (x-y)^2, (y-z)^2, (z-x)^2 ) \\leq \\frac{ \\sum x^2 }{2} $. Generalize it!\r\n\r\ncheers! :D :D", "Solution_1": "[quote=\"Lagrangia\"]If x,y,z are real numbers then:\n\nmin { (x-y)^2, (y-z)^2, (x-z)^2 } <=(sum(x^2))/2. Generalize it!\n\ncheers! :D :D[/quote]\r\n\r\nthus x²+y²-2xy or y²+z²-2yz or z²+x²-2xz\r\nto be compared to x²/2+y²/2+z²/2\r\n\r\nsuppose the first one smallest (parallel for the others)\r\nthen x+z \\leq 2y and y+z \\leq 2x thus\r\nx²/2+y²/2-2xy \\leq z²/2 ?\r\nz²/2 \\geq x²/2+y²/2+xy so yes this is true; \r\nxy is larger than -2xy, if x and y have the same sign. \r\n\r\nIn alike ways, one can prove the inequality starting from the fact that x should not have the same sign as y.", "Solution_2": "and it seems the pows aren't working...\r\n\r\n\u00b2 is supposed to be ^2", "Solution_3": "and now it does??? strange stuff :?" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Help me to solve the integral, thanks\r\n \r\n$ \\int\\limits_{\\minus{}\\frac{b}{2a}}^{\\plus{}\\infty }{\\frac{{{x}^{m}}.dx}{{{\\left( a{{x}^{2}}\\plus{}bx\\plus{}c \\right)}^{n}}}}$", "Solution_1": "hello, what do we know about the numbers $ m$ and $ n$?\r\nSonnhard.", "Solution_2": "I found these formulas on a book of mine:\r\n\r\nIf $ m\\ne 2n-1$ then \r\n\r\n\\[ \\int \\frac{x^m}{(ax^2+bx+c)^n}dx=-\\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}}-\\frac{(n-m)b}{(2n-m-1)a}\\int \\frac{x^{m-1}dx}{(ax^2+bx+c)^n}+\\frac{(m-1)c}{(2n-m-1)a} \\int \\frac{x^{m-2}dx}{(ax^2+bx+c)^n}.\\]\r\n\r\nand\r\n\\[{ \\int \\frac{x^m}{(ax^2+bx+c)^n}dx=\\frac{1}{a}\\int\\frac{x^{m-2}dx}{(ax^2+bx+c)^{n-1}}-\\frac{b}{a}\\int\\frac{x^{m-1}dx}{(ax^2+bx+c)^{n}}-}\\frac{c}{a}\\int\\frac{x^{m-2}dx}{(ax^2+bx+c)^{n}}.\\]", "Solution_3": "Please help me, I have solved this integral, but I want to check my result." } { "Tag": [ "symmetry", "linear algebra", "matrix", "probability", "expected value" ], "Problem": "There are three points on a circle: $A, B, \\text{ and }C$. A bug starts on $A$. At any moment, he may move to either point that he is not on. If he ever goes $ABCA$, he is squashed where he stands. What is the expected waiting time for his death?", "Solution_1": "So that the thread doesn't die...[hide=\"Hint\"]Let $A$, $B$, $C$, and $DEAD$ be states. [hide=\"Hint 2\"]Use Markov chains. [hide=\"Hint 3\"]Add the states $B'$ and $C'$ [/hide][/hide][/hide]", "Solution_2": "a solution that doesn't involve Markov chains\r\n\r\n[hide]\n\nLet $E_{A}$, $E_{B}$, $E_{C}$ be the expected values if the bug starts at $A$, $B$ or $C$, respectively. Let $E'_{B}$ be the expected value after the bug travels $AB$. After 3 moves there is a 1/8 chance the bug is dead, 1/8 chance the bug is back at $A$, a 1/8 chance the bug is at $B$ without previously visiting $A$, a 1/8 chance the bug is at $C$ without previously visiting $A$ a 1/4 chance the bug ends visiting $A$ then $B$, and a 1/4 chance the bug ends visiting $A$ then $C$. Hence, $E_{A}= \\frac18*3+\\frac18*(E_{A}+3)+\\frac18*(E_{B}+3)+\\frac38*(E_{C}+3)+\\frac14*(E'_{B}+3) = \\frac18(E_{A}+E_{B}+3E_{C}+2E'_{B}+24)$. \n\nNow if the bug starts at $B$ it either moves to $A$ or $C$, so it's expected value at $B$ is $E_{B}= \\frac12(E_{A}+1)+\\frac12(E_{C}+1)$, and similarly $E_{C}= \\frac12(E_{B}+1)+\\frac12(E_{A}+1)$. Adding these last two equations together yields $E_{B}+E_{C}= \\frac12(E_{B}+E_{C})+E_{A}+2$. Therefore $E_{B}+E_{C}= 2E_{A}+4$. \n\nBy the symmetry of the solution, we can conclude that $E_{B}= E_{C}= E_{A}+2$.\n\nIf the bug travels $AB$ then it has a 1/4 chance of dieing by traveling $BCA$, a 1/2 chance of ending at $B$ otherwise by traveling $BCB$ or $BAB$, and a 1/4 chance of ending at $C$ with $BAC$, so $E'_{B}= \\frac14*2+\\frac14(E_{C}+2)+\\frac14(E_{B}+2)+\\frac14(E'_{B}+2)$. Rearranging $E'B = \\frac23+\\frac13(E_{C}+E_{B}) = 4+\\frac23E_{A}$.\n\n\nPlugging these expressions into the expression for $E_{A}$ and solving for $E_{A}$ gives us $E_{A}= 24$. Hence it should take 24 moments for the bug to be squashed. \n\n[/hide]\r\n\r\nEdit: Fixed some major flaws.", "Solution_3": "[quote=\"calculuslover800\"][hide=\"Answer\"]$24$[/hide][/quote]I got a different answer.[hide=\"My answer\"]$17$[/hide]Which is correct?", "Solution_4": "[hide=\"I got the same answer as calculuslover\"]If the bug is both not on $A$ and has deviated from the $ABCA$ sequence, it will take him an expected two moves to get back to $A$ by a simple geometric sum.\n\nNow suppose the bug is on $A$ beginning a possible $ABCA$. He has a 1/2 chance of moving to $C$, and that adds 3 expected moves: one to move to $C$, expected two to get back to $A$.\n\nThe alternative is that he moved to $B$ with 1/2 chance. Now he has a 1/2 (1/4 cumulative) chance of moving back to $A$, adding 2 moves ($ABA$). Otherwise, he's still going.\n\nNow he's on $C$. There's a 1/2 (1/8 cumulative) chance of moving to $B$ and adding 5 moves ($ABCB$ plus the two moves to return to $A$). Else, he moves to $A$, gets squished, and finishes with 3 moves ($ABCA$).\n\nSo we have the following possibilities:\n$AC$: +3 moves (1/2 = 4/8 chance)\n$ABA$: +2 moves (1/4 = 2/8 chance)\n$ABCB$: +5 moves (1/8 chance)\n$ABCA$: +3 moves and finish (1/8 chance)\n\nSo the expected value is $3 \\cdot 4+2 \\cdot 2+5+3 = \\fbox{24}$[/hide]", "Solution_5": "[hide=\"My method\"]\nLet there be $6$ states: $A, B, C, D, B', C'$. $D$ represents dead, $B'$ represents $B$ coming from $A$, $C'$ represents $C$ coming from $B'$. $A$ can go to $B'$ or $C$, $B$ to $A$ or $C$, $C$ to $A$ or $B$, $B'$ to $A$ or $C'$, $C'$ to $D$ or $B$, and $D$ to $D$. Make a stochastic matrix with these states and multiply it by itself repeatedly and see how many steps it takes for $AD$ to be $> \\frac{1}{2}$. The answer I got using a calculator was $17$.\n[/hide]", "Solution_6": "miyomiyo, the thing you've cacluated isn't the expected value. If we have a random variable $X$ from which we take repeated samples and a possible outcome $x_{0}$, the expected number of samples necessary to get $x_{0}$ is not necessarily the same as the number of samples for which the probability of getting $x_{0}$ exceeds $\\frac{1}{2}$.\r\n\r\n24 is correct, incidentally." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "1) Prove that for $x,y,z>0$ then\r\n\\[ 2\\leq\\dfrac{x^2+y^2}{x^2+yz+xy}+\\dfrac{y^2+z^2}{y^2+xz+yz}+\\dfrac{z^2+x^2}{z^2+yx+zx}. \\]\r\n\r\n2) Prove that if $x,y,z\\geq0$ then\r\n\\[ \\dfrac{x^2+yz}{x^2+y^2}+\\dfrac{y^2+zx}{y^2+z^2}+\\dfrac{z^2+xy}{z^2+x^2}\\geq\\dfrac52. \\]", "Solution_1": "Part one: \r\n\\[ \\dfrac{x^2+y^2}{x^2+xy+y^2} \\geq \\dfrac{2}{3} \\]", "Solution_2": "your second ineq is not true\r\ntake z=0 , x=0.5,y=0.25" } { "Tag": [ "trigonometry" ], "Problem": "In right $\\triangle ABC$ with right $\\angle C$, point $D$ is on hypotenuse $\\overline{AB}$ such that $\\angle DAC \\cong \\angle DCA$. Given that $AC=7$ and $BC=24$, find $CD$.", "Solution_1": "[quote=\"lingomaniac88\"]In right $\\triangle ABC$ with right $\\angle C$, point $D$ is on hypotenuse $\\overline{AB}$ such that $\\angle DAC \\cong \\angle DCA$. Given that $AC=7$ and $BC=24$, find $CD$.[/quote]\r\n\r\n[hide]$CD$ is a median drawn to the hypotenuse of a right triangle and is therefore $\\frac{\\sqrt{7^{2}+24^{2}}}{2}=\\boxed{12.5}$[/hide]", "Solution_2": "[quote=\"lingomaniac88\"]In right $\\triangle ABC$ with right $\\angle C$, point $D$ is on hypotenuse $\\overline{AB}$ such that $\\angle DAC \\cong \\angle DCA$. Given that $AC=7$ and $BC=24$, find $CD$.[/quote]\r\n[hide]${\\angle A = \\tan^{-1}(\\frac{24}{7}}) = 73.7^\\circ$ \n\n$\\frac{CD}{\\sin(A)}= \\frac{AC}{\\sin(180-2A)}$\n\n$CD = \\boxed{12.5}$[/hide]", "Solution_3": "[hide]12.5[/hide]", "Solution_4": "[hide]This is just a median of the right triangle.\n\nSo since it is to the hypotenuse it is 12.5\n\n$12.5$[/hide]" } { "Tag": [ "geometry", "rectangle", "conics", "parabola" ], "Problem": "A variable rectangle [b]PQRS [/b]has its sides parallel to fixed directions. [b]Q[/b] and [b]S [/b]lie respectively on the line [b]x=b,x=-a [/b]and [b]P [/b]lies on the[b] x-axis.[/b] Then the locus of R is\r\n[b](a) a straight line\n(b) a circle\n(c) a parabola \n(d) none of these.\n[/b]\r\nThe answer is (a). Please explain the problem [u]if possible with figure[/u].Thanks for your help a lot.", "Solution_1": "you pretty much have a rectangle only defined by three lines, where the points touch on every line except for one. We want to find the possible value for the fourth point. Let's consider the rectangle as a square, so P is between a and b. D is right above point P. As the square is warped, the rectangle becomes elongated and point D is the opposite corner of the rectangle. When P is infinitely close to either a or b, then D will be exactly at the other side. Our second condition assures us that the height and width are held by the three lines, which only one possible height from D to the x-axis. Thus, the locus of D is a line.\r\n\r\nI hope my solution is of help, and I sincerely apologize for not being able to provide a diagram.", "Solution_2": "[quote=\"AMC 14\"]you pretty much have a rectangle only defined by three lines, where the points touch on every line except for one. We want to find the possible value for the fourth point. Let's consider the rectangle as a square, so P is between a and b. D is right above point P. As the square is warped, the rectangle becomes elongated and point D is the opposite corner of the rectangle. When P is infinitely close to either a or b, then D will be exactly at the other side. Our second condition assures us that the height and width are held by the three lines, which only one possible height from D to the x-axis. Thus, the locus of D is a line.\n\nI hope my solution is of help, and I sincerely apologize for not being able to provide a diagram.[/quote]\r\n\r\nAMC14 I can't understand your proof. Could you be more clear . :blush:" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Let $ \\sigma \\in S_n$ and $ \\alpha <2$. Evaluate$ \\displaystyle\\lim_{n\\to\\infty} \\displaystyle\\sum_{k\\equal{}1}^{n}\\frac{\\sigma (k)}{k^{\\alpha}}$.", "Solution_1": "$ A(n)\\equal{}f(1)\\plus{}...\\plus{}f(n)\\geq (n^ 2)/2$\r\n\r\nwhere f is in $ S_n$\r\n\r\n\r\ndenote u(n) the general term of the serie\r\nwith Abel tranformation \r\n\r\n$ u(n)\\geq \\sum_{k\\equal{}1}^{n\\minus{}1}\\frac{k^2}{2}(\\frac{1}{k^a}\\minus{}\\frac{1}{(k\\plus{}1)^a})$\r\n\r\nthe general in RHS is equivalent to $ \\frac{C}{k^{a\\minus{}1}}$ but a<2 it diverge \r\n\r\nthe require limit is +oo" } { "Tag": [ "geometry", "rectangle" ], "Problem": "Can someone explain to me how Carnot's Engine and the Second Law of Thermodynamics are linked. \r\n\r\nThanks.", "Solution_1": "[quote=\"BanishedTraitor\"]Can someone explain to me how Carnot's Engine and the Second Law of Thermodynamics are linked. \n\nThanks.[/quote]\r\n\r\nCarnot's engine is the most efficient engine possible, even in theory. On a S vs. T graph the Carnot cycle is a rectangle whose sides are two isothermal lines, and two adiabatic (constant entropy) lines. The net change in entropy for a Carnot cycle is zero. The efficiency of a Carnot engine is given by: $ \\eta \\equal{} 1 \\minus{} \\frac{T_C}{T_H}$, where $ T_C$ is the absolute temperature of the low temperature isotherm, and likewise $ T_H$ is the absolute temperature of the high temperature isotherm. Note: Even this efficiency is less than 100%, and it only occurs for an engine with the minimal increase in entropy, i.e. zero.\r\n\r\nThe Second Law of Thermodynamics implies that for a closed system, one where particles can neither enter nor leave, of which an engine is a primary example, entropy cannot decrease. Entropy is a measure of disorder. Increased entropy represents energy that has gone into making the system more chaotic rather than doing work. Thus increased entropy means decreased efficiency.", "Solution_2": "[quote=\"BanishedTraitor\"]Can someone explain to me how Carnot's Engine and the Second Law of Thermodynamics are linked.[/quote]\r\n\r\nStart by answering the following two questions:\r\n\r\n1) What is a Carnot's cycle?\r\n\r\n2) What statement of the second law are you considering?\r\n\r\nAnd don't forget that we are discussing Thermodynamics, so microscopic concepts like \"molecule\" or \"particle\" are not needed." } { "Tag": [ "probability", "combinatorics unsolved", "combinatorics" ], "Problem": "On a small island there are 25 inhabitants. One of these inhabitants starts a rumor which spreads around the isle. Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before. At that point, the person stops spreading it. What is the probability that all 25 inhabitants will eventually hear the rumor?\r\n\r\nGeneralize to any number of inhabitants.", "Solution_1": "There needs to be some model of how the people interact -- what's the probability space we're working over?", "Solution_2": "I guess, the most natural assumption is that people only meet in pairs, and at any point in time, the next \r\npair to meet is uniformly chosen. However, calculations with small numbers of inhabitants suggest\r\nthat in this case the answer will not be very nice." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Let ABC be a triangle and AD be the bisector of ABC. Denote I, J be the projections of D on AB, AC. Let U, V be respectively intersections of DI, DJ and (O) such that they lie on the small arcs AB and AC. Let K be the intersection of UJ and VI. Prove or disprove AK is perpendicular to BC. :maybe:", "Solution_1": "I drew a dynamic sketch, and it clearly appears not to be true.", "Solution_2": "Oh really! Poor me! :(" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Prove that if $4^n+2^n+1$ is a prime then $n$ is a power of $3$.", "Solution_1": "I'm sure it was posted at least two times before.\r\n\r\nOK I found them:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=69143\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=69141 (this one contains a lot of other links)\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=33850\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=27620\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=1489\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=291", "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=291\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=1489\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=27620 \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=33850\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=47701\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=69143\r\n\r\n :D\r\n\r\n :( Once more, I'm too slow... :( ;)", "Solution_3": "Thanks - I should have used search button in a first place..." } { "Tag": [ "search", "function" ], "Problem": "find all $ a$ and $ b$ in $ N$ such \r\n$ a^2\\plus{}a\\minus{}7\\equal{}b^2$", "Solution_1": "[quote=\"greatestmaths\"]find all $ a$ and $ b$ in $ N$ such \n$ a^2 \\plus{} a \\minus{} 7 \\equal{} b^2$[/quote]\r\n\r\nHint\r\n\r\n[hide]Multiply by 4 and find some way of gettin squares.[/hide]", "Solution_2": "[hide]$ 4a^2 + 4a + 1 - 29 = 4b^2$\n$ (2a + 1)^2 - (2b)^2 = 29$\n$ (2a + 2b + 1)(2a - 2b + 1) = 29$\n\n$ \\left \\{ \\begin{array}{c} 2a+2b+1=29 & 2a-2b+1=1 \\end{array} \\right \\; \\implies (a,b)= (7,7)$[/hide]", "Solution_3": "[quote=\"TZF\"][hide]$ 4a^2 \\plus{} 4a \\plus{} 1 \\minus{} 29 \\equal{} 4b^2$\n$ (2a \\plus{} 1)^2 \\minus{} (2b)^2 \\equal{} 29$\n$ (2a \\plus{} 2b \\plus{} 1)(2a \\minus{} 2b \\plus{} 1) \\equal{} 29$\n\n$ ( \\minus{} 8,7); \\; ( \\minus{} 8, \\minus{} 7) ; \\; (7,7); \\; (7, \\minus{} 7)$[/hide][/quote]\r\nAre you sure?", "Solution_4": "I am. Do you disagree with my solution?\r\n\r\nPlease share your thoughts :)", "Solution_5": "[quote=\"TZF\"]I am. Do you disagree with my solution?\n\nPlease share your thoughts :)[/quote]\r\n$ \\{a,b\\}\\subset\\mathbb N.$ :wink:", "Solution_6": "Hah oh yeah, thanks, fixed", "Solution_7": "Hmm well, I think that it might be a factoring problem. A^2 + A - 7 = B^2, but we can subtract B^2, subtract A and add 7 to get the equation; A^2 - B^2 = 7 - A, so we can factor the side on the left to get; \r\n(A + B)(A - B) = 7 - A. Hmm well I can't really do anything else from their, so that's all I can do. Well, tell me if it helped any.", "Solution_8": "It's already solved and if you don't have whole solution than don't post.", "Solution_9": "I suck at these kind of problems. How do you guys know how to solve them?", "Solution_10": "It's really easy to observe that if a is larger than 7, $ a^2 2a \\minus{} 2b \\plus{} 1$, since we want $ b$ to be positive.\r\n\r\nSo, $ 2a \\plus{} 2b \\plus{} 1 \\equal{} 29$ and $ 2a \\minus{} 2b \\plus{} 1 \\equal{} 1$ is the only system we care about.", "Solution_18": "Its already been solved.", "Solution_19": "Multiply both sides by 4:\r\n4a^2 + 4a - 28 = 4b^2\r\n(2a+1-2b)(2a+1+2b)=29\r\nConsider all divisors of 29 (2a+1+2b>2a+1-2b)\r\nTherefore a=7 and b=7 :lol:", "Solution_20": "$ a^2 \\plus{} a \\minus{} 7 \\equal{} b^2 \\leftrightarrow \\left( {a \\plus{} \\frac{1}{2}} \\right)^2 \\minus{} 7\\frac{1}{4} \\equal{} b^2 \\leftrightarrow \\left( {a \\plus{} \\frac{1}{2}} \\right)^2 \\minus{} b^2 \\equal{} \\frac{{29}}{4}$\r\n\r\n$ \\leftrightarrow \\left( {a \\plus{} \\frac{1}{2} \\plus{} b} \\right)\\left( {a \\plus{} \\frac{1}{2} \\minus{} b} \\right) \\equal{} \\frac{{29}}{4} \\leftrightarrow \\left( {2a \\plus{} 1 \\plus{} 2b} \\right)\\left( {2a \\plus{} 1 \\minus{} 2b} \\right) \\equal{} 29$\r\n\r\nWe feel 29 is a prime number. SO:\r\n$ \\left[ \\begin{array}{l}\r\n \\left\\{ \\begin{array}{l}\r\n 2a \\plus{} 1 \\plus{} 2b \\equal{} 29 \\\\ \r\n 2a \\plus{} 1 \\minus{} 2b \\equal{} 1 \\\\ \r\n \\end{array} \\right. \\\\ \r\n \\left\\{ \\begin{array}{l}\r\n 2a \\plus{} 1 \\plus{} 2b \\equal{} 1 \\\\ \r\n 2a \\plus{} 1 \\minus{} 2b \\equal{} 29 \\\\ \r\n \\end{array} \\right. \\\\ \r\n \\left\\{ \\begin{array}{l}\r\n 2a \\plus{} 1 \\plus{} 2b \\equal{} \\minus{} 1 \\\\ \r\n 2a \\plus{} 1 \\minus{} 2b \\equal{} \\minus{} 29 \\\\ \r\n \\end{array} \\right. \\\\ \r\n \\left\\{ \\begin{array}{l}\r\n 2a \\plus{} 1 \\plus{} 2b \\equal{} \\minus{} 29 \\\\ \r\n 2a \\plus{} 1 \\minus{} 2b \\equal{} \\minus{} 1 \\\\ \r\n \\end{array} \\right. \\\\ \r\n \\end{array} \\right.$\r\n\r\nI think that's right@@@ :wink:", "Solution_21": "Nice solution. Even one that I can understand. How do you guys get so good at problems like these?" } { "Tag": [], "Problem": "Alyssa has four marbles of different colors, red, blue, green and yellow. In how many\ndifferent ways can she form a set consisting of some of the marbles if she must choose at least one marble?", "Solution_1": "[hide=\"I don't think this is correct because\"] {2,3} is different from {3,2}, so the answer should be 64, but they get 15[/hide]", "Solution_2": "[hide]what's 2,3 and 3,2 if you mean like blue red and read blue, then yeah, that's the same. So each color can either be in or out. So 2^4=16, but it can't be all out so -1=15[/hide]", "Solution_3": "[hide=\"I am curious, wouldn't this be\"] 19 because we can make a chart\n\n\n1 marble chosen----- 4 choices\n2 marbles chosen--- 6 choices (4x3 divide by 2 because we don't care about the order)\n3 marbles chosen--- 8 choices (same as above but divide by our 3 ways we can choose)\n4 marbles chosen--- 1 choice (there is only one way for a marble to be chosen)[/hide]\n\n[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Hide your solutions.[/url][/color]", "Solution_4": "[quote=\"penny13\"][hide=\"I am curious, wouldn't this be\"] 19 because we can make a chart\n\n\n1 marble chosen----- 4 choices\n2 marbles chosen--- 6 choices (4x3 divide by 2 because we don't care about the order)\n3 marbles chosen--- 8 choices (same as above but divide by our 3 ways we can choose)\n4 marbles chosen--- 1 choice (there is only one way for a marble to be chosen)[/hide]\n\n[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Hide your solutions.[/url][/color][/quote]\nThat was my answer at first. The answer said 15 though.", "Solution_5": "[quote=\"penny13\"]\n3 marbles chosen--- [color=#BF0000]8[/color] choices (same as above but divide by our 3 ways we can choose)[/quote]\n\nThat should be $4$, and hence the answer is $15$.", "Solution_6": "It's simply $\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}=4+6+4+1=\\boxed{15}$\n\nSolution #56", "Solution_7": "[hide = \"Official Solution\"]Each of the four marbles can either be in Alyssa's set, or not. That gives two possibilities for each marble, for a total of $2^4=16$ possible sets. However, we are told the set must have at least one marble. We need to subtract one to eliminate the case of an empty set, which we counted. Our final answer is $16-1=\\boxed{15}$ sets.[/hide]", "Solution_8": "I agree with AoPS and Radioactive. Either way, its just 4 choose 1 plus 4 choose 2 plus 4 choose 3 plus 4 choose 4\n", "Solution_9": "Doesn't tell you enough information, I didn't know that you could only choose 2 marbles per set, so yeah.\n", "Solution_10": "this was my first post :)\ni am posting 8 or 9 months later :)\nnow i have 208 posts\n\n\n\n\n\nthis wasn't my first post... my first one was in my C&P class... but the thread is locked so i can't do anything ", "Solution_11": "[hide=process]Though you are supposed to do this with complimentary counting, with casework counting you could easily do it by saying one set marbles there are 4 two set marbles there are 6 3 set marbles there are 4 and 4 set marbles there are 1 and add to 15[/hide]", "Solution_12": "[hide=Hint]\nYou can do complementary counting (subtracting from total) or use combinatorics (adding to get total) to solve this. Also the color of the marbles has nothing to do with the solution. \n[/hide]\n[hide=Process of Combinations]\nWe are told that Alyssa must choose at least $1$ in $4$ marbles. So that means that she can choose $1$, $2$, $3$, or $4$ marbles from $4$ marbles. That's $\\binom{4}{4}+\\binom{4}{3}+\\binom{4}{2}+\\binom{4}{1}=1+4+6+4=\\boxed{15}$ ways she can form a set with the specifications. \n[/hide]" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Hi all, \r\n\r\nI need to show that for a space $ X$ the following three conditions are equivalent:\r\n(a) Every map $ \\phi: S^1 \\rightarrow X$ is homotopic to a constant map, with image a point.\r\n(b) Every map $ \\phi: S^1 \\rightarrow X$ can be extended to a map $ \\phi^*: D^2 \\rightarrow X$.\r\n(c) $ \\pi_1(X,x_0)\\equal{}0$ for every $ x_0 \\in X$.\r\nDeduce that a space $ X$ is simply-connected iff all maps $ S^1 \\rightarrow X$ are homotopic.\r\n\r\n\r\nFor $ (a) \\Rightarrow (b)$, am I not able to do the following?\r\nIf $ \\phi(S^1)\\equal{}x_0$, some $ x_0 \\in X$, then define $ \\phi^*(x)\\equal{}x_0$ for every $ x \\in D^2$. Then, $ \\phi^*$ agrees with $ \\phi$ on the subspace $ S^1 \\subset D^2$. Is there more to the definition of a map \"extended\" to $ D^2$?\r\n\r\nFor the rest, I'm pretty stuck. \r\n\r\nAny help or hints would be very much appreciated.", "Solution_1": "A map \"extended\" to $ D^2$ means a continuous map. The map you defined is not necessarily continuous.\r\n\r\nAnother way to think about extension is that if $ i: S^1 \\to D^2$ is the inclusion map, then $ \\phi \\equal{} \\phi^* \\circ i$.\r\n\r\nThink about the conceptual relationship between (1) and (2). Essentially, if a map can be extended to a map from $ D^2$, then the image of $ D^2$ can be thought of as the region enclosed by the image of $ S^1$. In other words, $ S^1$ actually bounds a region in $ X$. This means that $ S^1$ can be squished to a point through this region. This should help you with (2) to (1).\r\n\r\nTo get (1) to (2), notice that we try to deduce the existence of a map from a two-dimensional space $ D^2$ based on properties about a map from a one-dimensional space $ S^1$. This property, specifically, is the existence of a certain homotopy. How does this property help us create a map from a higher dimensional object? ;)\r\n\r\nTo understand why (1) and (3) are related, think about what an element of the fundamental group actually is." } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Let $k$ be a positive integer. A natural number $m$ is called [i]$k$-typical[/i] if each divisor of $m$ leaves the remainder $1$ when being divided by $k$.\r\n\r\nProve:\r\n\r\n[b]a)[/b] If the number of all divisors of a positive integer $n$ (including the divisors $1$ and $n$) is $k$-typical, then $n$ is the $k$-th power of an integer.\r\n\r\n[b]b)[/b] If $k > 2$, then the converse of the assertion [b]a)[/b] is not true.", "Solution_1": "If $n=p_1^{a_1}\\ldots p_t^{a_t}$, then the number of divisors is $(a_1+1)\\ldots(a_t+1)$. If this number is $k$-typical, then $a_i+1=M_k+1$, meaning that all $a_i$s are multiples of $k$, and this proves a). In order to prove b), just take numbers of the form $p^{k(k-2)}$. This is a $k$'th power, and the number of divisors is $(k-1)^2$, which is divisible by $k-1$, which is not $1\\pmod k$." } { "Tag": [ "algebra", "polynomial", "AMC", "USA(J)MO", "USAMO" ], "Problem": "This is from USAMO, but is appropriate for this level.\r\n\r\nSuppose $P(x)$ is a polynomial of degree $n\\geq1$ such that $P(k)=\\frac{k}{k+1}$ for $k=0,\\ 1,\\ 2,\\ ...,\\ n$. Find $P(n+1)$", "Solution_1": "[hide=\"hint\"]\nlook at $(k+1)p(k)-k$\n[hide=\"hint2\"]\nit has a degree of $n+1$, and $p(k)_{ k\\in\\{0,1,...,n\\}}=0$, so $(k+1)p(k)-k=c\\cdot k(k-1)...(k-n)$.\n[/hide]\n[/hide]", "Solution_2": "good hints, but are you going to solve the problem?", "Solution_3": "Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=289576&highlight=#289576]HERE[/url]. That problem's almost the same. You only need to do a little more algebra to solve it completely.", "Solution_4": "try solving it completely then", "Solution_5": "[hide=\"Here you go\"]So you build a polynomial [tex]Q(x)=(x+1)\\cdot P(x)-x[/tex] of degree [tex]n+1[/tex] with roots [tex]{0,1,...,n}[/tex]. \nThe polynomial has the form of:\n[tex]Q(x)=A\\cdot (x)\\cdot (x-1)\\cdot ...\\cdot (x-n)[/tex]. \nYou also know:\n[tex]Q(-1)=(-1+1)\\cdot P(-1)-(-1)=1[/tex] and hence:\n[tex]Q(-1)=A\\cdot (-1)\\cdot (-1-1)\\cdot ...\\cdot (-1-n)=A\\cdot (n+1)!\\cdot (-1)^{n+1}=1\\ \\Longrightarrow\\\\ \\Longrightarrow\\ A=\\frac{(-1)^{n+1}}{(n+1)!}[/tex]. \nNow you can compute:\n[tex]Q(n+1)=\\frac{(-1)^{n+1}}{(n+1)!}\\cdot (n+1)\\cdot (n+1-1)\\cdot ...\\cdot (n+1-n)=\\\\=\\frac{(-1)^{n+1}}{(n+1)!}\\cdot (n+1)!=(-1)^{n+1}[/tex]. \nTransforming the first expression you get:\n[tex]P(n+1)=\\frac{Q(n+1)+n+1}{n+2}=\\frac{n+1+(-1)^{n+1}}{n+2}[/tex]. \nIt means [tex]P(n+1)=1[/tex] for odd [tex]n[/tex] or [tex]P(n+1)=\\frac{n}{n+2}[/tex] for even [tex]n[/tex].[/hide]", "Solution_6": "good job, very nicely done, and CORRECT!" } { "Tag": [ "AMC", "AIME" ], "Problem": "so i am thinking about buying The Art and Craft of Problem Solving. i am wondering if this book has answers and what kinds of problems it provides the reader.", "Solution_1": "The problems are from hard AMC 12 up to Olympiad level. Most are around AIME to the easier side of the olympiad spectrum though.\r\n\r\nThere is a hint section at the end. But there is no solution manual available to the public. The best alternative is to post problems that you cannot figure out on these boards.", "Solution_2": "are there any books like this that have solutions manuals (other than AoPS), or are the hints very useful?", "Solution_3": "I think Problem Solving Strategies and Problem Solving Problems would work.", "Solution_4": "I've looked at the books fouier series mentioned, but they seem too dry. For Olympiadproblem solving materials,I am getting Math Oly Challenges and Treasures, which seem more interesting. I'll probably get PSS next year", "Solution_5": "does problem solving through problems come with solutions?", "Solution_6": "No, but you shouldn't really need a solution guide. If you aren't able to solve the problems, you're probably giving up too quickly. Post questions on the board if you really need help.", "Solution_7": "ACOPS is really good BECAUSE it doesn't have a solutions manual. It makes you think about problems, and keep thinking about them, too. I liked ACOPS as much as AoPS, and I am trying to buy (not borrow) a copy, but it's pretty expensive..." } { "Tag": [ "number theory", "relatively prime" ], "Problem": "A dart board has two concentric circles. Landing a dart in the outer ring will net you seven points. The number inside the inner circle is faded away and thus impossible to read, however, Albatross remembers that the highest score impossible to get is $5^3$ . What is the point value gained by landing a dart in the inner circle?", "Solution_1": "[hide]Use formula to get 7x - 7 - x = 125. Solve to get $\\boxed{22}$.[/hide]", "Solution_2": "[hide]$(x-1)(y-1)-1=125$\n$6(y-1)=126$\n$y-1=21$\n$y=22$\n\n$22$[/hide]", "Solution_3": "[hide]\n\n7x-x-7=125\n6x-7=125\n6x=132\nx=22\n\n22\n[/hide]\r\n\r\nAlbatross?", "Solution_4": "[hide]Apply Chicken McNugget, $7n-7-n=125$\n$6n=132$\n$n=22$.[/hide]", "Solution_5": "[hide]Chicken McNugget will give you 22. Checking that it is relatively prime to 7, which it is, then that is your answer.\n\n22.0 to the nearest tenth\n\n$22$[/hide]" } { "Tag": [ "search", "Pythagorean Theorem", "geometry", "geometry open" ], "Problem": "Hi!\r\nPlease help me to find problems which use the following well knownlemma\r\n\r\n $ AC\\perp BD$$ \\Longleftrightarrow$ $ AB^2 \\minus{} AD^2 \\equal{} CB^2 \\minus{} CD^2$.\r\n\r\nI spent 2 hours on search but I've found only a few problems.\r\n\r\nPlease give some links, if somebody remembers.\r\n\r\nThanks a lot!", "Solution_1": "hello, what do you know about the point $ D$?\r\nSonnhard.", "Solution_2": "I'm sorry. I've edited...", "Solution_3": "hello, if $ \\overline{AC}\\perp\\overline{BD}$, then let $ S$ the intersection point. Then we get after the Pythagorean theorem \r\n$ \\overline{AB}^2\\equal{}\\overline{BS}^2\\plus{}\\overline{AS}^2$ (1)\r\n$ \\overline{AD}^2\\equal{}\\overline{AS}^2\\plus{}\\overline{DS}^2$ (2)\r\nso we get from (1) and (2):\r\n$ \\overline{AB}^2\\minus{}\\overline{AD}^2\\equal{}\\overline{BS}^2\\minus{}\\overline{DS}^2$.\r\nAnalogously we get $ \\overline{CB}^2\\minus{}\\overline{CD}^2\\equal{}\\overline{BS}^2\\minus{}\\overline{DS}^2$. \r\nFrom here follows \r\n$ \\overline{AB}^2\\minus{}\\overline{AD}^2\\equal{}\\overline{CB}^2\\minus{}\\overline{CD}^2$.\r\nSonnhard.", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, if $ \\overline{AC}\\perp\\overline{BD}$, then let $ S$ the intersection point. Then we get after the Pythagorean theorem \n$ \\overline{AB}^2 \\equal{} \\overline{BS}^2 \\plus{} \\overline{AS}^2$ (1)\n$ \\overline{AD}^2 \\equal{} \\overline{AS}^2 \\plus{} \\overline{DS}^2$ (2)\nso we get from (1) and (2):\n$ \\overline{AB}^2 \\minus{} \\overline{AD}^2 \\equal{} \\overline{BS}^2 \\minus{} \\overline{DS}^2$.\nAnalogously we get $ \\overline{CB}^2 \\minus{} \\overline{CD}^2 \\equal{} \\overline{BS}^2 \\minus{} \\overline{DS}^2$. \nFrom here follows \n$ \\overline{AB}^2 \\minus{} \\overline{AD}^2 \\equal{} \\overline{CB}^2 \\minus{} \\overline{CD}^2$.\nSonnhard.[/quote]\r\n\r\nOf course.. But I asked about problems.. And not easy ones, if it's possible.", "Solution_5": "usually you can't solve hard problems with this lemma\r\nbut with this lemma you can prove the karno principle that has many hard problems" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find the smallest integer $ n>0$ such that $ n^2\\minus{}n\\plus{}11$ is the product of four primes (not necessarily distinct).", "Solution_1": "[hide=\"Impossible Prime divisors of n^2 - n + 11\"]Notice that neither prime from $ \\{2,3,5,7\\}$ can divide $ n^2 \\minus{} n \\plus{} 11$[/hide]\n[hide=\"First candidate\"]Checking $ 11^4$ that no integer solution to $ 11^4 \\equal{} n^2 \\minus{} n \\plus{} 11$ by evaluating $ n \\equal{} 11^2$ and $ n \\equal{} 11^2 \\plus{} 1$[/hide]\n[hide=\"Second Candidate (Lucky one)\"]Checking $ 11^3 \\times 13$ and noticing $ \\boxed{11^3 \\times 13 \\equal{} (11\\times 12)^2 \\minus{} (11\\times 12) \\plus{} 11}$[/hide]" } { "Tag": [ "equation", "RMO" ], "Problem": "Solve for real $x$ : \\[ (x^2 + x -2 )^3 + (2x^2 - x -1)^3 = 27(x^2 -1 )^3. \\]", "Solution_1": "$x^2 + x -2 = (x - 1)(x + 2)$\r\n$2*x^2 - x -1 = (x - 1)(2 x + 1)$\r\n$(x^2 + x -2 )^3 + (2*x^2 - x -1)^3 = 9 (x - 1)^3(x + 1) (x^2 + x + 1)$\r\n...\r\n\r\nSolution are $1,-1,-2,-1/2$", "Solution_2": "Or we can use fermats last theorem!", "Solution_3": "[quote=\"seamusoboyle\"]Or we can use fermats last theorem![/quote]\r\n\r\nThat's not the first thing I would have attempted. But keep in mind that it applies to natural numbers, not the reals. I guess it doesn't matter in this case.", "Solution_4": "Ah. Didnt see the real part. My mistake.", "Solution_5": "Well , the best way is this - standard method for this type of problems !\r\n\r\nWe write the equation in the form \r\n\r\n $(x^2 +x-2)^3 + (2x^2 -x-1)^3 +(3-3x^2)^3 = 0$\r\n\r\n Since the sum of the bases is $0$ , the equation becomes \r\n\r\n $3 (x^2 +x-2) (2x^2 -x-1) (3-3x^2) = 0$\r\n\r\n But this can be solved very easy and the solutions are $x = 1 , x = -1 , x = -2 , x= -\\frac{1}{2}$ , \r\n\r\n[b]Remember that if $a +b+c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ [/b]\r\n\r\n[u]Babis[/u]", "Solution_6": "If we simply factorize the two sides we will come to an equation : $9(x+2)(2x+1)(x+1)=0$..Earlier we got a solution $x=1$...Thus the other solutions are clearly $ -1,-2,\\frac{-1}{2}$.", "Solution_7": "-1 IQ solution :wallbash: " } { "Tag": [ "quadratics" ], "Problem": "The billionaire Dagobert Grinberg arranges his $m$ bag with gold coins. Surprised he realizes that the bag can be arranged in a triangular and quadratic form without residue. How many bags does Dagobert have, if $m \\in [1000,2000].$ There was even bigger surprise when Dagobert realized that he can arrange the gold coins in a triangular and quadratic way without residue too. How many coins does Dagobert have at least ? :)", "Solution_1": "Hmm, if you say me how much euros a gold coin is worth, I will see whether I will be happy with my $1225\\cdot 1$ (or $1225\\cdot 36$?) gold coins...\r\n\r\n Darij", "Solution_2": "Hey Darij,\r\n\r\nyou have 1225 bags with gold coins not 1225 gold coins !!! :D But you have not yet determined the number of coins. Maybe one gold coin for one pound sterling ? :)", "Solution_3": "I thought the number of gold coins was the number of gold coins in each bag... in any case, the number of gold coins can be only 1, 36, 1225, 41616 or etc. ([url=http://mathworld.wolfram.com/SquareTriangularNumber.html]square triangular numbers[/url]), but I don't know how little you assume your minimum to be...\r\n\r\n Darij", "Solution_4": "[quote=\"orl\"]The [b]billionaire[/b] Dagobert Grinberg arranges his $m$ bag with gold coins. [/quote]", "Solution_5": "Ah well, this is really a clue, but is it a German billionaire or an English one? I. e., how much is a billion?\r\n\r\n Isotomion (= the alter ego of Darij Grinberg, replacing the main ego when the latter has troubles with posting on ML during a firewall idiosyncracy of the internet room where he is posting from)", "Solution_6": "Well, in German it is a \"Milliardr\". So please find the smallest number $\\geq 10^9.$", "Solution_7": "$1 413 721 * 1 225=1 731 808 225$" } { "Tag": [ "calculus", "integration", "arithmetic sequence" ], "Problem": "The sum of all numbers of the form $ 2k + 1 $, where $ k $ takes on integral values from 1 to $ n $ is:\r\n\\[ \\text{(A)}\\ n^2 \\qquad \\text{(B)}\\ n(n+1) \\qquad \\text{(C)}\\ n(n+2) \\qquad \\text{(D)}\\ (n+1)^2 \\qquad \\text{(E)}\\ (n+1)(n+2) \\]", "Solution_1": "[hide=\"Sol\"]\n$ 2k+1 $\n1 will repeat n times so Sum of $ 1 $ is $ n $,\nNow, $ 2k $,\n$ 2[1+2+3+.....+n] $\nArithmetic sequence,\n$ 2n+n(n-1) $\n$ n^2+n $\nthen,\n$ n^2+2n $\n$ n(n+2) $\nAnswer=$ \\boxed{\\text{(C)}} $\n[/hide]", "Solution_2": "[hide]we see an arithmetic sequence 3, 7, 11, 15, etc., and it goes to $ 2n+1 $, and there are n terms. \n\n$ \\frac{(3+2n+1)(n)}{2}=n(n+2) $. [/hide]", "Solution_3": "[hide]Basically, we have to find the sum of odd integers from $ 3 $ to $ 2n+1 $\n\n$ 1+2+3+...+2n+1=(n+1)(2n+1) $\n\n$ 2+4+6+...+2n=2(1+2+3+...+n)=n(n+1) $\n\nThen $ 3+5+7+....+2n+1=(n+1)(2n+1)-n(n+1)-1=(n+1)^2-1=n(n+2) $\n\nAnswer: $ C $[/hide]" } { "Tag": [ "parameterization", "articles", "LaTeX" ], "Problem": "Is there a way to get this size?\r\n\r\n[img]http://img116.imageshack.us/img116/5891/holabz6.png[/img]\r\n\r\nI actually use 11pt but is it possible to set 10.5pt or the size of the image?", "Solution_1": "You can try[code]\\usepackage{fix-cm}[/code] then use [code]\\fontsize{10.5pt}{10pt}\\selectfont[/code]for where you want 10.5pt font. Change the second parameter 10pt to whatever line spacing looks best.", "Solution_2": "Oops... I have a doubt...\r\n\r\nIn the preamble I have\r\n\r\n\\documentclass{article}\r\n\\usepackage{fix-cm}\r\n\\fontsize{10.5pt}{10pt}\\selectfont\r\n\r\nWhat do I have to put after \\selectfont and \\documentclass[]{article} to get de 10.5pt size?", "Solution_3": "Here's a sample document\r\n[code]\\documentclass[12pt]{article}\n\\usepackage{fix-cm}\n\\begin{document}\nThis is in the size given by documentclass\n\\fontsize{10.5pt}{10pt}\\selectfont This is in 10.5pt\n\\end{document}[/code]", "Solution_4": "Ohhh great!!\r\n\r\nNow it worked!" } { "Tag": [ "function", "vector", "real analysis", "real analysis unsolved" ], "Problem": "$f: V \\to \\mathbb{R}$ a $C^{1}$ function where $V$ is a Banach space. Suppose that $m = \\inf_{x \\in V}f(x) >-\\infty$. Prove that for every $\\epsilon>0$ there exist $v \\in V$ such that:\r\n$c \\leq f(v) \\leq c+\\epsilon$ and $\\| Df(v) \\|_{V'}\\leq \\epsilon$", "Solution_1": "I assume that $c=m$ (otherwise the problem doesn't make much sense). Assume the contrary. Start with any point $x_{0}\\in V$ such that $f(x_{0})2 + 1/b 2 + 1/c 2 + 1/d 2 ) = (1/a + 1/b + 1/c + 1/d) 2 .\r\n\r\nPierre.", "Solution_2": "This was solved b4 on this forum:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=352[/url]\r\n\r\nI hope it's Ok. I don't know if ne1 has really checked it thoroughly :D" } { "Tag": [ "modular arithmetic", "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "prove [i]Lucas' Theorem[/i]:\r\n if $ m \\equal{} a_{0} \\plus{} a_{1}p \\plus{} ... \\plus{} a_{t}p^{t}$ with $ 0\\leq a_i\\frac{2-\\sqrt 2}{2}$.\r\nIf $\\theta_{k}<\\frac{2-\\sqrt 2 }{2}, k=m,m+1,...m+s$, then $n_{k+1}=2n_{k}-1, \\theta_{k+1}=2\\theta_{k}.$ Therefore exist s: $\\theta_{m+s}=2^{s}\\theta_{m}\\ge \\frac{2-\\sqrt 2}{2}.$", "Solution_2": "[quote=\"Rust\"] then $n_{k+1}=2n_{k}-1, \\theta_{k+1}=2\\theta_{k}.$ [/quote]\ni don not understand", "Solution_3": "If $\\theta_{k}<\\frac{2-\\sqrt 2 }{2},$ then \n\n$2^{k+1}=2n_{k}\\sqrt 2-2(1-\\theta_{k})\\sqrt 2$\n\n$=(2n_{k}-1)\\sqrt 2-(1-2\\theta_{k})\\sqrt 2$\n\n$ \\Longrightarrow n_{k+1}=2n_{k}-1, \\theta_{k+1}=2\\theta_{k}.$" } { "Tag": [ "geometry", "circumcircle" ], "Problem": "The [i]Simson theorem[/i] states that, for any point $P$ on the circumcircle of a triangle $ABC$, the orthogonal projections of the point $P$ on the lines $BC$, $CA$, $AB$ lie on one line. This line is called the [i]Simson line[/i] of the point $P$ with respect to triangle $ABC$.\r\n\r\n[b]a)[/b] Let $P$ and $Q$ be two points on the circumcircle of triangle $ABC$, and let $s$ and $t$ be the Simson lines of these points $P$ and $Q$ with respect to triangle $ABC$. Let $R$ be an arbitrary point on the circumcircle of triangle $ABC$. Prove that $\\measuredangle\\left(s;\\;t\\right)=\\measuredangle QRP$, where we use directed angles modulo $180^{\\circ}$.\r\n\r\n[Feel free to work with non-directed angles in the proof if you are more familiar with them.]\r\n\r\n[b]b)[/b] For any line $d$ in the plane, prove that there exists exactly one point on the circumcircle of triangle $ABC$ whose Simson line with respect to triangle $ABC$ is parallel to the line $d$. Construct this point.\r\n\r\n darij", "Solution_1": "[hide=\"a)\"]Let $Q_{1}, Q_{2}, Q_{3}$ be the feet of perpendiculars from $Q$ to $BC, AB, AC$ respectively.\n\nLemma 1\nLet $S$ be the intersection between the perpendicular from $Q$ to $BC$ and the circumcircle, then $AS \\| t$.\n\nWe have $\\angle QQ_{1}Q_{2}= 90^{\\circ}-\\angle Q_{2}Q_{1}B = \\angle B-\\angle QQ_{2}Q_{1}$.\nSince $QQ_{1}BQ_{2}$ is cyclic we have $\\angle QQ_{1}Q_{2}= \\angle B-QQ_{2}Q_{1}= \\angle QBA = \\angle QSA$.\nAlso, in the same way, $AR \\| s$.\n\nBack to the problem.\n$s$ intersect the circumcircle in $M, H$ and $t$ in $L, K$. Then $\\angle (HM, KL) = \\frac{\\smile HK+\\smile ML}{2}$. Let $S, R$ be the intersections between $QQ_{1}, PP_{1}$ and the circumcircle, respectively, where $P_{1}$ is the orthogonal projection of $P$ on the line $BC$. Then $\\smile PQ = \\smile SR$ and for Lemma 1, $\\smile AK = \\smile SL$ and $\\smile HA = \\smile MR$. Hence $\\angle (HM, KL) = \\angle (s, t) = \\frac{\\smile PQ}{2}$.[/hide]", "Solution_2": "First, sorry for a mistake in the problem. I have just fixed it (replaced < PRQ by < QRP).\r\n\r\nAndreas - your proof is correct, but hard to understand... e. g. you should have defined the point $P_{1}$ (as the orthogonal projection of the point P on the line BC) and the point R (as the point where the perpendicular to BC through P intersects the circumcircle of triangle ABC, apart from P).\r\n\r\n[hide=\"Anyway, here is my solution.\"][i]Solution.[/i] [b]a)[/b] Let X and Y be the orthogonal projections of the point P on the lines BC and CA. Then, the Simson line s of the point P with respect to triangle ABC is the line XY. Thus, < (s; CA) = < XYC. Now, since < CXP = 90\u00b0 and < CYP = 90\u00b0, the points X and Y lie on the circle with diameter CP; hence, < XYC = < XPC. Now, in the right-angled triangle PXC, we have < XPC = 90\u00b0 - < PCX. Thus, < (s; CA) = < XYC = < XPC = 90\u00b0 - < PCX = 90\u00b0 - < PCB. Similarly, < (t; CA) = 90\u00b0 - < QCB. Hence,\n\n$\\measuredangle\\left(s;\\;t\\right)=\\measuredangle\\left(s;\\;CA\\right)-\\measuredangle\\left(t;\\;CA\\right)=\\left(90^{\\circ}-\\measuredangle PCB\\right)-\\left(90^{\\circ}-\\measuredangle QCB\\right)$\n$=\\measuredangle QCB-\\measuredangle PCB=\\measuredangle QCP$.\n\nSince the points C, P, Q, R lie on one circle, < QCP = < QRP, so that < (s; t) = < QRP, and problem [b]a)[/b] is solved.\n\n[b]b)[/b] Fix an arbitrary point P on the circumcircle of triangle ABC, and let s be the Simson line of this point P with respect to triangle ABC. Consider a point Q on the circumcircle of triangle ABC, and the Simson line t of this point Q with respect to triangle ABC. We want to show that $t\\parallel d$ holds for exactly one position of the point Q on the circumcircle of triangle ABC, and construct this point Q.\n\nAccording to part [b]a)[/b], we have < (s; t) = < QCP (since C is a point on the circumcircle of triangle ABC). But $t\\parallel d$ holds if and only if < (s; t) = < (s; d). Since < (s; t) = < QCP, this is equivalent to < QCP = < (s; d). Now, C and P are fixed points and < (s; d) is a fixed angle; the locus of all points T in the plane such that < TCP = < (s; d) is a line through the point C. Hence, the equation < QCP = < (s; d) holds if and only if the point Q is the point of intersection of this line with the circumcircle of triangle ABC different from C; clearly, there is exactly one such point, and its construction is obvious. Thus, part [b]b)[/b] of the problem is solved.[/hide]\r\n\r\n darij" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ p$ be a prime and let $ G$ be an abelian group of order $ p^{n}$. Prove that the nilradical of the group ring $ \\mathbb{F}_pG$ is the augmentation ideal.\r\n\r\nUse : the augmentation ideal in the group ring $ RG$ is generated by {$ g$ - 1|$ g\\in G$}.", "Solution_1": "the augmentation ideal is per definition the kernel of a homomorphism $ \\mathbb{F}_p G \\to \\mathbb{F}_p$, hence a prime ideal. thus every nilpotent element lies in this ideal. conversely, $ (g \\minus{} 1)^{p^n} \\equal{} 0$ shows the other inclusion. $ G$ abelian just guarantees that $ \\mathbb{F}_p G$ is a commutative ring, so that the nilradical is, in fact, an ideal." } { "Tag": [], "Problem": "Set $ A$ contains all the two-digit integers that equal the product of their tens and units digits divided by the quotient of their tens and units digits. What is the square root of the product of all integers in set $ A$?", "Solution_1": "Let the number be $ 10a\\plus{}b$.\r\n\r\nThus, $ \\frac{ab}{\\frac{a}{b}}\\equal{}10a\\plus{}b \\implies b^2\\equal{}10a\\plus{}b$\r\n\r\nThe solutions are $ (2,5), (3,6)$.\r\n\r\nThus, $ \\sqrt{25 \\cdot 36}\\equal{}\\sqrt{5^2 \\cdot 6^2}\\equal{}5 \\cdot 6\\equal{}30$." } { "Tag": [], "Problem": "Thank you rjyanco for posting the State meet links :!: \r\n\r\nDoes anyone know which schools quallified for the New England meet :?: \r\n\r\nThere are different number of schools qualifying in each\r\nsection apparently based on the category (Large/Medium/Small)\r\n\r\nThis year I heard more schools attended in certain sections\r\nhence more qualifying spots... :maybe:", "Solution_1": "From my league (in Connecticut) Hopkins and Choate are going.", "Solution_2": "Worcester County, MA is sending $\\boxed{\\text{Shrewsbury HS}}$ into large division yay\r\n\r\nWe're gonna fail :(", "Solution_3": "In the Small Schools Division the two teams that will go from the Connecticut State Match will be:\r\n\r\nHopkins\r\nBrunswick School" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Let $x$ and $y$ be positive integers such that $xy$ divides $x^{2}+y^{2}+1$. Show that\r\n$\\frac{x^{2}+y^{2}+1}{xy}=3$.", "Solution_1": "Search function.", "Solution_2": "Set$x^{2}+y^{2}+1 =kxy.$\r\nChoose root$(x_{0},y_{0})$ of this equation so that $x_{0}+y_{0}$ min.\r\nUse Viet theorem , we have $k=3$(Easy to prove that)", "Solution_3": "Sorry thachpbc, do you can post the complete solution??\r\nI dont know Viet theorem... :oops:", "Solution_4": "This problem has been posted soooo many times :D\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=97705\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=71760\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=40207 \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=91590 \r\n\r\n :P", "Solution_5": "[quote=\"Yimin Ge\"]This problem has been posted soooo many times :D\n\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=97705\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=71760\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=40207 \nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=91590 \n\n :P[/quote]\r\nHow do you find so many links on it?\r\n :)" } { "Tag": [ "probability", "calculus", "geometry", "trigonometry", "number theory", "\\/closed" ], "Problem": "Hi I'm kinda new to AOPS, but I'm interested in the courses that it has to offer. I bought the Intro to Number Theory and Intro to Counting and Probability books, but I also want to take the class. Should I read them over first or wait until the class? Moreover, I'm taking pre-calculus next year. Does anyone have a good suggestion on the class I should take here that would prepare me for that?\r\n\r\nFurthermore, I've taken Algebra I, II and Geometry before, but I don't think the teacher covered it in depth. Should I also take the class and/or buy the book?\r\n\r\nThanks :lol:", "Solution_1": "Reading them before the class should help you with the harder problems presented in class and as homework, even just skimming through sections. If you've taken Algebra I, II, and Geometry in school, and understood the stuff they taught there, school pre-calculus should be fine.\r\n\r\nAs for your last question, buying AoPS Volume 1 and/or 2 might be the best (and more cost-effective) choice. Basically, they both cover a wide range of topics in the \"AoPS-style\", which is more in depth than what is taught at school", "Solution_2": "Thanks, do you think I should take Algebra III?", "Solution_3": "That would probably be a good idea, because the AoPS Alg 3 course covers not just Algebra 2 but also some pre-calculus(I think?) I'm taking the course this June, and going into pre-calc this fall.", "Solution_4": "K, cool. How many classes do you think I should take or can handle if I\"m also taking Biology over the summer?", "Solution_5": "Read ahead in the book. I always did this in college, and it made college *very* easy for me.\r\n\r\nAs for classes to take, Algebra 3 will get you well ahead of your class in non-trig precalc. We'll have a precalc class in the fall that will go with our precalc book. We haven't set the dates, but it will likely start just after the Algebra 3 class ends.\r\n\r\nYou can probably manage two classes during the summer, particularly if you get ahead early on. It's usually much tougher to carry two subject classes (as opposed to the problem series, which are less demanding time-wise) during the school year." } { "Tag": [], "Problem": "not", "Solution_1": "my zergling avatar", "Solution_2": "http://www.artofproblemsolving.com/Admin/latexrender/pictures/38d2e7af24be72e03dd245ceb8117594.png\r\nlol", "Solution_3": "", "Solution_4": "", "Solution_5": "dont look here :P" } { "Tag": [ "function", "inequalities", "algebra solved", "algebra" ], "Problem": "Prove that there does not exist any function f : [0, +\u221e) \u2192 (0, +\u221e) such that for all real x>= 0 we have f'(x) >=f(f(x)).", "Solution_1": "Pretty nice and easy. f is of course increasing. Thus, when we integrate from o to x the inequality we will have $ f(x)\\geq\\frac{x}{2}\\cdot f(f(\\frac{x}{2}))$. Thus, for all $ x>2$ we have $ f(\\frac{x}{2})1 $. Thus, using again the found inequality we obtain $ f(f(\\frac{x}{2}))<4 $ for $ x>2 $ and thus $ f(f(x))<4 $ for all $ x>1 $. But $ f(x)>\\frac{x}{2}\\cdot f(f(\\frac{x}{2})>\\frac{x}{2}\\cdot f(f(1))$ for all $ x>2$ and thus also $ 4> f(f(x))>\\frac{x}{2}\\cdot f(f(1)) $ for all $ x>2$ , false." } { "Tag": [ "number theory", "relatively prime", "prime factorization", "prime numbers", "number theory open" ], "Problem": "Let's try to find the number of primes less than 30=2*3*5. Since 30 < 49 = 7*7, all numbers less than 30 and relatively prime to it are prime, so the number of primes less than 30 is phi(30)-1+3 (accounting for 1,2,3,5) = 4*2+2 = 10.\r\n\r\nNow apply this method to 210=2*3*5*7. The only nontrivial numbers less than 210 that are relatively prime to it and composite are 11*11,11*13,11*17,11*19,13*13 (which is easy to figure out using a calculator with integer multiplication). Then 1 is relatively prime to 210 but composite, and 2,3,5,7 are not relatively prime to 210 but are prime. Then pi(210) = phi(210) - 1 - 5 + 4 = phi(210) - 2. Then phi(210) is easy to calculate using its prime factorization, for it is (2-1)(3-1)(5-1)(7-1) = 2*4*6 = 48. So pi(210) = 48-2=46 which is correct :)\r\n\r\nNotice that in that calculation, all we had to know was information about relatively small prime numbers (i.e. those < 20) and perform multiplication. This is a fairly simple way to find the number of primes less than 210. I'm not sure how well this works for other numbers, but are there any interesting implications of this method?", "Solution_1": "Computing $ \\pi(p_1\\cdot p_2\\cdot\\dots\\cdot p_n)$ by your method will be harder and harder as $ n$ grows. In particular, one would need to account all semi-primes with prime factors from the interval $ [p_n \\plus{} 1,\\sqrt {p_1\\cdot p_2\\cdot\\dots\\cdot p_n}]$ (of exponentially growing length).\r\n\r\nIf you are interested in efficient way to compute $ \\pi(x)$, take a look at this paper:\r\nM Deleglise, J Rivat [url=http://cr.yp.to/bib/1996/deleglise.pdf]Computing \u03c0 (x): the Meissel, Lehmer, Lagarias, Miller, Odlyzko method[/url]", "Solution_2": "Although this thread's a little old, [url=http://terrytao.wordpress.com/2007/06/05/open-question-the-parity-problem-in-sieve-theory/#more-89]Terence Tao[/url] has an interesting post related to this subject." } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO" ], "Problem": "The answers are now posted on AOPS. How did people do?\r\n\r\nI found this AIME to be slightly easier than last year's AIME 2, and much easier than last year's AIME 1.\r\n\r\n What do you think the cutoff will be for USAMO?", "Solution_1": "I did all right, got 11 so my index is 236 which should be enough I think.\r\nI believe it was easier, because there was not as many counting problems and ugly calculations.\r\nI think cutoff will be around 215-220.", "Solution_2": "I think I got 11 as well... this makes my index 227. I did pretty badly on AMC lol", "Solution_3": "I should get a 10 for an index of 236.5. Hopefully the USAMO will be kind to me.", "Solution_4": "Congratulations, guys. I messed up on AIME very badly, so I won't be writing USAMO, but our school do have two Juniors who will make it through the floor value. :) \r\n\r\nHow did your schools do?\r\n\r\nFoundation" } { "Tag": [ "function", "geometry", "ratio", "trigonometry", "probability", "limit", "search" ], "Problem": "In honor of today, pi day, let's make a list of problems whose answer is $\\pi$ and do [i]not[/i] mention $\\pi$ in them. Starting with:\r\n\r\n1. For the function $y = x[sin\\left(\\frac{180}{x}\\right)][cos\\left(\\frac{180}{x}\\right)]$, as $x$ goes to infinity, what does $y$ approach?\r\n\r\nEDIT: Also, feel free to do any solutions to people's problems.", "Solution_1": "Okay...\r\n\r\nCompute $\\sqrt{6\\sum_{i=1}^{\\infty}\\frac{1}{i^{2}}}$.\r\n\r\nLet $a$ be the area of the region in the first quadrant bounded by the graphs $y=0$, $x=0$, and $y=\\sqrt{1-x^{2}}$. Compute $4a$.", "Solution_2": "It's a nice idea. Here's one (though not intermediate level)\r\n\r\nFind the limit of $a_{n}$, as $n\\to \\infty$, where \\[a_{n}= \\sqrt{ 6 \\left( 1+\\frac 1{2^{2}}+\\frac 1{3^{2}}+\\cdots+\\frac 1{n^{2}}\\right) }.\\] :)", "Solution_3": "Calculate $\\sum^{\\infty}_{k=1}\\frac{4}{2k-1}\\cdot (-1)^{k+1}$", "Solution_4": "What's the ratio of a circle's length to its diameter? :D", "Solution_5": "[quote=\"Valentin Vornicu\"]What's the ratio of a circle's length to its diameter? :D[/quote]\r\n\r\nsuprisingly or not, this question was on a local contest...cough bad contest cough:\r\n\r\na circle passes through (0,0) and (3,1), what is the ratio of the circumference to the diameter of that circle", "Solution_6": "Haha Altheman NSML contest 5 last year right? I remember that, the sad thing is that only me and one other person from my school got it right.", "Solution_7": "Evaluate $\\frac{2}{\\cos \\frac{\\pi}{4}\\cos \\frac{\\pi}{8}\\cos \\frac{\\pi}{16}...}$\r\n\r\nEdit: Let $p$ be the probability that two randomly selected positive integers are relatively coprime. Evaluate $\\sqrt{ \\frac{6}{p}}$.\r\n\r\nEdit 2: Evaluate $\\frac{1}{2}\\cdot \\left( \\lim_{n \\to \\infty}\\frac{n!}{ n^{n+\\frac{1}{2}}e^{-n}}\\right)^{2}$", "Solution_8": "$\\frac{2}{p}$ where p is the probability that if you have two parallel cracks on the floor, and a stick that's the length of the distance between them, when you drop the stick it lands on a crack.", "Solution_9": "the area of a circle over the radius squared.", "Solution_10": "Does anybody want to show any proofs of these?", "Solution_11": "Search the forum ;)", "Solution_12": "$\\sqrt[4]{90 \\sum^{\\infty}_{x=1}x^{-4}$", "Solution_13": "[quote=\"diophantient\"]$\\sqrt[4]{90 \\sum^{\\infty}_{x=1}x^{-4}}$[/quote]\r\n\r\nThe message contains only quote and code tags. Please add some text of your own.", "Solution_14": "As long as we're still doing this, how about\r\n\r\n$\\frac{1}{2}\\cdot \\left( \\int_{-\\infty}^{\\infty}e^{ \\frac{-t^{2}}{2}}\\, dt \\right)^{2}$\r\n\r\nOr $\\sqrt{ 8 \\sum_{k=1}^{\\infty}\\frac{1}{(2k-1)^{2}}}$, which seems to have been neglected, the poor thing.\r\n\r\nOr how about the beautiful\r\n\r\n$2 \\cdot \\frac{2 \\cdot 2 \\cdot 4 \\cdot 4 \\cdot 6 \\cdot 6 \\cdot ... }{1 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 5 \\cdot 7 \\cdot ... }$", "Solution_15": "Find $\\lim_{n\\to\\infty}(1+\\frac{1}{n})^{n}$ :D", "Solution_16": "That's not pi!!!!! (read the topic...) :wink:", "Solution_17": "Sorry :oops: It is $e$ and $e\\neq \\pi$.", "Solution_18": "I'm going to do one that's not a formula...\r\n\r\n$n$ random numbers between 0 and 100 are chosen. We split them into $\\frac{n}{2}$ pairs of $2$. If the sum of the squares of a given pair is $< 10000$, we add $1$ to a given counter $c$. Otherwise, we add $1$ to a given counter $d$. Find $\\lim_{n\\to\\infty}\\frac{4c}{c+d}$.", "Solution_19": "The surface area of a sphere divided by the square of the diameter.\r\n\r\n\r\nThe volume of a sphere divided by 4/3 of the cube of the radius.\r\n\r\n\r\napprox. 22/7", "Solution_20": "To Mr. Vornicu's first expression:\r\nI wanted to attempt a proof.\r\n\r\nIt is well known that the sum of the inverses of the squares of the natural numbers is $\\frac{\\pi^{2}}{6}$ (the Basel Problem)\r\n\r\nThus, we have $a_{n}= \\sqrt{6 \\cdot \\frac{\\pi^{2}}{6}}$\r\n\r\nfrom which we have $a_{n}= \\sqrt{\\pi^{2}}$\r\n\r\nand finally, $a_{n}= \\pi$.", "Solution_21": "That hardly constitutes a proof. The content of the problem is the solution to the Basel Problem.\r\n\r\n[quote=\"1=2\"]approx. 22/7[/quote]\r\n\r\nIncidentally, there's a fascinating integral related to this. Calculate\r\n\r\n$\\frac{22}{7}-\\int_{0}^{1}\\frac{x^{4}(1-x)^{4}}{1+x^{2}}\\, dx$", "Solution_22": "Or you could just be like:\r\n\r\n$\\frac{\\ln{(-1)}}{i}$", "Solution_23": "I think we have to whip out [hide=\"a whip\"] $\\arcsin$ of a Taylor series. [/hide]", "Solution_24": "[quote=\"cincodemayo5590\"]I think we have to whip out [hide=\"a whip\"] $\\arcsin$ of a Taylor series. [/hide][/quote]\r\n\r\nBut being in radians implies pi.", "Solution_25": "Area of circle with diameter 2? ^^ Hopefully it hasn't be posted yet.", "Solution_26": "[quote=\"Karth\"]Or you could just be like:\n\n$\\frac{\\ln{(-1)}}{i}$[/quote]\r\n\r\nEven though that expression might equal $\\pi$, could it really be evaluated? Is $\\ln{(-1)}$ actually defined?", "Solution_27": "The problem with taking logs of negative numbers is that they're not uniquely defined. It's more correct to say that\r\n\r\n$e^{i(\\pi+2 \\pi k)}=-1$", "Solution_28": "[img]http://img011.photo.21cn.com/photos/user/24454028/20070412224157/o/8502198.jpg[/img]", "Solution_29": "Which of the methods listed so far converges fastest?", "Solution_30": "Bailey or Ramanujan are extremely fast convergers.", "Solution_31": "Ramanujan developed certain series which converge to pi that can generalize, then mathematicians with computers were later able to come up with formulas with huge numbers that converged super quick - 50 correct digits per iteration!!", "Solution_32": "Any with continued fractions?", "Solution_33": "$\\left( 1+\\frac{1}{2^{2}}\\right) \\left( 1+\\frac{1}{3^{2}}\\right) \\left( 1+\\frac{1}{5^{2}}\\right) ... = \\frac{6}{\\pi^{2}}$\r\nWhere $x$ for each term $\\left( 1+\\frac{1}{x^{2}}\\right)$ is prime.", "Solution_34": "[b]miyomiyo:[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?p=831263#831263\r\n\r\n[b]AstroPhys:[/b] http://mathworld.wolfram.com/EulerProduct.html", "Solution_35": "Wallpaper 1024*768\r\n[img]http://img012.photo.21cn.com/photos/user/24454028/20070226182623/o/7860697.jpg[/img]\r\n\r\nPi-Poster\r\n[img]http://img012.photo.21cn.com/photos/user/24454028/20070226190405/o/7861206.png[/img]", "Solution_36": "How many of these can be proved?", "Solution_37": "all of them - how else would they be known? although i would like to see a few more shown, as all of the ones i know have been linked to or stated.\r\n\r\nbut yeah some of the wallis formulas/etc. are really nasty.", "Solution_38": "Happy Pi Day!\r\n\r\nI feel a strange need to post as my school for some weird reason celebrated pi day today May 18 instead of March 14.", "Solution_39": "This year's wisconsin Mathcounts pin - wisconsin - pizza :)" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "One I see alot:\r\n\r\nFind the sum of all positive factors of 504.\r\n\r\nMy question is, is there a better way of solving this problem besides finding all the factors then adding them?", "Solution_1": "There's this really cool trick: Say you want to find the sum of the factors of 72. First, you prime factorize to get $ 2^3*3^2$. Then you add all the powers of 2 and powers of 3, and multiply: $ (1\\plus{}2\\plus{}4\\plus{}8)(1\\plus{}3\\plus{}9)\\equal{}195$.\r\n\r\nSimilarly, the prime factorization for 504 is $ 2^3*3^2*7^1$. $ (1\\plus{}2\\plus{}4\\plus{}8)(1\\plus{}3\\plus{}9)(1\\plus{}7)\\equal{}\\boxed{1560}$. I hope this helps! :)\r\n\r\nIf you don't understand how this product represents the sum of all the factors, distribute and you'll see this covers all possible combinations of the availible prime factors.", "Solution_2": "Hah, thanks. :]", "Solution_3": "Where can I find a proof for that trick?", "Solution_4": "You can prove it for yourself, just multiply it out (in Algebra of course) and you can see that all possible additions are covered." } { "Tag": [ "function", "algebra", "floor function", "calculus", "calculus computations" ], "Problem": "1). Prove that the limit (-1)^[x] as x approaches infinity does not exist. Bracket indicates floor function\r\n\r\n\r\n2).Suppose that the limit of f(x)=L as x approaches a and L is posotive. Show that there exists an interval I containing a such that f(x)>0 and x E I\r\n\r\n\r\nThank you :)", "Solution_1": "1) The possible values of $ (\\minus{}1)^{[x]}$ is only $ 1$ and $ \\minus{}1$, because $ [x]$ is always an integer regardless of $ x\\in\\mathbb R^\\plus{}$. So, if there exists the limit, we can find $ M>0$ such that $ x>M\\Rightarrow f(x)\\equal{}1$(without loss of generality, we assumed the value of the limit is $ 1$).By completeness property of $ \\mathbb R$, it is well known that the set of natural number $ \\mathbb N$ is not bounded above . So, we can find an odd natural number $ c$ larger than $ M$, this is contradiction because $ (\\minus{}1)^{[c]}\\equal{}\\minus{}1$.\r\n\r\n2)For all $ \\varepsilon >0$, there exists $ \\delta >0$ such that $ 0<|x\\minus{}a|<\\delta\\Rightarrow |f(x)\\minus{}L|<\\varepsilon$($ L>0$). The contradictive method also gives the way in this problem. Sorry, I have to go now!", "Solution_2": "@ J.Y. Choi, thank you so much for explaining it to me. I quite understand it now. :)" } { "Tag": [], "Problem": "The exterior wall of a building forms a right angle with the ground at its base. A 25 foot ladder is placed against the wall so that the foot of the ladder is 7 feet from the base of the wall. The ladder slips and its upper end slides 4 feet down from the wall. How many feet did the foot of the ladder slide along the ground?\r\n\r\n[hide]\nmy answer was 8. Is that right?[/hide]", "Solution_1": "yes. at the beginning it is a 7-24-25 right triangle. when it slides down 4 feet it becomes a x-20-25 right triangle, x, of course, being 15. so 15-7=8. your correct! hooray! give yourself a cookie! :)" } { "Tag": [ "email" ], "Problem": "[url=http://www.lunchlady.org/files/seizure.htm]i am recovering in the hospital[/url]\r\n\r\nI had a seizure! OMG then I passed out!\r\nit reminds me of a site which you were supposed to look realy closely to find differences in pictures and then a beast thing appeared and screamed real loudly after like a minute!\r\nBTW does anyone know which site that was? I redid my emails so i have no clue....", "Solution_1": "I laugh at your predicament.\r\n\r\n\r\n\r\nAnybody that posts a link to eBaumsWorld in this topic will be disemboweled.", "Solution_2": "Contrarily, anybody who posts a link to lunchlady will be reemboweled.", "Solution_3": "AND ANYONE who has had someone have a seizure right before their eyes and is joking like this after crying like a girl before is crazy for caring about their bowels.\r\nreally my head started to hurt after 10 min. of staring at it!", "Solution_4": "[quote=\"funcia\"]AND ANYONE who has had someone have a seizure right before their eyes and is joking like this after crying like a girl before is crazy for caring about their bowels.\nreally my head started to hurt after 10 min. of staring at it![/quote]\r\nme and my buddy made something like that during web design class at school and spent the whole period staring at it. good stuff.", "Solution_5": "yeah pretty sad. \r\nit hurt my eyes" } { "Tag": [], "Problem": "I've been writing this class, and I was wondering if I could simplify this using the switch statement. The problem is that I don't know how. Anyone want to help me?\r\n\r\n[code]\tprivate String message;\n\t\n\tpublic String getLetterGrade(int gradePercent){\n\t\tKeyboardReader reader = new KeyboardReader();\n\t\t\n\t\tif (gradePercent >= 96 && gradePercent <= 100){\n\t\t\tmessage = \"Grade is A+\";\n\t\t}\n\t\telse if (gradePercent >= 92 && gradePercent <= 95){\n\t\t\tmessage = \"Grade is A\";\n\t\t}\n\t\telse if (gradePercent >= 90 && gradePercent <= 91){\n\t\t\tmessage = \"Grade is A-\";\n\t\t}\n\t\telse if (gradePercent >= 86 && gradePercent <= 89){\n\t\t\tmessage = \"Grade is B+\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 82 && gradePercent <= 85){\n\t\t\tmessage = \"Grade is B\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 80 && gradePercent <= 81){\n\t\t\tmessage = \"Grade is B-\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 76 && gradePercent <= 79){\n\t\t\tmessage = \"Grade is C+\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 72 && gradePercent <= 75){\n\t\t\tmessage = \"Grade is C\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 70 && gradePercent <= 71){\n\t\t\tmessage = \"Grade is C-\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 66 && gradePercent <= 69){\n\t\t\tmessage = \"Grade is D+\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 62 && gradePercent <= 65){\n\t\t\tmessage = \"Grade is D\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 60 && gradePercent <= 61){\n\t\t\tmessage = \"Grade is D-\";\n\t\t}\n\t\t\n\t\telse if (gradePercent >= 0 && gradePercent <= 59){\n\t\t\tmessage = \"Grade is F\";\n\t\t}\t\t\n\t\treturn message;\n\t}[/code]", "Solution_1": "Switch only works where each case is a single value - so here, where you are testing ranges, you can't really use it.\r\n\r\nThere are a couple of things you could do to simplify it - firstly, assuming gradePercent is always going to be something between 0 and 100, a lot of those conditions are unnecessary - you can get rid of all the <= conditions (other than <= 100), since they are dealt with by the 'else if' part.\r\n\r\nBut the best way to simplify it would be to use an array/loop - I get the feeling that this code is something you'd write before you started using these at all though. (You could store all the grade cutoffs and grade letters in arrays; then just loop through the array, find the highest grade cutoff that is less than your mark, and use the respective letter).", "Solution_2": "You could let the case statements fall through:\r\n[code]\nswitch (gradePercent) {\n case 100:\n case 99:\n case 98:\n case 97:\n case 96:\n message = \"Grade is A+\";\n break;\n case 95:\n case 94:\n case 93:\n case 92:\n message = \"Grade is A\";\n break;\n ...\n}\n[/code]\r\nI don't know if it's simpler than what you already have though :D." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "algebra", "polynomial", "geometry", "ratio" ], "Problem": "Just received this information from Maria Monks, one of the coaches along with Allison Miller, at the China Girls Math Olympiad:\r\n\r\nCarolyn Kim with a score of 45 got a bronze\r\nJenny Jin with a score of 39 got a bronze\r\nIn Young Cho with a score of 54 got a bronze\r\nColleen Lee with a score of 48 got a bronze\r\nJoy Zheng with a score of 48 got a bronze\r\nWendy Mu with a score of 63 got a silver!\r\nJenny Iglesias with a score of 81 got a GOLD!!\r\nLynnelle Ye with a score of 87 got a GOLD!!!\r\n\r\nThat's a medal for each one on our team! \r\nCongratulations to all the team members (and their coaches!), great work!\r\n\r\nI am posting this in the AMC forum since the team members were selected on the basis of their USAMO scores and their training was at MOSP. Many regular readers will know them.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_1": "Congratulations to all! Just curious, how exactly is the CGMO scored (save the speculative answers...they're generally not too helpful)? Also, are the problems anywhere on the internet?", "Solution_2": "There are a couple problems in the olympiad section. Look in the Combinatorics unsolved.", "Solution_3": "Hopefully this isn't too speculative for you.\r\n\r\nhttp://www.msri.org/specials/gmo\r\n[quote]Sherry got first place with 114/120 points[/quote]\r\n\r\n20 points per problem, I suppose. This would make a score of $ 87$ quite impressive, as well as all of the others.", "Solution_4": "Wow nice Lynelle...", "Solution_5": "Aren't there 8 problems?\r\n\r\nWhen will the problems be officially released?\r\n\r\nEDIT: I have found 3 problems from CGMO 2008.\r\n7.\t[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221517[/url] \r\n5.\t[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221516[/url]\r\n1.\t[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1228708#1228708[/url]", "Solution_6": "Lot of AoPSers on that list, indeed, including one who, I believe, took the first two AoPS classes ever (over 5 years ago!)", "Solution_7": "[quote=\"rrusczyk\"]Lot of AoPSers on that list, indeed, including one who, I believe, took the first two AoPS classes ever (over 5 years ago!)[/quote]\r\n\r\nWoah. This site is really old. :D", "Solution_8": "Hmm I'm going to be a total hypocrite now. But it seems that those are scores are all multiples of 3. I find it unlikely that problems are scored by one person, out of 15 points, that seems rather inconvenient. I suspect 3 people grade the paper from 0 to 5. But this may be complete BS.", "Solution_9": "There are 8 problems, each worth 5 points each.\r\nThey just multiply your score by 3 to... I dunno make you feel good.\r\nUhh, I have the problems with me, but am I allowed to post them?", "Solution_10": "[quote=\"inyoung\"]There are 8 problems, each worth 5 points each.\nThey just multiply your score by 3 to... I dunno make you feel good.\nUhh, I have the problems with me, but am I allowed to post them?[/quote]\r\n\r\n3 have already been posted, so it should be okay. Everyone starts talking about the IMO problems like immediately after the test ends, too.", "Solution_11": "[url]http://www.msri.org/specials/gmo/2008[/url]\r\n[url]http://web.mit.edu/monks/www/Pictures.html[/url]\r\nInyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm", "Solution_12": "Congrats to the teams! Upward trend-- this is a good thing! :-)\r\n\r\n[quote=\"inyoung\"]There are 8 problems, each worth 5 points each.\nThey just multiply your score by 3 to... I dunno make you feel good.\nUhh, I have the problems with me, but am I allowed to post them?[/quote]\nLast year, Alison said that there were 5 steps, if you will, that you get to in the process of solving the problem. If you get the first step, you get 3 points; the second, 6, and so on, and if you solve the problem, you get 15. So it's easier to get partials than on, say, the USAMO.\n\nAnd the obligatory disclaimer that must come with everything one says on a public forum: If I've mangled her explanation and the scoring isn't actually done that way, I take full responsibility.\n\n[quote=\"minsoens\"]Inyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]\r\nHaha oh man, they made you do the aerobic dancing thing again! How'd you guys do in the competition? (the REAL competition you went to China for-- not the math thing, silly!)", "Solution_13": "[quote=\"mihail911\"][quote=\"rrusczyk\"]Lot of AoPSers on that list, indeed, including one who, I believe, took the first two AoPS classes ever (over 5 years ago!)[/quote]\nWoah. This site is really old. :D[/quote]\nwoah, I think I feel really old, haha. ):\n\n\n[quote=\"Piggypi314\"]Haha oh man, they made you do the aerobic dancing thing again! How'd you guys do in the competition? (the REAL competition you went to China for-- not the math thing, silly!)[/quote]\r\nWe got \"most energetic.\" :lol: \r\nAnd yeah, I might have spent more time doing aerobics than I spent doing math.", "Solution_14": "[quote=\"minsoens\"]Inyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]\r\n\r\nDude, it's a girl's olympiad. They like those things. :D", "Solution_15": "[quote=\"orl\"]Dude, it's a girl's olympiad. They like those things. :D[/quote]\r\nI, for one, didn't. I'm surprised I didn't singlehandedly disqualify our dance team from \"most energetic\".", "Solution_16": "Wait...dance? Is this actually part of the main event?", "Solution_17": "[quote=\"Piggypi314\"]\r\nAnd the obligatory disclaimer that must come with everything one says on a public forum: If I've mangled her explanation and the scoring isn't actually done that way, I take full responsibility.\r\n\r\n[end quote]\r\n\r\nWhat Piggypi314 explained is what I was told about scoring before I had ever gone to the competition. When I went to the competition, I found out that scoring works on a partial credit system similar to the USAMO/IMO (although not entirely the same: on one problem points were taken off for minor arithmetic errors on solutions that would have received full credit at the IMO).\r\n\r\nAnd yes, as said before: 4 hours for 4 problems each day, 2 days, problems graded on a score of 0-15 in multiples of 3. I've been told this is because the Chinese are used to tests graded out of 120...", "Solution_18": "[quote=\"Alison\"]\nI've been told this is because the Chinese are used to tests graded out of 120...[/quote]\r\nWhoa whoa whoa where did you hear that? It's out of 100, just like anywhere else...", "Solution_19": "Hmm...my parents generally took tests out of 120 back when they were high-schoolers in China...\r\n\r\n[quote]on one problem points were taken off for minor arithmetic errors on solutions that would have received full credit at the IMO[/quote]\r\nLike 4!=12 *sigh*\r\n\r\nI think the dance lessons/competition were just to keep us busy in place of having to organize more touring...", "Solution_20": "[quote=\"bookaholic\"]Hmm...my parents generally took tests out of 120 back when they were high-schoolers in China...\n\n[quote]on one problem points were taken off for minor arithmetic errors on solutions that would have received full credit at the IMO[/quote]\nLike 4!=12 *sigh*\n\nI think the dance lessons/competition were just to keep us busy in place of having to organize more touring...[/quote]\r\nWait you serious? My parents said they took tests out of 100. I've never heard of the 120 thing until now.", "Solution_21": "[quote=\"serialk11r\"]\nWait you serious? My parents said they took tests out of 100. I've never heard of the 120 thing until now.[/quote]\r\nYeah, i went to ask my parents, and they said 100 as well...", "Solution_22": "Well it probably differed from school to school then. Like, in every French book I've seen, it says tests in France are all out of 20, but I've heard of schools in France which use 100.", "Solution_23": "[quote=\"minsoens\"][url]http://www.msri.org/specials/gmo/2008[/url]\n[url]http://web.mit.edu/monks/www/Pictures.html[/url]\nInyoung, what's going on in [url=http://web.mit.edu/monks/www/ZhongShan1/images/p7310405.jpg]this[/url] picture? :huh: .... lol hmm[/quote]\r\n\r\nIs there a web-page which has the problems, possibly with solutions, and complete results?", "Solution_24": "Hello, these problems are from Inyoung:\r\n\r\n1. (a) Determine if the set $ \\{1,2,\\cdots,96\\}$ can be partitioned into 32 sets of equal size and equal sum.\r\n(b) Determine if the set $ \\{1,2,\\cdots,99\\}$ can be partitioned into 33 sets of equal size and equal sum.\r\n\r\n2. Let $ \\varphi(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d$ be a polynomial with real coefficients. Given that $ \\varphi(x)$ has three positive real roots and that $ \\varphi(0) < 0$, prove that\r\n\\[ 2b^3 \\plus{} 9a^2d \\minus{} 7abc \\leq 0.\r\n\\]\r\n3. Determine the least real number $ a$ greater than $ 1$ such that for any point $ P$ in the interior of the square $ ABCD$, the area ratio between two of the triangles $ PAB$, $ PBC$, $ PCD$, $ PDA$ lies in the interval $ \\left[\\frac {1}{a},a\\right]$.\r\n\r\n4. Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of\r\n\\[ \\frac {XY \\plus{} YW}{AC \\plus{} BD}.\r\n\\]\r\n5. In convex quadrilateral $ ABCD$, $ AB \\equal{} BC$ and $ AD \\equal{} DC$. Point $ E$ lies on segment $ AB$ and point $ F$ lies on segment $ AD$ such that $ B$, $ E$, $ F$, $ D$ lie on a circle. Point $ P$ is such that triangles $ DPE$ and $ ADC$ are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point $ Q$ is such that triangles $ BQF$ and $ ABC$ are similar and the corresponding vertices are in the same orientation. Prove that points $ A$, $ P$, $ Q$ are collinear.\r\n\r\n6. Let $ (x_1,x_2,\\cdots)$ be a sequence of positive numbers such that $ (8x_2 \\minus{} 7x_1)x_1^7 \\equal{} 8$ and\r\n\\[ x_{k \\plus{} 1}x_{k \\minus{} 1} \\minus{} x_k^2 \\equal{} \\frac {x_{k \\minus{} 1}^8 \\minus{} x_k^8}{x_k^7x_{k \\minus{} 1}^7} \\text{ for }k \\equal{} 2,3,\\cdots\r\n\\]\r\nDetermine real number $ a$ such that if $ x_1 > a$, then the sequence is monotonically decreasing, and if $ 0 < x_1 < a$, then the sequence is not monotonic.\r\n\r\n7. On a given $ 2008 \\times 2008$ chessboard, each unit square is colored in a different color. Every unit square is filled with one of the letters C, G, M, O. The resulting board is called [i]harmonic[/i] if every $ 2 \\times 2$ subsquare contains all four different letters. How many harmonic boards are there?\r\n\r\n8. For positive integers $ n$, $ f_n \\equal{} \\lfloor2^n\\sqrt {2008}\\rfloor \\plus{} \\lfloor2^n\\sqrt {2009}\\rfloor$. Prove there are infinitely many odd numbers and infinitely many even numbers in the sequence $ f_1,f_2,\\cdots$.", "Solution_25": "Just to be sure, for #4, are you sure it is XY+YW? I just want to be sure that the Z does not matter.", "Solution_26": "I'm fairly sure that it should involve Z, since I remember the numerator being the sum of the diagonals of XYZW (although I could be wrong here)", "Solution_27": "[quote]4. Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of\n\\[ \\frac {XY \\plus{} ZW}{AC \\plus{} BD}.\n\\]\n[/quote]yeah I think this makes sense. I can't edit my post though :(", "Solution_28": "Darn, Minseon, you suck!\r\n\r\nI already solved the problem as it was originally stated: XY+YW.\r\n\r\n(The answer is $ \\frac {1 \\plus{} \\sqrt {3}}{2}$.)\r\n\r\nDarn darn darn. Well is anyone certain about the correct statement of the problem?\r\n\r\n(And anyway, XY is not a diagonal? So perhaps you mean XZ+YW?)", "Solution_29": "So does anyone know the complete results? We are looking for the relative performance of the US team comparably to other nations, don't we? :)", "Solution_30": "[quote=\"gighiuhui\"]Darn, Minseon, you suck! I already solved the problem as it was originally stated: XY+YW. (The answer is $ \\frac {1 \\plus{} \\sqrt {3}}{2}$.)[/quote]good job im so proud of you[quote=\"gighiuhui\"]Darn darn darn. Well is anyone certain about the correct statement of the problem? (And anyway, XY is not a diagonal? So perhaps you mean XZ+YW?)[/quote]I don't know why I didn't think about actually drawing the diagram before I posted. XZ+YW makes sense, but if you want reassurance from someone who has more credibility than I now do, PM or IM Inyoung or Lynnelle or Joy.", "Solution_31": "Yeah, it's XZ+YW; I checked my problem sheet", "Solution_32": "[quote=\"minsoens\"][quote=\"gighiuhui\"]Darn, Minseon, you suck! I already solved the problem as it was originally stated: XY+YW. (The answer is $ \\frac {1 \\plus{} \\sqrt {3}}{2}$.)[/quote]good job im so proud of you[/quote]\r\n\r\nHey, hey no need for all that sarcasm. \r\n\r\nAnyway, it turns out the answer is the same either way, but whatever the equality case of XY+YW is a subset of that for XZ+YW. \r\n\r\nHrmm..", "Solution_33": "[quote=\"orl\"]So does anyone know the complete results? We are looking for the relative performance of the US team comparably to other nations, don't we? :)[/quote]\r\n\r\nWe had to leave before the closing ceremony, so we didn't get much information at all about the results (all we know is our own scores/medals, and the gold medal cutoff). Zuming might have the team results by now, but I'm not sure about that.", "Solution_34": "So are these problems appreciably easier compared to last year's?\r\n\r\nI recall struggling with a few of them last time, especially that ridiculous un-symmetric inequality that I never solved (and which Sherry said she Lagrange murderplied). \r\n\r\nThis year, I was just kind of like blah blah blah. Although perhaps this is a just a consequence of (most of) the problems being appreciably nicer than last year's if not easier." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "$(a)$There are two sequences of numbers, with $2003$ consecutive integers each, and a table of $2$ rows and $2003$ columns\n$\\begin{array}{|c|c|c|c|c|c|} \\hline\\ \\ &\\ &\\ &\\cdots\\cdots&\\ &\\ \\\\ \\hline \\ &\\ &\\ &\\cdots\\cdots&\\ &\\ \\\\ \\hline \\end{array}$\nIs it always possible to arrange the numbers in the first sequence in the first row and the second sequence in the second row, such that the sequence obtained of the $2003$ column-wise sums form a new sequence of $2003$ consecutive integers? \n$(b)$ What if $2003$ is replaced with $2004$?", "Solution_1": "a) since a sequence of 2003 consecutive integers may be written as $x+1, x+2, ..., x+2003$, it's enough to consider the case where both sequences are $1, 2, ..., 2003$. the following arrangement shows that it's possible:\r\n\r\n$\\left(\\begin{array} {ccccccccc} 1 & 1003 & 2 & 1004 & 3 & 1005 & {...} & 2003 & 1002 \\\\ 1002 & 1 & 1003 & 2 & 1004 & 3 & {...} &1001&2003 \\\\ \\end{array} \\right)$\r\n\r\nb) consider more generally the case of two sequences each of length $2k$. as in (a) we may assume both sequences are $1, 2, ..., 2k$. the sum of all the elements of both these sequences is $2\\cdot \\frac{2k(2k+1)}{2} = 2k(2k+1)$. on the other hand, assume they produced a set of $2k$ consecutive integers $m+1, m+2, ..., m+2k$. the sum of these numbers is $2km+k(2k+1) = k(2m+2k+1)$, and should equal the sum $2k(2k+1)$. but this is impossible because the latter has one more factor of 2 than the former." } { "Tag": [ "function", "calculus", "derivative", "trigonometry", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $f: \\mathbb{R}\\to \\mathbb{R}$ be a derivable function. Is the graph of $f'$ connected ?", "Solution_1": "No? One can construct a function with a derivative which has a jump discontinuity.", "Solution_2": "I'm talking about the set $G=(x; f'(x))$ as a subset of $\\mathbb{R}^{2}$. For example, the graph of $x^{2}\\sin(\\frac{1}{x})$ is connected..", "Solution_3": "Nevermind, I realized I am wrong here. Sorry to spam up your topic.", "Solution_4": ":huh: I don't get what you mean ?", "Solution_5": "I believe the graph is connected.\r\n\r\nLet $f: \\mathbb{R}\\to\\mathbb{R}$ be an everywhere differentiable function. Note that $f'$ is the limit of continuous functions. Assume the graph $G$ of $f'$ is not connected and let $A_{1}$,$A_{2}$ be a disconnection. Let $K_{j}$ be the projection of $G \\cap A_{j}$ onto the real line. $K_{j}$ are $F_\\sigma$ sets that fill up $\\mathbb{R}$. Hence they are $G_\\delta$ sets. Write $K_{1}= \\cap E_{n}$, $K_{2}= \\cap F_{n}$\r\n\r\nLet $x \\in \\partial K_{1}\\cap K_{1}$ and let $M \\subset \\mathbb{R}$ be an open set containing $x$. I claim that $M$ also intersects $\\partial K_{2}\\cap K_{2}$. The key to this is that $f'$ has the intermediate value property, so that if we have a ball $B(y,r)$ contained in $K_{2}$ then actually $\\overline{B(y,r)}$ is contained in $K_{2}$. I will not prove the claim unless you don't believe me. \r\n\r\nNow let $H = \\overline{(\\partial K_{1}\\cap K_{1}) \\cup (\\partial K_{2}\\cap K_{2})}$. By what has just been said, for any $n$ $E_{n}\\cap F_{n}\\cap H$ is an open and dense subset of $H$. By the Baire Category Theorem $K_{1}\\cap K_{2}\\cap H = \\emptyset$ is dense in $H$, so actually $K_{1}$ and $K_{2}$ are both open. However, using again that $B(y,r) \\subset K_{j}\\Rightarrow \\overline{B(y,r)}\\subset K_{j}$ we find that $K_{j}$ cannot be open unless it is all of $\\mathbb{R}$, a contradiction.", "Solution_6": "I would just like to add that the last contradiction follows easier by just observing that $\\mathbb{R}$ is connected. (I am not allowed to edit my post)." } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometry unsolved" ], "Problem": "A frieng of mine asked me a hard problem, which I can do nothing on it.\r\nGiven a triangle ABC, let S denote the area of the triangle, and let L be the length of a chord of the circumcircle of ABC that divides ABC into two parts with the same area. Prove that $L_{\\text{min}}\\leq\\sqrt{\\frac{2S}{\\sqrt3}}$.\r\nHope I didn't make any grammar mistakes. :)", "Solution_1": "the problem is attractive, I have spent a lot time on it,but it seems that it is beyond my knowledge. :?" } { "Tag": [], "Problem": "ex.\r\nA survey af 300 parks showed the fallowing:\r\n15 had only camping\r\n20 had only hiking trils\r\n35 had only picknicking\r\n185 had camping\r\n140 had camping and hiking trails\r\n125 had camping and picnicking.\r\n210 had hiking trails\r\nFind the number of parks that\r\na) had at least one of these features.\r\nb) had all three features.\r\nc)did not have any of these features.\r\nd)had exactly two of these features.", "Solution_1": "I think this should perhaps be moved to the Beginner's section.", "Solution_2": "Try using PIE. \r\n\r\nI'm not exactly sure on how to approach this problem, but I think that PIE might help.", "Solution_3": "[hide=\"hint\"]Draw a venn diagram and fill it in[/hide]", "Solution_4": "With the venn diagram: make sure you dont put things into the circles until you are sure they belong there (ie subtract correctly) otherwise you will overcount. write numbers (ie there are 25 in set A, this means at least in set A, could also be in other sets, so dont assume its just in A) outside the circles before you fill in." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "given a set S={1,2,3...2001} . determine the minimum value of m such that . with arbitrry subset m elements of S has 2 elements u,v such that u + v = $ 2^n$ with n $ \\ge$ 0", "Solution_1": "[hide=\"Hint for {1,2,..., 2^k-1}\"]Consider the following $ 2^{k\\minus{}1}\\minus{}1$ pairs: $ \\{1,2^k\\minus{}1\\},\\{2,2^k\\minus{}2\\},\\cdots,\\{2^{k\\minus{}1}\\minus{}1,2^{k\\minus{}1}\\plus{}1\\}$ [/hide]", "Solution_2": "oh, yes when I thought this problem , I thought about it (your hint) ! but it was not completed!\r\ncould you write your solution more clearly? :(" } { "Tag": [ "inequalities", "trigonometry", "inequalities unsolved" ], "Problem": "Let $ A$, $ B$, $ C$ be the three angles of the triangle $ ABC$. Show that for arbitrary positive reals $ x$, $ y$, $ z$, the following inequality holds:\r\n$ x\\sin{A}\\plus{}y\\sin{B}\\plus{}z\\sin{C}\\le\\frac{1}{2}\\left(xy\\plus{}yz\\plus{}zx\\right)\\sqrt{\\frac{x\\plus{}y\\plus{}z}{xyz}}$.", "Solution_1": "This is actually easier than it looks...\r\n\r\n[color=blue][b]Problem.[/b] For any triangle ABC with angles A, B, C, and any three nonnegative reals x, y, z, prove the inequality\n\n$ \\sqrt {xyz}\\cdot\\left(x\\sin A \\plus{} y\\sin B \\plus{} z\\sin C\\right)\\leq\\frac12\\left(yz \\plus{} zx \\plus{} xy\\right)\\sqrt {x \\plus{} y \\plus{} z}$.[/color]\r\n\r\n[i]Solution.[/i] Upon squaring, this inequality becomes\r\n\r\n[color=green][b](1)[/b][/color] $ xyz\\cdot\\left(x\\sin A \\plus{} y\\sin B \\plus{} z\\sin C\\right)^{2}\\leq\\frac14\\left(yz \\plus{} zx \\plus{} xy\\right)^{2}\\left(x \\plus{} y \\plus{} z\\right)$.\r\n\r\nBy the Cauchy-Schwarz inequality,\r\n\r\n$ \\left(x \\plus{} y \\plus{} z\\right)\\left(x\\sin^{2}A \\plus{} y\\sin^{2}B \\plus{} z\\sin^{2}C\\right)\\geq\\left(\\sqrt {x\\cdot x\\sin^{2}A} \\plus{} \\sqrt {y\\cdot y\\sin^{2}B} \\plus{} \\sqrt {z\\cdot z\\sin^{2}C}\\right)^{2}$\r\n$ \\equal{} \\left(x\\sin A \\plus{} y\\sin B \\plus{} z\\sin C\\right)^{2}$,\r\n\r\nso the inequality [color=green][b](1)[/b][/color] follows from\r\n\r\n$ xyz\\cdot\\left(x \\plus{} y \\plus{} z\\right)\\left(x\\sin^{2}A \\plus{} y\\sin^{2}B \\plus{} z\\sin^{2}C\\right)\\leq\\frac14\\left(yz \\plus{} zx \\plus{} xy\\right)^{2}\\left(x \\plus{} y \\plus{} z\\right)$.\r\n\r\nThis inequality simplifies to\r\n\r\n$ 4xyz\\cdot\\left(x\\sin^{2}A \\plus{} y\\sin^{2}B \\plus{} z\\sin^{2}C\\right)\\leq\\left(yz \\plus{} zx \\plus{} xy\\right)^{2}$,\r\n\r\nor, equivalently,\r\n\r\n$ 4\\left(zx\\cdot xy\\cdot\\sin^{2}A \\plus{} xy\\cdot yz\\cdot\\sin^{2}B \\plus{} yz\\cdot zx\\cdot\\sin^{2}C\\right)\\leq\\left(yz \\plus{} zx \\plus{} xy\\right)^{2}$.\r\n\r\nBut this follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=590661#590661]http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 post #11[/url] Theorem 2 for the reals yz, zx, xy and the angles A, B, C such that A + B + C is a multiple of 180\u00b0 (in fact, A + B + C = 180\u00b0 since A, B, C are angles of a triangle). Thus, the problem is solved.\r\n\r\n[b]PS.[/b]\r\n\r\n[color=blue][b]Equivalent version of the Problem.[/b] For any triangle ABC with angles A, B, C, and for any acute-angled triangle XYZ with angles X, Y, Z, prove the inequality\n\n$ \\tan X\\sin A \\plus{} \\tan Y\\sin B \\plus{} \\tan Z\\sin C\\leq\\frac12\\left(\\tan Y\\tan Z \\plus{} \\tan Z\\tan X \\plus{} \\tan X\\tan Y\\right)$.[/color]\r\n\r\n Darij" } { "Tag": [ "geometry", "inequalities", "trigonometry", "number theory", "complex numbers", "\\/closed" ], "Problem": "Which level of subject classes would you seriously consider taking during the Spring semester?\r\n\r\nIf you would like you can mention which classes you are most likely to take. We are more likely to offer classes that generate a high level of response.", "Solution_1": "By spring semester I would probably want to take an intermediate geometry, intermediate number theory (I don't know if you'll be able to offer these), or Olympiad courses in pretty much any other subject (algebra, inequalities, counting, etc.)", "Solution_2": "[quote=\"jeffrey\"]By spring semester I would probably want to take an intermediate geometry, intermediate number theory (I don't know if you'll be able to offer these), or Olympiad courses in pretty much any other subject (algebra, inequalities, counting, etc.)[/quote]\r\n\r\nWe may have an Intermediate Number Theory course prepared by the Spring. If not it will likely be prepared by the summer.", "Solution_3": "[quote=\"Rep123max\"]I will want to be taking Olympiad courses most likely, except for an Intermediate Geometry and Number Theory. For Olympiad courses, it would be Olympiad Problem Solving and will there be an Olympiad Algebra?[/quote]\r\n\r\nWe will eventually have an Olympiad Algebra course, but it is several courses back on our current list of classes to write.", "Solution_4": "Are there any plans to have an Intermediate Geometry course in the spring (or anytime soon)?", "Solution_5": "We hope to have one by the end of 2005, but it won't be in the spring, unfortunately.", "Solution_6": "This semester we were seriously interested in the Complex Numbers and Trigonometry course, now in progress, but took another course on the same evening. Having that one again in the spring (on any night but Thursday) would be wonderful.", "Solution_7": "We have scheduled the Spring 2005 Complex Numbers/Trigonometry class for Tuesdays. We will soon be making broader announcements about class listings." } { "Tag": [ "counting", "distinguishability" ], "Problem": "A nursery employee wishes to plant 2 Golden Delicious apple trees and 5 Bartlett pear trees in one row. How many distinct arrangements are possible?", "Solution_1": "This question should have a clause stating that trees of the same type are indistinguishable. Otherwise, the question is too confusing. :mad:", "Solution_2": "The way I see it is, if the trees were all distinguishable, then why would they take the time to specifically state that there are 2 Golden Delicious apple trees and 5 Bartlett pear trees? This is just my opinion; if there is some other rule that applies to this or if I am wrong, someone please post/correct me.", "Solution_3": "There are 7 places for the trees, and 2 of these 7 must be Golden Delicious apple trees. The answer is 7 choose 2.\r\n\r\n\r\n$ \\binom72\\equal{}\\frac{7\\times6}{2}\\equal{}\\boxed{21}$", "Solution_4": "The answer is 21.......", "Solution_5": "the answer is 9 + 10 = 21 ( do u get it )", "Solution_6": "[hide = Solution]There are seven trees total, and we have to choose where the 2 Golden Delicious apple trees go. So, there are $\\binom{7}{2} = \\boxed{21}$ ways.[/hide]", "Solution_7": "[hide=Alternate Solution]\nYou can first solve for organizing 7 trees in 7 spaces, so you get $7!$. However, there are 2 repeats for the apple trees, and 5 for the pear trees. So, you get $\\frac{7!}{5!2!}=\\boxed{21}$.\n[/hide]\n\nIn response to the posts about clarity, the Alcumus problem states the distinguishability of the trees: \n[quote=Alcumus]A nursery employee wishes to plant 2 identical Golden Delicious apple trees and 5 identical Bartlett pear trees in one row. How many distinct arrangements are possible? [/quote]\n\n", "Solution_8": "There are $7!$ ways to arrange the plants if they are distinguishable. Since the are not, we must divide by $2! \\cdot 5!$. Our answer is $\\frac{7!}{2!5!}=21$", "Solution_9": "[hide=Sol]Ans = (5+2)c5 or (5+2)c2 = 21[/hide]" } { "Tag": [ "limit", "trigonometry", "calculus", "derivative", "calculus computations" ], "Problem": "This is a problem my calc teacher made up.\r\n\r\nassume that $\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2}=\\frac{1}{2}$\r\nFind:\r\na) $\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x}$\r\n\r\na) $\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x^3}$\r\n\r\nThanks in advance.", "Solution_1": "Perhaps there is a way to evaluate them using the given limit. But using L'Hopital's Rule, we find the first limit is $0$, and using L'Hopital and Squeeze Theorem, we find the second limit is also $0$. Although I dont' think you need the squeeze theorem...", "Solution_2": "when you look at the top of this board, it says to not put calculus problems here, there is a special board for calculus that you can post calculus problems in", "Solution_3": "I asked Mr. Crawford and he said it was okay to post a limits problem.\r\nWhat is L'Hopital's Rule?\r\n\r\nMODERATORS:FEEL FREE TO MOVE TOPIC.", "Solution_4": "This limit problem seems to be more advanced than the type I think Mr. Crawford would allow to be posted here.\r\n\r\nBut... L'Hopital's Rule is as follows:\r\n\r\nIf $\\lim_{x \\to a} \\frac{f(x)}{g(x)}= \\frac{0}{0}$ or $=\\frac{\\infty}{\\infty}$, $\\lim_{x \\to a} \\frac{f(x)}{g(x)}= \\lim_{x \\to a} \\frac{f'(x)}{g'(x)}$\r\n\r\nIn other words, if a limit that you are evaluating is interdeterminate (check spelling) in both the numerator(f(x)) and the denominator(g(x)), you can evaluate the limit as the first derivative of the numerator divided by the first derivative of the denominator. This will give you the same answer.\r\n\r\n It's kinda hard to explain. If somebody else can do it in a more understandable way, please do so.", "Solution_5": "Moved from intermediate.", "Solution_6": "Multiply denominator and nominator with $1 + \\cos x$ and you can just use that $\\lim_{x\\to 0}\\frac{\\sin x}{x} = 1$.\r\n\r\nedit: Oh, you're supposed to use the thing you're supplied with. Well,\r\n$\\frac{1-\\cos x}{x} = x\\cdot\\frac{1 - \\cos x}{x^2}$ so it tends to zero.\r\n\r\n$\\frac{1-\\cos x}{x^3} = \\frac{1}{x}\\cdot\\frac{1 - \\cos x}{x^2}$ so the limit doesn't exist." } { "Tag": [], "Problem": "Is anyone else going?", "Solution_1": "No. My school doesn't even do Science Olympiad. :(", "Solution_2": "That sucks :( \r\n\r\nWe didn't get to go because we had a snow day Friday and our coaches said that if the roads were bad enough to close school it was too dangerous for us to drive to chicago!! We might go to another one though.." } { "Tag": [ "limit", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "I was given such a problem and still thinking how to solve:\r\n\r\nFind the limit ${\\lim_{n\\to +\\infty}({2\\over {\\ln(n+1)}} \\int_{x=0}^{\\pi\\over 2}{{\\sin^2 (n+1)x}\\over {\\sin x}}\\,dx )}$ :ninja:", "Solution_1": "Let $I_n=\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin ^ 2 (n+1)x}{\\sin x} dx,$ let's calculate $I_n-I_{n-1}.$", "Solution_2": "First step: $\\frac{\\sin (n+1)x}{\\sin x}=e^{-inx}+e^{i(2-n)x}+\\dots+e^{i(n-2)x}+e^{inx}$.\r\n\r\nNow integrate each term (multiplying by the other $\\sin ((n+1)x)$ and estimate." } { "Tag": [ "search", "\\/closed" ], "Problem": "Eh?\r\n :?:\r\nThe title says... all.", "Solution_1": "The search button is your friend a quick search for image signature revealed this\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709190#709190[/url]", "Solution_2": "This has been pointed out many times." } { "Tag": [ "vector", "real analysis", "real analysis unsolved" ], "Problem": "proved that (u x v) x w = u x (v x w) iff (u x w) x v=0\r\n\r\nusing only a geometric argument...\r\n\r\nu,v,w are all vectors.\r\nu x v => the cross product of u and v", "Solution_1": "Assume $u,v,w$ are unit vectors from the origin satisfying $[[u , w] , v] = 0$. We will prove \\[[[u,v], w] = [u , [v , w]] \\;\\;\\;\\;\\; \\text(1) \\]\r\n If $[[u , w] , v] = 0$ , which is equivalent to $v \\perp (u,w)$, then $[u , v] = u' \\subset (u,w) $ ( because $u' \\perp v$ ) and $ \\perp u$. Similarly we have $w'$. It is easy too see that the oriented angles $(w',w) =(u,u')$ , so the the resulting vectors in $(1)$ are the same .\r\n Now let \\[[[u , v] , w] = [u , [v , w]] = a \\] If $a = 0$ then $u,w,v$ are perpendicular to each other....\r\n It $a \\not= 0$, it is easy to see that $a \\perp (u,w).$ Then $[v , w] \\perp a,v,w$ then $a,v,w$ lie on the same plane. Similarly $a,v,u$ lie on the same plane. Therefore if $a $ is not parallel to $v$ then $a,v,w,u$ lie on the same plane which is absurd because $a \\perp (u,w)$. That implies $v \\perp u,w$ hence $[[u , w] , v] = 0$", "Solution_2": "Use Jacobi Identity $U\\times\\left(V\\times W\\right)+V\\times\\left(W\\times U\\right)+W\\times\\left(U\\times V\\right)=0$" } { "Tag": [], "Problem": "Two objects A and B of lengths 120 meters and 180 meters take 30 seconds to cross each when they travel in the same direction. However, if they were to travel in opposite directions, then they will cross each other in 7.5 seconds. If object A is faster than object B, then what is the speed of object A?\r\n\r\nKindly explain the solution. Thanks", "Solution_1": "I just set up a bunch of equations.\r\n\r\n$30(a - b) = 120$\r\n\r\n$7.5(a + b) = 300$\r\n\r\nSolving for $a$ I get $a= 22$.\r\n\r\nHopefully that's correct :|", "Solution_2": "Actually the answer is 90. I just dont know where did it come from. :blush:", "Solution_3": "I got 22 also. Maybe we misarranged the first equation or the answer key is wrong.", "Solution_4": "The question is from this site:\r\n\r\nhttp://free-quiz.4gmat.com/showquiz.php?quiz=22", "Solution_5": "What exactly does cross each other while they are travelling in the same direction mean ??", "Solution_6": "I think this is what it means \"cross each other\" means overtake since we are talking about the same direction( correct me if Im wrong). For example objects A and B are trains. If two trains start at the same place, one will eventually overtake the other if they have different speeds. So if the other train is longer, it will take sometime for the shorter train to overtake it.\r\n\r\nI think the problem itself is quite ambiguous.", "Solution_7": "Answer \r\n\r\n[hide](a-b)30 = 300 ==> a-b = 10\n(a+b)7.5=300 ==> a+b = 40\n\n2b+10=40\n==> b = 15\n==> a = 25 m/s\n\na = 25/1000 * 3600 = 90km/hr[/hide]", "Solution_8": "Could you explain how you got $30 (a - b) = 300$ ??", "Solution_9": "I can explain that. The two objects are going in the same direction. Therefore, the speed of the faster object decreased by the speed of the slower object gives you the rate at which the two objects are moving past each other. Since the two objects take 30 seconds to pass each other, 30 seconds multiplied by the rate of increase gives you 30(a-b)\r\n\r\nBecause they are moving parallel to each other, the distance the two objects must travel to pass each other is 300. Therefore, \r\n\r\n30(a-b) = 300" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A group of 6 students decided to make [i]study groups[/i] and [i]service activity groups[/i] according to the following principle:\n\nEach group must have exactly 3 members. For any pair of students, there are same number of study groups and service activity groups that both of the students are members.\n\nSupposing there are at least one group and no three students belong to the same study group and service activity group, find the minimum number of groups.", "Solution_1": "I think the answer is 8 groups.\r\n\r\nWhat about if there were $n$ students? (Let's assume that everybody has to be in at least one group) ;)", "Solution_2": "any full solution , please !", "Solution_3": "Clearly the number of groups is 2k: k study groups and k service activity groups.\nLet k=3. $G_1, G_2, G_3 ...$ - study groups, $J_1, J_2, J_3 ...$ - service activity groups.\nWLOG $G_1=\\{1;2;3\\}$. Then $J_1=\\{1;2;x\\} J_2=\\{1;3;y\\} J_3=\\{2;3;z\\}$ and it's all J-groups \nIf x,y and z are different then k>3. (Any two of pairs $\\{1;x\\}, \\{3;y\\}, \\{2;z\\}$ cann't be in one G-group, so it's at least 4 G-groups)\nIf x=y and x\u2260z then k>3. ($G_2=\\{1;x;v\\} G_3=\\{1;x;u\\}$ and $u,v$\u2260$z$)\nFinally if x=y=z it's also no solutions.\n8 groups are possibly:\n$G_1=\\{1;2;3\\}, G_2=\\{1;4;5\\}, G_3=\\{2;4;6\\}, G_4=\\{3;5;6\\}, J_1=\\{4;5;6\\}, J_2=\\{1;2;4\\}, J_3=\\{1;3;5\\}, J_4=\\{2;3;6\\}$", "Solution_4": "The answer is $\\boxed{8}.$ Let the students be labelled $1, 2, 3, 4, 5, 6.$ Then, taking the groups $(1, 2, 3), (3, 4, 5), (5, 6, 1), (2, 4, 6)$ to be study groups and $(2, 3, 4), (4, 5, 6), (6, 1, 2), (1, 3, 5)$ to be service activity groups works. \n\nLet's show that less than $8$ groups is not possible. Observe firstly that there must be the same name of groups. Clearly $2$ groups is impossible. It's also easy to see that $4$ groups is impossible. Let's now show that $6$ is impossible, i.e. when there are $3$ of each type of group (study or service activity).\n\nIf there is no pair of students who are in at least two study groups together, then the logic is rather easy. Indeed, two study groups must share a triangle, from which we can deduce that the other study group has student not in the first two, together with one student from each of the first two study groups. Then there clearly is no way to partition these people into service activities.\n\nOtherwise, if there is a pair of students who are in three study groups together, the logic is also easy.\n\nFinally, suppose there is a pair of students $(x, y)$ which are in exactly two study groups together. Then, the other study group can contain at most one of these two students. Hence, it's clear that any partition of these edges into service activity groups must have the same two triangle (which contain $(x, y)$). This is a contradiction.\n\n$\\square$ " } { "Tag": [ "AMC", "AIME" ], "Problem": "If you take the AMC 10/12 at a college you can still qualify for AIME right? Is it any different than taking it at a high school?", "Solution_1": "Yes, you can still qualify for the AIME as long as you take it on the correct day. It is the same as if you had taken it at a high school.", "Solution_2": "Colleges and Universities General Instructions and Information\r\nPage, 4, Item 4 d.\r\nhttp://www.unl.edu/amc/b-registration/b1-archive/2009-2010/Colleges/2010-College_Booklet.pdf\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "induction", "algebra", "polynomial" ], "Problem": "How do you evaluate the following summations?\r\n\r\n1. $ \\displaystyle\\sum_{k\\equal{}1}^{n}k^2\\binom{n}{k}$\r\n\r\n2. $ \\displaystyle\\sum_{k\\equal{}1}^{n}(\\minus{}1)^{k\\plus{}1}k^2\\binom{n}{k}$", "Solution_1": "$ \\boxed{k{}_nC{}_k=n_{n-1}C_{k-1}}$\r\n\r\n${ \\boxed{\\boxed{(1+x)^n=\\sum_{k=0}^n {}_nC{}_k x^k}}}$", "Solution_2": "hello, for the first sum we can prove by induction that\r\n$ \\sum_{k\\equal{}1}^{n}k^2\\binom{n}{k}\\equal{}2^{n\\minus{}1}n\\plus{}2^{n\\minus{}2}n(n\\minus{}1)$.\r\nSonnhard.", "Solution_3": "[quote=\"FantasyLover\"]How do you evaluate the following summations?\n\n1. $ \\displaystyle\\sum_{k \\equal{} 1}^{n}k^2\\binom{n}{k}$\n\n2. $ \\displaystyle\\sum_{k \\equal{} 1}^{n}( \\minus{} 1)^{k \\plus{} 1}k^2\\binom{n}{k}$[/quote]\r\n\r\n[b]Solution of 1[/b]\r\n\r\nconsider $ \\ (1\\plus{}x)^n\\equal{}\\sum_{k \\equal{} 0}^{n}\\binom{n}{k}x^k$\r\n\r\nand follow the following steps:\r\n\r\nStep 1: Differentiate both side with respect to $ \\ x$\r\n\r\nStep 2: Multiply both side by $ \\ x$\r\n\r\nStep 3: Differentiate both side with respect to $ \\ x$\r\n\r\nStep 4: Put $ \\ x\\equal{}1$ on both side and there goes your problem.\r\n\r\n[b]Solution of 2[/b]\r\n\r\nFollow the steps of first solution and in the final step 4:Put $ \\ x\\equal{} \\minus{}1$ on both side and there goes your problem.\r\n\r\nA generalization of problem 2\r\n\r\nProve that $ \\ \\displaystyle\\sum_{k \\equal{} 0}^{n}( \\minus{} 1)^{k}f(k)\\binom{n}{k}\\equal{}0$ where $ \\ f(k)$ is a polynomial of degree $ \\ < n .$", "Solution_4": "Wow, nice solution.\r\n\r\nThank you. :)" } { "Tag": [ "limit", "real analysis", "real analysis solved" ], "Problem": "Let a1 > 0 be a given real number. Define the sequence {an} recursively as follows :\r\n\r\nan+1 = an + an-2\r\n\r\nfor n = 1, 2, 3, ... . Prove that \r\n\r\nn\\lim\\infty ann-1/3\r\n\r\nexists and find it.", "Solution_1": "We will find \\lim a n 3/n which is the same as \\lim a n+1 3-a n 3 by Cesaro-Stolz. But a n+1 3-a n 3 =(a n+1- a n)( a n+1 2 +a n+1 a n +a n 2 )=(a n+1 /a n) 2 +a n+1 /a n+1, so we only have to find the limit of a n+1 /a n, which is 1 because a n->\\infty.", "Solution_2": "hi pcalin, you can have a better approximation\r\n\r\nan = (3n)1/3( 1 + ln(n)/(9n) + \\varepsilon n) \r\n\r\nwith \\lim \\varepsilon n = 0", "Solution_3": "An other approach :\r\n\r\na(n+1)^3 = a(n)^3 + 3 + 3/a(n)^3 + 1/a(n)^6\r\n\r\nb(n) = a(n)^3\r\nb(n+1) = b(n) + 3 + 3/b(n) + 3/b(n)\r\n\r\nb(n) >= b(1) + 3(n-1)\r\nb(n) <= b(1) + 3(n-1) + O(ln(n))\r\n\r\nThat's enough to prove b(n)/n -> 3" } { "Tag": [ "videos" ], "Problem": "I wanted to know what you people do for entertainment, excluding math and other academic excercises of course. Maybe you can give me some fun suggesstions?", "Solution_1": "Music, movie, and videogame (w/ some espn)", "Solution_2": "I usually play video games, DDR or watch some Anime.", "Solution_3": "Video games. Specifically, Resident Evil 4 and FIFA 2006 (PS2). So addictive...\r\nI also play the violin some... I watch tv sometimes (only when there's good stuff on).", "Solution_4": "But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?", "Solution_5": "[quote=\"hello\"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]\r\n\r\nI go out and eat...that's fun.\r\nUnfortunately, I don't live in the city like some of my friends...I live in Hastings on Hudson, a little sleepy town in Westchester. So even when I have time to go places, it's not that easy. Occasionally, I go watch an awaited movie with friends or invite them over and go to the park. But it's not that often...\r\n\r\nDuring the week, I'm generally too busy for much entertainment. But school is fun, so meh.", "Solution_6": "I go for a walk....\r\nlisten to or play music...\r\nmeet friends.... \r\n\r\n :)", "Solution_7": "[quote=\"hello\"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]\r\n\r\nOh. Me and my friends go to the mall to hang out every once in a while. Or the arcade. =P", "Solution_8": "[quote=\"hello\"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]\r\n\r\nWell, I'm not really able to do a lot of these things, but you could go to concerts, movies, malls, restaurants, plays, and lots of other stuff.", "Solution_9": "I like to hang out with my friends most of the time. Otherwise, I swim, and I fence.", "Solution_10": "i drink, smoke hang out with palz, listen to music play rpg's :D :D", "Solution_11": "I generally go out and play basketball.......... really don't have a habit of hanging around anywhere though........ :D", "Solution_12": "I read, write nonsense, watch UEFA Champion's League, and play PC strategy games", "Solution_13": "Any one of you by any chance every played paintball? I am interested in playing that so I am curious. So you people rarely do non studying things, right?", "Solution_14": "[quote=\"hello\"]Any one of you by any chance every played paintball? I am interested in playing that so I am curious. So you people rarely do non studying things, right?[/quote]\r\n\r\nWow, what a generalization. \r\n\r\nI'm ALWAYS doing non-studying things. Juggling clubs and hobbies for years. Not to mention a social life too.\r\n\r\nNever tried paintballing, but I'm told its a lot of fun. I go laser-tagging more often though, its closer to home for me.", "Solution_15": "Paintballing is loads of fun, it doesn't really hurt that much either.", "Solution_16": "Err.........what's paintball ?? I really haven't quite heard of it before... :oops:", "Solution_17": "Must be ballpainting.\r\n\r\nOr if you are really imaginative, you really can play around balls using paint brushes ;)", "Solution_18": "I read sci-fi and fantasy, I juggle, I play video games, I go on AOPS, and I IM people.\r\nThat's all I can think of right now, I will probably come up with some more things later.", "Solution_19": "I sleep a lot.", "Solution_20": "Ok, juggling. But do you people do things outdoors?", "Solution_21": "Well, you could go camping with friends, or go backpacking, or canoeing, or kayaking, or go biking.", "Solution_22": "[quote=\"hello\"]Ok, juggling. But do you people do things outdoors?[/quote]\r\n\r\nOutside?!? But that's where the (gulp) Sun lives! \r\n(Just kidding. Although I am currently sitting in my room with the lights off and the shades closed in the middle of the day... :ninja: )\r\n\r\nYes, in reality I do do things outside- cross-country skiing in the winter (which is about 7 months of the year) and rowing or soccer the other five months.", "Solution_23": "i do baseball, soccer, and tennis, and play a lot of basketball at home", "Solution_24": "Mr Potter, where do you do kayaking and canoeing, since you live in Pennsylvania. Do you go to the poconos? I am interested in going there, so I am trying to find out.", "Solution_25": "Uh, I live in Ohio. But I was just listing ideas for what you could do, I've only gone conoeing once, and that was in Canada, kind of near Toronto.", "Solution_26": "I play alot of basketball. I also generally go chill with a couple friends in the evening after I'm completely exhausted from doing math in the mornings (this is more applicable to spring break right now, hah).\r\n\r\nYea, video games too", "Solution_27": "Watch T.V(Friends,Simpsons,soccer,tennis,nickelodeon-specificallyKenan n Kel,jimmyneutron,spongebob)I'm really into T.V\r\n\r\nI'm not a bookaholic \r\n\r\nAoPS(though now i'm trying to come on AoPS for doing math :roll: )\r\n\r\nPlay Tennis,soccer,table tennis,etc(I'm the best outdoor sportsman you can ever get)\r\n\r\nI like to Just sit back and $Imagine$", "Solution_28": "[quote=\"hello\"]Ok, juggling. But do you people do things outdoors?[/quote]\r\n\r\nLol, when I said juggling clubs and hobbies, i meant that I've been balancing a huge number of clubs and committments. I just realized that sounded funny. No, I don't juggle, lol.", "Solution_29": "OK, any more suggesstions here? For those that tried paintball, can anyone explain in more detail?", "Solution_30": "[quote=\"bubka\"]Must be ballpainting.\n\nOr if you are really imaginative, you really can play around balls using paint brushes ;)[/quote]\r\n\r\nL. O. L.\r\nMy gosh, that got me laughing, no offense. You're kidding? Or are you serious?\r\n\r\nPaintballing is when you go to these places where you can play. You put on like protective gear and get a paintball gun, and buy however many paintballs (balls filled with paint) you want. Then you go in the field or arena or whatever, and like... shoot people. \r\nIt's fun. I've only done it once, but it. is. FUN.\r\nLaser tag is fun too. \r\nBoth Paintball and laser tag are I guess \"easier\" for smart people, not sure exactly why.\r\nNo stereotyping intended.\r\n\r\nNowadays I have little time, so I play basketball whenever my neighbor's friends are over.", "Solution_31": "For entertainment I have a very peculiar activity [hide=\"my peculiar activity\"][hide=\"my very peculiar activity\"][hide=\"my very very peculiar activity\"]Post mesages on an odd website called: [url=http://www.artofproblemsolving.com/Forum/index.php?f=139]a very peculliar website[/url][/hide][/hide][/hide]\r\n\r\nNo but seriously my outdoor hobies are playing Table tennis (I play almost everyday), tennis, and basketball", "Solution_32": "I've done paintballing several times. Essentially like a battle, but using paintballs. Quite fun. The key is not to hide...I used to do so and shoot people from afar if I could when I was young, being scared :D \r\nNow, I just run through. Really, getting shot might hurt a bit, but with all of the adrenaline pumping, the pain is unnoticeable.", "Solution_33": "i spend most of my time playing video games and playing tennis.", "Solution_34": "i play tennis, soccer, ride bikes, and try to play on my pc as much as i can.", "Solution_35": "Music, movie, computer games and graphics, reading, playing quitar, surfing (internet), skiing", "Solution_36": "[quote=\"shadysaysurspammed\"]Watch T.V(Friends,Simpsons,soccer,tennis,nickelodeon-specificallyKenan n Kel,jimmyneutron,spongebob)I'm really into T.V\n\nI'm not a bookaholic \n\nAoPS(though now i'm trying to come on AoPS for doing math :roll: )\n\nPlay Tennis,soccer,table tennis,etc(I'm the best outdoor sportsman you can ever get)\n\nI like to Just sit back and $Imagine$[/quote]\r\n\r\nI LUV the simpsons! Remember the time when maggie shot Mr. Burns?" } { "Tag": [ "MIT", "college" ], "Problem": "Hello,\r\n\r\nI have some questions that the answers have eluded me so far. Let's say an intelligent person studies on his/her own at home, yet he/she does not do school work due to boredom, et cetera, and thus gets fairly low marks. If this person wants to continue his studies or research, what should this person do without being able to go to university? Are there any facilities that this person can make use of? Thanks in advance.\r\n\r\nMasoud Zargar", "Solution_1": "MIT OpenCourseWare?", "Solution_2": "I can't imagine that \"this person\" has low enough marks to be unable to go to any university. Also, somewhat high test scores ought to help.", "Solution_3": "It doesn't matter if you get bored or not, you have to finish school anyway.......", "Solution_4": "Well, school is the prison of the small men ;)\r\n\r\nAs I know there are exceptions possible (here, don't know systems in other countries).", "Solution_5": "See, this is what I don't get. I was bored at my old school- beyond bored, my lowest grade was in PE (and PE is always a giveaway grade). So I [i]changed schools[/i]. I took the opportunities available to me, and don't tell me there are no opportunities available to you. Off the top of my head, I can name three top public high schools in my state (admission based on talent, not geography), and I live in a state infamous for its education funding. I highly doubt that colleges will accept boredom as an excuse. They expect you to find a way to not be bored.", "Solution_6": "when i read the first message in this topic, the first thing that came to my mind was my own status at school. :roll: just like u, i have an interest in math but get bad grades at school. in fact, my lowest grade was in math. :P :blush:", "Solution_7": "Those who knew me can probably attest to my TERRIBLE grades during HS (and college for that matter). \r\n\r\nHaving gone through the experience, I highly suggest you take your GPA far more seriously than you have. \r\n\r\nBleumoose is definitely right about colleges not accepting boredom as an excuse. However, colleges are willing to make exceptions if you show that you are very good at what you are passionate about. Consider doing USACO if you are somewhat interested in CS, etc.", "Solution_8": "My Suggestion is to suround your self with subjects you like then the ones you dont like dosent seem as bad (if you have more that you like then dislike) for example next year im taking 4 math classes Physics 2 and Economics.", "Solution_9": "that happened to me...\r\n\r\ni got good grades my freshman year, but each year my grades basically went down a letter grade.\r\n\r\nso right now i'm going to a community college that's non-competitve, meaning there was no admission process, when i applied they weren't full so i got in. right now i'm trying to get good grades and i'll transfer.", "Solution_10": "I teach, and am an academic advisor, at a university where admissions is not highly selective, and is fairly automatic based on test scores and GPA. I've seen combinations like a 1510 SAT (the old version) with a 2.7 high school GPA, and I've seen combinations like a 960 SAT with a 4.0 GPA.\r\n\r\nThe students with the high grades and the low test scores do better in our university than do the students with the low grades and the high test scores.\r\n\r\nIf I see that combination - high test scores with low grades - I see two possibilities:\r\n\r\n1. The person is a misunderstood genius, whose brilliance cannot be captured within the narrow confines of high school, or\r\n\r\n2. The person is a flake and a lazy bum.\r\n\r\nI'd estimate that there are at least 99 flakes for every misunderstood genius. The bad grades are a bad indicator. And a flake might be able to pass some courses without working at them just on native intelligence - but sooner or later, there will come a course that he can't pass without some serious work.\r\n\r\nBut life does come with second chances, and I have seen ex-flakes who discovered both their true interest and some useful self-discipline while in college.", "Solution_11": "Academics mean a lot to students these days though\r\nNow, GPA is just as important as SAT or ACT scores.\r\n-sigh-\r\nthat's a bad thing for me", "Solution_12": "[quote=\"Aryth\"]Now, GPA is just as important as SAT or ACT scores.[/quote]\r\n\r\nYou say that as if there was a time when the opposite was true. (In fact, grades are probably more important to colleges than any single standardized test score, as they should be.)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Among the points corresponding to number $1,2,...,2n$ on the real line, $n$ are colored in blue and $n$ in red. Let $a_1,a_2,...,a_n$ be the blue points and $b_1,b_2,...,b_n$ be the red points. Prove that the sum $\\mid a_1-b_1\\mid+...+\\mid a_n-b_n\\mid$ does not depend on coloring , and compute its value. :roll:", "Solution_1": "This is called Proizvolov's identity, I believe. You forgot to mention that $a_1,\\ldots,a_n$ are the numbers of the first set [b]in increasing order[/b], while $b_1,\\ldots,b_n$ are those in the second set [b]in decreasing order[/b] (or vice-versa, of course; the idea is that the sequences are ordered in opposite directions).\r\n\r\nJust observe that in the pairs $(a_1,b_1),\\ldots,(a_n,b_n)$, each numebr $\\le n$ is paired up with a number $\\ge n+1$, so the sum is, in fact, $(n+1)+\\ldots+2n-(1+2+\\ldots+n)=n^2$.\r\n\r\nMaybe this observaton isn't that trivial, but I don't have the time to go into details right now, sorry :)." } { "Tag": [ "function", "inequalities unsolved", "inequalities" ], "Problem": "Let $x_{k} \\in [0,1]$ $ \\forall k=1,n$. Prove that: $x_{1}+x_{2}+ \\ldots + x_{n} - (x_{1}x_{2}+x_{2}x_{3}+ \\ldots +x_{n}x_{1})$ is less or equal to: a) 1 for n=3; b) 2 for n=4; c) $[ \\frac{n}{2} ]$, for $n \\geq 3$.", "Solution_1": "Easy problem :D", "Solution_2": "this is bulgaria 95 ,i think.and it uses keeping all other values constant except one the function will be linear so minima will be reached when all variables take the boundary values (either of them)" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Consider the circumference sorrounding the isoceles triangle ABC (with AC=BC). In the arc BC opossite from A we choose a point D. Let E be a point in AD such that CE and AD are perpendicular. prove that AE=BD+DE.\r\n\r\n it has a lot of diferent solutions", "Solution_1": "By Ptolemy's theorem, we have $(DA-DB)\\cdot AC=DC\\cdot AB$. We have to show that $DA-DB=2DE$, so we have to prove that $\\frac{CD}{CB}=\\frac{CE}{CM}$, where $M$ is the midpoint of $BC$, which is clear since $CAM,CDE$ are similar.", "Solution_2": "nice\u00a1\u00a1\u00a1\u00a1 :D", "Solution_3": "Is this not the Archimedes Midpoint Theorem? http://mathworld.wolfram.com/ArchimedesMidpointTheorem.html" } { "Tag": [ "MATHCOUNTS" ], "Problem": "How many positive three-digit integers less than 400 satisfy each of the following characteristics simultaneously? \r\n\r\n-Each of the three digits is prime. \r\n-The sum of the three digits is prime. \r\n-The number is prime.", "Solution_1": "[hide]2\n\n233\n\n323[/hide]", "Solution_2": "2+3+3 isn't prime :)", "Solution_3": "Also 2 is not a three-digit number... :)", "Solution_4": "I think the 2 was his answer to your question ;)", "Solution_5": "Oh I see...\n\nHint: [hide]consider the three things given in this order:\n\n1.) Each of the three digits is prime. \n\n2.) The number is prime.\n\n3.) The sum of the three digits is prime.[/hide]", "Solution_6": "oops... \n\n[hide]1\n\n223[/hide]", "Solution_7": "I'm afraid there's far more solutions than that...unless you do this systematically and organize your work, sooner or later you'll lose count since there's so many...", "Solution_8": "exactly why i did not attempt this problem", "Solution_9": "ok... then I have no idea.", "Solution_10": "Think about what the ones digit[b] can't[/b] be. That should reduce the possibilities significantly. Then I guess it's just brute force/trial and error...if anyone have any elegant way of attacking this problem it would be really helpful since it's really annoying by trial and error, especially to check and see if the number is a prime or not...yeck! :lol:", "Solution_11": "OoOoOoOo....\r\nI was only using 2's and 3's.... you can use the other ones... oops.", "Solution_12": "I found that it was simpler to check for the sum of the digits being prime first. So that gives you that the sum of the digits is 19, 17, 13, 11, or 7. (It's bounded because each digit is between 2 and 7.) This gives us the following possibilities:\n19 = 7+7+5\n17 = 7+7+3 = 7+5+5\n13 = 7+3+3 = 5+5+3\n11 = 7+2+2 = 5+3+3\n7 = 3+2+2\n\nSo, checking these out, we have to try\n577 757 377 737\n773 557 337 373\n733 553 227 533\n353 223\n\nAnd you only have to check divisibility by primes up to 29 at most for any of them . . . I mean, it's tedious, but I don't see there being an easier way.", "Solution_13": "The original problem wants all of those primes LESS than 400. furthermore, 377,737,553,and 533 are all composite.", "Solution_14": "So that shortens it down to 337 373 227 353 223.\nNote, that if there was no 400 bound, this would be tedious, if not for a very slick way: [hide]Write a program in Java and let the computer calculate it! :rotfl:[/hide]", "Solution_15": "Well, logically speaking, the 3 digits can either be 2,3,5, or 7, and the last digit cannot be 2 or 5, or otherwise, the number itself wouldn't be prime. This should narrow down some cases..." } { "Tag": [ "articles", "Stanford", "college" ], "Problem": "http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences\r\n\r\nThoughts? My own view is essentially that of Tegmark's - I hadn't even thought deeply on this before and more or less implicitly assumed such a statement. \r\n\r\nhey also guys did you know godel had a completeness theorem\r\n\r\nthat was kind of irrelevant oh well I just felt like sharing\r\n\r\nguys if you want to discuss Tibet please make another topic", "Solution_1": "What does Tibet have to do with it? Is it just a trick to attract more attention to the post? :spam:", "Solution_2": "no, it had a much more germane title initially. the main thrust is the original post -- I think it's an interesting thing which I haven't thought about much.", "Solution_3": "Why should math not apply and be effective in the sciences? This goes back to the whole \"math is pure\" thing, but math is so abstract and based on the simplest axioms that it would only make sense for it to be true and effective in the sciences. The only parts of math \"made up\" are those basic axioms which are obviously based on intuition and what we see around us, so math would naturally describe it.", "Solution_4": "see that's what I implicitly assumed for quite a while. but if you step back and think about it, it seems that your argument really only applies insofar as explaining why math is true with respect to the world around us. it doesn't go further and explain why math is so perfect at explaining almost every single thing in the world around us. (kind of like showing something is necessary but not sufficient). by the way, that same wikipedia article has a link further down on the surprising ineffectiveness of math in biology, the cognitive sciences and so forth.", "Solution_5": "Okay, but if we wonder why math can explain almost everything around us, then we could ask what else there would be to explain it. I don't know. Nor do I know why math would be ineffective at biology and the cognitive sciences. Possibly those are simply far more complicated? eg. biology is made of cells which has tons of different parts made of molecules working together in ridiculous ways. Perhaps they would be explained at math if we could find a way to examine those sciences so simply? But then we would be back down to, like examining how individual molecules interact, and we get into a different science, so I now I'm not precisely sure what I think about it.", "Solution_6": "Well, isn't physics made upon axioms itself? I have not studied quantum theory, so I am not sure of this statement, but one can easily imagine a world without gravity. It would probably be more like a sea of materials, but more importantly, it seems possible that an absense of an axiom in physics is somewhat plausible, so it is not mathematical related.\r\n\r\nI dont' think I know what I am talkign abotu at all, so I will read more on this later :)", "Solution_7": "Well, if there was no gravity, then we would develop a mathematical system to describe physics that doesn't involve math. It would be different, but still described by math. \r\n\r\nAnyway, what I was really trying to say in my last post is this, and I don't know if it makes sense or if I know what I'm talking about either, but:\r\n\r\nThe field of biology is complicated. Again, the cell has a lot of different parts working together. Therefore, it is not easy at all to come up with mathematical rules to govern the biological world. Take this analogy: the primes. The primes are very complicated, and we need very very advanced math to describe their distribution well, (Riemann Hypothesis level math, for instance). However, despite how complex it is, we can still do it. So, maybe biology can be described by math, it is just so complicated that it seems on the surface that math cannot describe it [i]effectively[/i].", "Solution_8": "[quote=\"fedja\"]What does Tibet have to do with it? Is it just a trick to attract more attention to the post? :spam:[/quote]\r\n\r\nThe O.P. seems to believe that rules of netiquette don't apply to him. :spam:, indeed.", "Solution_9": "[quote=\"JBL\"]The O.P. seems to believe that rules of netiquette don't apply to him. [/quote]\r\n\r\nhah, this is another topic for another time. clearly not spam, anyway (dudes totally don't know the definition of spam anymore).\r\n\r\n@hance see, it seems very easy to convince ourselves that math should indeed be able to accurately model anything it wants to. like, this line of thought makes perfect sense to me as well. that's why I was kinda surprised to read the wikipedia article (and then some of the papers) in the first place -- people were investigating what seemed to me to be a tautology. so I'm trying to step back and say, well, is there any sense in which this line of thought (math describes everything) might fail? well, is there? or are people just hatin' on math for no reason? I'm not sure (this isn't a topic where I have a specific point I'm arguing, more a topic where I'm like hey this is kind of weird guys)\r\n\r\n@anax blehh sorry didn't really understand your point there", "Solution_10": "http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Fictionalism", "Solution_11": "[quote=\"MysticTerminator\"]http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Fictionalism[/quote]\r\n\r\nWhoa what\r\n\r\nI really don't understand that. It's quite difficult to comment on because I don't entirely understand how is \"science without mathematics\" works. But to me, it looks like he designed his physical system without mathematics, and then said that we can use math even though math is just fiction or something -- what?\r\n\r\nI mean what is up with saying \"mathematics is a reliable process whose physical applications are all true, even though its own statements are false\"?", "Solution_12": "tjhance: fictionalism is essentially an anti-platonist interpretation of mathematics. that is, fictionalism claims that mathematical objects are disembodied abstractions completely disconnected from the space and time, and thus, statements referring to them cannot possibly be true. to a fictionalist, the statement \"14 is divisible by 2\" is false for the same reason that the statement \"santa claus is jolly\" is false. santa claus does not exist, and hence, ascribing characteristics to santa claus is absurd. similarly, \"14\" and \"2\" do not actually exist, so statements that try to say something about them are necessarily false. (side note: is it necessarily the case that the fictionalist approach simply inverts the convention we use for describing vacuous truths?) fictionalism, however, concedes that mathematics may be a useful fiction that can help point us towards physical reality, but it can never, in and of itself, contain physical truth.\r\n\r\nEDIT: stanford encyclopedia of philosophy strikes again: http://plato.stanford.edu/entries/fictionalism-mathematics/", "Solution_13": "Well okay, that makes a bit more sense. But what exactly is the difference between an object being \"abstract\" (as in Platonism) and an object begin \"fictitious\"? Is there a difference between looking at the properties of a \"real\" (what does this mean?) abstract object and looking at the abstract object which does not exist except in a fictitious system which still is able to describe the physical universe? (I don't know what I'm arguing for here; I just don't understand what the point is or what this means entirely.)", "Solution_14": "Can I make an analogy without people going crazy? It is just too perfect with some of the recent threads.\r\n\r\nMathematical fictionalism says that math is like religion. It may be useful, but this says nothing about whether what it uses is real or not. In fact, since they are made up, we can almost with certainty say that what is described in them don't exist.\r\n\r\nAs for what the difference between a \"mathematical form\" or whatever in Platonistic terms and a \"fictitious\" object, my guess would be that the philosophical purpose is to eliminate some philosophical problems. Platonists need a way for us to come in contact with or come to know the object that actually exists. These objects also have to exist somewhere. Fictionalists don't have to describe what that process is or where they exist, because they say they don't exist. \r\n\r\nMathematically the difference is probably very subtle. It is probably only in attitude. Platonists' thinking is probably more along the lines of looking for an objective truth and fictionalists are probably more likely to manipulate and change the rules of the game. I don't know, I'm just making stuff up now so I'll stop." } { "Tag": [ "trigonometry", "geometry", "parallelogram" ], "Problem": "Two consecutive sides of a paralellagram have lengths 13 and 6. Find the length of the longest diagonal.", "Solution_1": "#1: It's \"parallelogram.\"\r\n#2: Do you mean the longest possible diagonal? As there is not enough information given to determine a unique parallelogram.", "Solution_2": "Yeah, I agree, and if you specified some ugly angles it would turn into a trig problem.\r\n\r\nThe longest possible diagonal would be some number infinitesimally smaller than 19.", "Solution_3": "sorry. It was the longest possible diagonal, and the diagonals were integers. I've already solved the problem.\r\n\r\n[hide=\"solution\"]The diagonal is less than 13+6=19, and greater than rt(169+36)=rt205. Therefore, the diagonal is either 15, 16, 17, or 18. We can draw a regular parallelogram ABCD, label AB and CD 13, and BC and DA 6. We also draw height BE, and label it h. Make sure that BE is on the outside of the parallelogram. Now draw EC, and label it x. We now know that\n\n$x^{2}+h^{2}=36$\n\n$169+26x+x^{2}+h^{2}=long_{diagonal}^{2}$\n\n$205+26x=long_{diagonal}^{2}$\n\nWe now draw heeight AF, label it h, and label FD x. Now\n\n$169-2x+x^{2}+h^{2}=short_{diagonal}^{2}$\n\n$205-26x=short_{diagonal}^{2}$\n\nWe need to find the long diagonal such that the shortest diagonal is also an integer. The only number that works is 17. Therefore, 17 is the longest diagonal that can be achieved.[/hide]" } { "Tag": [ "calculus", "integration", "logarithms", "trigonometry", "function", "calculus computations" ], "Problem": "i need the integral (and some explanation) of 1n((1+x)^(1/2) + (1-x)^(1/2)) from 0 to 1.", "Solution_1": "This should be moved to Calculus Computation Tutrials.\r\n\r\nJapanese Communities Moderator\r\n\r\nkunny", "Solution_2": "is this what you mean:\r\n\r\n$ \\int^1_0 \\ln (\\sqrt {1\\plus{}x} \\plus{} \\sqrt{1\\minus{}x})$ $ dx$\r\n\r\nusing integration by parts, with $ u$ $ \\equal{}$ $ \\ln (\\sqrt {1\\plus{}x} \\plus{} \\sqrt{1\\minus{}x})$ and $ dv$ $ \\equal{}$ $ 1$, \r\n\r\n$ du$ $ \\equal{}$ $ \\frac {\\sqrt {1\\minus{}x}\\minus{}\\sqrt {1\\plus{}x}}{\\sqrt {1\\plus{}x}\\plus{}\\sqrt {1\\minus{}x}}$ $ \\frac{1}{2 \\sqrt{1\\minus{}x^2}}$ \r\n\r\n= $ \\frac {2 \\minus{} 2\\sqrt {1\\minus{}x^2}}{4x\\sqrt{1\\minus{}x^2}}$ (multiplying up and down with $ \\sqrt {1\\plus{}x}\\plus{}\\sqrt {1\\minus{}x}$ )\r\n\r\n= $ \\frac {1 \\minus{} \\sqrt{1\\minus{}x^2}}{2x\\sqrt{1\\minus{}x^2}}$\r\n\r\nthe original integral is now\r\n\r\n$ x \\ln (\\sqrt{1\\plus{}x}\\plus{}\\sqrt{1\\minus{}x}) \\minus{} \\int x \\frac {1 \\minus{} \\sqrt{1\\minus{}x^2}}{2x\\sqrt{1\\minus{}x^2}} dx$\r\n\r\n= $ x \\ln (\\sqrt{1\\plus{}x}\\plus{}\\sqrt{1\\minus{}x}) \\minus{} \\int \\frac {1 \\minus{} \\sqrt{1\\minus{}x^2}}{2\\sqrt{1\\minus{}x^2}} dx$\r\n\r\n= $ x \\ln (\\sqrt{1\\plus{}x}\\plus{}\\sqrt{1\\minus{}x}) \\minus{} \\frac {1}{2} \\int \\frac {1}{\\sqrt {1\\minus{}x^2}} \\minus{} 1 dx$\r\n\r\nfor $ \\int \\frac {1}{\\sqrt {1\\minus{}x^2}} dx$, substitute $ u \\equal{} \\sin \\theta$\r\n$ du \\equal{} \\cos \\theta d\\theta$\r\n\r\n\r\n$ \\int \\frac {1}{\\sqrt {1\\minus{}x^2}} dx$ = $ \\int d\\theta$ \r\n\r\n = $ \\theta$ \r\n\r\n = $ \\arcsin x$\r\n\r\n\r\nso the original integral is \r\n\r\n$ \\int \\ln (\\sqrt {1\\plus{}x} \\plus{} \\sqrt{1\\minus{}x})$ $ dx$ = $ x \\ln (\\sqrt{1\\plus{}x}\\plus{}\\sqrt{1\\minus{}x}) \\minus{} \\frac {\\arcsin x}{2} \\plus{} \\frac {x}{2}$ + $ C$\r\n\r\nnow just substitute the limits. you'll get:\r\n\r\n$ \\int^1_0 \\ln (\\sqrt {1\\plus{}x} \\plus{} \\sqrt{1\\minus{}x})$ $ dx$ $ \\equal{}$ $ \\frac {1}{2} \\ln 2 \\minus{} \\frac {\\pi}{4} \\plus{} \\frac {1}{2}$", "Solution_3": "[quote=\"anonym115\"]i need the integral (and some explanation) of 1n((1+x)^(1/2) + (1-x)^(1/2)) from 0 to 1.[/quote]\r\n\r\nIf you want $ \\int^1_0 \\ln (\\sqrt {1\\plus{}x} \\plus{} \\sqrt{1\\minus{}x})$\r\nans is$ \\frac{(\\minus{}2 \\plus{} \\Pi\\plus{} ln 4)}{4}$", "Solution_4": "Isn't this subject the same as\r\n[b]integral of a certain natural log expression[/b]?", "Solution_5": "Let $ x \\equal{} \\sin \\theta$, \r\n\r\n$ \\int_0^1 \\ln (\\sqrt {1 \\plus{} x} \\plus{} \\sqrt {1 \\minus{} x})dx \\equal{} \\int _0^{\\frac {\\pi}{4}} \\ln (\\sqrt {1 \\plus{} \\sin 2\\theta} \\plus{} \\sqrt {1 \\minus{} \\sin 2\\theta})\\cdot 2\\cos 2\\theta \\ d\\theta$\r\n\r\n$ \\equal{} 2\\int_0^{\\frac {\\pi}{4}} \\cos 2\\theta \\cdot \\ln (|\\sin \\theta \\plus{} \\cos \\theta| \\plus{} |\\sin \\theta \\minus{} \\cos \\theta|)d\\theta$\r\n\r\n$ \\equal{} 2\\int_0^{\\frac {\\pi}{4}} \\cos 2\\theta \\ln (\\sin \\theta \\plus{} \\cos \\theta \\plus{} \\cos \\theta \\minus{} \\sin \\theta)d\\theta$\r\n\r\n$ \\equal{} \\int_0^{\\frac {\\pi}{4}} 2\\cos 2\\theta \\cdot \\ln (2\\cos \\theta)d\\theta$\r\n\r\n$ \\equal{} \\left[\\sin 2\\theta \\cdot \\ln (2\\cos \\theta)\\right]_0^{\\frac {\\pi}{4}} \\plus{} \\int_0^{\\frac {\\pi}{4}} \\sin 2\\theta \\cdot \\frac {\\sin \\theta}{\\cos \\theta}d\\theta$\r\n\r\n$ \\equal{} \\ln \\sqrt {2} \\plus{} 2\\int_0^{\\frac {\\pi}{4}} \\sin ^ 2 \\theta d\\theta$\r\n\r\n$ \\equal{} \\ln \\sqrt {2} \\plus{} \\int_0^{\\frac {\\pi}{4}} (1 \\minus{} \\cos 2\\theta)d\\theta$\r\n\r\n$ \\equal{} \\ln \\sqrt {2} \\plus{} \\left[\\theta \\minus{} \\frac {1}{2}\\sin 2\\theta\\right]_0^{\\frac {\\pi}{4}}$\r\n\r\n$ \\equal{} \\boxed{\\frac {1}{2}\\ln 2 \\plus{} \\frac {\\pi}{4} \\minus{} \\frac {1}{2}}$", "Solution_6": "Lemma ${ \\boxed{\\int_0^1 \\ln (1 + \\sqrt {1 - x^2})dx = \\frac {\\pi}{2} - 1}}$\r\n\r\nProof: Let $ x = \\sin \\theta$, \r\n\r\n$ \\int_0^1 \\ln (1 + \\sqrt {1 - x^2})dx = \\int_0^{\\frac {\\pi}{2}} \\ln (1 + \\cos \\theta)\\cdot \\cos \\theta d\\theta$\r\n\r\n$ = \\left[\\sin \\theta \\cdot \\ln (1 + \\cos \\theta)\\right]_0^{\\frac {\\pi}{2}} + \\int_0^{\\frac {\\pi}{2}} \\sin \\theta \\cdot \\frac {\\sin \\theta}{1 + \\cos \\theta}d\\theta$\r\n\r\n$ = \\int_0^{\\frac {\\pi}{2}} (1 - \\cos \\theta)d\\theta = \\frac {\\pi}{2} - 1$ Q.E.D.\r\n\r\nNow let denote $ f(x) = \\int_{ - x}^x \\ln (\\sqrt {1 + t} + \\sqrt {1 - t})dt$, we have $ f'(x) = \\ln (\\sqrt {1 + x} + \\sqrt {1 - x}) + \\ln (\\sqrt {1 - x} + \\sqrt {1 + x})$\r\n\r\n$ = 2\\ln (\\sqrt {1 + x} + \\sqrt {1 - x}) = \\ln (\\sqrt {1 + x} + \\sqrt {1 - x})^2 = \\ln 2 + \\ln (1 + \\sqrt {1 - x^2})$, thus by $ f(1) = 0$, \r\n\r\n$ \\therefore f(1) = f(1) - f(0) = [f(x)]_0^1 = \\int_0^1 f'(x)dx$\r\n\r\n$ = \\int_0^1 \\{\\ln 2 + \\ln (1 + \\sqrt {1 - x^2})\\}dx = \\ln 2 + \\int_0^1 \\ln (1 + \\sqrt {1 - x^2})dx$\r\n\r\n$ = \\ln 2 + \\frac {\\pi}{2} - 1$, since $ \\ln (\\sqrt {1 + t} + \\sqrt {1 - t})$ is an even fuction,\r\n\r\n$ \\int_0^1 \\ln (\\sqrt {1 + x} + \\sqrt {1 - x})dx = \\frac {1}{2}f(1) = \\boxed{\\frac {1}{2}\\ln 2 + \\frac {\\pi}{4} - \\frac {1}{2}}$" } { "Tag": [ "algorithm", "invariant", "IMO Shortlist", "combinatorics", "chip-firing" ], "Problem": "A finite number of coins are placed on an infinite row of squares. A sequence of moves is performed as follows: at each stage a square containing more than one coin is chosen. Two coins are taken from this square; one of them is placed on the square immediately to the left while the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one coin on each square. Given some initial configuration, show that any legal sequence of moves will terminate after the same number of steps and with the same final configuration.", "Solution_1": "There exists $ n \\in \\mathbb{N}_0$ and $ (a_1,a_2,\\ldots,a_n) \\in \\mathbb{N}^n$ such that $ \\min\\{a_1,a_n\\} > 0$ and such that if $ a_1$ represents the (initial) number of coins in the square $ x$ then $ a_i$ represents number of coins in the square $ x \\plus{} i \\minus{} 1$. Define also $ y: \\equal{} \\sum_{i \\equal{} 1}^n{a_i}$. Obviously all coins stay always in squares $ \\ge x \\minus{} y$ and always in squares $ \\le x \\plus{} n \\minus{} 1 \\plus{} y$, and we can wlog assume $ x \\minus{} y$ the is the square number 0, so all moves is done in a finite number of [i]squares[/i]. So for each time there exist a $ (n \\plus{} 2y \\plus{} 2)$-upla of non negative integer $ (a_0,a_1,a_2,\\ldots,a_{n \\plus{} 2y \\plus{} 1})$ with sum $ y$ such that $ a_i$ represents number of coins in the square $ i$. Now the number $ \\sum_{i \\equal{} 0}^{n \\plus{} 2y \\plus{} 1}{i^2a_i}$ increases at each step by 2, and it is always less of $ y(n \\plus{} 2l \\plus{} 1)^2$: it shows that the algorithm finishes in a finite number of [i]steps[/i]. It remains only to show that we'll have always the same number of moves and the final configuration is invariant by the algorithm chosen. Denote with $ h_i$ the number of moves done at the end of the game in the square $ i$. Then the final number of coins in the square $ i$ is $ a_i \\plus{} h_{i \\plus{} 1} \\minus{} 2h_i \\plus{} h_{i \\minus{} 1} \\in \\{0,1\\}$. Assume by contradiction that there exists two distinct sequence $ \\{h_i\\}$ and $ \\{H_i\\}$ that satisfies the previous relation. Then we can choose $ \\{a_i\\}_{0 \\le i \\le n \\plus{} 2y \\plus{} 1}$ s.t. $ S: \\equal{} \\min\\{\\sum{h_i},\\sum{H_i}\\}$ reaches minimum. Since the initial configuration is the same by assumption, then there exist $ 0 < j < n \\plus{} 2y \\minus{} 1$ s.t. $ a_j > 1$, but now the $ n \\plus{} 2y \\plus{} 2$ upla of non negative integers $ (a_0,a_1,\\ldots,a_{j \\minus{} 1} \\plus{} 1,a_j \\minus{} 2,a_{j \\plus{} 1} \\plus{} 1,\\ldots,a_{n \\plus{} 2y \\minus{} 1})$ reaches a new minimum in S, contradiction.", "Solution_2": "[quote=\"bboypa\"]There exists $ n \\in \\mathbb{N}_0$ and $ (a_1,a_2,\\ldots,a_n) \\in \\mathbb{N}^n$ such that $ \\min\\{a_1,a_n\\} > 0$ and such that if $ a_1$ represents the (initial) number of coins in the square $ x$ then $ a_i$ represents number of coins in the square $ x \\plus{} i \\minus{} 1$. Define also $ y: \\equal{} \\sum_{i \\equal{} 1}^n{a_i}$. Obviously all coins stay always in squares $ \\ge x \\minus{} y$ and always in squares $ \\le x \\plus{} n \\minus{} 1 \\plus{} y$, and we can wlog assume $ x \\minus{} y$ the is the square number 0, so all moves is done in a finite number of [i]squares[/i]. So for each time there exist a $ (n \\plus{} 2y \\plus{} 2)$-upla of non negative integer $ (a_0,a_1,a_2,\\ldots,a_{n \\plus{} 2y \\plus{} 1})$ with sum $ y$ such that $ a_i$ represents number of coins in the square $ i$. Now the number $ \\sum_{i \\equal{} 0}^{n \\plus{} 2y \\plus{} 1}{i^2a_i}$ increases at each step by 2, and it is always less of $ y(n \\plus{} 2l \\plus{} 1)^2$: it shows that the algorithm finishes in a finite number of [i]steps[/i]. It remains only to show that we'll have always the same number of moves and the final configuration is invariant by the algorithm chosen. Denote with $ h_i$ the number of moves done at the end of the game in the square $ i$. Then the final number of coins in the square $ i$ is $ a_i \\plus{} h_{i \\plus{} 1} \\minus{} 2h_i \\plus{} h_{i \\minus{} 1} \\in \\{0,1\\}$. Assume by contradiction that there exists two distinct sequence $ \\{h_i\\}$ and $ \\{H_i\\}$ that satisfies the previous relation. Then we can choose $ \\{a_i\\}_{0 \\le i \\le n \\plus{} 2y \\plus{} 1}$ s.t. $ S: \\equal{} \\min\\{\\sum{h_i},\\sum{H_i}\\}$ reaches minimum. Since the initial configuration is the same by assumption, then there exist $ 0 < j < n \\plus{} 2y \\minus{} 1$ s.t. $ a_j > 1$, but now the $ n \\plus{} 2y \\plus{} 2$ upla of non negative integers $ (a_0,a_1,\\ldots,a_{j \\minus{} 1} \\plus{} 1,a_j \\minus{} 2,a_{j \\plus{} 1} \\plus{} 1,\\ldots,a_{n \\plus{} 2y \\minus{} 1})$ reaches a new minimum in S, contradiction.[/quote]\n\nI do not understand the lasst step. How is this a contradiction? More over, even if it is, then why does this imply that they must end in the same configuration?", "Solution_3": "[hide=\"English mode\"]Note that moves are commutative- that is, the final configuration depends only on which moves we make, not on what order we make them in.\n\nFurthermore, any square with at least two coins on it must be moved at some point.\n\nLook at all the moves legal from the starting position- all of them must be made at some point. No move we make now can block off a different move in the future, so we might as well make every initially legal move simultaneously.\n\nBut of course we can continue with this logic, making every legal move simultaneously from every subsequent position; so we will be done if we can show the process must end at some point.\n\nNo move can create a gap between two piles of coins wider than any gap that already exists, so the breadth of the group of coins is limited, and therefore there are a finite number of positions the coins can be in. In addition, each move strictly increases the total rotational inertia of the coins around their center of mass, so no position can be repeated- the process must terminate, and we are done.[/hide]", "Solution_4": "[quote=\"lokitos\"]Look at all the moves legal from the starting position- all of them must be made at some point. No move we make now can block off a different move in the future, so we might as well make every initially legal move simultaneously.[/quote]\nFor someone like me with terrible intuition, here's an easier way to see this (basically induction): suppose that we have two sequences of moves $(r_1,\\ldots,r_m)$ and $s=(s_1,\\ldots,s_n)$. Then initially, square $r_1$ must have at least $2$ coins, so there must exist $1\\le i\\le n$ such that $s_i=r_1$. WLOG $s_1=r_1$. (If not, then letting $i_0>0$ denote the smallest such $i$, it's easy to verify that the sequence $s'=(s_{i_0}=r_1,s_1,\\ldots,s_{i_0-1},s_{i_0+1},\\ldots,s_n)$ is equivalent to the sequence $s$.) Now we can just operate on $r_1$ and then induct to show that $r$ is a permutation of $s$.", "Solution_5": "It seems that I don't understand any of the solutions posted here... But combined together, they make sense.\n\n[quote=\"bboypa\"]There exists $ n \\in \\mathbb{N}_0$ and $ (a_1,a_2,\\ldots,a_n) \\in \\mathbb{N}^n$ such that $ \\min\\{a_1,a_n\\} > 0$ and such that if $ a_1$ represents the (initial) number of coins in the square $ x$ then $ a_i$ represents number of coins in the square $ x \\plus{} i \\minus{} 1$. Define also $ y: \\equal{} \\sum_{i \\equal{} 1}^n{a_i}$. Obviously all coins stay always in squares $ \\ge x \\minus{} y$ and always in squares $ \\le x \\plus{} n \\minus{} 1 \\plus{} y$, and we can wlog assume $ x \\minus{} y$ the is the square number 0, so all moves is done in a finite number of [i]squares[/i].[/quote]\n\nSorry to say, but I don't understand any of this. What is $n$? What are the $a_i$? Why are all moves done in a finite number of squares?\n\n[quote=\"lokitos\"]Note that moves are commutative- that is, the final configuration depends only on which moves we make, not on what order we make them in.[/quote]\n\nWhy? (I am aware that this is true, but I only know how to prove it using the stabilization property, i. e., the fact that final configurations always exist. How do you show this at this point, way before you have proven the stabilization property?)\n\n[quote=\"lokitos\"]No move can create a gap between two piles of coins wider than any gap that already exists,[/quote]\n\nOkay, I finally understood what you mean here with the help of [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=17461]Naphthalin[/url]. The claim is that the \"maximum gap\" (i. e., the highest distance $j-i$ between two distinct integers $i$ and $j$ such that $i0$. Fortunately, this is okay because all we want is that this \"maximum gap\" is bounded a-priori.\n\nWe are now discussing a more complex version of this problem at https://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=428298 . My post there (after some minor modifications) also gives a solution of this problem.", "Solution_6": "[quote=\"darij grinberg\"]It seems that I don't understand any of the solutions posted here... But combined together, they make sense.\n\nSorry to say, but I don't understand any of this. What is $n$? What are the $a_i$? Why are all moves done in a finite number of squares?[/quote]\n\n$a_1>0$ is the number of coins in the first [i]non empty[/i] square, called $x$; $a_i \\ge 0$ is the number of coins in the square $x+i-1$, for all $i=1,2,\\ldots,n-1$; $a_n>0$ is the number of coins in the last[i] non empty[/i] square, placed in $x+n-1$. why finite number of squares? Just think about the best and the worst case..", "Solution_7": "Oh, I see. You're talking about the [b]initial[/b] configuration. I still don't see why this:\n\n[quote=\"bboypa\"]Obviously all coins stay always in squares $ \\ge x \\minus{} y$ and always in squares $ \\le x \\plus{} n \\minus{} 1 \\plus{} y$[/quote]\n\nis \"obvious\". (I agree that this is true and easy to show after the problem is solved...)", "Solution_8": "[quote=\"bboypa\"]Then we can choose $ \\{a_i\\}_{0 \\le i \\le n \\plus{} 2y \\plus{} 1}$ s.t. $ S: \\equal{} \\min\\{\\sum{h_i},\\sum{H_i}\\}$ reaches minimum. Since the initial configuration is the same by assumption, then there exist $ 0 < j < n \\plus{} 2y \\minus{} 1$ s.t. $ a_j > 1$, but now the $ n \\plus{} 2y \\plus{} 2$ upla of non negative integers $ (a_0,a_1,\\ldots,a_{j \\minus{} 1} \\plus{} 1,a_j \\minus{} 2,a_{j \\plus{} 1} \\plus{} 1,\\ldots,a_{n \\plus{} 2y \\minus{} 1})$ reaches a new minimum in S, contradiction.[/quote]\n\nWe need to prove: If $\\sum {{H_i}} = 0$,there not exist no zero sequence $\\left\\{ {{h_i}} \\right\\}$ satisfies ${a_i} + {h_{i + 1}} - 2{h_i} + {h_{i - 1}} \\in \\left\\{ {0,1} \\right\\}$.", "Solution_9": "A terribly long, but rather straightforward approach may be found [url=http://artofproblemsolving.com/community/c6h1458641p9083581]here[/url].", "Solution_10": "I have also outlined a solution of the $2$-dimensional analogue of this exercise (i.e., coins are placed on the integer lattice $\\mathbb{Z}^2$ rather than on $\\mathbb{Z}$, and accordingly at each step we take $4$ coins from a point and distribute them among its $3$ neighboring squares) in [url=http://www.cip.ifi.lmu.de/~grinberg/t/17s/hw5s.pdf]Math 5707 Spring 2017 homework set #5 exercise 5[/url]. There are a bunch of illustrations in a recent popular article: [url=http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile]Jordan Ellenberg, *The Amazing, Autotuning Sandpile*, Nautilus[/url].", "Solution_11": "Solution from [i]Twitch Solves ISL[/i]:\n\n[b]First part[/b]: We start by proving the procedure always terminates. Index the squares by ${\\mathbb Z}$.\n\n[b][color=red]Claim:[/color][/b] \tFor any starting configuration, \tthere exists an interval $[A,B]$ such that \tno coin may ever exit $B$.\n[i]Proof.[/i] \tLet $n$ be the largest index of a square with any coins. \tNote that for any square $m > n$, \tthe following property is true: \tif $m$ ever gains a coin, \tthen forever after, either $m$ or $m+1$ has a coin. \tThis follows since any move affecting either $m$ or $m+1$ \twill add a coin to at least one of them.\nThus, we make take $B = m + 2c$ where $c$ is the number of coins. \tThe choice of $A$ is similar. $\\blacksquare$\nNow note that the sum of squares of indices of coins increases each step. This shows that configurations may never repeat; but there are finitely many configurations, ensuring termination.\n\n[b]Second part[/b]: Suppose $S = (x_1, \\dots, x_n)$ is a valid sequence of moves that leads to an end state. We perform the following procedure for $i = 1, \\dots, n$. [list] \t[*]Before the $i$'th move $s_i$, \tlook at the leftmost square which has more than one coin. \tIt must be operated on eventually, \tsince that is the only way it can lose coins. \tLet $x_j$ be the first such move on this square.\n[*]Rearrange the sequence so that this move $x_j$ comes next instead. \tThat is, apply the change \t\\[ (x_i, x_{i+1}, \\dots, x_{j-1}, x_j) \t\t\\longmapsto \\left( x_j, x_i, x_{i+1}, \\dots, x_{j-1} \\right). \\] \tNote that validity of the whole operation is preserved. [/list] In this way, any valid sequence can be rearranged to a certain [i]canonical sequence[/i] where one always operates on the leftmost possible square. This implies that the lengths of all valid sequences are the same (actually, they are permutations of each other), and also that the ending states match.", "Solution_12": "First we claim that there exists an interval $[A,B]$ which the coins can never leave, say there doesn't exist such $B$, now if there doesn't exist such $B$, we can reach any square on the right hand side, but that's a contradiction because like consider the rightmost square in the initial configuration, note that if you start from moving $k$ coins towards the right, the number of coins moving towards the right decreases each step, it'll never be $k$ again (due to symmetry of distribution), atmost it can be of the form $k/2 + k/8 ... k/2^{n}$ but that is never equal to $k$ for some finite $k$, so eventually at some point we'll not be able to move on the right hand side. Similar argument holds for the left hand side. Now note that the sum of the square of the coin's indices increase each step so the process terminates. Now no matter how you move, you'll always have the same permutation of moves, so we end up in the same final configuration again." } { "Tag": [], "Problem": "Find ${ a_{1},a_{2},\\ldots,a_{12} \\in \\mathbb{Z}^{ + }}$ such that:\r\n\\[ a_{1} + a_{2} + \\ldots + a_{12} = a_{1}.a_{2}...a_{12}\\]", "Solution_1": "This question has multiple answers, I'm relatively certain. \r\n\r\n[hide=\"One Solution\"]$ a_1 \\equal{} a_2 \\equal{} ... \\equal{} a_8 \\equal{} 1$, and $ a_9 \\equal{} a_{10} \\equal{} a_{11} \\equal{} a_{12} \\equal{} 2$.[/hide]", "Solution_2": "I don't understand. Help me, please!!!!!!!!!!!!!" } { "Tag": [ "search", "function", "number theory proposed", "number theory" ], "Problem": "Sorry if the following problem has been posted before ,but search function didnt help me . \r\n\r\nFind all positive integers $ x,y$ such that $ x^{2} \\plus{} y^{2} \\equal{} 1997(x \\minus{} y)$", "Solution_1": "since $ x \\minus{} y|x^2 \\plus{} y^2 \\implies x \\minus{} y |2xy \\implies (x \\minus{} y)k \\equal{} 2xy \\implies \\frac {1}{2y} \\equal{} \\frac {1}{2x} \\plus{} \\frac {1}{k}$ so there exists integers like $ a,b,d$ with $ \\gcd(a,b) \\equal{} 1$ and $ 2y \\equal{} dab$ and $ 2x \\equal{} da(a \\plus{} b)$\r\nnow just use these equalities and the fact that $ 1997$ is a prime!", "Solution_2": "[quote=\"srinath.r\"]Sorry if the following problem has been posted before ,but search function didnt help me . \n\nFind all positive integers $ x,y$ such that $ x^{2} \\plus{} y^{2} \\equal{} 1997(x \\minus{} y)$[/quote]\r\n\r\nLet $ x\\equal{}au$ and $ y\\equal{}av$ with $ \\gcd(u,v)\\equal{}1$\r\n\r\nThe equation may be written $ a(u\\minus{}v)^2\\plus{}2auv\\equal{}1997(u\\minus{}v)$\r\n\r\n$ u\\ne v$ and so $ u\\minus{}v|2auv$ and so $ u\\minus{}v|2a$ and let $ 2a\\equal{}k(u\\minus{}v)$. We get $ a(u\\minus{}v)\\equal{}1997\\minus{}kuv$ and also $ k(u\\minus{}v)^2\\equal{}3994\\minus{}2kuv$ and so $ k|3994$\r\n\r\nLet then $ 3994\\equal{}kw$ and we get $ (u\\minus{}v)^2\\equal{}w\\minus{}2uv$ and so $ w\\equal{}u^2\\plus{}v^2$, hence the solutions :\r\n\r\n$ (k,w)\\equal{}(1,3994)$ $ \\implies$ $ 3994\\equal{}u^2\\plus{}v^2$ $ \\implies$ $ (u,v)\\in\\{(5,63),(63,5)\\}$ and so $ (x,y)\\equal{}(1827,145)$\r\n$ (k,w)\\equal{}(2,1997)$ $ \\implies$ $ 1997\\equal{}u^2\\plus{}v^2$ $ \\implies$ $ (u,v)\\in\\{(29,34),(34,29)\\}$ and so $ (x,y)\\equal{}(170,145)$\r\n$ (k,w)\\equal{}(1997,2)$ $ \\implies$ $ 2\\equal{}u^2\\plus{}v^2$ $ \\implies$ $ (u,v)\\in\\{(1,1)\\}$ and so no positive integer solution\r\n$ (k,w)\\equal{}(3994,1)$ $ \\implies$ $ 1\\equal{}u^2\\plus{}v^2$ $ \\implies$ $ (u,v)\\in\\{(1,0),(0,1)\\}$ and so no positive integer solution\r\n\r\nHence the only two solutions $ (1827,145)$ and $ (170,145)$", "Solution_3": "sorry dear mathlinkers i can't find the year but u can see it [url=http://www.mathlinks.ro/viewtopic.php?search_id=1326770934&t=239405]here[/url]!", "Solution_4": "Yes shoki ,it is bulgaria 1998...I tried to find it using search function but couldnt find it . here is my solution ,\r\nGiven ,$ x^{2} \\minus{} y^{2} \\equal{} 1997x \\minus{} 1997y \\implies x(x \\minus{} 1997) \\equal{} y( \\minus{} y \\minus{} 1997)$ . \r\nIt is clearly evident that $ x \\neq y$ and leave out the trivial cases where $ x \\minus{} 1997 \\equal{} 0$ or $ \\minus{}y \\minus{} 1997 \\equal{} 0$ which are not possible . \r\n\r\nSo we have $ \\frac{x}{y} \\equal{} \\frac{1997 \\plus{} y}{1997 \\minus{} x}$\r\n\r\n$ \\implies \\frac{x \\plus{} y}{x \\minus{} y} \\equal{} \\frac{3994 \\minus{} (x \\plus{} y)}{x \\plus{} y} \\implies ( x \\plus{} y)^{2} \\equal{}\r\n( x \\minus{} y)(3994 \\minus{} ( x \\minus{} y))$ . \r\n\r\nLet $ x \\minus{} y \\equal{} a$ and $ 3994 \\minus{} (x \\minus{} y) \\equal{} b$ ;$ a \\plus{} b \\equal{} 3994$ . \r\nLet the $ \\gcd(a , b) \\equal{} d$ ; $ a \\equal{}da'$ and $ b \\equal{} db'$ where $ a'$ and $ b'$ are coprime .\r\n\r\nSo we have $ d ( a' \\plus{} b' ) \\equal{} 3994 \\implies d \\mid 3994$. \r\nSince $ 1997$ is a prime , $ d \\equal{}1$ or $ d \\equal{} 2$ or $ d \\equal{} 1997$ or $ d \\equal{} 3994$ ,it is easy to see that $ d \\neq 1997 ,3994$ . \r\nIf $ d \\equal{} 2$ \r\n$ a' \\plus{} b' \\equal{} 1997$ and $ x \\minus{} y \\equal{} 2a'$ and $ 3994 \\minus{} (x \\minus{} y) \\equal{} 2b'$ \r\nSince $ (x \\minus{} y)(3994 \\minus{} (x \\minus{} y) \\equal{} (x \\plus{} y)^{2} \\implies 4a'b' \\equal{} (x \\plus{} y)^{2}$ ,we easily observe that $ a'b'$ is a perfect square , but since $ \\gcd(a' ,b') \\equal{} 1 \\implies a' \\equal{} m^{2}$ and $ b \\equal{} n^{2}$ where $ m ,n \\in \\mathbb{N}$ \r\n\r\nWe get that $ m^{2} \\plus{} n^{2} \\equal{} 1997$ ,solving this diophantine (it is easy :D ) we get that the non trivial case to be $ m \\equal{}29$ and $ n \\equal{} 34$ \r\n\r\n$ \\implies 2a' \\equal{} a \\equal{} x \\minus{} y \\equal{} 2 * 29^{2}$ , $ 3994 \\minus{} (x \\minus{} y) \\equal{} 2 * 34^{2}$ ,using these two we get \r\n\r\n$ x \\plus{} y \\equal{} 2*34*29$ ,from these conditions we can obtain that $ (x ,y ) \\equal{} (1827 ,145)$ . \r\n\r\n[b]If[/b] $ d \\equal{} 1$ , we have $ a \\plus{} b \\equal{} 3994$ ,repeating the whole story again we can obtain that, \r\n$ a \\equal{} m^{2}$ and $ b \\equal{} n^{2}$ and solving this diophantine , $ m^{2} \\plus{} n^{2} \\equal{} 3994$ by doing some work with $ \\mod 7$ ,we get the only non trivial solution which is likely to be possible is $ m \\equal{} 5$ and $ n \\equal{} 63$ . Thus ,using this we can find $ x , y$ with $ m , n$ and we finally get \r\n$ x \\equal{} 170$ and $ y \\equal{} 145$ \r\nThus the only solutions $ ( x ,y ) \\equal{} (1827 ,145 )$ or $ (x ,y ) \\equal{} (170 ,145)$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let a, b, c be positive real numbers. Prove that\r\n\\frac {a^2 + a - c^2}{b + c} + \\frac {b^2 + 4b - a^2}{c + a} + \\frac{c^2 + 9c - b^2}{a + b}\\ge\\ 4", "Solution_1": "hello, do you meant this\r\nLet $ a, b, c$ be positive real numbers. Prove that \r\n$ \\frac {a^2 \\plus{} a \\minus{} c^2}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} 4b \\minus{} a^2}{c \\plus{} a} \\plus{} \\frac{c^2 \\plus{} 9c \\minus{} b^2}{a \\plus{} b}\\ge\\ 4$?\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, do you meant this\nLet $ a, b, c$ be positive real numbers. Prove that \n$ \\frac {a^2 \\plus{} a \\minus{} c^2}{b \\plus{} c} \\plus{} \\frac {b^2 \\plus{} 4b \\minus{} a^2}{c \\plus{} a} \\plus{} \\frac {c^2 \\plus{} 9c \\minus{} b^2}{a \\plus{} b}\\ge\\ 4$?\nSonnhard.[/quote]\r\n\r\nWe have $ \\frac {a^2 \\minus{} c^2}{b \\plus{} c} \\plus{} \\frac {b^2 \\minus{} a^2}{a \\plus{} c} \\plus{} \\frac {c^2 \\minus{} b^2}{a \\plus{} b} \\equal{} \\frac {a^4 \\plus{} b^4 \\plus{} c^4 \\minus{} a^2b^2 \\minus{} b^2c^2 \\minus{} c^2a^2}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq 0$\r\n\r\n$ \\frac {a}{b \\plus{} c} \\plus{} \\frac {4b}{a \\plus{} c} \\plus{} \\frac {9c}{a \\plus{} b} \\geq \\frac {(a \\plus{} 2b \\plus{} 3c)^2}{2(ab \\plus{} bc \\plus{} ca)}$\r\nBut\r\n$ \\frac {(a \\plus{} 2b \\plus{} 3c)^2}{2(ab \\plus{} bc \\plus{} ca)} \\minus{} 4 \\equal{} \\frac {(a \\minus{} 2b \\minus{} c)^2 \\plus{} 8c^2}{2(ab \\plus{} bc \\plus{} ca)} \\geq 0$" } { "Tag": [], "Problem": "Al, Bob, Carl, David, Earl, and Frank sit down to eat dinner with their mother and father on a circular table. How many distinct ways are there to seat the six children and their parents provided that the mother and father sit across from one another?", "Solution_1": "[hide]Position the mother, then the father has to sit across from her. The other six seats are optional. So it's $6!=720$ ways.[/hide]", "Solution_2": "[hide]After seating mother and father, there are 6!=720 remaining ways to seat the children.[/hide]", "Solution_3": "[quote=\"1234567890\"]Al, Bob, Carl, David, Earl, and Frank sit down to eat dinner with their mother and father on a circular table. How many distinct ways are there to seat the six children and their parents provided that the mother and father sit across from one another?[/quote]\r\n[hide=\"unsure\"]We can use the mother and father as fixed reference points. There are $\\binom{6}{3}=20$ ways to pick three people for the top above them. There are 6*6=36 ways to order the people once the divisions have been assigned. So there are 720 ways, unless you can flip the people, (I'm pretty sure this is overcounted) so I think the answer is $\\frac{720}{2}=360$[/hide]" } { "Tag": [ "FTW" ], "Problem": "I will start a tournament but instead of being countdown to determine the winner like all the other tournaments are, I will make this tournament normal games.\r\n\r\nYou will be put into a group and play 3 games of 10 questions 22 seconds and timed. The winners (maybe more depending on the number of groups) of each group will get to play one final game of 20 questions 11 seconds and ranking to determine the overall winner.\r\n\r\nSign up now. I will use lowest rating to determine seeding. :lol:\r\n\r\nmewto55555(1)\r\nernie(1056)\r\ndistracted523(1172)\r\nRomanianGenius(1198)\r\ntinytim\r\n(^_^)\r\nthdanh90\r\nhurdler\r\nmath154\r\nra5249(joined even though you don't want too)", "Solution_1": "ummmm...this is uncomfortable...because of glitch day, my lowest rating is 1...", "Solution_2": "i will join and 1056 (the day i bombed), but if u want my real lowest rating- its 1113.", "Solution_3": "Ill join. :lol:", "Solution_4": "i'll join", "Solution_5": "Don't miss me. :blush: :blush:", "Solution_6": "I'll join.\r\n\r\nNow I wish I hadn't kept my lowest rating so high.....\r\n\r\nThere sure have been quite a few tourneys started over the last few days.", "Solution_7": "Low: 1198 I join", "Solution_8": "Oh, hi [b]--[/b]...!!! lol!!! (mewto55555's \"brother\").", "Solution_9": "Hmm, I guess I'll join. Low: 1172", "Solution_10": "Please tell me your lowest ratings if you have not yet.", "Solution_11": "Sorry, forgot to mention my lowest is 1190.", "Solution_12": "mewto55555(1)\r\nernie(1056)\r\n(^_^)(1057)\r\nmath154(1131)\r\ndistracted523(1172)\r\nhurdler(1190)\r\nRomanianGenius(1198)\r\nra5249(1200)\r\ntinytim\r\nthdanh90", "Solution_13": "I'll join.\r\n\r\nOops, my lowest rating is 1000... :P", "Solution_14": "fine, lowest 1200", "Solution_15": "THDANH90 vs. (^_^), a.k.a. Johny Fasee Waisee\r\n90-75\r\n153-63\r\n\r\nHe got about 3 blanks in the 2nd games; and this results wouldn't interpret anything about his excellent mind.", "Solution_16": "group 1\r\nra5249 vs BOGTRO\r\n\r\ngroup 1337\r\ndannyhamtx vs. tinytim\r\n\r\nAdvanced\r\nmewto55555(1)\r\nRomanianGenius(2)\r\nYongyi781(1)\r\n\r\nFinals\r\nthdanh90\r\n\r\nNeeds to get lucky to advance\r\nernie\r\ndistracted523\r\nhurdler\r\n\r\nOut\r\n(^_^)", "Solution_17": "lowest rating is 1088 (tanked)", "Solution_18": "um it already started", "Solution_19": "[quote=\"budi713\"]Out\n(^_^)[/quote]\r\n\r\nOh, that's sad...and a bit unfair. If you're going to have some consolation advancers, why not make it double-elimination and get it over with?", "Solution_20": "um i did really understand what you were saying\r\n\r\nbut this tournament is special since i made if so unfair :D", "Solution_21": "BOGTRO DEFEATS ra5246 BY FORFEIT\r\n\r\nFrom: ra5249 \r\nTo: BOGTRO \r\nPosted: Today, at 8:15 am \r\nSubject: Re: FTW \r\num just tell budi that i withdraw and that you win by withdrawal because i'm at camp and i can't play", "Solution_22": "group 1337\r\ndannyhamtx vs. tinytim\r\n\r\nAdvanced\r\nmewto55555(1)\r\nRomanianGenius(2)\r\nYongyi781(1)\r\nBOGTRO(1)\r\nhurdler(2)\r\n\r\n\r\nFinals\r\nthdanh90", "Solution_23": "*gasp* you got my name wrong SHUN", "Solution_24": "SHUN! SHUN!", "Solution_25": "DId this tournament just like...die?", "Solution_26": "[quote=\"ash-Pokemon\"]lowest rating is 1088 (tanked)[/quote]\r\n\r\nTHE ASHTER? :rotfl: :rotfl: :rotfl:", "Solution_27": "[quote=\"ernie\"][quote=\"ash-Pokemon\"]lowest rating is 1088 (tanked)[/quote]\n\nTHE ASHTER? :rotfl: :rotfl: :rotfl:[/quote]\r\n\r\nernie, don't u have [i]anything[/i] better to do than respond 2 every single post there is??", "Solution_28": "[quote=\"vallon22\"][quote=\"ernie\"][quote=\"ash-Pokemon\"]lowest rating is 1088 (tanked)[/quote]\n\nTHE ASHTER? :rotfl: :rotfl: :rotfl:[/quote]\n\nernie, don't u have [i]anything[/i] better to do than respond 2 every single post there is??[/quote]\r\n\r\nactually, no. :rotfl: \r\n\r\njk\r\n\r\ni was just expressing my laughter on the [b]ashter[/b]\r\n\r\nu hav 2 admit, it is pretty funny, but so is \"ernie\"...", "Solution_29": "[quote]i was just expressing my laughter on the ashter \n\nu hav 2 admit, it is pretty funny, but so is \"ernie\"...[/quote]\r\n\r\n\r\ni completely agree :D :rotfl:" } { "Tag": [ "MATHCOUNTS", "Perfect Squares" ], "Problem": "What is the smallest integer that can be expressed as the sum of two different perfect squares in two different ways?\r\n\r\nI answered 25, 0 + 25 & 16 + 9. The given answer was 65, 64 +1, & 49 + 16.\r\n\r\nShould one consider 0 as a perfect square?", "Solution_1": "Yes, 0 is a perfect square. So your answer is correct. What contest did this question come from?", "Solution_2": "It is true that zero considered as a perfect square and it is...\r\nBut in MATHCOUNTS question, they don't include the zero in the answer for perfect square question, from what I know so...\r\nSo, that's why they have that answer in the question..", "Solution_3": "[quote=\"Ravi B\"]Yes, 0 is a perfect square. So your answer is correct. What contest did this question come from?[/quote]\r\n\r\nI remember seeing this on warmup7", "Solution_4": "Question is from a 1993-1994 Handbook Warm-up worksheet.\r\n\r\nSo, I am a little confused. 0 is to be considered a perfect square except in Mathcounts questions?", "Solution_5": "Perhaps the issue here that influenced Mathcounts scoring is that zero is the additive identity element, so that counting zero as fair game for SUMMING perfect squares leaves no distinction between the smallest number that is the sum of perfect squares in one way and the smallest number that is a sum in two ways. \r\n\r\nI think another issue that is in view is that 0 times ANYTHING is 0, so while one dictionary I consulted listed 1 as a perfect square, none that that I found with a list gave 0 as an example in defining what a perfect square is. \r\n\r\nHere's a Mathcounts forum thread that raises but doesn't definitively answer the question: \r\n\r\nhttp://mathcounts.org/meeting/Upper_Forum_DB/Queries/Active_Forum.taf?function=CombinedDetail&ParentID=6185 \r\n\r\n(After edit:) See also the thread on the Getting Started forum about the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]definition of a perfect square[/url]. The short answer is that zero is NOT a perfect square, because it is not a figurate number at all.", "Solution_6": "I believe the Mathcounts definition of perfect square is \"square of a positive integer,\" not \"square of an integer.\"", "Solution_7": "I remeber 1 team round we did and we got it wrong cuz we didn't use 0 as a perfect square.", "Solution_8": "Check for what the handbook says, and then protest it at the contest if something like this ever happens.", "Solution_9": "Is 0 an even number? Is -2 an even number? Is -1 a factor of 5?", "Solution_10": "O is a definitely even number.. It is considered as even number because numbers are in pattern of odd, even, odd, even, etc..\r\nSo, 1 is odd and 0 is before 1, meaning it's even..\r\nAnd I heard theone853's team got wrong because they didn't use the zero as a perfect squre.. Can you post the problem so we can look at?", "Solution_11": "On the MathCounts web site, they have the following \"toolbox\": [url]http://www.mathcounts.org/Program/toolbox.pdf[/url]. There it says that the perfect squares are 1, 4, 9, etc. In other words, they don't include 0. That's not how I would have defined the term, but I guess if you are taking a MathCounts competition, then you should follow their definitions.", "Solution_12": "theone were u on our team when we did that or were you on a different team? so is 0 a perfect square or not?", "Solution_13": "For mathcounts, no.", "Solution_14": "but wut about that weird team problem where theone got wrong cause he forgot 0 as a perfect square?", "Solution_15": "We can't answer directly without seeing the problem..\r\nProblem must mention some special words in it.. Because as I said, zero doesn't count on usual perfect square question on Mathcounts..", "Solution_16": "This is always an interesting sort of topic. Personally, I am of the opinion that more inclusive definitions are usually better.\r\n\r\nOne of the basic properties of mathematics that students must master is that of generalization. If you create too many special cases where generalization is not allowed (simply due to definition), a student may become reluctant to generalize.\r\n\r\nIs a square a special rectangle? Is a rhombus a special parallelogram? Is a parallelogram a special trapezoid? Is an equilateral triangle a special isoceles triangle? Different books and different mathematicians give different answers.\r\n\r\nAs I said before, I tend to like the more inclusive definitions as long as properties that apply to the specific case also apply to the general case.\r\n\r\nTo Ravi B: This is the first year I recall seeing MATHCOUNTS provide a glossary of mathematical terms. If MATHCOUNTS has published a definition, then you should probably assume the question-writers will be using the official MATHCOUNTS definitions. However, Section III of the new MATHCOUNTS Toolbox does not define \"Perfect Square\". It simply lists some sample values. Should we assume 676 is not a perfect square because the table ends with 25^2 = 625? Of course not! So I'm not sure you should conclude that MATHCOUNTS has defined \"perfect square\" to exclude zero.\r\n\r\nTo tokenadult: I believe most sources that give a written definition say that a perfect square is the square of an \"integer\". That would seem to include 0. I don't believe I've ever seen a written definition say a perfect square is the square of a \"counting number\". On the other hand, most sources that list perfect squares begin with 1.\r\n\r\nThere is also the possible distinction between a \"perfect square\" and a \"square number\". I don't think anyone would argue that the first 4 \"triangular numbers\" were 1, 3, 6, and 10. Or that the third triangular number is 6. But if you asked what is the zeroeth triangular number, I think most mathematicians would respond \"it's zero, of course\". A few might claim it is undefined. So is zero a triangular number? Is zero a square number? Is zero a perfect square?\r\n\r\nThe best answer is that it depends on the context.\r\n\r\nA good question-writer will avoid the ambiguities. He won't refer to a quadrilateral as a \"rectangle\" if it is also a square. If he uses the term \"perfect square\", he will either use it in a way that it must be positive or he will add the qualifier \"non-zero\" or \"positive\" or \"including zero\" to the problem-statement, or he will define the term.\r\n\r\nA good question-taker will consider the possible ambiguities. He will consider the possibility that the \"rectangle\" refered to in the problem *might* actually be a square. He will consider that the \"perfect square\" *might* be zero. He will consider that sqrt(17) *might* be negative. If he finds these ambiguities exist, the good question-taker will reexamine the problem to see if there is an extra constraint the eliminates the ambiguity. If one case is trivial and one case is non-trivial, he will make note of the trivial case, but solve the non-trivial case.\r\n\r\nIf ambiguities remain that cannot be resolved, a good math teacher will appreciate that the student noticed the defect in the problem. In a competition, a student should note the ambiguity on his test and bring the ambiguity to the attention of his or her coach.\r\n\r\nJust my opinions. :-)", "Solution_17": "I agree with your opinion in general, although I would like to nit-pick some of the particulars. If one were to list the perfect squares, starting with 1 and continuing to 625, in increasing order, it is quite reasonable to assume that the definition being used does not include 0, since the list could easily have been extended backwards to 0 but could not have been extended back any further. Thus, the choice to start the list with the number 1 seems quite signficant in this case. Also, non-mathematical definitions of mathematical terms are consistantly awful. With all respect to Webster's dictionary and tokenadult and anyone else who refered to such a reference, I would never turn to that kind of source for this kind of question, because the people who write them are not being held to a mathematical level of precision.\r\n\r\nFrom a mathematical standpoint, none of this really matters. Any of the definitions you listed are completely irrelevant to the mathematical truths associated with them. It is nice to have standard, comprehensible definitions, and it is usually nice to have them include the widest reasonable array of things possible. But the issue here isn't really from a mathematical viewpoint; it's from the perspective of math competitions. In competitions, it matters very much what the rules are, because even if you can solve the question when 0 is a perfect square and when 0 isn't, you only get to put down one answer. Very careful problem writers [i]should[/i] always define ambiguous terms in their questions, but they don't. So the default contest rules are quite significant, even given their total ambiguity.", "Solution_18": "The problem goes something like this:\r\n\r\n1 is less than or equal to x is less than 100.\r\n\r\nhow many x are there that can be expressed as a62+b^2, where a and b integers?", "Solution_19": "But that question never uses the term \"perfect square\". It just says b2 where b is an integer. Because 0 is an integer, 02 is okay for this problem.", "Solution_20": "that confuses me...", "Solution_21": "Well, it says to express something as a^2 + b^2 where a and b are integers.. so a and b can be {.. -3, -2, -1, 0, 1, 2, 3, ..}. \r\nNow if it said to express it as the sum of two perfect squares, then we may have a problem. but it doesn't. this is just what Ravi B said but I don't know how else to explain it.. which part confused you?", "Solution_22": "Well this is the problem i had trouble with:\r\n\r\nHow many numbers less than 100 can be expressed as the sum of two squares.", "Solution_23": "Are you sure you're quoting the question [i]exactly[/i] as it was written? Because if the question is the way Starry wrote it, there isn't any problem at all -- it'd be a reading issue, not a definition issue.\r\n\r\n\r\nAnd as much as I enjoy getting credit for such things, I think you (TripleM) meant to say Ravi B., not JBL, right?", "Solution_24": "Doh. Fixed :)", "Solution_25": "[quote=\"rcv\"]. . . . \n\nTo tokenadult: I believe most sources that give a written definition say that a perfect square is the square of an \"integer\". That would seem to include 0. I don't believe I've ever seen a written definition say a perfect square is the square of a \"counting number\". On the other hand, most sources that list perfect squares begin with 1. \n. . . . \nJust my opinions. :-)[/quote]\r\n\r\nI asked about what the mathematical sources say in the thread on the Getting Started Forum \"[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]How is 'perfect square' defined in books?[/url]\" And after seeing the answer posted to that thread, I have no doubt at all that 0 is NOT a perfect square, and has never been regarded as such by the mathematicians who investigate number theory professionally (who would be the arbiters of usage on such a question). \r\n\r\nIt appears that this thread developed from a competition problem that was misremembered, or related to another person without sufficient detail, or something like that. Getting the right answer to a mathematical problem always involves, first of all, figuring out what the problem is about. The problem in this thread, it turns out, is not defining the term \"perfect square\" (which is the proper \"problem\" posed by the thread I posted) but finding a correct answer given a specified set of constraints, in which it appears the term \"perfect square\" didn't appear at all (if other participants here are relating the story correctly).", "Solution_26": "[quote=\"JBL\"]I agree with your opinion in general, although I would like to nit-pick some of the particulars. . . . With all respect to Webster's dictionary and tokenadult and anyone else who refered to such a reference, I would never turn to that kind of source for this kind of question, because the people who write them are not being held to a mathematical level of precision.[/quote]\r\n\r\nNow it's my turn to nitpick. :) Where did you ever get the idea I would turn to a general English dictionary, or to any nonmathematical source, for the definition of a mathematical term? I notice the absence of quoted text from any of my posts in your discussion excerpted above. I asked, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]in the thread I posted[/url], for definitions from mathematically acceptable sources (and offered some from sources I have at hand) and it's clear to me now that the mathematicians who work with the issue day by day DO NOT call zero a perfect square (for some of the same reasons that they don't call one a prime number). \r\n\r\nThis, of course, is all in the spirit of \"I agree with your opinion in general, although I would like to nit-pick some of the particulars.\"", "Solution_27": "[quote=\"tokenadult\"]\nI asked about what the mathematical sources say in the thread on the Getting Started Forum \"[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]How is 'perfect square' defined in books?[/url]\" And after seeing the answer posted to that thread, I have no doubt at all that 0 is NOT a perfect square, and has never been regarded as such by the mathematicians who investigate number theory professionally (who would be the arbiters of usage on such a question). \n[/quote]\r\n\r\nThe famous theorem that says that a number can be written as the sum of two squares only if any prime factor that has the form 4k-1 has an even multiplicity.\r\n\r\nIn this case, 0 is a square. Of course it doesn't say perfect square, but really this is pretty similar to this problem..", "Solution_28": "Just for clarification purposes, the question which originated the thread was verbatim from the 1993-94 Mathcounts handbook, warm-up 18, #3. It is stated early on in the thread.\r\n\r\nAnother question brought up was theone's team test question, that we aren't sure of the exact wording.\r\n\r\nI now imagine me taking the state test on Saturday in great fear that every question will involve determining whether or not I should consider 0 as a square or a perfect square! :lol:", "Solution_29": "Assume that 0 is not a perfect square. If you get the problem wrong because of that, challenge their answer by showing them the MathCounts Toolbox.", "Solution_30": "We'll do, & really thanks.", "Solution_31": "Yes a perfect square but not an integer", "Solution_32": "0 is in fact an integer.", "Solution_33": "[url=https://en.wikipedia.org/wiki/Square_number]According to Wikipedia[/url] zero is a perfect square", "Solution_34": "yea cause 0 * 0 = 0" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all integers $ x$ such that $ x(x\\plus{}1)(x\\plus{}7)(x\\plus{}8)$ is a perfect square\r\nIt's a nice problem ...hope you enjoy it!\r\n\r\nDaniel", "Solution_1": "[quote=\"lambruscokid\"]Find all integers $ x$ such that $ x(x \\plus{} 1)(x \\plus{} 7)(x \\plus{} 8)$ is a perfect square\nIt's a nice problem ...hope you enjoy it!\n\nDaniel[/quote]\r\n\r\n\r\n\r\n$ x(x \\plus{} 1)(x \\plus{} 7)(x \\plus{} 8)\\equal{})(x^{2}\\plus{}8x)(x^{2}\\plus{}8x\\plus{}7)\\equal{}b(b\\plus{}7)\\equal{}a^{2}$\r\n$ b\\equal{}x^{2}\\plus{}8x$\r\n$ 4b(b\\plus{}7)\\plus{}49\\minus{}49\\equal{}(2b\\plus{}7)^{2}\\minus{}49\\equal{}4a^{2}$\r\n----->\r\n\r\n$ (2b\\plus{}7\\minus{}2a)(2b\\plus{}7\\plus{}2a)\\equal{}49$\r\n\r\n$ 2b\\plus{}7\\minus{}2a\\equal{}1;7;49;\\minus{}1;\\minus{}7;\\minus{}49$\r\n$ 2b\\plus{}7\\plus{}2a\\equal{}49;7;1;\\minus{}49;\\minus{}7;\\minus{}1$\r\n$ 4b\\plus{}14\\equal{}50;17;50;\\minus{}50;\\minus{}14;\\minus{}50$-----> $ 4b\\equal{}36;0;36;\\minus{}64;\\minus{}28;\\minus{}64$----> $ b\\equal{}9;0;9;\\minus{}16;\\minus{}7;\\minus{}16$\r\n\r\n\r\n$ b\\equal{}9$---> $ x\\equal{}\\minus{}9;1$\r\n$ b\\equal{}0$---> $ x\\equal{}\\minus{}8;0$\r\n$ b\\equal{}\\minus{}16$--->$ x\\equal{}4$\r\n$ b\\equal{}\\minus{}7$---> $ x\\equal{}\\minus{}7;\\minus{}1$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Minimize (a+b)(c+b) subject to abc(a+b+c)=1.", "Solution_1": "For $ a>0,\\ b>0,\\ c>0$, $ (a\\plus{}b)(b\\plus{}c)\\equal{}ca\\plus{}\\frac{1}{ca}\\geq 2$ by AM-GM.", "Solution_2": "I didn't assume a, b, c > 0. I expanded (a+b)(c+b) and then completed the square to get a minimum of -((a-c)^2)/4 for b = -(a+c)/2 but am having problems relating this to the condition abc(a+b+c)= 1." } { "Tag": [ "ratio", "geometry", "similar triangles" ], "Problem": "A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of\r\nthe vertices of the square lie on the hypotenuse and the other two vertices of the\r\nsquare lie on the legs of the triangle, what is the length of a side of the square?\r\nExpress your answer as a common fraction.", "Solution_1": "EDIT: I found a mistake in my solution. Sorry!\r\n\r\nHowever:\r\n\r\n[hide=\"HINT\"]\nSimilar triangles, similar triangles, similar triangles!!\n[/hide]", "Solution_2": "Can you give another hint? :blush:", "Solution_3": "[hide=\"Another hint\"]\nWhich similar triangle ratios do you want? What are we trying to solve for? [/hide]", "Solution_4": "Oh wow!! I never noticed those triangles were similar!!! :rotfl:", "Solution_5": "yes but how would you relate the side lengths to set up a proportion?", "Solution_6": "OK I figured it out now.\r\n\r\nI labeled the parts of the hypotenuse a,b,c with b being side of square, so a+b+c=17\r\n\r\na/b=15/8\r\nc/b=8/15\r\n\r\na=15b/8\r\nc=8b/15\r\n\r\nso 15b/8+b+8b/15=17\r\n409b/120=17\r\nb=2040/409?", "Solution_7": "Yes, that is right :)", "Solution_8": "I feel stupid now...so easy" } { "Tag": [ "number theory", "least common multiple" ], "Problem": "Four distinct 3-digit integers have the same hundredth digit and 3 of the numbers are divisors of the sum of the 4 numbers. Find all possible such 4 numbers.", "Solution_1": "[hide]400a+10b+10c+10d+10e+f+g+h+j is the total sum.\nWithout loss of generality, let's assume 100a+10b+f, 100a+10c+g, 100a+10d+h are the three numbers that divide the total sum. By setting up actual divison, we can cancel out the 400a. This gives us a 4. Of course, there's a bunch of remainder terms. 10(b+c+d+e) could possibly add up to 360. f+g+h+j could add up to 36 but in order for that to happen, b doesn't equal c doesn't equal d doesn't equal e and for 360, similar requirements are placed on f+g+h+j. So the maximum is 360+9+8+7+6 or 390. The minimum would have 10(b+c+d+e)=0 and have f+g+h+j=0+1+2+3=6. So the range is between 400a+6 to 400a+390. We can prove that for a>2 the quotients are 3 and 4 only. We first prove that a quotient of 2 is impossible. 200a+{0,2,...198}=400a+{6,7,...390}. 200a+{6,7,...390}={0,2,...198} which is obviously not true since a is a digit. Now, we prove that 5 and numbers greater than 5 will fail. We pick the smallest a that's greater than 2; this is 3. 300+400(3)=1500 which equals 1500/300=5. But I can't have three numbers in the 300s average together to equal 400. The best I can do with those 3 numbers is 399+398+397. If I was to increase the 300, that would be increasing my lowest number that's increasing my sum by 1. But for any fraction p/q greater than one, (p+1)/(q+1)

0,xy+yz+zx=1$\r\n\\[\\frac{x(y+z)^{2}}{(1+yz)^{2}}+\\frac{y(z+x)^{2}}{(1+zx)^{2}}+\\frac{z(x+y)^{2}}{(1+xy)^{2}}\\ge \\frac{3\\sqrt{3}}{4}\\]\r\n:)", "Solution_1": "Please use English in name of this topic! Locked!" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $ AA_1$ \u03b7 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c2 \u03bf\u03be\u03c5\u03b3\u03c9\u03bd\u03af\u03bf \u03c4\u03c1\u03b3\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$, \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ M$ \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf\u03bd \u03c4\u03bf\u03c5 $ AA_1$. \u03a0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ P$ \u03c3\u03c4\u03b7\u03bd $ BM$ \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 $ \\angle APC \\equal{} 90^{\\circ}$ \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ Q$ \u03c3\u03c4\u03b7\u03bd $ CM$ \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ \\angle AQB \\equal{} 90^{\\circ}$. \r\n\u039d\u03b4\u03bf \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ A_1$, $ P$, $ M$ and $ Q$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac .\r\n\r\n\u03a0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03c4\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b1\u03c2 :)", "Solution_1": "[quote=\"silouan\"]\u0388\u03c3\u03c4\u03c9 $ AA_1$ \u03b7 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c2 \u03bf\u03be\u03c5\u03b3\u03c9\u03bd\u03af\u03bf\u03c5 \u03c4\u03c1\u03b3\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ ABC$, \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ M$ \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf\u03bd \u03c4\u03bf\u03c5 $ AA_1$. \u03a0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ P$ \u03c3\u03c4\u03b7\u03bd $ BM$ \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 $ \\angle APC = 90^{\\circ}$ \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ Q$ \u03c3\u03c4\u03b7\u03bd $ CM$ \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 $ \\angle AQB = 90^{\\circ}$. \n\u039d\u03b4\u03bf \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ A_1$, $ P$, $ M$ and $ Q$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac .\n\n\u03a0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03c4\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b1\u03c2 :)[/quote]\r\n\r\n[color=red][b]\u039b\u03a5\u03a3\u0397[/b][/color]\r\n\r\n\u0395\u03cd\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03af\u03c7\u03bd\u03bf\u03c2 $ D$ \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 $ h_a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03c9\u03bd \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd $ AQB$ \u03ba\u03b1\u03b9 $ APC$. \u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ MD = MA = MA_{1}$ \u03b1\u03c0\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c3\u03bf\u03c5 \u03c3\u03b5 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf. \u0395\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc \u03cc\u03c4\u03b9 $ \\angle{DAA_{1} = \\theta = \\frac {|\\angle{B} - \\angle{C}|}{2}}$ \u03c9\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03cd\u03c8\u03bf\u03c5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c5.\r\n\r\n[b]\u039b\u0397\u039c\u039c\u0391 1.[/b]\u039f\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03c4\u03c9\u03bd \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03c9\u03bd $ AQC$,$ QCA_1$,$ BPA_1$ \u03ba\u03b1\u03b9 $ APB$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 $ AA_1$.\r\n\r\n[i]\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7.[/i] \u0391\u03c2 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ Q'\\in CM$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03ce\u03c3\u03c4\u03b5 \u03bf \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ AQ'C$ \u03bd\u03b1 \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 $ AA_1$. \u03a3\u03b7\u03bc\u03b5\u03b9\u03ce\u03c3\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ac \u03b1\u03c0\u03bb\u03cc \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03b9 \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c0\u03c9\u03c2 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7 $ CM$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc.\r\n\u0388\u03c7\u03bf\u03c5\u03bc\u03b5 $ MQ'\\cdot MC = MA^2 = MD^2$. \u0391\u03c5\u03c4\u03cc \u03b4\u03b5\u03af\u03c7\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03bf \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 $ Q'CD$ \u03b5\u03c6\u03ac\u03c0\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c4\u03b7\u03c2 $ MD$. \u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b4\u03b5\u03ad\u03b9 \u03c4\u03b7\u03bd \u03c5\u03c0\u03bf \u03c7\u03bf\u03c1\u03b4\u03ae\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b3\u03c9\u03bd\u03af\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03b7 \u03c0\u03bf\u03c5 \u03b2\u03b1\u03af\u03bd\u03b5\u03b9 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c4\u03cc\u03be\u03bf, \u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b1\u03ba\u03cc\u03bb\u03bf\u03c5\u03b8\u03b5\u03c2 \u03b3\u03c9\u03bd\u03b9\u03b1\u03ba\u03ad\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2:\r\n\r\n$ \\angle{MQ'A} = \\frac {\\angle{A}}{2}$\r\n\r\n$ \\angle{MQ'D} = \\angle{MDA_{1}} = \\angle{MA_{1}D = 90 - \\theta}$\r\n\u03a3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $ \\angle{B} + \\angle{AQ'D} = 180$, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c4\u03bf $ Q'$ \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf $ AQB$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 $ Q' = Q$.\r\n\u038c\u03bc\u03bf\u03b9\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf $ P$.\r\n\r\n\r\nT\u03ce\u03c1\u03b1, \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03cc\u03c4\u03b9 $ \\angle{PBA_{1}} = \\angle{PA_{1}A}$ \u03ba\u03b1\u03b9 $ \\angle{MCA_{1}} = \\angle{MA_{1}C}$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b5\u03be\u03ae\u03c2 \u03b3\u03c9\u03bd\u03b9\u03b1\u03ba\u03ad\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03bd\u03cd\u03bf\u03c5\u03bd \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03b1 $ MPDA_{1}$ \u03ba\u03b1\u03b9 $ MDA_{1}Q$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03b3\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03b1:\r\n\r\n$ \\angle {MPA_{1}} = \\angle{PBA_1} + \\angle{PA_{1}B} = \\angle{MA_{1}D} = \\angle{MDA_1}$ \u03ba\u03b1\u03b9 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 $ \\angle{MQD} = \\angle{MA_{1}D}$.\r\n\r\n\u03a3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b1\u03c3\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac, \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ M,P,D,A_{1},Q$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac.\r\n\r\n[i]\u03a0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7.[/i] \u0391\u03bd \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 $ B$ \u03ba\u03b1\u03b9 $ C$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c4\u03c1\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae $ A$ \u03bd\u03b1 \u03ba\u03b9\u03bd\u03b5\u03af\u03c4\u03b1\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 $ \\angle{BAC}$ \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bc\u03ad\u03bd\u03b5\u03b9 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ae \u03c4\u03cc\u03c4\u03b5 \u03bf \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03cc\u03c2 \u03c4\u03cc\u03c0\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ba\u03ad\u03bd\u03c4\u03c1\u03bf\u03c5 \u03c4\u03c9\u03bd \u03c0\u03c1\u03bf\u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03ad\u03bd\u03c4\u03c9\u03bd \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ce\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03c9\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf\u03c5 $ T$, \u03cc\u03c0\u03bf\u03c5 \u03c4\u03bf $ T$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 \u03b5\u03bb\u03ac\u03c3\u03c3\u03bf\u03bd\u03bf\u03c2 \u03c4\u03cc\u03be\u03bf\u03c5 $ \\overarc{BC}$.\r\n\r\n\u039d\u03af\u03ba\u03bf\u03c2 \u03a1\u03ac\u03c0\u03b1\u03bd\u03bf\u03c2", "Solution_2": "\u03a9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u039d\u03af\u03ba\u03bf!\r\n\r\n\u039c\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03c0\u03c1\u03bf\u03c3\u03ad\u03b3\u03b3\u03b9\u03c3\u03b7 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03c6\u03bf\u03cd \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $ MQ\\cdot MC\\equal{}MA_1^2$ \u03bd\u03b1 \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03ae \u03bc\u03b5 \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf $ M$ \u03ba\u03b9 \u03b1\u03ba\u03c4\u03af\u03bd\u03b1 $ MA_1$. \u0395\u03c4\u03c3\u03b9 \u03c4\u03b1 $ P,Q$ \u03b8\u03b1 \u03c0\u03ae\u03b3\u03b1\u03b9\u03bd\u03b1\u03bd \u03c3\u03c4\u03b1 $ B,C$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03ba\u03b1\u03b9 \u03b1\u03c6\u03bf\u03cd \u03c4\u03bf $ A_1$ \u03b8\u03b1 \u03ad\u03bc\u03b5\u03bd\u03b5 \u03c3\u03c4\u03b7 \u03b8\u03ad\u03c3\u03b7 \u03c4\u03bf\u03c5, \u03b8\u03b1 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 $ P',A_1,Q'$ \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ba\u03b9 \u03ac\u03c1\u03b1 $ M, P, A_1, Q$ \u03bf\u03bc\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac :wink:" } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "$x^{5}-6x^{4}-24x^{3}+134x^{2}-105x$\r\n\r\nhow do i factor this?", "Solution_1": "Well, one of the factors is x.\r\n\r\nAnother one may be found by rational root theorem.", "Solution_2": "well, by the RRT, you can find that 1 is a root, but I dont know how to solve anymore for that third degree polynomial-try graphing it(on a calculator) and finding the zeros..", "Solution_3": "I graphed it and the roots were appaent. It factors as [hide]$(x+5)x(x-1)(x-3)(x-7)$[/hide]\r\n\r\nBut yeah, if you can't use a graphing calc., go RRT.", "Solution_4": "Not sure if this is the most efficient way, but here goes.\r\nFirst we know that $0$ is a root, since $x$ divides the polynomial. We now know one root out of five.\r\nA good next step is to see what $x=1$ yields, since it is easy to evaluate. In this case, it works as well. Two roots out of five.\r\nWe divide the polynomial by $x$ and $(x-1)$ which gives us $x^{3}-5x^{2}-29x+105$.\r\nWe use the Rational Root Theorem to find that our next root must divide $105$. We try $-5$, and it works. We can continue guessing according to RRT, or we can simply just...\r\nDivide the polynomial we got by $(x+5)$, and we get $x^{2}-10x+21$. The remaining roots are $3$ and $7$.\r\n\r\nOf course, we could've just started using RRT after dividing by x. This is just how I did it.", "Solution_5": "o wow..hmm im stupid, when I tested out 7, it didnt work.. and i was using a calculator" } { "Tag": [ "geometry", "function", "algebra", "domain", "perimeter", "inequalities" ], "Problem": "A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped like a circle.\r\n\r\n(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.\r\n\r\n(b) What is the domain of A?", "Solution_1": "I am not sure if this is correct:\r\n[hide] \na. Since the side length of the equilateral triangle is x, we have that the perimeter of the triangle is 3x. Therefore, it takes 3x meters of wire to make the equilateral triangle. The area of the equilateral triangle thus formed is $\\frac{x^{2}\\sqrt{3}}{4}$. Now, there is 10-3x meters of wire left for the circle. Say that r is the radius of the circle. $2r\\pi=10-3x$, so $r=\\frac{10-3x}{2\\pi}$. This implies that the area of the circle formed is $r^{2}\\pi=\\frac{100+9x^{2}-60x}{4\\pi}$. Therefore, the total area of the triangle and circle is $\\frac{x^{2}\\pi\\sqrt{3}+100+9x^{2}-60x}{4\\pi}$.\nb. Since 3x cannot exceed 10 and 3x cannot be less zero, we have the domain of x is $0\\geq(x)\\geq\\frac{10}{3}$, so the domain of A is $\\frac{100}{4\\pi}\\geq(A)\\geq\\frac{25\\sqrt{3}}{9}$. [/hide]", "Solution_2": "I thank you for your help.", "Solution_3": "[quote=\"The QuattoMaster 6000\"]I am not sure if this is correct:\n[hide] \na. Since the side length of the equilateral triangle is x, we have that the perimeter of the triangle is 3x. Therefore, it takes 3x meters of wire to make the equilateral triangle. The area of the equilateral triangle thus formed is $\\frac{x^{2}\\sqrt{3}}{4}$. Now, there is 10-3x meters of wire left for the circle. Say that r is the radius of the circle. $2r\\pi=10-3x$, so $r=\\frac{10-3x}{2\\pi}$. This implies that the area of the circle formed is $r^{2}\\pi=\\frac{100+9x^{2}-60x}{4\\pi}$. Therefore, the total area of the triangle and circle is $\\frac{x^{2}\\pi\\sqrt{3}+100+9x^{2}-60x}{4\\pi}$.\nb. Since 3x cannot exceed 10 and 3x cannot be less zero, we have the domain of x is $0\\geq(x)\\geq\\frac{10}{3}$, so the domain of A is $\\frac{100}{4\\pi}\\geq(A)\\geq\\frac{25\\sqrt{3}}{9}$. [/hide][/quote]\n\n[hide=\"Pretty much, but...\"]\nYou've got a mistake in part b. First of all, $3x$ and $10-3x$ cannot be $0$ either because you must make both a circle and a triangle. Also, you meant $\\leq$ where you put $\\geq$. So it should be:\n\n$0 < x < \\frac{10}{3}$\n\nAlso note that the domain of $A(x)$ is the possible values of $x$, not the possible values of $A$. So you computed (with your reversed inequalities) the range of $A$, not its domain. The interval I just described, $(0 , \\frac{10}{3})$ is the domain of $A$.[/hide]", "Solution_4": ":blush: Oh, yes. You are right, sorry about that.", "Solution_5": "Great math notes for me to review and study.\r\n\r\nThanks!" } { "Tag": [ "function", "algebra", "polynomial", "inequalities", "LaTeX", "quadratics", "AMC" ], "Problem": "Given the proportion $ \\frac{x\\plus{}y}{3}\\equal{}\\frac{y\\plus{}z}{4}\\equal{}\\frac{x\\plus{}z}{5}$, determine the value of $ \\frac{xy\\plus{}yz\\plus{}xz}{x^2\\plus{}y^2\\plus{}z^2}$", "Solution_1": "Note that the thing you want to compute is homogeneous in $ x, y, z$, so you can just set $ x \\plus{} y \\equal{} 3, y \\plus{} z \\equal{} 4, x \\plus{} z \\equal{} 5$.", "Solution_2": "what does homogeneous mean?", "Solution_3": "A function $ f(x, y, z)$ is homogeneous of degree $ k$ if $ f(tx, ty, tz) \\equal{} t^k f(x, y, z)$. The usual example is [url=http://en.wikipedia.org/wiki/Homogeneous_polynomial]homogeneous polynomials[/url]; here we have a quotient of two homogeneous polynomials of the same degree.\r\n\r\nIf you don't like buzzwords, let $ \\frac {x \\plus{} y}{3} \\equal{} ... \\equal{} d$ and set $ a \\equal{} \\frac {x}{d}, b \\equal{} \\frac {y}{d}, c \\equal{} \\frac {z}{d}$. Then $ \\frac {xy \\plus{} yz \\plus{} zx}{x^2 \\plus{} y^2 \\plus{} z^2} \\equal{} \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}$.", "Solution_4": "so $ (x\\plus{}y)(y\\plus{}z)\\plus{}(y\\plus{}z)(z\\plus{}x)\\plus{}(z\\plus{}x)(x\\plus{}y) \\equal{} xy\\plus{}zx\\plus{}yz\\plus{}y^2 \\plus{}yz\\plus{}yx\\plus{}xz\\plus{}z^2 \\plus{} zx\\plus{}zy\\plus{}yx\\plus{}x^2 \\equal{} 3xy\\plus{}3yz\\plus{}3xz\\plus{}y^2\\plus{}x^2\\plus{}z^2 \\equal{} p$... perhaps not so useful.... yet.\r\n\r\n$ (x\\plus{}y)^2\\plus{}(x\\plus{}z)^2\\plus{}(y\\plus{}z)^2 \\equal{} 2x^2\\plus{}2y^2\\plus{}2z^2 \\plus{} 2xy\\plus{}2yz\\plus{}2xz \\equal{} q$\r\n\r\nNow $ q\\minus{}2p \\equal{} \\minus{}4(xy\\plus{}xz\\plus{}yz) \\implies \\frac{2p\\minus{}q}{4} \\equal{} xy\\plus{}yz\\plus{}xz$ and $ 2p\\minus{}3q \\equal{} \\minus{}4(x^2\\plus{}y^2\\plus{}z^2) \\implies \\frac{3q\\minus{}2p}{4} \\equal{} x^2\\plus{}y^2\\plus{}z^2$.\r\n\r\nNow it follows for t0r's remarks that we can let $ x\\plus{}y\\equal{}3$, $ y\\plus{}z\\equal{}4$, and $ x\\plus{}z\\equal{}5$. Then we get $ p\\equal{}3*4\\plus{}4*5\\plus{}5*3 \\equal{} 12\\plus{}20\\plus{}15 \\equal{} 47$ and $ q\\equal{}3^2\\plus{}4^2\\plus{}5^2 \\equal{} 9\\plus{}16\\plus{}25 \\equal{} 50$. Then we have $ xy\\plus{}yz\\plus{}xz \\equal{} (2p\\minus{}q)/4 \\equal{} (2*47\\minus{}50)/4\\equal{}11$ and $ x^2\\plus{}y^2\\plus{}z^2 \\equal{} \\frac{3*50\\minus{}2*47}{4} \\equal{} 14$, so our answer is $ \\frac{11}{14}$.\r\n\r\ncomments:\r\n1) pretty much I just found the two symmetric polynomials that I could knowing that I could set $ x\\plus{}y$ and such to whatever I like. The only polynomials with a degree of 2 were $ \\sum_{cyc} (x\\plus{}y)(y\\plus{}z)$ and $ \\sum_{cyc} (x\\plus{}y)^2$, and then I saw they were both linear combinations of the two I wanted to find.\r\n2) by rearrangment inequality, it is clear our answer will be less than 1.", "Solution_5": "[hide=\"Or...\"] $ x \\equal{} 2, y \\equal{} 1, z \\equal{} 3$. [/hide]", "Solution_6": "Ooh. Hmm. You guys make it sound so easy. And that is beastly LaTeX people. Wow... Thanks! :D", "Solution_7": "[quote=\"jonathanchou711\"]Given the proportion $ \\frac {x \\plus{} y}{3} \\equal{} \\frac {y \\plus{} z}{4} \\equal{} \\frac {x \\plus{} z}{5}$, determine the value of $ \\frac {xy \\plus{} yz \\plus{} xz}{x^2 \\plus{} y^2 \\plus{} z^2}$[/quote]\r\n\r\nThis is easy for Japanese high school students.\r\n\r\nLet $ \\frac {x \\plus{} y}{3} \\equal{} \\frac {y \\plus{} z}{4} \\equal{} \\frac {x \\plus{} z}{5}\\equal{}k$, we have $ x\\plus{}y\\equal{}3k,\\ y\\plus{}z\\equal{}4k,\\ z\\plus{}x\\equal{}5k$.\r\nAdding these gives $ 2(x\\plus{}y\\plus{}z)\\equal{}12k\\Longleftrightarrow x\\plus{}y\\plus{}z\\equal{}6k\\Longrightarrow x\\equal{}(x\\plus{}y\\plus{}z)\\minus{}(y\\plus{}z)\\equal{}2k$, yielding $ y\\equal{}k,\\ z\\equal{}3k$.\r\nSubstitue these for the given expression, we have $ \\frac{xy\\plus{}yz\\plus{}zx}{x^2\\plus{}y^2\\plus{}z^2}\\equal{}\\frac{(2*1\\plus{}1*3\\plus{}3*2)k^2}{(2^2\\plus{}1^2\\plus{}3^2)k^2}\\equal{}\\boxed{\\frac{11}{14}}$", "Solution_8": "[quote=\"t0rajir0u\"][hide=\"Or...\"] $ x \\equal{} 2, y \\equal{} 1, z \\equal{} 3$. [/hide][/quote]\r\n\r\n(EDIT: seems kunny beat me to it.)\r\nOf course, I missed the simplest way. note we could also let $ d$ be the common value, then solve the system in terms of $ d$ to get $ x\\equal{}2d$, $ y\\equal{}d$ and $ z\\equal{}3d$. The $ d$'s would cancel out in the end.\r\n\r\nby the way... I did not know if my method would work out at the beginning, I just suspected. Strongly. The fundamental theorem of symmetric polynomials says there is a way for any symmetric polynomial to be expressed in terms of elementary symmetric polynomials (for example, for three variables $ x\\plus{}y\\plus{}z$, $ xy\\plus{}yz\\plus{}xz$, and $ xyz$ are the elementary symmetric polynomials). So it seems it could be possible to write non-elementary polynomials in terms of other polynomials.", "Solution_9": "Practice 1\r\n\r\nFind the value of $ p$ such that $ \\frac{a}{b\\plus{}c}\\equal{}\\frac{b}{c\\plus{}a}\\equal{}\\frac{c}{a\\plus{}b}\\equal{}p$.\r\n\r\n1985 Tama Art University entrance exam/Architecture", "Solution_10": "[quote=\"kunny\"][quote=\"jonathanchou711\"]Given the proportion $ \\frac {x \\plus{} y}{3} \\equal{} \\frac {y \\plus{} z}{4} \\equal{} \\frac {x \\plus{} z}{5}$, determine the value of $ \\frac {xy \\plus{} yz \\plus{} xz}{x^2 \\plus{} y^2 \\plus{} z^2}$[/quote]\n\nThis is easy for Japanese high school students.\n\nLet $ \\frac {x \\plus{} y}{3} \\equal{} \\frac {y \\plus{} z}{4} \\equal{} \\frac {x \\plus{} z}{5} \\equal{} k$, we have $ x \\plus{} y \\equal{} 3k,\\ y \\plus{} z \\equal{} 4k,\\ z \\plus{} x \\equal{} 5k$.\nAdding these gives $ 2(x \\plus{} y \\plus{} z) \\equal{} 12k\\Longleftrightarrow x \\plus{} y \\plus{} z \\equal{} 6k\\Longrightarrow x \\equal{} (x \\plus{} y \\plus{} z) \\minus{} (y \\plus{} z) \\equal{} 2k$, yielding $ y \\equal{} k,\\ z \\equal{} 3k$.\nSubstitue these for the given expression, we have $ \\frac {xy \\plus{} yz \\plus{} zx}{x^2 \\plus{} y^2 \\plus{} z^2} \\equal{} \\frac {(2*1 \\plus{} 1*3 \\plus{} 3*2)k^2}{(2^2 \\plus{} 1^2 \\plus{} 3^2)k^2} \\equal{} \\boxed{\\frac {11}{14}}$[/quote]\r\n\r\n\r\nOk you know what? Don't be bragging about your \"super awesome skillz\"! I'm only in 7th grade! And I have to do this crap! K? You probably only finished pre-algebra or something in 7th grade! :mad: Don't give yourself compliments! The real skilled waits for the compliments and earns them! :mad:\r\n\r\nBTW: Thnx for the practice anyways...", "Solution_11": "My solution is not a super awsome skills, just put the common value of three fractional expressions as $ k$.\r\n\r\nWhat's pre- algebra?\r\n\r\nt0rajir0u and facis made the solution more difficult, don't they?", "Solution_12": "[quote=\"kunny\"]t0rajir0u and facis made the solution more difficult, don't they?[/quote]\r\n$ x \\equal{} 1, y \\equal{} 2, z \\equal{} 3$ is difficult? All I claimed is that the answer is independent of the choice of $ x, y, z$, and all that's left to do is justify that.", "Solution_13": "Students less than junior highschool students usually don't notice $ homogeneous$.\r\nBy the way, what's level of 7 th grade?", "Solution_14": "[quote=\"kunny\"]Practice 1\n\nFind the value of $ p$ such that $ \\frac {a}{b + c} = \\frac {b}{c + a} = \\frac {c}{a + b} = p$.\n\n1985 Tama Art University entrance exam/Architecture[/quote]\r\n\r\nCross multiplying, we clearly have \r\n\\begin{align*}\r\na(a+c) &= b(b+c) \\\\\r\na(a+b) &= c(b+c). \r\n\\end{align*}\r\nDividing the two, we have \r\n\\begin{align*}\r\n\\frac{a+c}{a+b} &= 1\\\\\r\na+c &= b+c \\\\\r\nb&=c. \r\n\\end{align*}\r\nSo it follows that $ a^2 + ab - 2b^2 = 0$ so $ a=b$ or $ a=-2b$. Thus $ p$ is equal to $ \\frac12$ or $ -1$.\r\n\r\nKunny: Normal 7th grade differs from school to school, but the stuff taught in public schools is usually quadratic equations, maybe a little bit of Pythagorean theorem. (That's why I spend so much time on AoPS :P .)", "Solution_15": "Uh... 7th grade? 7th grade should be Pre-Algebra. You don't know what Pre-Algebra is??? Uh... http://en.wikipedia.org/wiki/Pre-algebra It's wiki, and it's not that reliable, but it's good enough. But I'm in Geometry. That's 2 levels up. But what I posted wasn't even Geometry. That was like Algebra 2! I'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?", "Solution_16": "[quote=\"ThinkFlow\"][quote=\"kunny\"]Practice 1\n\nFind the value of $ p$ such that $ \\frac {a}{b + c} = \\frac {b}{c + a} = \\frac {c}{a + b} = p$.\n\n1985 Tama Art University entrance exam/Architecture[/quote]\n\nCross multiplying, we clearly have\n\\begin{align*} a(a + c) & = b(b + c) \\\\\na(a + b) & = c(b + c). \\end{align*}\nDividing the two, we have\n\\begin{align*} \\frac {a + c}{a + b} & = 1 \\\\\na + c & = b + c \\\\\nb & = c. \\end{align*}\nSo it follows that $ a^2 + ab - 2b^2 = 0$ so $ a = b$ or $ a = - 2b$. Thus $ p$ is equal to $ \\frac12$ or $ - 1$.\n\nKunny: Normal 7th grade differs from school to school, but the stuff taught in public schools is usually quadratic equations, maybe a little bit of Pythagorean theorem. (That's why I spend so much time on AoPS :P .)[/quote]\r\n\r\n\r\nThank you for the explanation.\r\n\r\nAlternative Solution:\r\n\r\n$ \\frac {a}{b + c} = \\frac {b}{c + a} = \\frac {c}{a + b} = p\\Longrightarrow a = p(b + c),\\ b = p(c + a),\\ c = p(a + b)$\r\n\r\nAdding both sides gives $ a + b + c = 2p(a + b + c)$.\r\n\r\nCase 1. $ a + b + c\\neq 0$, we have $ 1 = 2p$, yielding $ p = \\boxed{\\frac {1}{2}}$.\r\n\r\nCase 2. $ a + b + c\\neq 0$, we have $ p = \\frac {c}{a + b} = \\frac {c}{ - c} = \\boxed{ - 1}$.\r\n\r\nGenellary it is holds that $ \\boxed{\\frac {a}{b} = \\frac {c}{d} = \\frac {e}{f} = \\frac {pa + qc + re}{pb + qd + rf}}$, which is called \"componend\" by Mathematics Japanese English dictionary.\r\n\r\nUse the fact, for $ a + b + c\\neq 0$, we have $ \\frac {a}{b + c} = \\frac {b}{c + a} = \\frac {c}{a + b} = \\frac {a + b + c}{(b + c) + (c + a) + (a + b)} = \\frac {1}{2}$", "Solution_17": "[quote=\"jonathanchou711\"]I'm in Geometry. That's 2 levels up. But what I posted wasn't even Geometry. That was like Algebra 2! I'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?[/quote]\nYou're being [b]extremely[/b] rude. It should be obvious that kunny is from Japan and if you aren't aware of how much more difficult the math curriculum is in most countries, including Japan, than it is here, you're in for a surprise. And I don't know why you would expect either that middle-school math levels line up within the country or that they would line up [b]outside[/b] the country. \n\n[quote=\"jonathanchou711\"]You probably only finished pre-algebra or something in 7th grade! :mad:[/quote]\r\nAnd not only is this ad hominem likely to be totally incorrect, it reveals a lot more about your character than about kunny's.", "Solution_18": "[quote=\"jonathanchou711\"]Uh... 7th grade? 7th grade should be Pre-Algebra. You don't know what Pre-Algebra is??? Uh... http://en.wikipedia.org/wiki/Pre-algebra It's wiki, and it's not that reliable, but it's good enough. [/quote]\n\nWhere do you live? The system of education has different style by country.\n\n[quote=\"jonathanchou711\"]\n\nI'm in Kumon: http://www.kumon.org And that was hard for me. Have you never been a 7th grader before?[/quote]\r\n\r\nI know Kumon well, in my opinion, Kumon is effective for rapid calculation, but it is just repeating skills for solving, but I don't like the style.\r\n\r\nI think that my ideal mathematics education is \"Guess and Checking\", say, U.S.A.", "Solution_19": "Practice 2\r\n\r\nLet $ a,\\ b,\\ c$ be non-zero real numbers such that $ \\frac{\\minus{}a\\plus{}b\\plus{}c}{a}\\equal{}\\frac{a\\minus{}b\\plus{}c}{b}\\equal{}\\frac{a\\plus{}b\\minus{}c}{c}$.\r\n\r\nFind the value of $ \\frac{(b\\plus{}c)(c\\plus{}a)(a\\plus{}b)}{abc}$.\r\n\r\n1985 Tokyo Women University entrance exam/Mathematics", "Solution_20": "Using the componend, we find that the three fractions equal one. Thus, we find\r\n\\begin{align*}\r\nb+c &= 2a \\\\\r\na+c &= 2b \\\\\r\na+b &= 2c\r\n\\end{align*}\r\nif we set the fractions equal to one and cross-multiply. Solving, we find that $ a=b=c$, so the desired value is $ 2^3 = 8$. Thanks for the practice. :)", "Solution_21": "You have almost solved it, but not perfect.", "Solution_22": "[hide=\"Solution\"]\nLet $ k\\equal{}\\frac{\\minus{}a\\plus{}b\\plus{}c}a\\equal{}\\frac{a\\minus{}b\\plus{}c}b\\equal{}\\frac{a\\plus{}b\\minus{}c}c$, so that $ b\\plus{}c\\equal{}a(k\\plus{}1)$, $ a\\plus{}c\\equal{}b(k\\plus{}1)$, and $ a\\plus{}b\\equal{}c(k\\plus{}1)$, so adding yields $ 2(a\\plus{}b\\plus{}c)\\equal{}(a\\plus{}b\\plus{}c)(k\\plus{}1)\\implies(a\\plus{}b\\plus{}c)(k\\minus{}1)\\equal{}0$. If $ a\\plus{}b\\plus{}c\\equal{}0$, then substitute $ b\\plus{}c\\equal{}\\minus{}a$, $ c\\plus{}a\\equal{}\\minus{}b$, and $ a\\plus{}b\\equal{}\\minus{}c$ to get $ \\frac{\\minus{}a\\cdot\\minus{}b\\cdot \\minus{}c}{abc}\\equal{}\\minus{}1$. If $ a\\plus{}b\\plus{}c\\ne0$, then $ k\\equal{}1$, so multiplying the second group of equations yields $ \\frac{(b\\plus{}c)(c\\plus{}a)(a\\plus{}b)}{abc}\\equal{}\\frac{abc(k\\plus{}1)^3}{abc}\\equal{}2^3\\equal{}8$.\n[/hide]", "Solution_23": "Different Solution:\r\n\r\nI see, I forgot the case when $ a+b+c=0$. If $ a+b+c=0$, then $ a+b=-c$ so \r\n\\begin{align*}\r\n\\frac{(c+b)(c+a)(a+b)}{abc} &= \\frac{-c(c^2 + (a+bc + ab))}{abc} \\\\\r\n&= \\frac{c^2-c^2 + ab}{-ab} \\\\\r\n&= -1,\r\n\\end{align*}\r\nand our values are $ 8$ and $ -1$.", "Solution_24": "[quote=\"math154\"]\nLet $ k = \\frac { - a + b + c}a = \\frac {a - b + c}b = \\frac {a + b - c}c$, so that $ b + c = a(k + 1)$, $ a + c = b(k + 1)$, and $ a + b = c(k + 1)$, so adding yields $ 2(a + b + c) = (a + b + c)(k + 1)\\implies(a + b + c)(k - 1) = 0$. If $ a + b + c = 0$, then substitute $ b + c = - a$, $ c + a = - b$, and $ a + b = - c$ to get $ \\frac { - a\\cdot - b\\cdot - c}{abc} = - 1$. If $ a + b + c\\ne0$, then $ k = 1$, so multiplying the second group of equations yields $ \\frac {(b + c)(c + a)(a + b)}{abc} = \\frac {abc(k + 1)^3}{abc} = 2^3 = 8$.[/quote]\n\nMy solution is same as yours, but you need to examine the case $ a + b + c = 0$.\n\n[quote=\"ThinkFlow\"]Different Solution:\n\nI see, I forgot the case when $ a + b + c = 0$. If $ a + b + c = 0$, then $ a + b = - c$ so\n\\begin{align*} \\frac {(c + b)(c + a)(a + b)}{abc} & = \\frac { - c(c^2 + (a + bc + ab))}{abc} \\\\\n& = \\frac {c^2 - c^2 + ab}{ - ab} \\\\\n& = - 1, \\end{align*}\nand our values are $ 8$ and $ - 1$.[/quote]\r\n\r\nNow We have just finished Practice 2 :lol: .", "Solution_25": "[hide=\"Or...\"] [b]Lemma:[/b] $ \\frac {a}{b} \\equal{} \\frac {c}{d} \\Rightarrow \\frac {a}{b} \\equal{} \\frac {a \\plus{} c}{b \\plus{} d}$.\n\nSumming pairwise, the original three fractions are equal to $ \\frac {2a}{b \\plus{} c} \\equal{} \\frac {2b}{c \\plus{} a} \\equal{} \\frac {2c}{a \\plus{} b}$ and summing all three, the original three fractions are equal to $ 1$ (after the provision about the $ a \\plus{} b \\plus{} c \\equal{} 0$ case, where we can write\n\n$ \\frac{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{abc} \\equal{} \\frac{\\minus{}abc}{abc} \\equal{} \\minus{}1$\n\nand in the $ a \\plus{} b \\plus{} c \\neq 0$ case none of $ a \\plus{} b, b \\plus{} c, c \\plus{} a$ are equal to zero either.) [/hide]", "Solution_26": "[quote=\"kunny\"]Practice 2\n\nLet $ a,\\ b,\\ c$ be non-zero real numbers such that $ \\frac { \\minus{} a \\plus{} b \\plus{} c}{a} \\equal{} \\frac {a \\minus{} b \\plus{} c}{b} \\equal{} \\frac {a \\plus{} b \\minus{} c}{c}$.\n\nFind the value of $ \\frac {(b \\plus{} c)(c \\plus{} a)(a \\plus{} b)}{abc}$.\n\n1985 Tokyo Women University entrance exam/Mathematics[/quote]\r\n\r\nAdd two to each of those fractions to get $ \\frac{a\\plus{}b\\plus{}c}{a}\\equal{}\\frac{a\\plus{}b\\plus{}c}{b}\\equal{}\\frac{a\\plus{}b\\plus{}c}{c}$.\r\n\r\nIf $ a\\plus{}b\\plus{}c\\equal{}0$, $ b\\plus{}c\\equal{}\\minus{}a$ a.s.o. and the fraction is $ \\minus{}1$. If $ a\\plus{}b\\plus{}c\\ne 0$, then $ a\\equal{}b\\equal{}c\\ne 0$ and we get $ 8$.", "Solution_27": "@altheman: Dang almost exactly what I did, except I didn't immediately realize I should add 2 so I went through a little work.\r\n\r\n@kunny:\r\nSorry to ruin the math but...\r\n\r\n[quote=\"kunny\"]I think that my ideal mathematics education is \"Guess and Checking\", say, U.S.A.[/quote]\n\nWas this a joke or were you being serious? I'm not totally sure what you mean (I want to know, whatever it turns out to be)[/quote]", "Solution_28": "[hide][quote=\"jonathanchou711\"]Given the proportion $ \\frac {x \\plus{} y}{3} \\equal{} \\frac {y \\plus{} z}{4} \\equal{} \\frac {x \\plus{} z}{5}$, determine the value of $ \\frac {xy \\plus{} yz \\plus{} xz}{x^2 \\plus{} y^2 \\plus{} z^2}$[/quote]\nThis seems quite straightforward if you note the homogenuity (hinted by t0rajir0u)\n\n$ x \\plus{} y \\equal{} 3; y \\plus{} z \\equal{} 4; x \\plus{} z \\equal{} 5 \\implies x \\plus{} y \\plus{} z \\equal{} (3 \\plus{} 4 \\plus{} 5)/2 \\equal{} 6$\n$ \\implies(x \\plus{} y \\plus{} z)^2 \\equal{} x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 2xy \\plus{} 2yz \\plus{} 2zx \\equal{} 36$\n\n$ x^2 \\plus{} 2xy \\plus{} y^2 \\equal{} 9;y^2 \\plus{} 2yz \\plus{} z^2 \\equal{} 16;x^2 \\plus{} 2xz \\plus{} z^2 \\equal{} 25$\n\n$ \\implies 2(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} xy \\plus{} yz \\plus{} zx) \\equal{} 9 \\plus{} 16 \\plus{} 25 \\equal{} 50$\n\nSo we have \n\n$ x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 2xy \\plus{} 2yz \\plus{} 2zx \\equal{} 36 ...eqn(1)$\n\n$ x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} xy \\plus{} yz \\plus{} zx \\equal{} 25 ...eqn(2)$\n\n$ eqn(1) \\minus{} eqn(2)$ yields $ xy \\plus{} yz \\plus{} zx \\equal{} 11$\n\n$ 2*eqn(2) \\minus{} eqn(1)$ yields $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 14$.\n\nso the desired answer is $ 11/14$\n\n... by the way, don't judge your math ability to regular school curriculums... they aren't great tests for your creativity and problem solving abilities... (not to brag, but guess what... I self-studied up to diff eqns before 9th grade but sometimes I cannot even solve an AIME or AMC problem :blush: )[/hide]", "Solution_29": "[quote=\"kunny\"]Practice 2\n\nLet $ a,\\ b,\\ c$ be non-zero real numbers such that $ \\frac { - a + b + c}{a} = \\frac {a - b + c}{b} = \\frac {a + b - c}{c}$.\n\nFind the value of $ \\frac {(b + c)(c + a)(a + b)}{abc}$.\n\n1985 Tokyo Women University entrance exam/Mathematics[/quote]\r\n\r\nHere is my solution:\r\n\r\n$ \\frac { - a + b + c}{a} = - 1 + \\frac {b + c}{a}$, we have \r\n\r\n$ \\frac { - a + b + c}{a} = \\frac {a - b + c}{b} = \\frac {a + b - c}{c}$\r\n\r\n$ \\Longleftrightarrow \\frac {b + c}{a} = \\frac {c + a}{b} = \\frac {a + b}{c} = k\\Longrightarrow b + c = ak, \\c + a = bk,\\ a + b = ck$.\r\n\r\nAdding these both sides gives $ 2(a + b + c) = k(a + b + c)\\ \\cdots [*]$\r\n\r\n$ \\therefore \\frac { - a + b + c}{a} = \\frac {a - b + c}{b} = \\frac {a + b - c}{c} = \\frac {ak\\ \\cdot bk\\cdot ck}{abc} = k^3$.\r\n\r\nThus, from $ [*]$, the answer is for $ a + b + c\\neq 0$, yielding $ = 2^3 = \\boxed{8}$, for $ a + b + c = 0$, yielding $ k = \\frac { - c}{c} = - 1\\Longrightarrow ( - 1)^3 = \\boxed{ - 1}$.\r\n\r\nOf course, I have the solution using the componend.", "Solution_30": "[quote=\"kunny\"][quote=\"math154\"]\nLet $ k \\equal{} \\frac { \\minus{} a \\plus{} b \\plus{} c}a \\equal{} \\frac {a \\minus{} b \\plus{} c}b \\equal{} \\frac {a \\plus{} b \\minus{} c}c$, so that $ b \\plus{} c \\equal{} a(k \\plus{} 1)$, $ a \\plus{} c \\equal{} b(k \\plus{} 1)$, and $ a \\plus{} b \\equal{} c(k \\plus{} 1)$, so adding yields $ 2(a \\plus{} b \\plus{} c) \\equal{} (a \\plus{} b \\plus{} c)(k \\plus{} 1)\\implies(a \\plus{} b \\plus{} c)(k \\minus{} 1) \\equal{} 0$. If $ a \\plus{} b \\plus{} c \\equal{} 0$, then substitute $ b \\plus{} c \\equal{} \\minus{} a$, $ c \\plus{} a \\equal{} \\minus{} b$, and $ a \\plus{} b \\equal{} \\minus{} c$ to get $ \\frac { \\minus{} a\\cdot \\minus{} b\\cdot \\minus{} c}{abc} \\equal{} \\minus{} 1$. If $ a \\plus{} b \\plus{} c\\ne0$, then $ k \\equal{} 1$, so multiplying the second group of equations yields $ \\frac {(b \\plus{} c)(c \\plus{} a)(a \\plus{} b)}{abc} \\equal{} \\frac {abc(k \\plus{} 1)^3}{abc} \\equal{} 2^3 \\equal{} 8$.[/quote]\n\nMy solution is same as yours, but you need to examine the case $ a \\plus{} b \\plus{} c \\equal{} 0$.[/quote]\r\n\r\nI think I did examine that case...", "Solution_31": "Sorry, I missed the context.", "Solution_32": "Practice 3\r\n\r\nLet $ a,\\ b,\\ c$ be non-zero real numbers. Find the values of $ k,\\ l$ such that $ \\frac {a^2 \\plus{} b^2}{c^2} \\equal{} \\frac {b^2 \\plus{} c^2}{a^2} \\equal{} \\frac {c^2 \\plus{} a^2}{b^2} \\equal{} k$, \r\n$ \\frac {a^3 \\plus{} b^3}{c^3} \\equal{} \\frac {b^3 \\plus{} c^3}{a^3} \\equal{} \\frac {c^3 \\plus{} a^3}{b^3} \\equal{} l$.\r\n\r\n2002 Tokai University entrance exam/Humanities", "Solution_33": "[quote=\"ThinkFlow\"][quote=\"kunny\"]Practice 1\n\nFind the value of $ p$ such that $ \\frac {a}{b + c} = \\frac {b}{c + a} = \\frac {c}{a + b} = p$.\n\n1985 Tama Art University entrance exam/Architecture[/quote]\n\nCross multiplying, we clearly have\n\\begin{align*} a(a + c) & = b(b + c) \\\\\na(a + b) & = c(b + c). \\end{align*}\nDividing the two, we have\n\\begin{align*} \\frac {a + c}{a + b} & = 1 \\\\\n\\end{align*}\n[/quote]Shouldn't the RHS be $ b/c$ instead of $ 1$ (of course $ b = c$ but we do not know that yet at this point of the solution...) :huh:", "Solution_34": "[quote=\"kunny\"]Practice 3\n\nLet $ a,\\ b,\\ c$ be non-zero real numbers. Find the values of $ k,\\ l$ such that $ \\frac {a^2 \\plus{} b^2}{c^2} \\equal{} \\frac {b^2 \\plus{} c^2}{a^2} \\equal{} \\frac {c^2 \\plus{} a^2}{b^2} \\equal{} k$, \n$ \\frac {a^3 \\plus{} b^3}{c^3} \\equal{} \\frac {b^3 \\plus{} c^3}{a^3} \\equal{} \\frac {c^3 \\plus{} a^3}{b^3} \\equal{} l$.\n\n2002 Tokai University entrance exam/Humanities[/quote]\r\nWe have $ \\frac {a^2 \\plus{} b^2 \\plus{} b^2 \\plus{} c^2 \\plus{} c^2 \\plus{} a^2}{c^2 \\plus{} a^2 \\plus{} b^2} \\equal{} 2(a^2 \\plus{} b^2 \\plus{} c^2)/(a^2 \\plus{} b^2 \\plus{} c^2)$\r\n\r\nIf $ a^2 \\plus{} b^2 \\plus{} c^2\\neq 0$ the answer is $ 2$.\r\n\r\nIf $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 0$, then $ b^2 \\plus{} c^2 \\equal{} \\minus{} a^2$ etc\r\n\r\nSo we are left with $ \\minus{} c^2/c^2 \\minus{} a^2/a^2 \\minus{} b^2/b^2 \\equal{} \\minus{} 3$ (of course $ a,b,c\\neq0$)\r\n\r\nThe second problem can be solved similarly.\r\n\r\nEDIT: Yeah, forgot to mention that -3 won't work...", "Solution_35": "Yes, $ k\\equal{}2$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Does anybody know how did they do at Northwest chapter in CT?\r\nIt was supposed to be last Chapter competition in CT, right?", "Solution_1": "I'm from Litchfield County! But I wasn't in Mathcounts that year, sorry!" } { "Tag": [ "group theory", "abstract algebra", "vector", "search", "superior algebra", "superior algebra unsolved" ], "Problem": "Since I am taking a course in group & rings this semester I am revising the beginner's course in algebra that I took 1.5 years ago since I've forgotten all of it. I have a problem from a past exam which is probably very simple to you. I found it moderately difficult, but there is a detail in my solution that I feel uneasy about, even though it's correct.\r\n\r\n\"Prove that every element $ z$ in a finite field $ F$ can be expressed $ z \\equal{} a^2 \\plus{} b^2$ with $ a,b \\in F$.\"\r\n\r\nIf the characteristic of $ F$ is two then every element is a square so we are done.\r\n\r\nIf the characteristic is larger, then let the order of $ F$ be $ r$. $ r$ is odd. We consider how many squares there are in $ F$. Squaring $ 0$ gives $ 0$ and after that we choose a remaining element and it's (distinct) additive inverse; these give the same square. Then we keep choosing two elements at a time like that (each time we get a new square because $ x^2 \\equal{} c$ has at most two solutions), and so we realize there are $ \\frac{r\\plus{}1}{2}$ squares in $ F$. \r\n\r\nNow comes the part where I am \"uneasy\": What feels natural to me is to prove that this now implies that when going through $ a^2 \\plus{} b^2$ for all $ a, b \\in F$ we must get $ r$ distinct elements. However, the only argument I can find is this: Choose a $ z \\in F$. Varying $ a$ in $ F$, $ a^2$ assumes $ \\frac{r\\plus{}1}{2}$ different values, and varying $ b$, $ z \\minus{} b^2$ also assumes $ \\frac{r\\plus{}1}{2}$ different values. Since there cannot be $ r\\plus{}1$ distinct elements in $ F$ we get a solution to $ z \\equal{} a^2 \\plus{} b^2$.\r\n\r\nHow can I do the last bit more directly, i.e. directly showing that $ a^2\\plus{}b^2$ gives $ r$ distinct elements? Maybe a weird question that does not have too much to do with mathematics, but I feel the current solution is too tricky :P", "Solution_1": "This is the only way I know how to prove the theorem.", "Solution_2": "I believe that the solution you proposed is essentially the most direct and simple way to prove this (at least that I am aware of). You can shorten your argument slightly by looking at cosets, I think. But I'm not sure.", "Solution_3": "There is one cool proof:\r\n\r\nLet $ F^\\plus{}\\equal{}F$ and $ F^*\\equal{}F\\backslash \\{0\\}$ be the additive and multiplicative subgroups of $ F$, respectively.\r\n\r\nWe consider the set $ S$ of that nonzero elements that are a sum of two squares.\r\nIf $ x\\equal{}a^2\\plus{}b^2, y\\equal{}c^2\\plus{}d^2 \\in S$, then so is $ xy\\equal{}(a^2\\plus{}b^2)(c^2\\plus{}d^2)\\equal{}(ac\\minus{}bd)^2\\plus{}(ad\\plus{}bc)^2$. Thus we have a subset of a finite group $ F^*$ that is closed under multiplication, so $ S$ is a subgroup of $ F^*$.\r\nAs $ S$ contains all squares, it has index $ \\leq 2$ in $ F^*$.\r\nIf index $ \\equal{}1$, then we are done.\r\nOtherwise, index $ \\equal{}2$.\r\nThen every sum of squares is a square, thus $ S \\cup \\{0\\}$ is closed under addition. But now we got a subgroup of $ F^\\plus{}$ having $ \\frac{|F|\\minus{}1}{2}\\plus{}1 > \\frac{|F|}{2}$ elements, so this set must be the full $ F^\\plus{}$. Again, we are done.", "Solution_4": "[quote=\"ZetaX\"]\nAs $ S$ contains all squares, it has index $ \\leq 2$ in $ F^*$.\n[/quote]\r\nWhy is this? Sorry, I don't get it. Index is the number of cosets, right?", "Solution_5": "The squares $ T$ have index $ 2$ (or $ 1$, but then $ |F|$ is a power of two; but we never need to make this case work, every step works independent from this). We have $ T \\subset S \\subset F^*$, thus the index of $ S$ is at most that of $ T$, being $ 2$ (every coset $ xS$ of $ S$ contains a coset $ xT$ of $ T$).\r\nTo see that $ T$ has at most index $ 2$:\r\nThe squaring map $ f(x)\\equal{}x^2$ has kernel $ \\mathrm{ker}(f) \\equal{} \\{ \\pm 1\\}$. As $ |T| \\equal{} |\\mathrm{im}(f)| \\equal{} \\frac{|F^*|}{\\mathrm{ker}(f)} \\geq \\frac{|F^*|}2$, we get what we wanted to show.", "Solution_6": "uhm.. why should this $ S\\cup\\{0\\}$ be a subgroup of $ F^\\plus{}$?\r\n\r\nanyway, i think there's a linear way to get to the conclusion: $ S\\cup\\{0\\}$ is a $ \\mathbb{F}_p$-vector subspace of $ F$, since every element in $ \\mathbb{F}_p$ is a sum of two squares (pigeonhole).\r\nbut it can't be contained in any proper subspace, because of its cardinality, so we are done.\r\n\r\n(and this would help in case of two-squares-like formulas, if there's any for higher powers)", "Solution_7": "[quote=\"ma_go\"]uhm.. why should this $ S\\cup\\{0\\}$ be a subgroup of $ F^ \\plus{}$?[/quote]\r\nAs the sentence before that part says: the sum of two squares is a square.\r\n\r\nAnd my goal was to give a proof based purely on the groups.", "Solution_8": "sorry, you're right :) i didn't quite follow... and, by the way, nice solution ;)\r\n\r\ndo you know any higher-degree identity along the line of the two- and four-squares ones? i'm too lazy to search for it on wikipedia ;)" } { "Tag": [ "algebra", "difference of squares", "special factorizations" ], "Problem": "Ok, Ive been given this algebraic fraction to simplify\r\n\r\nx^2 - 9\r\n______\r\nx^2+x-6\r\n\r\nI think I need to do this but Im not sure what else to do.\r\n\r\nx^2 - 9\r\n______\r\n(x-2)(x+3)\r\n\r\nIf im right what can I do next, if not where have I gone wrong, please don't only post the answer. Thankyou", "Solution_1": "[hide=\"\"Difference of squares--so CLICK ME if you still want to know\"\"]$a^2-b^2=(a-b)(a+b)$[/hide]", "Solution_2": "Yes -- you've factored the bottom, now you need to factor the top.", "Solution_3": "Just wondering, isn't this should be on Getting Started Forum? ;)", "Solution_4": "Should it be in getting started? Probably. But as a rule of thumb I generally don't move topics down the ladder.", "Solution_5": "A *very good* rule of thumb. Occasionally easier than expected level is much better than what happens when posts move down regularly - lots of stuff that's too hard.", "Solution_6": "well don't bother about where it should be now. Once you factor the top using a difference of squares problem, cross out the common factors in the top and bottom. note 9=3^2." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "[b][u]The author of this posting is: Megus.[/u][/b]\r\n_____________________________________________________________\r\n\r\nGiven is function $f(x) = \\frac{1}{x+1}$ where $x > 0$. For all natural $n$ we know that:\r\n\r\n$g_n(x) = x+f(x)+f(f(x))+\\ldots+f(f((\\cdots f(x)))$ ($n$ times iterated)\r\n\r\nProve:\r\n(a) function $g_n$ is strictly increasing\r\n(b) $g_n(1)=\\frac{F_1}{F_2}+\\frac{F_2}{F_3}+\\cdots +\\frac{F_{n+1}}{F_{n+2}}$ where $F_0 = F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for $n \\geq 1$.\r\n_________________\r\nPrzemyslaw Chojecki", "Solution_1": "[b][u]The author of this posting is: Myth.[/u][/b]\r\n_____________________________________________________________\r\n\r\n(b) is a well known property of Fibonacci numbers:\r\n\r\n\\[{\\frac{F_{n+1}}{F_n} = 1+\\frac{1}{1+\\frac{\\ldots}{1+\\frac{1}{1}}}}.\\]\r\n\r\n_________________\r\nAmicus Plato, sed magis amica est veritas", "Solution_2": "[b][u]The author of this posting is: Myth.[/u][/b]\r\n_____________________________________________________________\r\n\r\nFor (a). \r\nIt is easy to show that for positive increasing function $h(x)$:\r\n\r\n1) ${h(x)+f(h(x)}$ is increasing\r\n2) $f(f(h(x)))$ is increasing\r\n\r\n_________________\r\nAmicus Plato, sed magis amica est veritas" } { "Tag": [], "Problem": "Four basketball players, whose heights in inches are 60, 62, 64 and 70, are standing on the court. Their teammate Natalie, who is 74 inches tall, walks out onto the court, and the mean height increases by $ m$ inches while the median height increases by $ n$ inches. What is the positive difference between $ m$ and $ n$ ?", "Solution_1": "Hi;\r\n\r\nm - n = 1", "Solution_2": "To elaborate:\n\nOriginal heights: ${60}, {62}, {64}, {70}$ <---Median is $\\frac{62+64}{2}$, or $63$\n\nMean is $\\frac{60+62+64+70}{4}$, or $64$.\n\nAfter the new height is added: Median=middle number--> now $64$\n\nMean: $\\frac{60+62+64+70+74}{5}={66}$\n\nMedian increases by 1.\nMean increases by 2.\n\n${2}-{1}={1}$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c>o prove that\r\n\r\n\r\n\\[ \\frac{1}{{5(a^{2}+b^{2})-ab}}+\\frac{1}{{5(a^{2}+c^{2})-ac}}+\\frac{1}{{5(b^{2}+c^{2})-bc}}\\ge \\frac{1}{{a^{2}+b^{2}+c^{2}}}\\]", "Solution_1": "[quote=\"Alisher\"]a,b,c>o prove that\n\\[ \\frac{1}{{5(a^{2}+b^{2})-ab}}+\\frac{1}{{5(a^{2}+c^{2})-ac}}+\\frac{1}{{5(b^{2}+c^{2})-bc}}\\ge \\frac{1}{{a^{2}+b^{2}+c^{2}}}\\]\n[/quote]\r\nEasy or hard?\r\nI think, it's transparent. SOS kills it easy enough. :wink:", "Solution_2": "[quote=\"arqady\"][quote=\"Alisher\"]a,b,c>o prove that\n\\[ \\frac{1}{{5(a^{2}+b^{2})-ab}}+\\frac{1}{{5(a^{2}+c^{2})-ac}}+\\frac{1}{{5(b^{2}+c^{2})-bc}}\\ge \\frac{1}{{a^{2}+b^{2}+c^{2}}}\\]\n[/quote]\nEasy or hard?\nI think, it's transparent. SOS kills it easy enough. :wink:[/quote]\r\n\r\n Hi! I use SOS it becomes that \r\n\\[ S_{c}= (a+b)^{2}(a^{2}+b^{2})-\\frac{{(a^{2}+b^{2}+c^{2})(a^{2}+c^{2})(b^{2}+c^{2})}}{{5(a^{2}+b^{2})-ab}}\\]\r\n i proof \r\n\\[ S_{c},S_{b}\\] are non-negative but i can't prove \r\n\\[ S_{a}+S_{b}\\ge 0 \\]", "Solution_3": "Utkirstudios, I have obtained $ S_{c}=\\frac{3(a+b)(5a+5b-c)}{(5a^{2}-ac+5c^{2})(5b^{2}-bc+5c^{2})}-\\frac{1}{5a^{2}-ab+5b^{2}}.$ :D", "Solution_4": "$ \\frac{1}{{5(a^{2}+b^{2})-ab}}+\\frac{1}{{5(a^{2}+c^{2})-ac}}+\\frac{1}{{5(b^{2}+c^{2})-bc}}\\geq\\frac{(1+1+1)^{2}}{{10(a^{2}+b^{2}+c^{2})-(ab+bc+ca)}}$\r\n\r\nI used here CBS:$ \\frac{x^{2}}{a}+\\frac{y^{2}}{b}+\\frac{z^{2}}{c}\\geq\\frac{(x+y+z)^{2}}{a+b+c}$\r\nfor positive numbers.\r\n\r\n\r\n\r\nUnfortunately we cannot use now:$ a^{2}+b^{2}+c^{2}\\geq ab+bc+ca$ in the initial inequality.\r\nIt's no good using this method :blush:", "Solution_5": "Yes, Cyberjunkie. It does not work.", "Solution_6": "[quote=\"Alisher\"]a,b,c>o prove that\n\\[ \\frac{1}{{5(a^{2}+b^{2})-ab}}+\\frac{1}{{5(a^{2}+c^{2})-ac}}+\\frac{1}{{5(b^{2}+c^{2})-bc}}\\ge \\frac{1}{{a^{2}+b^{2}+c^{2}}}\\]\n[/quote]\r\nWe have:\r\n$ \\frac{1}{5(a^{2}+b^{2})-ab}+\\frac{1}{5(b^{2}+c^{2})-bc}+\\frac{1}{5(c^{2}+a^{2})-ca}\\ge \\frac{1}{a^{2}+b^{2}+c^{2}}\\leftrightarrow \\sum (\\frac{1}{5(a^{2}+b^{2})-ab}-\\frac{1}{3(a^{2}+b^{2}+c^{2})}) \\ge 0$\r\n$ \\leftrightarrow \\sum \\frac{3(c^{2}-a^{2}+c^{2}-b^{2})-(a-b)^{2}}{5(a^{2}+b^{2})-ab}\\ge 0$\r\nOr:\r\n$ \\sum S_{c}(a-b)^{2}\\ge 0$\r\n$ S_{c}=3(a+b)(5(a+b)-c)(5(a^{2}+b^{2})-ab)-(5(c^{2}+a^{2})-ac)(5(c^{2}+b^{2})-cb)$\r\n$ S_{b}=3(c+a)(5(c+a)-b)(5(c^{2}+a^{2})-ca)-(5(b^{2}+a^{2})-ab)(5(b^{2}+c^{2})-bc)$\r\n$ S_{a}=3(b+c)(5(b+c)-a)(5(b^{2}+c^{2})-bc)-(5(a^{2}+b^{2})-ab)(5(a^{2}+c^{2})-ac)$\r\nWe may assume that $ a \\ge b \\ge c$\r\nBecause $ 5(c^{2}+a^{2})-ca \\ge (5(b^{2}+c^{2})-bc)$ and $ 3(c+a)(5(c+a)-b)-5(a^{2}+b^{2})+ab >0$ so $ S_{b}\\ge 0$\r\nWe have:\r\n$ S_{a}+S_{b}=3(c+a)(5(c+a)-b)(5(c^{2}+a^{2})-ca)+3(b+c)(5(b+c)-a)(5(b^{2}+c^{2})-bc)-(5(b^{2}+a^{2})-ab)(5(b^{2}+c^{2})-bc)-(5(a^{2}+b^{2})-ab)(5(a^{2}+c^{2})-ac)$\r\n$ =(5(a^{2}+c^{2})-ac)(10a^{2}+15c^{2}+30ac-5b^{2}-2ab-3bc)+(5(b^{2}+c^{2})-bc)(10b^{2}+15c^{2}+30bc-2ab-3ac-5a^{2})$\r\n$ \\ge (5(b^{2}+c^{2})-bc)(5a^{2}+30c^{2}+27ac+27bc-4ab+5b^{2}) \\ge 0$\r\nSimilarly $ S_{b}+S_{c}\\ge 0$\r\nIt follows that: $ S_{a}(b-c)^{2}+S_{b}(c-a)^{2}+S_{c}(a-b)^{2}\\ge (S_{a}+S_{b})(b-c)^{2}+(S_{b}+S_{c})(a-b)^{2}\\ge 0$", "Solution_7": "[quote=\"NguyenDungTN\"]\nWe have:\n$ \\frac{1}{5(a^{2}+b^{2})-ab}+\\frac{1}{5(b^{2}+c^{2})-bc}+\\frac{1}{5(c^{2}+a^{2})-ca}\\ge \\frac{1}{a^{2}+b^{2}+c^{2}}\\leftrightarrow \\sum (\\frac{1}{5(a^{2}+b^{2})-ab}-\\frac{1}{3(a^{2}+b^{2}+c^{2})}) \\ge 0$\n$ \\leftrightarrow \\sum \\frac{3(c^{2}-a^{2}+c^{2}-b^{2})+(a-b)^{2}}{5(a^{2}+b^{2})-ab}\\ge 0$\n[/quote]\r\nMaybe $ \\frac{3(c^{2}-a^{2}+c^{2}-b^{2})-(a-b)^{2}}{5(a^{2}+b^{2})-ab}\\ge 0$ $ ?$ :wink:", "Solution_8": "[quote=\"arqady\"][quote=\"NguyenDungTN\"]\nWe have:\n$ \\frac{1}{5(a^{2}+b^{2})-ab}+\\frac{1}{5(b^{2}+c^{2})-bc}+\\frac{1}{5(c^{2}+a^{2})-ca}\\ge \\frac{1}{a^{2}+b^{2}+c^{2}}\\leftrightarrow \\sum (\\frac{1}{5(a^{2}+b^{2})-ab}-\\frac{1}{3(a^{2}+b^{2}+c^{2})}) \\ge 0$\n$ \\leftrightarrow \\sum \\frac{3(c^{2}-a^{2}+c^{2}-b^{2})+(a-b)^{2}}{5(a^{2}+b^{2})-ab}\\ge 0$\n[/quote]\nMaybe $ \\frac{3(c^{2}-a^{2}+c^{2}-b^{2})-(a-b)^{2}}{5(a^{2}+b^{2})-ab}\\ge 0$ $ ?$ :wink:[/quote]\r\nI'm sorry, i have fixed my post!", "Solution_9": "[quote=\"Alisher\"]Let $ a,b,c>0,$ prove that\n\n$ \\frac {1}{5b^2\\minus{} bc\\plus{}5c^2} \\plus{} \\frac {1}{5c^2\\minus{}ca\\plus{}5a^2} \\plus{} \\frac {1}{5a^2\\minus{}ab\\plus{}5b^2}\\ge \\frac {1}{a^2 \\plus{} b^2\\plus{} c^2}.$[/quote]$ \\sum{\\frac {1}{5b^2\\minus{} bc\\plus{}5c^2}}\\minus{}\\frac {1}{a^2 \\plus{} b^2\\plus{} c^2}$\r\n\r\n$ \\equal{}\\sum{\\frac{(b\\minus{}c)^2\\left[(b\\plus{}c\\minus{}a)^2\\left(2a^2\\plus{}40b^2\\plus{}23bc\\plus{}40c^2\\right)\\plus{}31a^2bc\\plus{}10\\left(b^4\\plus{}c^4\\right)\\plus{}57bc\\left(b^2\\plus{}c^2\\right)\\right]}{4\\left(a^2\\plus{}b^2\\plus{}c^2\\right)\\left(5b^2\\minus{}bc\\plus{}5c^2\\right)\\left(5c^2\\minus{}ca\\plus{}5a^2\\right)\\left(5a^2\\minus{}ab\\plus{}5b^2\\right)}}$\r\n\r\n$ \\geq0.$\r\n\r\nBy the way, this inequality holds for [b]real[/b] $ a,b,c$ with $ 9bc\\plus{}a^2>0,9ca\\plus{}b^2>0,9ab\\plus{}c^2>0.$", "Solution_10": "Ji Chen, did you use computer to find this hard formula?\r\n\r\n\r\nThank you very much. :)", "Solution_11": "Yes, manlio. Another formula relate to Vasc's inequality\r\n\r\n$ \\frac {7}{4b^2 \\minus{} bc \\plus{} 4c^2} \\plus{} \\frac {7}{4c^2 \\minus{} ca \\plus{} 4a^2} \\plus{} \\frac {7}{4a^2 \\minus{} ab \\plus{} 4b^2}\\ge \\frac {9}{a^2 \\plus{} b^2 \\plus{} c^2}$\r\n\r\nsee : http://www.mathlinks.ro/Forum/viewtopic.php?t=185227" } { "Tag": [], "Problem": "A woman stands on a scale in a moving elevator. Her mass is 60.0kg, and the combined mass of the elevator and the scale is an additional 815kg. Starting from rest, the elevator accelerates upward. During the acceleration, there is tension of 9410N in the hoisting cable. What does the scale read during the acceleration?", "Solution_1": "[color=blue]Denote woman's mass $m$ and the mass of the elevator and the scale $M$.\n$T_0$ is the tension of the hoisting cable and $T$ is the scale read during the acceleration.\n$a$ is the acceleration of the system\nWe have these equations\n$\\large T_0-(m+M)g=(m+M)a$\n$T-mg=ma$\nFrom these two equations, you can get $T=mg+\\frac{m}{m+M}[T_0-(m+M)g]$[/color]" } { "Tag": [], "Problem": "p&q are prime\r\n[b]pql(5^p-2^p)(5^q-2^q)[/b]", "Solution_1": "Can you explain clearly,please?" } { "Tag": [ "geometry", "perimeter" ], "Problem": "To fulfil the demand of light speed :D , i ll post some problems that is too hard for me.\r\n\r\n1. 10 sqs. each with sides of 1in, arranged in a plain to form a figure so that no sq has all four edges on the perimeter of the figure, no 2 sq. overlap, and where two sq share a point they share edge....(thats the original problem i m serious)\r\n\r\n2.A model train is made by connecting cars with possible lengths of 2in and 5 in, what is the greatest unachieveable length?(easy!)", "Solution_1": "1.[hide]Don't know lol[/hide]\n\n2.[hide]using the formula ab-a-b, you get \n2*5-2-5\n10-7\n=3[/hide]\r\n3. Thanks for fulfilling my demands lol :D", "Solution_2": "i dun even understant wats N.O1 asking for \r\n\r\nAnswer is \r\n\r\n[hide]18[/hide]", "Solution_3": "Did I get number two right though?", "Solution_4": "lets c, yes u got it right", "Solution_5": "Yay!!! I'm so happy now :D", "Solution_6": "here is a tricky one\r\n\r\nThe purity of gold is measured in karats. 24 karat gold is puregold and k karat gold is k/24 by mass. one piece of 16-karat gold jewelery has a mass of 30 grams . Another piece of 10-karat gold has a mass of 40 gram. Find the numbe of grams in the positive difference between the amount of pure gold in each piece of jewelry. express answer as a mixed number\r\n\r\nanswer:\r\n\r\n[hide]3(1/3)[/hide]", "Solution_7": "[hide=\"question 1\"] This is a problem? They don't even ask a question! Where did you get this from?[/hide]\n\n[hide=\"question above\"]\n16/24*30=2/3*30=20\n10/24*40=5/12*40=50/3\n20-50/3=60/3-50/3=10/3=3 1/3\n>>3 1/3<<[/hide]\r\nChecks his answer...YES!\r\nBTW, kidwithshirt, the way you express the answer to some say multiply. better remove the parenthesis and add a space instead." } { "Tag": [ "geometry", "incenter", "trapezoid", "geometric transformation", "reflection", "symmetry", "circumcircle" ], "Problem": "Dear Mathlinkers, \r\nlet ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC, \r\nP, Q the meetpoints of the parallel to AD passing through X with BI, AC, \r\nS, R the meetpoints of the parallel to AD passing through Y with AI, BD. \r\nProve that the center of the circle passing through X, Y, Q, R lies on PS. \r\nSincerely \r\nJean-Louis", "Solution_1": "a is the square side, E intersection of its diagonals. x,y are inradii of triangles MAD, MBC. MD = m, MC = n. \r\n\r\n$ x \\equal{} \\frac {a \\plus{} m \\minus{} \\sqrt {a^2 \\plus{} m^2}}{2} \\Longrightarrow m \\equal{} \\frac {2x(a \\minus{} x)}{a \\minus{} 2x},\\ n \\equal{} \\frac {2y(a \\minus{} y)}{a \\minus{} 2y}$\r\n\r\n$ a \\equal{} m \\plus{} n \\equal{} \\frac {2x(a \\minus{} x)}{a \\minus{} 2x} \\plus{} \\frac {2y(a \\minus{} y)}{a \\minus{} 2y}.$\r\n\r\n$ \\triangle BPX \\sim \\triangle BYC \\Longrightarrow PX \\equal{} \\frac {YC \\cdot BX}{BC} \\equal{} \\frac {2y(a \\minus{} x)}{a}$\r\n$ \\triangle ASY \\sim \\triangle AXD \\Longrightarrow SY \\equal{} \\frac {XD \\cdot AY}{AD} \\equal{} \\frac {2x(a \\minus{} y)}{a}$\r\n\r\nK, L are midpoints of XQ, YR. G is midpoint of KL. PS || QR cuts KL at O. WLOG, m < n => x < y.\r\n\r\n$ KO \\equal{} \\frac {2 \\cdot KL \\cdot KP}{XP \\minus{} YS} \\equal{} \\frac {(a \\minus{} x \\minus{} y)}{y \\minus{} x} \\cdot \\left(\\frac {2y(a \\minus{} x)}{a} \\minus{} \\frac {a \\minus{} 2x}{2}\\right) \\equal{}$\r\n\r\n$ \\equal{} \\frac {(a \\minus{} x \\minus{} y)}{y \\minus{} x} \\cdot \\left(\\frac {a(x \\plus{} y) \\minus{} 2xy}{a} \\minus{} \\frac {a \\minus{} 2y}{2}\\right) \\equal{}$\r\n\r\n$ \\equal{} \\frac {1}{y \\minus{} x} \\cdot \\left(\\frac {x(a \\minus{} x)(a \\minus{} 2y) \\plus{} y(a \\minus{} y)(a \\minus{} 2x)}{a} \\minus{} \\frac {(a \\minus{} x \\minus{} y)(a \\minus{} 2y)}{2} \\right) \\equal{}$\r\n\r\n$ \\equal{} \\frac {1}{y \\minus{} x} \\cdot \\left(\\frac {(a \\minus{} 2x)(a \\minus{} 2y)}{2} \\minus{} \\frac {(a \\minus{} x \\minus{} y)(a \\minus{} 2y)}{2} \\right) \\equal{}$\r\n\r\n$ \\equal{} \\frac {1}{y \\minus{} x} \\cdot \\frac {(y \\minus{} x)(a \\minus{} 2y)}{2} \\equal{} \\frac {a \\minus{} 2y}{2} \\equal{} LY \\equal{} LR \\equal{} LE$\r\n\r\nE is also diagonal intersection of isosceles trapezoid XQRY with perpendicular diagonals => E is its anticenter. Reflection O of E in its centroid G is its circumcenter.\r\n________________________________\r\n\r\nEDIT:\r\n\r\nPerpendicular to $ CD$ through $ M$ cuts symmetry axis $ KL$ of trapezoid $ XYRQ$ at $ N.$ $ QR$ cuts $ AB$ at $ T.$ $ \\triangle MQR$ is isosceles with $ MQ \\equal{} MR$ $ \\Longleftrightarrow$ perpendicular bisector of $ QR$ goes through $ M$ $ \\Longleftrightarrow$ $ \\angle QTA \\equal{} \\angle OMN$ $ \\Longleftrightarrow$\r\n\r\n$ \\frac {y \\minus{} x}{a \\minus{} x \\minus{} y} \\equal{} \\frac {KO \\minus{} KN}{MN} \\equal{} \\frac {(a \\minus{} 2y) \\minus{} 2(m \\minus{} x)}{a} \\equal{} \\frac {(a \\minus{} 2y)(a \\minus{} 2x) \\minus{} 2ax}{a(a \\minus{} 2x)}$\r\n\r\n$ \\Longleftrightarrow$ (multiplying this through and rearranging)\r\n\r\n$ a(a \\minus{} 2x)(y \\minus{} x) \\plus{} 2ax(a \\minus{} x \\minus{} y) \\plus{} (x \\plus{} y)(a \\minus{} 2x)(a \\minus{} 2y) \\equal{} a(a \\minus{} 2x)(a \\minus{} 2y)\\ \\Longleftrightarrow$\r\n\r\n$ 2[a(x \\plus{} y) \\minus{} 2xy] (a \\minus{} x \\minus{} y) \\equal{} a(a \\minus{} 2x)(a \\minus{} 2y)\\ \\Longleftrightarrow$\r\n\r\n$ 2[x(a \\minus{} x)(a \\minus{} 2y) \\plus{} y(a \\minus{} y)(a \\minus{} 2x)] \\equal{} a(a \\minus{} 2x)(a \\minus{} 2y),$\r\n\r\nwhich is an obvious identity, in view of $ a \\equal{} m \\plus{} n \\equal{} ...$ (see above).", "Solution_2": "Dear Yetti and Mathlinkers,\r\nit seems that we can say more about this figure :\r\nfor example,\r\nthat the center of the circle passing through X, Y, Q, R is the orthopole of the triangle PXY wrt AB.\r\nSincerely\r\nJean-Louis", "Solution_3": "[quote=\"jayme\"]... the center of the circle passing through X, Y, Q, R is the orthopole of the triangle PXY wrt AB.[/quote]\nYou mean orthopole of the $ \\triangle MXY,$ I trust. This is easy to prove and on top of it, it gives synthetic proof of the $ \\triangle MQR$ being isosceles.\n\nLet the incircles $ (X), (Y)$ touch $ CD$ at $ U, V,$ let $ M', U', V' \\in AB$ be feet of perpendiculars from $ M, U, V$ to $ AB.$ Otherwise, keeping the same notation as above: $ E$ is diagonal intersection of the square $ ABCD$ and of the isosceles trapezoid $ XYRQ.$ $ K, L$ are midpoints of $ XQ, YR.$ The right $ \\triangle MUX, \\triangle UKE$ with corresponding sides $ UM \\perp KU, UX \\perp KE$ are similar by SAS, because $ \\frac {UM}{UX} \\equal{} \\frac {a}{a \\minus{} 2x} \\equal{} \\frac {KU}{KE}$ $ \\Longrightarrow$ $ MX \\perp UE$ $ \\Longrightarrow$ reflection of $ UE$ in the trapezoid centroid $ G$ (common midpoint of $ KL, OE$) goes through $ V'$ and through the trapezoid circumcenter $ O$ and $ (V'O \\parallel UE) \\perp MX.$ Similarly, $ (U'O \\parallel VE) \\perp MY.$ Then $ O$ is orthopole of $ AB$ WRT $ \\triangle MXY$ $ \\Longrightarrow$ perpendicular to $ XY$ from $ M'$ goes through $ O$ $ \\Longrightarrow$ $ M'O$ is perpendicular bisector of $ XY$ $ \\Longrightarrow$ $ \\triangle M'XY$ is isosceles with $ M'X \\equal{} M'Y.$ The same holds for $ \\triangle MQR,$ who is reflection of $ \\triangle M'XY$ in the trapezoid symmetry axis $ KL.$\n\n\n_________________________________________\n\nEDIT: Using the two problems about inradii (see [url]http://www.mathlinks.ro/viewtopic.php?t=282316[/url]), the above metric proof of the trapezoid $ XQRY$ circumcenter on $ PS$ can be re-written as follows:\n\nAs before, $ a$ is the square side, $ E$ intersection of its diagonals. $ x, y, r$ are inradii of triangles $ \\triangle MAD, \\triangle MBC, \\triangle MAB.$ $ K, L$ are midpoints of $ XQ, YR.$ $ G$ is midpoint of $ KL.$ $ PS \\parallel QR$ cuts $ KL$ at $ O.$\n\n$ \\left(1 \\minus{} \\frac {2x}{a}\\right)\\left(1 \\minus{} \\frac {2y}{a}\\right) \\equal{} 1 \\minus{} \\frac {2r}{a}\\ \\Longrightarrow\\ r \\equal{} x \\plus{} y \\minus{} \\frac {2xy}{a},$\n\n$ \\frac {1}{r} \\equal{} \\frac {1}{a \\minus{} 2x} \\plus{} \\frac {1}{a \\minus{} 2y} \\equal{} \\frac {2\\ KL}{XQ \\cdot YR}.$\n\n[quote=\"yetti\"]$ \\triangle BPX \\sim \\triangle BYC \\Longrightarrow PX \\equal{} \\frac {YC \\cdot BX}{BC} \\equal{} \\frac {2y(a \\minus{} x)}{a}$\n$ \\triangle ASY \\sim \\triangle AXD \\Longrightarrow SY \\equal{} \\frac {XD \\cdot AY}{AD} \\equal{} \\frac {2x(a \\minus{} y)}{a}$.[/quote]\n\n$ \\Longrightarrow PQ \\equal{} SR \\equal{} a \\minus{} \\left(2x \\plus{} 2y \\minus{} \\frac {2xy}{a}\\right) \\equal{} a \\minus{} (x \\plus{} y \\plus{} r)$\n\n$ KO \\equal{} \\frac {2\\ KL \\cdot KP}{XP \\minus{} YS} \\equal{} \\frac {KL \\cdot (XQ \\minus{} 2\\ PQ)}{XQ \\minus{} YR} \\equal{} \\frac {KL\\ (2r \\minus{} YR)}{XQ \\minus{} YR} \\equal{}$\n\n$ \\equal{} \\frac {YR\\ (XQ \\minus{} KL)}{XQ \\minus{} YR} \\equal{} \\frac {_1}{^2} YR \\equal{} LR \\equal{} LY \\equal{} LE$\n\n[quote=\"yetti\"]$ E$ is also diagonal intersection of isosceles trapezoid $ XQRY$ with perpendicular diagonals $ \\Longrightarrow$ $ E$ is its anticenter. Reflection $ O$ of $ E$ in its centroid $ G$ is its circumcenter.[/quote]", "Solution_4": "Dear Mathlinkers,\nfor a proof see: \nhttp://perso.orange.fr/jl.ayme vol.5, le r\u00e9sultat de Larrosa Canestro, p.\nSincerely\nJean-Louis" } { "Tag": [ "ratio", "Support" ], "Problem": "[i]Moderator's note: This was split from the discussion about who should win the American presidential race because it was silly and irrelevant. --JBL\n[/i]\r\n\r\nCJH, why must you always stir up politics.\r\nYou even have taiwan politics in your sig", "Solution_1": "That started with TokenAdult posting pro-Chen Sui Bian comments in some Taiwan article. (and not telling the dark side of it like what Lee Teng Hui did). In this Democratic World we like to express our own opinions. That is why we have nice discussions in these forums and carefully written facts of our views. \r\n\r\nPlus my skateboard guy looks cool. I'm going to have it for a while.", "Solution_2": "[quote=\"churchilljrhigh\"]That started with TokenAdult posting pro-Chen Sui Bian comments in some Taiwan article. (and not telling the dark side of it like what Lee Teng Hui did).[/quote]\r\n\r\nWell, that's not an accurate statement, as anyone who looks at the board can verify. Moreover, there isn't anyone I have mentioned whose name is Chen Sui Bian. \r\n\r\nTaking care to read carefully and avoid careless mistakes is helpful in math contests and helpful in discussions of public policy.", "Solution_3": "[quote=\"tokenadult\"][quote=\"churchilljrhigh\"]That started with TokenAdult posting pro-Chen Sui Bian comments in some Taiwan article. (and not telling the dark side of it like what Lee Teng Hui did).[/quote]\n\nWell, that's not an accurate statement, as anyone who looks at the board can verify. Moreover, there isn't anyone I have mentioned whose name is Chen Sui Bian. \n\nTaking care to read carefully and avoid careless mistakes is helpful in math contests and helpful in discussions of public policy.[/quote]\n\nWell I can verify that you had many pro pan-green statements and made a statement about neck to neck which was the Taiwan Election 2000 involving Pan Green candidate Chen Sui Bian, \n\n[quote=\"tokenadult\"]I'm not a citizen of that country, so I don't get a vote over there, but I would like to see the incumbent (the \"pan green\" candidate) get reelected. From what I hear about polling data (which tends to be skewed somewhat by who takes a particular poll), the race is neck-and-neck, with a statistically insignificant lead enjoyed by the challenger, the KMT-PFP coalition (\"pan blue\") candidate.[/quote]\r\n\r\nAs you can see Tokenadult publicly stated his views on which side he was on and gave alot of posts supporitng his cause (just visit the thread). Looking at the statement your statement is not very accurate either, thus undermining the credibility of your argument.\r\n\r\nFinally the whole Taiwan thread was started by you talking about the Great Wall of People in Taiwan. This was organized by Chen Sui Bian and the Pan-Green Coalition. Although you haven't mentioned his name you have mentioned his party countless times and have firmly established your views. I notice that you are an very educated person on the case of Taiwan and I am also too. It was good to have these political discussions.", "Solution_4": "That signature is so stupid churchilljrhigh.", "Solution_5": "but Chen was stupid enough so anything could happen (remember the time when he revealed a letter about China trying to compromise the political issue?)", "Solution_6": "joel: why didn't you just delete it as a whole. I don't foresee anything fruitful coming out of this discussion.", "Solution_7": "In general, I'd rather not delete anything with actual content. If no one says anything, it will eventually die a lonely death at the bottom of the forum.", "Solution_8": "Answerseeker95... Tell me the story!", "Solution_9": "[quote=\"theone853\"]CJH, why must you always stir up politics.\nYou even have taiwan politics in your sig[/quote]\r\n\r\n[i]exactly.[/i] ur location too. \r\n\r\ni agree with cats delete this i bet im not the only person that's offended", "Solution_10": "I can't believe everybody is making a big fuss about being Taiwanese or Chinese. I was at the National Geography Bee last year and I met a guy who was from Taiwan, but he accepted that he was chinese when he was talking to other people and never made refutes that he was Taiwanese. I mean all these \"Radical Taiwanis\" are forgetting their culture and ancestors because of some stupid political division that was started by two crazy dictators who killed many. Do we math students have to care about this that was started in the past and elsewhere and won't effect us because most of us on this forum are in U.S? [b](Fact 98% of Taiwanese are Han Chinese)[/b] Boy these radical pan green party need to get less liberal and have some conservatism! \r\n\r\nMore than 330,000 votes were declared invalid in the presidential election\r\n\r\nThere is a Vote-Rigging and a Assasination Attempt Questioning which gives basis to my signature. \r\n\r\nBTW The U.S. is the greatest country in the world so what they say (like if they wanted to invade Iraq they have the right to) is true so we have to listen them.\r\n\r\nPropositions from two presidents\r\n\r\nPresident Clinton pledged \"three nos\" - no to Taiwan independence, no to two Chinas, and no to Taiwan joining international organisations that need statehood for membership.\r\n\r\nThe administration of George W Bush has remained careful not to encourage Taiwan's President Chen Shui-bian, warning him against any actions which risked altering the status quo.", "Solution_11": "Just wondering: Why is the ratio of discussion between Taiwanese politics and U.S. presidental election like 5:1? :lol:", "Solution_12": "the chen sui bien guy is kinda illogical.\r\nchina has already pledged to keep taiwan under its control and will use military force if neccesary.\r\nthats why the usa doesnt support taiwan rifting from china.\r\nhas the election ended yet?", "Solution_13": "Should Singapore be under the control of the P.R.C. because of the ethnicity of its citizens? Should Tibet NOT be under the control of the P.R.C. because of the ethnicity of its citizens? How about Eastern Turkestan?", "Solution_14": "[quote=\"Scrambled\"]\nhas the election ended yet?[/quote]\r\n\r\nI think the electron ended, Chen won. But they think there are about erm. 300,000 or something number like that invalid votes and Chen only won by like 30,000. Some numbers like that. So they're doing a recount.", "Solution_15": "[quote=\"fubu\"][quote=\"Scrambled\"]\nhas the election ended yet?[/quote]\n\nI think the electron ended, Chen won. But they think there are about erm. 300,000 or something number like that invalid votes and Chen only won by like 30,000. Some numbers like that. So they're doing a recount.[/quote]\r\n\r\nYeah, that's what some Houston Chinese newspapers are saying. The difference was too little.", "Solution_16": "I think we should do what the Taiwanese wish to do. Perhaps have some sort of vote (sponsored by the UN or some independent third party) asking the Taiwanese themselves what their opinion is, namely, to truly grant themselves independent nation status or to be a part of China. Should they choose to be their own nation succession would occur and I guess it'd get a little more complicated. Personally I recognize Taiwan as independent from China, I think the olympiads do so as well, though I am unsure of this." } { "Tag": [ "function", "\\/closed" ], "Problem": "Critical Error\r\n\r\nmessage_die() was called multiple times.\r\n Error #2\r\nSQL requests not achieved\r\n\r\nLine : 566\r\nFile : /var/www/Forum/includes/functions_valentin.php\r\nSQL : SELECT c.class_id FROM classes c, enrollments e WHERE e.user_id = 1430 AND c.start_date <= DATE_ADD(NOW(),INTERVAL 7 DAY) AND c.end_date > DATE_ADD(NOW(),INTERVAL -50 DAY) AND e.dropped = 0\r\n \r\nError #1\r\nSQL requests not achieved\r\n\r\nLine : 661\r\nFile : /var/www/Forum/viewtopic.php\r\nSQL : SELECT fav_id FROM phpbb_favorites WHERE user_id = 1430 AND topic_id = 173438 ORDER BY topic_id ASC LIMIT 1\r\n \r\n \r\nPlease, contact the webmaster. Thank you.", "Solution_1": "My functions always causing trouble ..." } { "Tag": [ "trigonometry", "LaTeX", "articles" ], "Problem": "If tanx*tany= (a-b)/(a+b).\r\nshow that\r\n(a-bcos2x)*(a-bcos2y) is independent of x and y.", "Solution_1": "PLease post in $ LaTeX$.", "Solution_2": "[hide]\n$ \\tan x \\cdot \\tan y = \\frac {a - b}{a + b}$\n\n$ \\frac {\\sin x \\cdot \\sin y}{\\cos x \\cdot \\cos y} = \\frac {\\frac {1}{2}(\\cos (x - y) - \\cos (x + y))}{\\frac {1}{2}(\\cos (x - y) + \\cos (x + y))}$\n\nWe can assume:\n\n$ a = \\cos (x - y)$\n$ b = \\cos (x + y)$\n\n$ (a - b\\cos 2x)(a - b \\cos 2y) = a^2 - ab(\\cos 2x + \\cos 2y) + b^2(\\cos 2x \\cdot \\cos 2y)$\n\n$ \\cos 2x + \\cos 2y = 2 \\cos \\frac {2x + 2y}{2} \\cos \\frac {2x - 2y}{2} = 2ab$\n\n$ \\documentclass{article} \\usepackage{amsmath} \\begin{document} \\begin{align*} \\cos 2x \\cdot \\cos 2y & = \\frac {1}{2}(\\cos 2(x - y) + \\cos 2(x + y)) \\\\\n& = \\frac {1}{2}(\\cos^2 (x - y) - \\sin^2 (x - y) + \\cos^2(x + y) - \\sin^2 (x + y)) \\\\\n& = \\frac {1}{2}(2 \\cos^2 (x - y) - 1 + 2 \\cos^2(x + y) - 1) \\\\\n& = \\frac {1}{2}(2a^2 - 1 + 2b^2 - 1) \\\\\n& = a^2 + b^2 - 1 \\end{align*} \\end{document}$\n\nSo the beginning expression is independent of $ x$ and $ y$ but this is somewhat awkward to say. It would be more appropriate to say that it depends only on $ a$ and $ b$.\n\n\nBut there is something that should be added to this solution. We actually should have assumed that\n\n$ \\cos(x - y) = ka$\n$ \\cos(x + y) = kb$\n\nwhere $ k$ can be any real number. So:\n\n$ \\cos 2x + \\cos 2y = 2k^2ab$\n\n$ \\cos 2x \\cdot \\cos 2y = k^2(a^2 + b^2) - 1$\n\nSo when we put it in the expression we get that it equals to\n\n$ (a^2 - b^2)(1 - k^2b^2)$\n\nwhere $ k$ is \"stabilizing\" factor such that $ ka$ and $ kb$ are in interval $ [ - 1,1]$. That way we can input any real $ a$ and $ b$ to get valid solution. But then again, it doesn't depend on $ a$ and $ b$ only. Could someone help me with this?[/hide]", "Solution_3": "[quote=\"Flame\"][hide]\n$ \\tan x \\cdot \\tan y = \\frac {a - b}{a + b}$\n\n$ \\frac {\\sin x \\cdot \\sin y}{\\cos x \\cdot \\cos y} = \\frac {\\frac {1}{2}(\\cos (x - y) - \\cos (x + y))}{\\frac {1}{2}(\\cos (x - y) + \\cos (x + y))}$\n\nWe can assume:\n\n$ a = \\cos (x - y)$\n$ b = \\cos (x + y)$\n\n$ (a - b\\cos 2x)(a - b \\cos 2y) = a^2 - ab(\\cos 2x + \\cos 2y) + b^2(\\cos 2x \\cdot \\cos 2y)$\n\n$ \\cos 2x + \\cos 2y = 2 \\cos \\frac {2x + 2y}{2} \\cos \\frac {2x - 2y}{2} = 2ab$\n\n$ \\documentclass{article} \\usepackage{amsmath} \\begin{document} \\begin{align*} \\cos 2x \\cdot \\cos 2y & = \\frac {1}{2}(\\cos 2(x - y) + \\cos 2(x + y)) \\\\\n& = \\frac {1}{2}(\\cos^2 (x - y) - \\sin^2 (x - y) + \\cos^2(x + y) - \\sin^2 (x + y)) \\\\\n& = \\frac {1}{2}(2 \\cos^2 (x - y) - 1 + 2 \\cos^2(x + y) - 1) \\\\\n& = \\frac {1}{2}(2a^2 - 1 + 2b^2 - 1) \\\\\n& = a^2 + b^2 - 1 \\end{align*} \\end{document}$\n\nSo the beginning expression is independent of $ x$ and $ y$ but this is somewhat awkward to say. It would be more appropriate to say that it depends only on $ a$ and $ b$.\n\n\nBut there is something that should be added to this solution. We actually should have assumed that\n\n$ \\cos(x - y) = ka$\n$ \\cos(x + y) = kb$\n\nwhere $ k$ can be any real number. So:\n\n$ \\cos 2x + \\cos 2y = 2k^2ab$\n\n$ \\cos 2x \\cdot \\cos 2y = k^2(a^2 + b^2) - 1$\n\nSo when we put it in the expression we get that it equals to\n\n$ (a^2 - b^2)(1 - k^2b^2)$\n\nwhere $ k$ is \"stabilizing\" factor such that $ ka$ and $ kb$ are in interval $ [ - 1,1]$. That way we can input any real $ a$ and $ b$ to get valid solution. But then again, it doesn't depend on $ a$ and $ b$ only. Could someone help me with this?[/hide][/quote]\r\n\r\nHmm try this for root of (a-b)/(a+b).\r\nMaybe then we can work it out.", "Solution_4": "Where did you get this problem? Probably my approach is wrong.", "Solution_5": "[quote=\"Flame\"]But then again, it doesn't depend on $ a$ and $ b$ only. Could someone help me with this?[/quote]\r\n\r\nThat's fine; it's still independent of $ x,y$. (as the original expression is homogeneous in $ a,b$, the ending expression can be scaled to reach many different values)", "Solution_6": "Its in my 11th grade test player." } { "Tag": [ "function", "quadratics", "calculus", "derivative", "algebra", "domain", "analytic geometry" ], "Problem": "Find all functions $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ such that:\r\n$f(x-f(y))=f(x)+f(f(y))-axf(y)-bf(y)-c$ $\\forall x,y\\in \\mathbb{R}$\r\n(a,b,c are given real numbers)", "Solution_1": "may i ask what level of math this is from?", "Solution_2": "it is the generalisation of a number 6 problem in IMO, it is thus difficult. The solution uses elementary mathematics (maybe one can call this technical ,though). Looking at your signature I think you should try some easier ones :roll: \r\nPs: next time use PM-s please", "Solution_3": "[hide]\n[b]As Albanian Eagle correctly pointed out, it was not mentioned that the function is continuous and differentiable. Redone with finite differences.[/b]\n\n---------------\n\n[i]Not too sure about this, but... here's a swing... :maybe: [/i]\n\nLet $y$ be any real constant, and $f(y)$ therefore be any constant in the range of $f(x)$.\n\n$f(x+1-f(y))=f(x+1)+f(f(y))-axf(y)-af(y)-bf(y)-c$ and $f(x-f(y))=f(x)+f(f(y))-axf(y)-af(y)-bf(y)-c$ \u2192 $(f(x+1-f(y))-f(x-f(y)))=(f(x+1)-f(x))-af(y)$\n\nLet us use $f_{1}(x)=f(x+1)-f(x)$ represent the first row of finite differences:\n$f_{1}(x-f(y))=f_{1}(x)-af(y)$\u2192$f_{1}(x)-f_{1}(x-f(y))=af(y)$\n\nOver a distance of $f(y)$ in the $x$ direction, $f_{1}(x)$ changes by $af(y)$. Second level of finite differences $f_{2}(x)$ is proportional to change in $x$ by proportion constant $a$ \u2192 $f_{2}(x)=a$.\n\n$f_{1}(x)=ax+k+p(x)$ where $k$ is some unknown constant and $p(x)$ is some periodic function with period that makes some number of complete cycles over an $x$ distance of $f(y)$.\n$f(y)$ can be any number in the range of $f$.\nLet $m$ be a rational value in the range of $f$\nLet $n$ be a trascendental value on that range\n$p(x)$ must make some integer number of complete cycles across the $x$ interval $[x,x+m]$ and across the $x$ interval $[x,x+n]$.\nThis is impossible unless $p(x)$ is constant.\nTherefore, $p(x)$ is a constant, and $f_{1}(x)=ax+k+p(x)=ax+b$ for all x.\nIt follows that $f(x)$ is a quadratic with an $x^{2}$ coefficient of $\\frac{a}{2}$\n\n$f(x)=\\frac{a}{2}x^{2}+Bx+C$\nUse $x=0$ and $z=f(y)$ \u2192 $f(-z)=f(0)+f(z)-bz-c$\n\n$f(z)-f(-z)=bz+c-f(0)$ \u2192 $2Bz=bz+c-C$ \u2192 $(2B-b)z=c-C$\n$z$ is a variable and $c-C$ is not, so equality occurs only when both sides are 0.\n$2B-b=0$ \u2192 $B=\\frac{b}{2}$\n$c-C=0$ \u2192 $C=c$.\n\nThe final equation is now:\n$f(x)=\\frac{a}{2}x^{2}+\\frac{b}{2}x+c$\n[/hide]", "Solution_4": "no one mentioned conditions of differentiability or even continuity... so your solution is not acceptable but you did find the answer :P", "Solution_5": ":oops_sign: Good point.\r\n\r\nYou can arrive at the same conclusion using finite differences instead of derivatives, and finite differences don't require those conditions, right?\r\n\r\nHmm....\r\n[hide]$f(x+1-f(y))=f(x+1)+f(f(y))-axf(y)-af(y)-bf(y)-c$\n$f(x-f(y))=f(x)+f(f(y))-axf(y)-af(y)-bf(y)-c$\n\nSubtracting,\n$(f(x+1-f(y))-f(x-f(y)))=(f(x+1)-f(x))-af(y)$\n\nLet us use $f_{1}(x)=f(x+1)-f(x)$ represent the first row of finite differences:\n$f_{1}(x-f(y))=f_{1}(x)-af(y)$\n\n$f_{1}(x)-f_{1}(x-f(y))=af(y)$\n\nSo, the over a distance of $f(y)$ in the $x$ direction, $f_{1}(x)$ changes $af(y)$. Therefore, the second level of finite differences $f_{2}(x)$ is proportional to change in $x$ by a proportion constant of $a$, and so $f_{2}(x)=a$.\n\nIt follows that $f(x)$ is a quadratic with an $x^{2}$ coefficient of $\\frac{a}{2}$[/hide]\r\nDoes that work better?\r\n\r\nI'm trying to avoid using $x==f(x)$ to avoid limiting the domain to the range.", "Solution_6": "here is some things that are true : :lol: \r\n(let f(y0)=0 for some y0. then f(x) = f(x) + f(0) -c\r\nso f(0) = c if we know that f=0 for some y.) \r\n\r\nNow let x= f(y) then f(0) = 2f(f(y)) -af(y)^2 -bf(y) - c\r\n\r\nso f(f(y)) = a/2 f(y)^2 + b/2 f(y) + [c+f(0)]/2\r\n\r\nso f(z) = a/2 z^2 + b/2 z + k for all z in range of f.\r\nNow let x -> f(y)/2 - b/2a\r\nthen we see that f(-b/2a + z) = f(-b/2a -z ) for all z in range of f.\r\n\r\nAlso we can easily see that f(-z) = a/2 z^2 - b/2 z + f(0) for all z in range of f.", "Solution_7": "I am not sure that u actually have the right idea here, because using your finite difference ideas you won't be able to prove that f(x) is a quadratic for all x. rather u can only show it for all points in range of f(x).", "Solution_8": "Finite differences = slope of secant lines.\r\n\r\nWhat it does is show that $f_{1}(x)=ax+k+p(x)$ where $k$ is some unknown constant and $p(x)$ is some periodic function with period that makes some number of complete cycles over an $x$ distance of $f(y)$ . However, $f(y)$ can be any number in the range of $f$. Let $m$ be a rational value in the range of $f$ and $n$ be a trascendental value on that range. Thus, $p(x)$ must make some integer number of complete cycles across the $x$ interval $[x,x+m]$ and across the $x$ interval $[x,x+n]$. This is impossible unless $p(x)$ is constant. Therefore, $p(x)$ is a constant, and $f_{1}(x)=ax+k+p(x)=ax+b$ for all x.", "Solution_9": "Sorry but you can't conclude that p(x)=0 so try some other argument :(", "Solution_10": "I did this in about 5 minutes over a coffee, knowing it to be rubbish but I was astounded when this completely daft approach coincided with the vastly superior solution given about. \r\n\r\nlet x = f(y). \r\nThen f(x) = [f(0) + ax^2 + bx + c]/2 for all x in the range of f.\r\nAssume continuity from range to the rest of the domain. \r\nx = 0 lies in the domain hence c = f(0). \r\nHence f(x) = [ax^2 + bx] + c.\r\nDirect substition verifies the solution.\r\n\r\nThis makes me suspect the prescence of some principle to some extent justifying these gung-ho shortcuts. Does anyone know of any general results governing the connections between the behaviour of a function evaluated at all points in its range and the extension of the function to the superset consisting of the whole domain. What restrictions need to be placed on: If for all y = f(x), f(y) = g(y), then f(x) = g(x) for all x in the domain? It fails for constant polynomials but otherwise it is true for all polynomials defined on the whole of R. (A polynomial is completely determined by any segment of itself consisting of more than one point, and the range of a polynomial always contains more than one point unless the polynomial is constant). \r\n\r\nIs there any criterion for a functional equation to have a polynomial solution?", "Solution_11": "If I understood correctly the post above :\r\nYeah it's quite easier in this case to guess the answer to the problem. But as you see there is still no solution. \r\n\r\nAs for your questions, well they are a bit general :P" } { "Tag": [], "Problem": "$ 1)$.Find all primes $ p$ and $ q$ such that $ p \\plus{} q \\equal{} (p \\minus{} q)^3$ (easy)\r\n\r\n$ 2)$.Let $ a,b$ be dinstict positive integers such that $ ab(a \\plus{} b)$ is divisible by $ a^2 \\plus{} ab \\plus{} b^2$.Prove that $ |a \\minus{} b| > \\sqrt [3]{ab}$ (normal)", "Solution_1": "[hide]\u03a9\u03c1\u03b1\u03b9\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03b1\u03ba\u03b9\u03b1..\n\n\u03b3\u03b9\u03b1 \u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf, \u03c0\u03b1\u03c1\u03b5 $ gcd(a,b) \\equal{} d$ \n\u0391\u03c1\u03b1 $ a \\equal{} d*a_1, b \\equal{} d*a_2$ \u03bc\u03b5 $ gcd(a_1,a_2) \\equal{} 1$\n \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 \u03b5\u03ba\u03bc\u03b5\u03c4\u03b1\u03bb\u03bb\u03b5\u03c5\u03c3\u03bf\u03c5 \u03bf\u03c4\u03b9 $ a_1a_2 < a_1^2 \\plus{} a_1a_2 \\plus{} a_2^2 | d$ \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b6\u03b7\u03c4\u03bf\u03c5\u03bc\u03b5\u03bd\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1.\n\n\n\u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf, \u03b5\u03b9\u03bd\u03b1\u03b9 $ p \\equal{} \\frac {(p \\minus{} q)( (p \\minus{} q)^2 \\plus{} 1)}{2}$ \u03ba\u03b1\u03b9 $ q \\equal{} \\frac {(p \\minus{} q)( (p \\minus{} q)^2 \\minus{} 1)}{2}$\n\u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03b7 p,q \u03c0\u03c1\u03c9\u03c4\u03bf\u03b9 $ p \\minus{} q \\equal{} 2$[/hide]", "Solution_2": "\u0393\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03af\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7:\r\n\u0388\u03c3\u03c4\u03c9 $ p\\minus{}q\\equal{}x$\r\n\u03a4\u03cc\u03c4\u03b5 \r\n$ p\\plus{}q\\equal{}x^3$\r\n$ p\\minus{}q\\equal{}x$\r\n\u039c\u03b5 \u03c0\u03c1\u03cc\u03c3\u03b8\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03ac \u03bc\u03ad\u03bb\u03b7 \r\n$ x|2p$,but $ x \\not| p$,so $ x|2 \\Leftrightarrow x\\in \\{1,2\\}$\r\nA\u03bb\u03bb\u03ac $ p\\minus{}q\\equal{}1$ \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c4\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 $ p\\equal{}3$ \u03ba\u03b1\u03b9 $ q\\equal{}2$ which is impossible" } { "Tag": [ "ARML", "AMC", "AIME", "MATHCOUNTS", "Harvard", "college", "MIT" ], "Problem": "Is ARML comparable to AIME? It has some same concepts, the AIME is harder but ARML lots less time.\r\n\r\nSo it seems that doing AIME won't help much with ARML because the formats are drastically different. What about MC National Target?", "Solution_1": "I don't see how AIME won't help with ARML, because practicing AIMEs helps with your problem solving, and problem solving is used in ARML (obviously). It might not be super effective because of drastically different formats, but having good problem solving skills will help on any math competition (obviously again). I agree that the ARML is not comparable to AIME though.\r\n\r\nMC National Target is nowhere near as hard as ARML. I'd say that only 2-3 problems each year in the individual part of ARML are easy enough to be in mathcounts, and even the hardest mathcounts problems would probably only be about average difficulty in ARML (although I don't remember much of the mathcounts tests I practiced, so I could be wrong).", "Solution_2": "Seriously? A mere six threads down in this forum is a 24-post thread titled \"ARML and AIME,\" subtitled \"whats easier,\" and you decided to create a thread about how the difficulty of AIME and ARML compare?", "Solution_3": "JBL makes a good point: look around a little before posting duplicate topics. \r\n\r\nHowever, let me throw in my two cents. The very fact that people frequently compare AIME and ARML difficulty levels suggests that there are plenty of people who lean one way or the other. I've had this very discussion with fellow AIME and ARML problem-writers, and the lack of a clear consensus leads me to believe that the problem-by-problem difficulty levels are very comparable. \r\n\r\nMaybe the easy problems in ARML are easier than the easy problems in AIME; maybe #15 on the AIME is harder than I-10 at ARML, but those are outlier problems. By and large, both exams cover the same topics, so to any given student it'll come down to taste. If you thrive in one exam setting you'll find that exam easier than the other, simple as that.", "Solution_4": "I you handed the competitors all 10 individual problems at once and gave them 50 minutes to do the lot, allowing for highly variable per-problem time allocations, then the number of people getting 2009 #10 would have been more than 8.", "Solution_5": "Sorry about the repeated thread:blush:\r\nGot to be more careful later.\r\n\r\nWhat I meant was I need to get used to ARML format sort of, since the AIME has a drastically different format, I think I should just do ARML problems.\r\n\r\nHow does this compare to MC national target?", "Solution_6": "IMHO, ARML and AIME are very comparable in difficulty, but they have very different styles of problems. ARML problems tend to have a solution that's a little more straightforward (assuming you can find it), whereas AIME problems often require a bit more slogging in the solution.\r\n\r\nFor problems very similar in style to ARML, try Harvard-MIT or Stanford/Rice Math Tournament problems. I can't speak for HMMT, but when I was writing for SMT, we tried to write very ARML-like problems.", "Solution_7": "[quote=\"AwesomeToad\"]How does this compare to MC national target?[/quote]\r\n\r\nAlthough the format of targets is more similar to that of ARMLs, the questions are much too easy to be of any good practice. You even have a calculator on targets, so there is a lot more room for calculator-bashing in problems. \r\nI tend to think that ARML problems aren't so much about speed as they are about insight. If you get used to the kinds of ideas that ARML problems use often, you will most likely be able to answer a set of 2 questions in well under the 10 min time allotment. (Of course, I-10 last year was just mean :P )", "Solution_8": "[quote=\"cyberspace\"](Of course, I-10 last year was just mean :P )[/quote]\r\nAbout 1900 students (I can't be more exact than that since not all teams were full teams) participated in ARML. Here are how many got each individual question:\r\n\\[ \\begin{tabular}{rr}1. &1672\\\\\r\n2. &1256\\\\\r\n3. &1011\\\\\r\n4. & 470\\\\\r\n5. &749\\\\\r\n6. &263\\\\\r\n7. &541\\\\\r\n8. &284\\\\\r\n9. &1172\\\\\r\n10. & 8\\end{tabular}\\]\r\nBut, as I said above, more time would have made a considerable difference with that problem." } { "Tag": [ "geometry", "rectangle" ], "Problem": "In the grid below, place one digit in each of 24 non-shaded squares such that 14 numbers occurs below either vertically or horizontally in adjacent squares. What is the number A,BCD,EFG where each letter denotes the digit in the so lettered square?\r\n\r\n[img]7507[/img]\r\n\r\nThe answer key says that it is 2228657 but I got 8222657 :maybe:", "Solution_1": "66236 has to go into the 5-digit rectangle and use logic for everything else." } { "Tag": [ "floor function", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ X$ and $ Y$ be two disjoint finite set, $ \\mathcal{S} \\equal{} \\{A\\subseteq X\\cup Y| A\\cap X\\neq\\emptyset\\}$.\r\n\r\nLet $ \\mathcal{M}$ be a subset of $ \\mathcal{S}$ satisfying: \r\n\r\n1) there exists $ A\\in\\mathcal{M}$ such that $ A\\subseteq X$\r\n2) if $ A, B\\in\\mathcal{M}$, neither $ A\\subseteq B$ nor $ B\\subseteq A$. \r\n\r\nThen what's the maximal cardinality of $ \\mathcal{M}$ given $ |X| \\equal{} n$, $ |Y| \\equal{} m$.\r\n\r\nI know if $ |X| \\equal{} n$, then any subset $ \\mathcal{M}$ of $ 2^X$ satisfying 2) must satisfying $ |\\mathcal{M}|\\leq C(n,[n/2])$. But how to solve this new problem or estimate a tighter upper bound than just $ C(n\\plus{}m, [(n\\plus{}m)/2]$.\r\n\r\nAny hints are welcomed. Thanks!", "Solution_1": "Well, if $ n>m$, the bound stated by you is correct, since any set in the Sperner antichain, having cardinality $ \\lfloor (n\\plus{}m)/2 \\rfloor$ has nonempty intersection with $ X$, and there exists one included in $ X$.\r\n\r\nIf $ n \\leq m\\minus{}2$, the Sperner antichain will both contain sets with empty intersection with $ X$, and none included in $ X$. The issue of calculating the exact length of a maximal antichain might be arduous. There may be some considerations in an excellent book of Anderson, Combinatorics of sets, dedicated to antichains in sets environment.", "Solution_2": "[quote=\"mavropnevma\"]Well, if $ n > m$, the bound stated by you is correct, since any set in the Sperner antichain, having cardinality $ \\lfloor (n \\plus{} m)/2 \\rfloor$ has nonempty intersection with $ X$, and there exists one included in $ X$.\n\nIf $ n \\leq m \\minus{} 2$, the Sperner antichain will both contain sets with empty intersection with $ X$, and none included in $ X$. The issue of calculating the exact length of a maximal antichain might be arduous. There may be some considerations in an excellent book of Anderson, Combinatorics of sets, dedicated to antichains in sets environment.[/quote]\r\n\r\nThanks. Thanks for your recoomendation of that book. I think I have to spend a lot of time to work on it.\r\n\r\nCan we prove $ C(n\\plus{}m, [(n\\plus{}m)/2])\\minus{}C(m, [(n\\plus{}m)/2])\\plus{}1$ is a upper bound?\r\n(let $ C(m, [(n\\plus{}m)/2])\\equal{}0$ if $ m<[(n\\plus{}m)/2]$). I guess the first term is the maximal lenth of antichain in $ X\\cup Y$, the second term is just to remove the set with $ [(n\\plus{}m)/2]$ element but included in $ Y$?\r\n\r\nI can't prove it....." } { "Tag": [ "calculus", "integration", "limit", "calculus computations" ], "Problem": "For positive integers $n,$ let \r\n\r\n$S_{n}=\\frac{1}{\\sqrt{1}}+\\frac{1}{\\sqrt{2}}+\\cdots\\cdots+\\frac{1}{\\sqrt{n}},\\ T_{n}=\\frac{1}{\\sqrt{1+\\frac{1}{2}}}+\\frac{1}{\\sqrt{2+\\frac{1}{2}}}+\\cdots\\cdots+\\frac{1}{\\sqrt{n+\\frac{1}{2}}}.$\r\n\r\nFind $\\lim_{n\\to\\infty}\\frac{T_{n}}{S_{n}}.$", "Solution_1": "Use Cesaro Stolts lemma.\r\nWe have that \r\n$\\lim\\frac{T_{n}}{S_{n}}=\\lim\\frac{T_{n+1}-T_{n}}{S_{n+1}-S_{n}}=\\lim\\frac{\\sqrt{n+1}}{\\sqrt{n+3/2}}=1$.", "Solution_2": "Your answer is correct, but I would like to see a basic solution.", "Solution_3": "I do not know what you mean by a more basic solution, however another approach of this problem, without making use of Cesaro-Stolz lemma, would be this:\r\n\r\nLet $x_{n}=1+\\frac{1}{\\sqrt{2}}+\\cdots+\\frac{1}{\\sqrt{n}}-2\\sqrt{n}$, \r\n\r\nand \r\n\r\n$y_{n}=\\frac{1}{\\sqrt{1+\\frac{1}{2}}}+\\cdots+\\frac{1}{\\sqrt{n+\\frac{1}{2}}}-2\\sqrt{n+\\frac{1}{2}}$.\r\nThen it is not hard to prove that these two sequences are convergent, say $x_{n}\\rightarrow a$ and $y_{n}\\rightarrow b$.\r\nThen we see that $S_{n}=x_{n}+2\\sqrt{n}$ and $T_{n}=y_{n}+2\\sqrt{n+\\frac{1}{2}}$.\r\n\r\nThus $\\frac{T_{n}}{S_{n}}=\\frac{y_{n}+2\\sqrt{n+\\frac{1}{2}}}{x_{n}+2\\sqrt{n}}=\\frac{\\sqrt{n+\\frac{1}{2}}}{\\sqrt{n}}\\frac{\\frac{y_{n}}{2\\sqrt{n+\\frac{1}{2}}}+1}{\\frac{x_{n}}{2\\sqrt{n}}+1}\\rightarrow 1$", "Solution_4": "[b]Is it a \"basic solution\" ?[/b] \r\n\r\n[b]Lemma.[/b] $|\\alpha |\\le 1$ $\\implies$ $\\lim_{n\\to\\infty}\\frac{1}{n}\\sum_{k=1}^{n}f(\\frac{k+\\alpha }{n})=\\int_{0}^{1}f(x)\\ dx\\ .$\r\n\r\n$\\{\\begin{array}{c}f(x)=\\frac{1}{\\sqrt x}\\\\\\\\ \\{t,\\alpha ,\\beta \\}\\subset [-1,1]\\\\\\\\ S_{n}(t)=\\sum_{k=1}^{n}f(k+t)\\\\\\\\ \\end{array}$ $\\implies$ $\\{\\begin{array}{c}\\int_{0}^{1}f(x)\\ dx=2\\\\\\\\I_{n}\\equiv\\frac{1}{n}\\cdot\\sum^{n}_{k=1}f(\\frac{k+\\alpha}{n})\\rightarrow 2\\\\\\\\ J_{n}\\equiv\\frac{1}{n}\\cdot\\sum^{n}_{k=1}f(\\frac{k+\\beta}{n})\\rightarrow 2\\end{array}$ $\\implies$ $\\frac{S_{n}(\\alpha )}{S_{n}(\\beta )}=\\frac{I_{n}}{J_{n}}\\rightarrow 1\\ .$" } { "Tag": [ "Mafia", "USAMTS" ], "Problem": "Is there gonna be a funeral or something?\r\n\r\nI have to feel sorry for him, even though I don't know him very well - the poor guy died before he could even finish a game of Mafia! (I can only hope his replacement wins it for him.)", "Solution_1": "He like actually died in real life? :o", "Solution_2": "Wait is this a joke or something? :o", "Solution_3": "I am still trying to grasp what this means. :( \r\nI mean, what kind of a perverted joke is this huh? :mad: \r\nI doubt I'll be able to sleep tonight. The mental peace seems to be in no mood to return till I get a reply to the PM I just sent to him. :(", "Solution_4": "Who's perfect628?\r\n\r\nAnd why is he dead? :maybe:", "Solution_5": "http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=21169\r\n\r\nwait is this a sick joke or did an AoPSer really die? how d'you know?", "Solution_6": "How would we know if he was dead? He had no way to tell us he was dead. :|", "Solution_7": "Last post by perfect628:\r\n\r\nYesterday, at 1:37 pm\r\n\r\n\r\nIf true, how did you find out so fast?", "Solution_8": "The topic for mafia game 23 says Mafia Game 23-Day 5 Perfect628 died\r\n\r\n :roll:", "Solution_9": "[quote=\"mustafa\"]The topic for mafia game 23 says Mafia Game 23-Day 5 Perfect628 died\n\n :roll:[/quote]\r\n\r\nyeah i think zeb misenterpreted that :D", "Solution_10": "Yeeeah...it's just a Mafia game. :P", "Solution_11": "I [i]hope[/i] Zeb misinterpreted it.\r\nAnd if that is the case, perfect628 won't be happy to find this topic out.\r\nBut why hasn't he posted yet after his last post about six hours before Zeb's post?", "Solution_12": "perfect628 was in the USAMTS math jam last night. Not dead, just was killed in mafia.\r\n\r\n...", "Solution_13": "That's reassuring.\r\nWhat sort of an irresponsibel idiot are you zeb huh? :mad: \r\nDon't you see how serious this can get?", "Solution_14": "Yeah, making jokes about this is no laughing matter. I'll give zeb another chance to explain himself before this topic is locked...", "Solution_15": "[quote=\"zeb\"]Is there gonna be a funeral or something?\n\nI have to feel sorry for him, even though I don't know him very well - the poor guy died before he could even finish a game of Mafia! (I can only hope his replacement wins it for him.)[/quote]\r\n\r\nI am not dead. I died in mafia #23. And what are you talking about, I was not in the USAMTS math jam.", "Solution_16": "[quote=\"perfect628\"]I am not dead. I died in mafia #23. And what are you talking about, I was not in the USAMTS math jam.[/quote]\r\n\r\nproof enough :D \r\n\r\nzeb, explain yourself please", "Solution_17": "[quote=\"besttate\"]perfect628 was in the USAMTS math jam last night. [/quote]\r\nnope, that was me. I am not perfect628.", "Solution_18": "[quote=\"perfectnumber628\"][quote=\"besttate\"]perfect628 was in the USAMTS math jam last night. [/quote]\nnope, that was me. I am not perfect628.[/quote]\r\n :rotfl:", "Solution_19": "Wow, this is getting confusing... :rotfl:", "Solution_20": "Roflmao. I thought someone actually made a thread to talk about a person's death. That would be sick.", "Solution_21": "[quote=\"xxazurewrathxx\"]Roflmao. I thought someone actually made a thread to talk about a person's death. That would be sick.[/quote]\r\n\r\nThere was a thread about Anna Nicole Smith's death :roll:", "Solution_22": "I meant about how we have perfect628 and perfectnumber628. I had a feeling that zeb misinterpreted the topic title from the Mafia game, saying that \"perfect628 died\". He's just out of that game; he's not really dead.", "Solution_23": "Or this whole thread was just a joke, and people didn't want to point it out just in case perfect628 really had died.\r\n\r\nBut now that we know he didn't, I can say this:\r\n\r\nIf you die in Mafia, you die in real life! *Sinister grin*", "Solution_24": "imo, it's idiotic that this topic has gotten 2 pages of posts...", "Solution_25": "Okay cheers everyone! This topic has got a happy ending, and that's all that matters. But zeb should be more careful in future. :D", "Solution_26": "And if I was dead, I would hope that the topic wouldn't go in GFF...", "Solution_27": "CAN I HAVE HIS ACCOUNT", "Solution_28": "/|\\\r\n |\r\n |\r\n\r\nlol" } { "Tag": [], "Problem": "$ \\frac {9x^2}{({1 \\minus{} \\sqrt {1 \\plus{} 3x}})^2}$ is less than $ 3x \\plus{} 5$\r\n\r\nfind the range of real values of $ x$", "Solution_1": "[hide]$ \\minus{}\\frac{1}{3}\\leq x <\\frac{5}{12}, x\\neq0$[/hide]", "Solution_2": "[quote=\"123456789\"][hide]$ \\minus{} \\frac {1}{3}\\leq x < \\frac {5}{12}, x\\neq0$[/hide][/quote]\r\n\r\ncan you show the steps involved?", "Solution_3": "[hide]$ \\frac{9x^2}{(1\\minus{}\\sqrt{3x\\plus{}1})^2}\\equal{}(1\\plus{}\\sqrt{3x\\plus{}1})^2$, and $ x\\neq 0$\n$ (1\\plus{}\\sqrt{3x\\plus{}1})^2<3x\\plus{}5$\n$ 3x\\plus{}2\\plus{}2\\sqrt{3x\\plus{}1}<3x\\plus{}5$\n$ 2\\sqrt{3x\\plus{}1}<3$\n$ 4(3x\\plus{}1)<9$, and $ x\\geq\\minus{}\\frac{1}{3}$\n$ x<\\frac{5}{12}$\nSo the range is $ \\minus{}\\frac{1}{3}\\leq x < \\frac{5}{12}$, $ x\\neq 0$[/hide]" } { "Tag": [], "Problem": "Find all primes $ p$ such that $ \\frac{p\\plus{}1}{2}$ and $ \\frac{p^2\\plus{}1}{2}$ are both squares.", "Solution_1": "[hide]Let $ \\frac {p \\plus{} 1}{2} \\equal{} m^2, \\frac {p^2 \\plus{} 1}{2} \\equal{} n^2.$ The second equation implies that $ p^2 \\equal{} 2n^2 \\minus{} 1 > n^2$, so $ p > n$, and similarly $ p > m$. Subtract the two equations to obtain $ p(p \\minus{} 1) \\equal{} 2(n^2 \\minus{} m^2) \\equal{} 2(n \\minus{} m)(n \\plus{} m).$ $ p$ is an odd prime, so either $ p|(n \\minus{} m)$ or $ p|(n \\plus{} m).$ If $ p|(n \\minus{} m)$, since $ p < n$ we get that $ m \\equal{} n$, but then $ p \\equal{} 0$ or $ 1$, a contradiction, so $ p|(n \\plus{} m).$ In addition, since $ n, m < p$ then $ n \\plus{} m < 2p$ and thus $ n \\plus{} m \\equal{} p$. Substituting this into the equation yields $ m \\equal{} \\frac {p \\plus{} 1}{4}$, thus $ \\frac {p \\plus{} 1}{2} \\equal{} (\\frac {p \\plus{} 1}{4})^2$ with solution $ p \\equal{} 7.$[/hide]" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Let's work in $\\mathbb{R}^{2006}$. Find the biggest positive integer $n$ for which there exist linear subspaces $V_1,V_2,\\,\\cdots,V_n$ of $\\mathbb{R}^{2006}$ such that $\\dim V_k = k$ for all $k$ and such that $V_{i + j} = V_i + V_j$ for all $i, j$ satisfying $i + j \\leq n$.", "Solution_1": "Sorry, I was confused, this is rubbish :D", "Solution_2": "so strange... is it 1?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that \r\n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} > 2 \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}$\r\nwhere n is an integer and not less than 2 and a,b,c >0", "Solution_1": "[quote=\"NkMAsTeR\"]Prove that \n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} > 2 \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}$\nwhere n is an integer and not less than 2 and a,b,c >0[/quote]\r\nI think i prove that when a,b,c is sides of a triangle the ineq holds:\r\n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}}\\geq \\sqrt[2]{\\frac{a^2}{a(b+c)}} +\\sqrt[2]{\\frac{b^2}{b(c+a)}} +\\sqrt[2]{\\frac{c^2}{c(a+b)}}\\geq \\sum \\sqrt \\frac{a^2}{(\\frac{a+b+c}{2})^2}=2$", "Solution_2": "[quote=\"Hawk Tiger\"][quote=\"NkMAsTeR\"]Prove that \n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} > 2 \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}$\nwhere n is an integer and not less than 2 and a,b,c >0[/quote]\nI think i prove that when a,b,c is sides of a triangle the ineq holds:\n[b]$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}}=\\sqrt[2]{\\frac{a^2}{a(b+c)}} +\\sqrt[2]{\\frac{b^2}{b(c+a)}} +\\sqrt[2]{\\frac{c^2}{c(a+b)}}[/b: 6bf2465f5f]\\geq \\sum \\sqrt \\frac{a^2}{(\\frac{a+b+c}{2})^2}=2$[/quote]\r\nCan you explain this point ? But I think it true where a,b,c be positive reals.", "Solution_3": "sorry the first $=$should be $\\geq$\r\nand i have only found a ugly solution when $a\\geq b+c$\r\nbut i don't want to post it", "Solution_4": "[quote=\"NkMAsTeR\"]Prove that \n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} > 2 \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}$\nwhere n is an integer and not less than 2 and a,b,c >0[/quote]\r\n\r\n*This is inequality true:\r\n $\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}>2$", "Solution_5": "[quote=\"Frey\"][quote=\"NkMAsTeR\"]Prove that \n$\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} > 2 \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}$\nwhere n is an integer and not less than 2 and a,b,c >0[/quote]\n\n*This is inequality true:\n $\\sqrt[n]{\\frac{a}{b+c}} +\\sqrt[n]{\\frac{b}{c+a}} +\\sqrt[n]{\\frac{c}{a+b}} \\geq \\frac{n \\sqrt[n]{n-1}}{n-1}>2$[/quote]\r\nI'm sorry but I think my Inequalitity is true! You can try with some values of n\r\nEx:\r\n$A= \\frac{n \\sqrt[n]{n-1}}{n-1} If n=2 then A=2 If n=3 then A \\approx 1.89 ..$", "Solution_6": "We have $\\frac{a}{b+c} \\geq \\frac{4a^2}{(a+b+c)^2}$ in fact we have:\r\n$a(a+b+c)^2 \\geq 4a^2(b+c)$ and so $a(a-b-c)^2 \\geq 0$ that it's true. So we have:\r\n\r\n$\\sum \\sqrt[n] {\\frac{a}{b+c}} \\geq \\sum \\sqrt{\\frac{a}{b+c}} > \\sum \\frac{2a}{a+b+c} =2$\r\n\r\nFor the second part we have:\r\n\r\n$\\left ( 1+ \\frac{n-2}n \\right) ^n \\geq 1+ \\frac{n-2}n * n = n-1$ for bernoulli\r\n\r\nTaking $n$-th root we have:\r\n\r\n$\\frac{2n-2}n \\geq \\sqrt[n]{n-1}$ and thus : $2 \\geq \\frac{n \\sqrt [n] {n-1} } {n-1}$", "Solution_7": "[quote=\"Simo_the_Wolf\"]\n$\\sum \\sqrt[n] {\\frac{a}{b+c}} \\geq \\sum \\sqrt{\\frac{a}{b+c}}$[/quote]\r\n\r\nWhy is this true ? :?:", "Solution_8": "The above holds if $b+c>a$ and I think for $a>b+c$ it is trivial ;)", "Solution_9": "[quote=\"silouan\"]The above holds if $b+c>a$ and I think for $a>b+c$ it is trivial ;)[/quote]\r\n\r\nif $a\\geq b\\geq c$ and $b+c\\geq a$ then the inequality is true .\r\n\r\nBut if $a>b+c$ , when $a\\rightarrow\\infty$ and fix $b+c=1$ then it becomes\r\n\r\n$\\sqrt[n]{a}>\\sqrt{a}$\r\n\r\nwhich is not true :?:", "Solution_10": "I'm sorry but why nobody prove it?", "Solution_11": "Lemma:\r\n$\\frac{a^{1/n}}{\\left(b+c\\right)^{1/n}}\\ge \\frac{2a^{2/n}}{a^{2/n}+b^{2/n}+c^{2/n}}$(n \\ge 2)\r\nProve:\r\njust notice:$b^{2/n}+c^{2/n}\\ge (b+c)^{2/n}$" } { "Tag": [ "probability", "expected value", "combinatorics unsolved", "combinatorics" ], "Problem": "Suppose a computer can generate a whole number from 1 to k in one instance randomly.\r\n\r\nWhat is the expected number of instances, n required for each given k such that a complete set of 1 to k has been produced? Ie, the first time where you get all of the numbers from 1 to k.\r\n\r\nHow do I calculate Prob(set first completed in exactly nth instance) for all n?\r\n\r\nThanks all!", "Solution_1": "the expected number is $\\sum_{i=1}^k \\frac{k}{i}$\r\n\r\nonly have the proof in german. dont know the specific termology in english.\r\ncould somebody else prove it plz?", "Solution_2": "just keep computing the conditional expectations of the waiting time for arrival of the next new symbol given that we have already gotten $i$ of the symbols from $\\{1,2,\\ldots,k\\}$ and finally add them all up." } { "Tag": [ "AMC", "AIME", "\\/closed" ], "Problem": "Could someone help me by telling me where I can find the 2006 AIME II solutions?\r\nThanks. :D", "Solution_1": "Click the number next to a particular problem. That will bring you to a thread. Then scroll down to find the solution to the problem (someone should have posted it)." } { "Tag": [ "Princeton", "college", "IMC", "college contests" ], "Problem": "we may post here some information about who is going at the imc as it is done elsewhere in the forum it may be also important for some to know how to get to macedonia (i've heard for example that you cant go there from bulgaria by train)\r\ngenerally feel free to submit any information", "Solution_1": "I am going to drive a Romanian Team to the IMC this year. Probably Horia is going to join me with his car, because we are about 8 people wanting to come (amfulger, claude, kappa, phoria, ana caraiani, tanase raluca, a friend of mine and me). \r\n\r\nAre you going to come qw?", "Solution_2": "btw, the train information is acurate. Also you can't travel by air from Romania to Skopje direclty. You have to take the plane to Milano (or something) and then back to Skopje ... which is kind of weird ...", "Solution_3": "well if all goes ok i'll come our team will probably consist of 4 people but we are struggling to find a means on transport.it's up to the university's authorities but i have completely no idea what will they do. it may finish with hiking :)", "Solution_4": "our univ. decide not to pay for our being there, so it's totaly a private thingy :) (except for Ana Caraiani, for which Princeton is paying) we did not find any other (safe) way to get than, besided driving a car, so we'll just do that. :)", "Solution_5": "oh, and i suppose everybody knows that we have high competition fee this year as far as i remember it was 225 euro for the \"poorer countries\"", "Solution_6": "the fee we (the romanians) are requied to meet is smaller than that ... so I guess we are the poor-est country :)", "Solution_7": "Hi!\r\nI'm going to IMC too. :) \r\n\r\n[quote=\"Valentin Vornicu\"]the fee we (the romanians) are requied to meet is smaller than that ... so I guess we are the poor-est country :)[/quote]\r\nI don't think so because we also got a discount :)", "Solution_8": "cool, so we can make this topic like the one for IMO \"who's coming to\" :)", "Solution_9": "maja do you know how to reach macedonia from your location are there any buses or trains? i would be grateful for any information", "Solution_10": "[quote=\"qw\"]maja do you know how to reach macedonia from your location are there any buses or trains? i would be grateful for any information[/quote]\r\n\r\nI think there are regular lines (both bus and train lines) from Serbia to Macedonia. In next few days, Im going to look for more information about this and then Ill tell you what I found out.\r\n\r\n...", "Solution_11": "OK, here is what I found out.\r\nFrom Novi Sad to Skopje, one can go only by bus which departs at 6:10 a.m. and 10:20 a.m. every day. The trip takes about 12 hours (at least I was told so). The ticket costs about 15 EURO and it should be booked in advanced (3-5 days). \r\nIf you would go from Belgrade, then there is an additional option to go by train (which I dont prefer because trains are very dirty here). Anyway, it departs at 7:00, 13:50, 17:45, 23:20 every day and the ticket costs about 25 EURO, which is kind of strange because train transportation is usually cheaper here. \r\nAlso I believe that the daily number of buses which goes from Belgrade to Skopje is probably greater that two, but I havent checked that.\r\n\r\nMaja" } { "Tag": [ "trigonometry", "quadratics", "algebra", "geometry proposed", "geometry" ], "Problem": "Fin the angles of $ \\triangle ABC$ if $ a^2\\equal{}R^2\\plus{}b^2\\plus{}c^2$(usual notations)", "Solution_1": "$ a^2 \\equal{} R^2 \\plus{} b^2 \\plus{} c^2 > b^2 \\plus{} c^2 \\Rightarrow A > 90^0$.\r\nand:\r\n$ a^2 \\equal{} R^2 \\plus{} b^2 \\plus{} c^2\\Rightarrow b^2 \\plus{} c^2 \\equal{} a^2 \\minus{} R^2 \\Rightarrow \\\\\r\n\\Rightarrow 0 < 9.|AG|^2 \\equal{} 2.(b^2 \\plus{} c^2) \\minus{} a^2 \\equal{} 2(a^2 \\minus{} R^2) \\minus{} a^2 \\equal{} a^2 \\minus{} 2.R^2 \\Rightarrow \\\\\r\n\\Rightarrow \\sqrt {2}.R< a \\Rightarrow A < 145^0.$\r\n____________________________________________________________________\r\n\r\nIf b=c and $ a \\equal{} 120^0 \\Rightarrow b \\equal{} c \\equal{} R$ and $ a \\equal{} R.\\sqrt {3}\\Rightarrow b^2 \\plus{} c^2 \\plus{} R^2 \\equal{} 3.R^2 \\equal{} a^2.$", "Solution_2": "I don't understand your solution.:(...the idea to compose this pb was from this relation:$ \\cos AsinBsinC\\equal{}\\minus{}\\frac{1}{8}$, which is equivalwnt with the ip.\r\nBut $ sinBsinC\\equal{}\\frac{1}{2}\\left(\\cos(B\\minus{}C) \\plus{}cosA \\right)$ and we will obtain a quadratic equation in $ cosA$, with $ \\Delta\\equal{}16(\\cos^2(B\\minus{}C)\\minus{}1)\\leq 0$, so $ B\\equal{}C$.The only solution is $ A\\equal{}120,B\\equal{}C\\equal{}30$" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "all Lipschitz functions are absolutely continuous? Do you look at the length of the interval? And choose $ \\delta$ based on that?", "Solution_1": "State the definition of absolute continuity. Write the Lipschitz condition right below it. Stare at them for a minute. Then tell what exactly your difficulty is.", "Solution_2": "Suppose $ f: X \\to Y$ is a function. Let $ k$ be a positive integer. If $ d_{Y}(f(a), f(b)) \\leq k d_{X}(a,b)$ for all $ a,b \\in X$ then $ f$ is Lipschitz.\r\n\r\nA function $ I \\to X$ is absolutely continuous on $ I$ if for every positive number $ \\varepsilon$, there is a positive number $ \\delta$ such that whenever a sequence of pairwise disjoint sub-intervals $ [x_{k}, y_{k}]$ of $ I$ satisfies $ \\sum_{k} |y_{k}\\minus{}x_{k}| < \\delta$ then $ \\sum_{k} d(f(y_{k}), f(x_{k})) < \\varepsilon$.\r\n\r\nSince $ I \\subseteq \\mathbb{R}$ it has a finite diameter. Call this $ p$. Let $ \\varepsilon > 0$. Choose $ \\delta \\equal{} \\frac{\\varepsilon}{kp}$. Then:\r\n\r\n$ \\sum_{k} |y_{k}\\minus{}x_{k}| < \\delta \\implies \\sum_{k} d(f(y_{k}), f(x_{k})) \\leq kd_{X}(a,b) < \\frac{\\varepsilon}{p}$. Hence $ f$ is absolutely continuous.\r\n\r\nIs this correct?", "Solution_3": "you are using the same k :) \r\n$ \\sum_{k} |y_{k}\\minus{}x_{k}| < \\delta \\implies \\sum_{k} d(f(y_{k}), f(x_{k})) \\leq \\sum_{k} Kd(y_{k}, x_{k}) \\equal{} K \\sum_{k} d(y_{k}, x_{k})$\r\n\r\nthen choose $ \\delta\\equal{}\\epsilon/K$\r\n\r\nI don't know why you have to assume I to be an interval?", "Solution_4": "Would my $ \\delta$ work?", "Solution_5": "[quote=\"1234567a\"]you are using the same k :) \n$ \\sum_{k} |y_{k} \\minus{} x_{k}| < \\delta \\implies \\sum_{k} d(f(y_{k}), f(x_{k})) \\leq \\sum_{k} Kd(y_{k}, x_{k}) \\equal{} K \\sum_{k} d(y_{k}, x_{k})$\n\nthen choose $ \\delta \\equal{} \\epsilon/K$\n\nI don't know why you have to assume I to be an interval?[/quote]\r\n\r\nI think you are proving uniform continuity. I am trying to prove absolute continuity.", "Solution_6": "No, 1234567a is proving exactly what is needed: absolute continuity (though, of course, uniform continuity is proved in the same way and even with the same $ \\delta$ in the end).", "Solution_7": "does my $ \\delta$ work?", "Solution_8": "No, generally speaking, it does not." } { "Tag": [ "Asymptote", "LaTeX" ], "Problem": "I've just installed asymptote on my new Windows Vista (premium 64bit) machine having used it in the past on my old one. I've got ghostscript and GSView installed and checked that they are working by opening a .ps file. Next I installed asymptote thought I'd check it was OK by trying to produce a simple output in interactive mode:\r\n\r\n>asy\r\n>draw((0,0)--(100,100));\r\n\r\nbut I just get the told:\r\n\r\nCannot write to out.eps\r\n\r\nAny ideas?", "Solution_1": "Vista puts restrictions on the Program Files directory which some but not all programs can deal with. In particular it won't let programs save files there and expect them to save them elsewhere. As a workaround (there are probably better ways) re-install Asymptote to another directory outside Program Files.", "Solution_2": "Thanks a lot. I just saved it directly onto my C: drive. Also, I had to change the default path in asymptote to look for gsview64.exe rather than gsview32.exe.\r\n\r\nNow to configure with TeXnicCenter (fingers crossed).", "Solution_3": "I've just tried to configure aysmptote to work with TeXnicCenter but I'm having a slight difficulty:\r\n\r\nI set up asymptote by going to Tools-> Customise->Tools then creating \"Asyptote\" before entering the following info into the three fields:\r\n\r\n\"C:\\Asymptote\\asy.exe\"\r\n\r\n-batchView -tex \"pdflatex\" -align C -f \"pdf\" %tc\r\n\r\n%dc\r\n\r\nwhich is the same setup as on my other machine (except, of course, the path).\r\n\r\nWhen I try to build a document involving the asymptote environment it gets stuck straight away in the package-loading stage and says:\r\n\r\n[code]! LaTeX Error: File 'asymptote.sty' not found.[/code]\n\nIncidentally, when I execute the same command on my XP machine, the corresponding line in the dialogue box is:\n\n [code](C:\\Program Files\\MiKTeX 2.5\\tex\\latex\\Asymptote\\ayymptote.sty\")[/code]\r\n\r\nwhence it proceeds to build the document with no problems whatsoever.\r\n\r\nNow I suspect this is related to not having asymptote installed in Program Files but I'm out of ideas. I've tried copying the folder \"Asymptote\" which contains asymptote.sty and two other .sty files from the XP machine into various folders within TeXnicCenter without joy.", "Solution_4": "You've forgotten to do the bit that tells LaTeX all about asymptote - you have to put some asymptote files where LaTeX can find them. You'll find the details at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]Using Asymptote in LaTeX[/url]." } { "Tag": [ "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "nice one:\r\nassume that $G$ is p-subgroup of $S_n$ .and $n p $. It is well known\r\nthat every $ p $ - group has nontrivial center. Take some element\r\n$ a $ , which lies in the center of $ G $. Then it has a\r\ndecomposition to nonintersecting cycles, which consist of $ 1 $ or\r\n$ p $ elements. Denote by $ C_{1},C_{2}, \\ldots , C_{m} $ the\r\ncycles which consist of $ p $ elements ( by $ C_{j} $ I mean the\r\nset of elements in corresponding cycle). We have that $ m\r\n\\geqslant 1 $, because $ a $ is nontrivial. Now, for each $ x \\in\r\nG $, we have that $ xa = ax $, therefore $ x $ maps each cycle $\r\nC_{i} $ to some cycle $ C_{j} $. Therefore every $ x \\in G $ acts\r\non the set of cycles $ \\{ C_{1}, C_{2} , \\ldots , C_{m} \\} $, so $\r\nG $ acts on the set $ \\{ C_{1}, C_{2} , \\ldots , C_{m} \\} $. Each\r\norbit of $ G $, acting on $ \\{ C_{1}, C_{2} , \\ldots , C_{m} \\} $\r\nconsists of $ p^{j} $ elements, because it's order must divide $\r\n|G| = p^{k} $. Therefore if we take the union of $ C_{i} $ lying\r\nin this orbit, then it has $ p^{j+1} $ elements. But again, by $ n\r\n< p^{2} $ we have that $ j=0 $, so this orbit consists of only $ 1\r\n$ element. So we have that every element $ x \\in G $ maps $ C_{j}\r\n$ to itself for every $ j $. Take $ A = C_{1} $, $ B = \\{\r\n1,2,\\ldots,n \\} \\backslash C_{1} $. Then $ |A| = p $, $ |B| = n-p\r\n$, and $ G $ acts separately on $ A $ and on $ B $. If we denote\r\nby $ H $ the group of permutations on $ A $, which comes from $ G\r\n$, and by $ K $ the group of permutations on $ B $, which comes\r\nfrom $ G $, then by induction hypothesis we have that $ H,K $ are\r\ncommutative, hence $ G $ is commutative.\r\n\r\nI hope it is correct." } { "Tag": [ "function", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Suppose that the function $ f: [a,b] \\to [c,d]$ is absolutely continuous and bijection.\r\nIs it true that $ f^{\\minus{}1}: [c,d] \\to [a,b]$ is absolutely continuous?\r\nCan we weaker the assumption?, \r\ni.e. what is necessary and sufficient for an absolutely continuous function to have inverse image is also absolutely continuous?", "Solution_1": "Here's a counterexample: let $ g$ be the Cantor devil's staircase function on $ [0,1]$ (increasing, not absolutely continuous) and let $ f^{\\minus{}1}(x)\\equal{}g(x)\\plus{}x$. $ f$ then maps $ [0,2]$ to $ [0,1]$ bijectively, and is absolutely continuous since it satisfies a Lipschitz condition $ |f(x)\\minus{}f(y)|\\le |x\\minus{}y|$.", "Solution_2": "Thanks Jmerry,", "Solution_3": "The necessary and sufficient condition to be added is that $ f'\\ne 0$ almost everywhere. (Which means either $ f'>0$ a.e. or $ f'<0$ a.e., since $ f$ is monotone.)\r\n\r\nIn higher dimensions this is a topic of current research. There the analogue of the class of absolutely continuous functions is the Sobolev space $ W^{1,1}$ of mappings $ f\\colon \\mathbb R^n\\to\\mathbb R^n$. In two dimensions the situation is now [url=http://adsabs.harvard.edu/abs/2006ArRMA.180...75H]clear[/url]: if such $ f$ is a homeomorphism and has positive Jacobian determinant a.e., then its inverse is also in $ W^{1,1}$. When $ n>2$, things get more complicated." } { "Tag": [], "Problem": "Hello\r\n\r\nIs there people here which speaks spanish? (or at least write it)", "Solution_1": "Well, the \"Spanish Communities\" forum is reasonably active. It's too bad that AoPS/Mathlinks stopped posting flags beside posters' screen names, but I think most of the participants in that forum live in the Americas, from Mexico through Argentina and Chile, scattered across quite a few of the countries in between.", "Solution_2": "Okay\r\n\r\nWell actually I posted this 'cause I want to see if there're some people (excluding Spanish-speaking countries) which at least writes in spanish :), not necessary to talk the language.", "Solution_3": "Yo se un poco de espanol. Pero es muy muy muy mal.", "Solution_4": "yo voy a estudiar espanol en mi collegio en la ciudad de los angeles.", "Solution_5": "S\u00ed, habla espa\u00f1ol porque apprendo espa\u00f1ol en mi escuela en Canad\u00e1... Me gusta espa\u00f1ol mucho, pero mi espa\u00f1ol es muy mal :( \u00a1Viva Espa\u00f1ol!)", "Solution_6": "Yo hablo espa\u00f1ol, soy de San Luis Potosi, Mexico. Me llamo Ernesto H. Escobedo. No se cual es la comunidad para hispanoparlantes aqui en MathForums, o una comunidad para la olimpiada Mexicana de Matematicas", "Solution_7": "[b][color=red]Creo que la comunidad en Espa\u00f1ol se encuentra por ac\u00e1:[/color][/b]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/index.php?f=255]*****[/url]\r\n\r\n[color=red][b]Hasta la vista, [i]baby[/i].[/b][/color]", "Solution_8": "\u00a1Hola! Soy de El Salvador y hablo espa\u00f1ol. :D", "Solution_9": "\u00a1Hola!\r\n \r\n\r\n Estoy bien. [color=cyan]:lol: [/color]", "Solution_10": "Hola, yo puedo hablar un poco espanol. Buena suerte en sus estudias de matematicas.", "Solution_11": "Hola Como Estas\r\n\r\nI forgot most of what I learned....shoot....thats no good\r\n\r\nI have soo much espanol tarea :( \r\n\r\nI remember this though\r\n\r\nNos Vemos!![/b]", "Solution_12": "Yo soy de Puetro Rico y quiero un forum para Puerto Rico", "Solution_13": "Yo soy Louisville, you hablo un poco espan^~ol y mucho americano. Me gusta deportiste y matematicas. Yo estudio para los exams mucho. Yo (Aprendo pret\u00e9ritos en espa\u00f1ol. ~ Utilic\u00e9 a un traductor para las partes en el ()).", "Solution_14": "Hablo un poco espanol.\n\nSoy de los Estados Unidos.\n\nTengo nada hablar (esta fraze tiene razon? No se?).", "Solution_15": "No debo hablar?", "Solution_16": "The $\\text{Spanish Communities}$ is not very active. \n\nLas comunidades en espa\u00f1ol no son muy activas, creo que se debe a los moderadores del mismo, pues desde hace tiempo ya no se muestra alguna actividad. Espero que los participantes de foro que hablen o lean en el idioma espa\u00f1ol se manifieste.", "Solution_17": "You hablo y escribo espa\u00f1ol porque nac\u00ed y crec\u00ed en Uruguay (South America). All\u00ed estudi\u00e9 qu\u00edmica en la universidad.\nTambi\u00e9n hablo y escribo en ingl\u00e9s porque hace m\u00e1s de 30 a\u00f1os que vivo en EEUU (USA).\n\n\"Tengo nada hablar\" no tiene sentido (means nothing). I don't know what was meant by that. If I knew what was meant, I could translate.", "Solution_18": "!Hola, al\u00f3, jal\u00f3, bueno, al, diga \n\n\n\n??", "Solution_19": "I was trying to say that I have nothing to say.", "Solution_20": "Absolutamente nada que decir, se\u00f1orita???\n\nIgnore the last word.", "Solution_21": "!hola, Los Zapotas...!mui bien!", "Solution_22": "All I can say is:\n\nEl coche es grande\n\nEl coche es pequenio\n\nEl coche es nuevo\n\nEl coche es viecho\n\nLa camionetta es grande\n\nLa camionetta es pequenia\n\nLa camionetta es nueva\n\nLa camionetta es viecha\n\nDid I spell everything correctly?", "Solution_23": "[quote=\"3333\"]All I can say is:\n\nEl coche es grande\n\nEl coche es pequenio\n\nEl coche es nuevo\n\nEl coche es viecho\n\nLa camionetta es grande\n\nLa camionetta es pequenia\n\nLa camionetta es nueva\n\nLa camionetta es viecha\n\nDid I spell everything correctly?[/quote]\n\nThere are some mistakes\n\nEl coche es pequenio $\\Rightarrow$ El coche es peque\u00f1o.\n\nEl coche es viecho $\\Rightarrow$ El coche es viejo.\n\nLa camionetta es grande $\\Rightarrow$ La camioneta es grande.\n\nLa camionetta es pequenia $\\Rightarrow$ La camioneta es peque\u00f1a.\n\nLa camionetta es viecha $\\Rightarrow$ La camioneta es vieja.\n\nSaludos y qu\u00e9 bueno que est\u00e9s aprendiendo es idioma espa\u00f1ol (spanish)", "Solution_24": "what the heck...\nI only took half a year of Spanish, and I found spanish so uninteresting. \nHola\nBueno Dias----> good morning\nBueno Tardes----> good afternoon\nBueno Noches----> good night\nPapel----> paper\nBoligrafo---> Pen i think...\nMarcadora----> Marker\nPizara----> board", "Solution_25": "[quote=\"carlosmath\"][quote=\"3333\"]All I can say is:\n\nEl coche es grande\n\nEl coche es pequenio\n\nEl coche es nuevo\n\nEl coche es viecho\n\nLa camionetta es grande\n\nLa camionetta es pequenia\n\nLa camionetta es nueva\n\nLa camionetta es viecha\n\nDid I spell everything correctly?[/quote]\n\nThere are some mistakes\n\nEl coche es pequenio $\\Rightarrow$ El coche es peque\u00f1o.\n\nEl coche es viecho $\\Rightarrow$ El coche es viejo.\n\nLa camionetta es grande $\\Rightarrow$ La camioneta es grande.\n\nLa camionetta es pequenia $\\Rightarrow$ La camioneta es peque\u00f1a.\n\nLa camionetta es viecha $\\Rightarrow$ La camioneta es vieja.\n\nSaludos y qu\u00e9 bueno que est\u00e9s aprendiendo es idioma espa\u00f1ol (spanish)[/quote]\n\nMeh\n\n[quote]Bueno Noches----> good night[/quote]\nThat is awesome because in Russian, night is \u043d\u043e\u0447\u044c which is pronounced \"noch.\"\nWouldn't it be buenas noches though instead of bueno?", "Solution_26": "[quote=\"yugrey\"]I was trying to say that I have nothing to say.[/quote]\nNo tengo nade que decir. = I have nothing to say.\nNo tengo nade de que hablar. = I have nothing to talk about.\nNo s\u00e9 que decir. = I don't know what to say.\n(I used \"alt 0233\" for the \u00e9).", "Solution_27": "Hola\n\nComo estas?\n\nDe nada.\n\nGracias." } { "Tag": [ "number theory", "greatest common divisor", "algebra proposed", "algebra" ], "Problem": "[color=blue]resolve this exercice[/color]", "Solution_1": "it' anyone here give us a solution \r\n\r\nit' so easy :?", "Solution_2": "If one of the 18 cosecutive numbers is 999, we are done. If not, at least one of the 18 cosecutive numbers is divisible by 18. Hence the sum of its digits must be 9 or 18, and our proof is complete.", "Solution_3": "[quote=\"k2c901_1\"]If one of the 18 cosecutive numbers is 27, we are done. If not, at least one of the 18 cosecutive numbers is divisible by 18. Hence the sum of its digits must be 9 or 18, and our proof is complete.[/quote]\r\n\r\nIsnt the question asking for three digit consecutive number ? So i doubt 27 will be one of them . \r\n\r\nSince we know that if a number is divisible by 3 , the sum of its digit must also be a multiple of three . So in any 18 consecutive number , there will always be at least one of it which can be divisible by 3 .Or stronger one is at least 6 number which can be divisible by its own sum of digit .", "Solution_4": "A correction has been posted. As for your arguement, I don't think it's correct, as having the sum of the digits of a certain number divisible by 3 does not guarantee that the it is divisible by the sum of its digits. Take, for example, 984.", "Solution_5": "Ya , i agree with you k2c901_1 and I think I have mess up with the question itself .My arguement can only take out the gcd of the number itself and its sum of digit. :blush: :blush: I first thought that we can see it through divisibility of 9 and then come to divibility of 3 . I guess divisibilty of 18 is more accurate here .... :P ." } { "Tag": [ "modular arithmetic", "number theory", "prime factorization" ], "Problem": "Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.", "Solution_1": "[quote=\"N.T.TUAN\"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]\r\n\r\n[hide=\"Would it make the most sense to...\"]\nprove that any number $A$ is divisible [i]not only[/i] by a number in the set $\\{ 3,5,7,9 \\}$? (That is, in the prime factorization of $A$, those numbers are not the only primes.) [/hide]", "Solution_2": "[quote=\"cincodemayo5590\"][quote=\"N.T.TUAN\"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]\n\n[hide=\"Would it make the most sense to...\"]\nprove that any number $A$ is divisible [i]not only[/i] by a number in the set $\\{ 3,5,7,9 \\}$? (That is, in the prime factorization of $A$, those numbers are not the only primes.) [/hide][/quote]\r\n\r\nI don't think that would work; an even number would work by your condition, but not by his, since $2$ is a prime less than $11$, but it is not in that set.", "Solution_3": "[quote=\"Hikaru79\"][quote=\"cincodemayo5590\"][quote=\"N.T.TUAN\"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]\n\n[hide=\"Would it make the most sense to...\"]\nprove that any number $A$ is divisible [i]not only[/i] by a number in the set $\\{ 3,5,7,9 \\}$? (That is, in the prime factorization of $A$, those numbers are not the only primes.) [/hide][/quote]\n\nI don't think that would work; an even number would work by your condition, but not by his, since $2$ is a prime less than $11$, but it is not in that set.[/quote]\r\n\r\nBut $A$ can't be a multiple of $2$ with these digits of 1,3,7,9....", "Solution_4": "if A is a prime number then A is divisible by itself (we easily have $A \\geq 11$)\r\nif A is a composit number , assume that A has no divisot greater than 11 , we obtain $A$ max $=3.7=21$ , but $11;13;17;19$ are all primes , so A min $=21$ , BUT $21$ IS NOT CONTRACDITION !!", "Solution_5": "[quote=\"HTA\"]if A is a prime number then A is divisible by itself (we easily have $A \\geq 11$)\nif A is a composit number , assume that A has no divisot greater than 11 , we obtain $A$ max $=3.7=21$ , but $11;13;17;19$ are all primes , so A min $=21$ , BUT $21$ IS NOT CONTRACDITION !![/quote]\r\nIt's not that easy, A is not necessarily squarefree.\r\n[hide]\nClearly $A$ can't be divisible by 2 or 5, so suppose all of its prime factors are less than 11. Then $A$ is a power of 3 multiplied by a power of 7, that is, $3^{a}\\cdot 7^{b}$. We now prove that the tens digit of $A$ is a problem (since $A>9$, it actually has a tens digit).\n\nTo see this, consider parity of $a$ and $b$. If the same, then $A\\equiv 1\\pmod{4}$ and $A\\equiv \\pm 1\\pmod{5}$. If different, then $A\\equiv 3\\pmod{4}$ and $A\\equiv \\pm 2\\pmod{5}$. (This is so because $A\\equiv 3^{a+b}\\pmod{4}$ and $A\\equiv 3^{a-b}\\pmod{5}$.) In particular, $A$ must be congruent to one of 1,3,7,9 modulo 20, making its tens digit even, a contradiction.\n[/hide]", "Solution_6": "[quote=\"HTA\"]if A is a prime number then A is divisible by itself (we easily have $A \\geq 11$)\nif A is a composit number , assume that A has no divisot greater than 11 , we obtain $A$ max $=3.7=21$ , but $11;13;17;19$ are all primes , so A min $=21$ , BUT $21$ IS NOT CONTRACDITION !![/quote]\r\noh , i'm sorry , i read the question wrong :D", "Solution_7": "[quote=\"N.T.TUAN\"]Let $A$ be a positive integer number greater than $9$, written in decimal system with digits $1,3,7,9$. Prove that $A$ has at least a prime divisor not less than $11$.[/quote]\r\nif we assume that A has no prime divisors greater than 11 then A=3^x*7^y (checking that other cant be divisors)\r\n\r\nbut A=10^a_1+3*10^a_1+7*10^a_1+9*10^a_4\r\na_i are distinct arbitary integers.\r\nso,3^x*7^y=10^a_1+3*10^a_2+7*10^a_3+9*10^a_4\r\n\r\n3^x*7(7^{y-1}-10^a_3)=10^a_1+3*10^a_2+9*10^a_4\r\ntaking mod 3 left hand side is 0 mod 3 while rhs=1 mod 3\r\nnd its the contradiction we desired.", "Solution_8": "What about the possibility of repeated digits? Also your numbers allow for 0 to be among the digits.", "Solution_9": "Oups...I forgot to mention that a_i=(0,1,2,3) and distinct.\r\n\r\nBut do i need to prove it for repeated digits?\r\nIf so then it is really hard to seek for contradiction :(\r\nbcz thr is 1 so u may take as many as u want.So i thought that probably reapeated digits are not allowed ;)\r\nwhatever, if repeated digits are allowed then i need to think again!" } { "Tag": [ "function" ], "Problem": "For a function f: Z-->R, the following statement is true:\r\n\r\n$ f(z) \\equal{} z \\minus{} 10$ for $ z > 100$\r\n and $ f(z)\\equal{}f(f(z \\plus{} 11))$ for $ z\\le11$\r\n\r\nProve that for all $ z\\le100, f(z) \\equal{} 91$.", "Solution_1": "[quote=\"Smartguy\"]For a function f: Z-->R, the following statement is true:\n\n$ f(z) \\equal{} z \\minus{} 10$ for $ z > 100$\n and $ f(f(z \\plus{} 11))$ for $ z\\le11$\n\nProve that for all $ z\\le100, f(z) \\equal{} 91$.[/quote]\r\ncould you please double-check your problem :idea:", "Solution_2": "well i copied it exactly from a source--> Bratislava Correspondence Seminar, Fall 1998 3rd series", "Solution_3": "$ \\blacktriangleright$ Did you copy this from 8.28 on page 8 of [url=http://mathcircle.berkeley.edu/recur.pdf]this presentation handout[/url]?\r\n$ \\blacktriangleright$ Perhaps the source meant $ f(z) \\equal{} f(f(z \\plus{} 11))$ for $ z \\leq 101$?\r\n\r\n\r\n[b]Objection to problem as-is:[/b]\r\n\r\nConsider $ f(z) \\equal{} 1 \\; \\; \\forall \\; \\; z \\leq 22$. This is internally consistent because:\r\n\r\n1. $ z \\leq 11 \\implies z \\leq 22 \\implies f(z) \\equal{} 1$\r\n2. $ z \\leq 11 \\implies z \\plus{} 11 \\leq 22 \\implies f(z \\plus{} 11) \\equal{} 1 \\implies f(f(z \\plus{} 11)) \\equal{} f(1) \\equal{} 1$", "Solution_4": "[quote=\"TZF\"]$ \\blacktriangleright$ Did you copy this from 8.28 on page 8 of [/quote]\r\n\r\nIndeed, I did.", "Solution_5": "Well, I emailed the guy that made those notes -- let's see if he says anything (or if someone else figures this out).", "Solution_6": "So I emailed the author of the lecture notes and he gave me the link to the actual competition problems. Indeed, there was a typo.\r\n\r\n[url=http://kms.sk/bkms/starezadania/9798/az979813.html]See here[/url].\r\n\r\nThe question should read:\r\n\r\nFor a function $ f \\; : \\; \\mathbb{Z} \\to \\mathbb{R}$, the following statement holds:\r\n\r\n$ f(z) \\equal{} z\\minus{}10$, for $ z \\geq 101$\r\n$ f(z) \\equal{} f(f(z\\plus{}11))$, for $ z \\leq 100$\r\n\r\nProve that for all $ z \\leq 100$, $ f(z) \\equal{} 91$.", "Solution_7": "f(90)=f(f(90+11))=f(101)=101-10=91\r\n+) for 100 $ \\geq x \\geq 90 f(x) \\equal{} f(f(x\\plus{}11)) \\equal{} f(x\\plus{}11\\minus{}10)\\equal{}f(x\\plus{}1) \\equal{} 91$\r\n\r\n+)for x<100 there exist k where $ 100 \\geq x\\plus{}11k \\geq 90$\r\n$ f(x) \\equal{} f(f(...f(x\\plus{}11k))..) \\equal{} f(..f(90).. ) \\equal{} 91$" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": " is not isomorphic to \r\n is not isomorphic to \r\n\r\nThanks", "Solution_1": "$ \\mathbb{Q}^\\times\\le\\mathbb{R}^\\times$ contain the element $ \\minus{}1$ of finite order $ 2$.\r\nBut an element $ x$ of $ \\mathbb{Q}$ or $ \\mathbb{R}$ of order $ 2$ would satisfy $ 2x\\equal{}0\\not\\equal{}x$." } { "Tag": [], "Problem": "Of the following five statements, $I$ to $V$, about the binary operation of averaging(arithmetic mean),\r\n\r\nI. Averaging is associative\r\nII. Averaging is commutative\r\nII. Averaging distributes over addition\r\nIV. Addition distributes over averaging\r\nV. Averaging has an identity element\r\n\r\nthose of which are always true are\r\n\r\n(A) All\r\n(B) I and II only\r\n(C) II and III only\r\n(D) II and IV only\r\n(E) II and V only", "Solution_1": "Let's denote $\\frac{x+y}{2}=x*y$\r\n$I$ is definetly false.\r\n$II$ is true.\r\n$III$ is false $x*(a+b)=\\frac{x+a+b}{2}$ while $x*a+x*b=x+\\frac{a+b}{2}$.\r\n$IV$ is true $x+(a*b)=x+\\frac{a+b}{2}$ and $(x+a)*(x+b)=\\frac{2x+a+b}{2}=x+\\frac{a+b}{2}$.\r\n$V$ is false , if $x0$ and $ abcd\\equal{}1$, we define a real function $ f$ as $ f(x)\\equal{}a^x\\plus{}b^x\\plus{}c^x\\plus{}d^x$\r\n\r\n(a). For two positive reals $ x$ and $ y$, prove that $ x\\ge y$ implies $ f(x)\\ge f(y)$\r\n\r\n(b). Prove or disprove the converse statement.", "Solution_1": "$ ya^x + (x - y).1\\geq xa^y$\r\n$ \\Rightarrow\\sum ya^x\\geq\\sum xa^y - 4(x - y) = y\\sum a^y + (x - y)(%Error. \"suma\" is a bad command.\n^y - 4)$\r\nAnd $ \\sum a^y\\geq 4\\Rightarrow f(x)\\geq f(y)$", "Solution_2": "I don't understand your solution..\r\n\r\nPart (b), please ? I got stuck..", "Solution_3": "Oh,I only use AM-GM generaliza:D", "Solution_4": "My solution for part (a) is inspired by [b]nayel[/b]..\r\n\r\nbecause $ x\\ge y$, we may set $ x \\minus{} y \\equal{} k,k\\ge 0$\r\n\r\nUsing Chebyshev, we see that :\r\n\r\n$ a^x \\plus{} b^x \\plus{} c^x \\plus{} d^x \\ge \\frac {(a^k \\plus{} b^k \\plus{} c^k \\plus{} d^k)(a^y \\plus{} b^y \\plus{} c^y \\plus{} d^y)}{4} \\ge a^y \\plus{} b^y \\plus{} c^y \\plus{} d^y$\r\n\r\nBecause $ a^k \\plus{} b^k \\plus{} c^k \\plus{} d^k\\ge 4\\sqrt [4]{(abcd)^k} \\equal{} 4$", "Solution_5": "[quote=\"Ronald Widjojo\"]For $ a,b,c,d > 0$ and $ abcd = 1$, we define a real function $ f$ as $ f(x) = a^x + b^x + c^x + d^x$\n\n(a). For two positive reals $ x$ and $ y$, prove that $ x\\ge y$ implies $ f(x)\\ge f(y)$\n\n(b). Prove or disprove the converse statement.[/quote]\r\n\r\nWe have $ f(x) = a^x + b^x + c^x + d^x$\r\n\r\n$ \\rightarrow f '(x) = a^x.lnx + b^x.lnb + c^x.lnc + d^x.lnd$\r\n\r\nFollow Chebyshev we have \r\n\r\n$ f '(x) \\ge\\ \\frac {1}{3}(a^x + b^x + c^x + d^x)(lna + lnb + lnc + lnd) \n= \\frac {1}{3}(a^x + b^x + c^x + d^x)ln(abcd) = 0$\r\n\r\nSo with $ x \\ge\\ y$ we have $ f(x) \\ge\\ f(y)$", "Solution_6": "Yeah, the (a) part is easy, but for part (b) ?\r\n\r\nI can't solve it, though i can make it.. :oops:", "Solution_7": "We have $ \\frac{f(x)\\minus{}f(y)}{x\\minus{}y}\\equal{}f'(\\xi)\\ge0$ for a certain $ \\xi\\in[y,x]$ (mean value theorem). So it only remains to show $ f(x)\\equal{}f(y)\\Rightarrow x\\equal{}y$ which is easy again." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that for any four real numbers $a$, $b$, $c$, $d$, the inequality\r\n\\[ \\left(a-b\\right)\\left(b-c\\right)\\left(c-d\\right)\\left(d-a\\right)+\\left(a-c\\right)^2\\left(b-d\\right)^2\\geq 0 \\]\r\nholds.\r\n\r\n[hide=\"comment\"]This is inequality (350) in: Mihai Onucu Drimbe, [i]Inegalitati, idei si metode[/i], Zalau: Gil, 2003.\n\nPosted here only for the sake of completeness; in fact, it is more or less the same as http://www.mathlinks.ro/Forum/viewtopic.php?t=3152 .[/hide]\r\n\r\n Darij", "Solution_1": "We need to prove : $-xyz(x+y+z)+(x+z)^2(y+z)^2 \\ge 0$\r\n\r\n$\\Leftrightarrow (z^2+xy+yz+zx)^2 \\ge xyz(x+y+z)=xy[z(x+y)+z^2]$\r\n\r\nWhich this one is obvious true!\r\n\r\nAnd I think It's still true:\r\n\r\n$2(x-y)(y-z)(z-t)(t-x)+(x-z)^2(y-t)^2 \\ge 0$" } { "Tag": [], "Problem": "i was solving some problems the other day and saw this thing called \r\nDescartes Theorum\r\nsomething to do with simplifying polymonials\r\nthe book did a horrible job of explaining it\r\nand i have no idea what to do\r\n\r\ncan someone explain Descartes Theorum to me?\r\nin easy words?", "Solution_1": "Are you saying about Decartes sign rule?", "Solution_2": "I'd say he is, but it's definitely not going to be used in MOEMS :wink: \r\n[url]http://www.purplemath.com/modules/drofsign.htm[/url]", "Solution_3": ":D i get it\r\nthank you so much xpmath" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Solve\r\n$\\frac{dy}{dx} = \\frac{3x^2+2y^2-xy}{3x^2+2y^2+xy}$", "Solution_1": "rewrite as\r\n $ \\frac{dy}{dx} = \\frac{(3\\frac{x}{y}+2\\frac{y}{x}-1)}{(3\\frac{x}{y}+2\\frac{y}{x}+1)}$\r\nmake the subtitution $y = vx$ ..change of variable\r\nso $ \\frac{dy}{dx}=v+x\\frac{dv}{dx}$\r\n\r\nfrom then on it easy to seperate it and integrate to find v as a function of x and then multiply by x to find y the orginal function.", "Solution_2": "$\\frac{dy}{dx}=\\frac{3-z+z^2}{3+z+2z^2},\\ z=xy$", "Solution_3": "hello, with $ y\\equal{}xv$ i have got\r\n$ \\frac{3\\plus{}2v^2\\plus{}v}{3\\plus{}v^2\\minus{}4v\\minus{}2v^3}\\,dv\\equal{}\\frac{dx}{x}$\r\nSonnhard." } { "Tag": [], "Problem": "find the product\r\ncatechol + chloroform(KOH) ->A + $ \\ce{CH2I2} / \\ce{KOH}$ ->B", "Solution_1": "the product is piperonal", "Solution_2": "yes the mechanism :)", "Solution_3": "obviously you know the first part( how cho came to the para position). in the second you dont get a carbene as it is unstable. (it is not a dihalocarbene) you get a carbocation(CH2I+)( also bcoz I- is a better leaving group as compared to cl- so removal of I- takes place b4 removal of h) which is attacked by oh. similarly the other oh attacks replacing i giving piperonal.\r\nalternative mechanism:\r\nCH2I2+base->CH2(OH)2->HCHO which on reaction with substituted catechol gives piperonal", "Solution_4": "Thanks how -CHO came to the para posistion?", "Solution_5": "Doesn't your book explain that? Does your book also says that the yield of A is only about 11%, and so that reaction product is just a curiosity?", "Solution_6": "no Carcul please explain the entire thing", "Solution_7": "The only explanation that comes to my mind is a statistical reason: most of the product obtained from catechol comes from the normal Reimer-Tiemann reaction*, with formyl in position 3, since this is the most favoured position (probably due to some kind of complexation); however, some dichlorocarbene may also attack position 4, since it is also electronically favoured.\r\n\r\n\r\n* Actually, the Reimer-Tiemann reaction cannot in general be made on catechols because of the ease with which they are oxidized, specially under the basic condictions of the reaction. As far as I know, the biggest yield of 4-formylcatechol from catechol with the Reimer-Tiemann reaction that was obtained was a 29% (and only with a special technique)." } { "Tag": [], "Problem": "The cost of five pencils and one pen is $ \\$2.50$, and the cost of one\npencil and two pens is $ \\$1.85$. What is the cost of two pencils and\none pen?", "Solution_1": "Let a pencil value=x and a pen=y.\r\nAdd the 2 equations together and 6x+3y=4.35.\r\nDivide by 3 and get 2x+y=1.45." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let K include {Rational Numbeers} /\\ (0,1) \r\nProve if Points of K partition (0,1) into adjacent open intervals \r\n[b]==> K is CLOSE [/b]", "Solution_1": "Hmm... The phrase \"Points of $K$ partition $(0,1)$ into adjacent open intervals\" may only mean that $(0,1)\\setminus K$ is a union of open intervals and, therefore, open. Thus, $K$ is relatively closed in $(0,1)$ (but not necessarily closed: take the sequence $1,\\frac12,\\frac13,\\dots$; it partitions $(0,1)$ into adjacent open intervals in any reasonable sense but doesn't form a closed set). In any case, the problem, as stated, is either very simple or wrong. Perhaps, I misunderstand something :?" } { "Tag": [ "LaTeX" ], "Problem": "If I want to make a Latex file (I am later going to convert it to PDF) and I want to make a nice PDF, how do I skip multiple lines between the title and the first line? The \\par command only allows me to skip one line, no matter how many times I type it. Is there something I can do to skip multiple lines?", "Solution_1": "The command \\vspace allows you to skip space (v=vertical, \\hspace skips horizontal space). For example, \\vspace{2.3in} skips 2.3 inches. You can also use other units, such as cm, pt, etc. A particularly useful one to know is \\baselineskip, which is the height of a line of text. So \\vspace{2\\baselineskip} skips 2 lines.\r\n\r\nWarning: don't overuse it! The automatic formatting on LaTeX is designed to be aesthetically pleasing, and artificial formatting commands such as this can make the document look less appealing. I usually only use this when positioning a picture on the page or in similar circumstances.", "Solution_2": "Also, if you want to end a page and have the text stop until the next page, you can use the \\pagebreak command." } { "Tag": [ "probability", "percent" ], "Problem": "Hey this is a fairly simple question I think but I don't know how to do it...\r\n\r\nIf I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?\r\n\r\nThanks.\r\n\r\nP.S. At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...", "Solution_1": "[hide]if you mean the percent chance that you will lose 75 times in a row, it's 1/655^75[/hide]", "Solution_2": "no i meant what is the chance i will lose at all in the 75 times i play\r\n\r\nwould it be\r\n\r\n(1/655)^75\r\n\r\nto get around a 10% chance?????", "Solution_3": "I you mean the probabiliity you'll lose only once, then\r\n[hide]$\\frac{1}{655}$ chance you'll lose once, and then to win 74 times, the probability is $(\\frac{654}{655})^{74}$. However, there are 75 places to put this one loss. \nSo the total is $\\frac{1}{655}\\cdot(\\frac{654}{655})^{74}\\cdot75\\approx.102$[/hide]", "Solution_4": "I don't think he means that, either.", "Solution_5": "[quote=\"mathbeginner\"]Hey this is a fairly simple question I think but I don't know how to do it...\n\nIf I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?\n\nThanks.\n\nP.S. At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...[/quote]\r\n\r\n The probability that you will lose at least once, while playing 75 times is:\r\n\r\n$1-(\\frac{654}{655})^{75}= \\approx.01$", "Solution_6": "[quote=\"G-UNIT\"][quote=\"mathbeginner\"]Hey this is a fairly simple question I think but I don't know how to do it...\n\nIf I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?\n\nThanks.\n\nP.S. At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...[/quote]\n\n The probability that you will lose at least once, while playing 75 times is:\n\n$1-(\\frac{654}{655})^{75}= \\approx.01$[/quote]Explanation of his solution:\r\nNote that if you have two probabilities $P(A)$ and $P(\\text{not }A)$ then $P(A)+P(\\text{not }A)=1$. \r\n\r\nIn this case $P(A)$ is the probability that lose at least once\r\nSo $P(\\text{not }A)$ is the probability that you don't lose at least once, in other words the probability that you win every single game.\r\n\r\nSo if you find the probability that he wins all $75$ games: it is $\\left(\\frac{654}{655}\\right)^{75}$ In this case basically you just found $P(\\text{not }\r\nA)$. So to find $P(A)$ you want $1-\\left(\\frac{654}{655}\\right)^{75}$.", "Solution_7": "OK its done, i obviously am not good with words and its hard to explain but i figured that its going to be near 10% and thats all i wanted to know, thx for the help everyone", "Solution_8": "I think my answer comes out to be about 10 percent, saying that you will lose once and [i]only[/i] once.", "Solution_9": "[quote=\"mathbeginner\"]Hey this is a fairly simple question I think but I don't know how to do it...\n\nIf I'm playing a game and I have a 1/655 chance to lose, what is the % chance that I will lose if I play 75 times?\n\nThanks.\n\nP.S. At first I thought to just say 75/655, but that is wrong because if that were true then if I played 655 games I would have a 100% chance to lose, which is obviously incorrect...[/quote]\r\n\r\n[hide]\n1:655::x:75\n655x=75\nx=75/655=15/131\n15/31|65\n15/31*1/65=15/2015=3/403$\\approx$.7%?\n[/hide]", "Solution_10": "[hide]$1-\\left( \\frac{654}{655} \\right) ^{75} \\approx 11 \\%$[/hide]", "Solution_11": "whoa there's been about 5 answers all different, which one is correct? \r\n\r\nI guess the wording needs to be improved." } { "Tag": [], "Problem": "\u4ece\u672c\u7ad9\u4e0b\u8f7d\u7684 MathLinks Contest \u4e0a\u7684\u9898\u5e94\u8be5\u7528\u4ec0\u4e48\u201c\u6253\u5f00\u65b9\u5f0f\u201d\u6765\u9605\u8bfb\uff1f", "Solution_1": "\u90a3\u662f pdf \u6a94\uff0c\u4f60\u8981\u7528 Adobe Reader \u4f86\u6253\u958b\u3002 ;) \r\n\u4f60\u53ef\u5230\u4ee5\u4e0b\u7db2\u5740\u4e0b\u8f09\u3002\r\n[url=http://www.chinese-s.adobe.com/main.html]http://www.chinese-s.adobe.com/main.html[/url]" } { "Tag": [ "algebra", "polynomial", "modular arithmetic", "coefficients", "IMO Shortlist" ], "Problem": "For a polynomial $ P$ of degree 2000 with distinct real coefficients let $ M(P)$ be the set of all polynomials that can be produced from $ P$ by permutation of its coefficients. A polynomial $ P$ will be called [b]$ n$-independent[/b] if $ P(n) \\equal{} 0$ and we can get from any $ Q \\in M(P)$ a polynomial $ Q_1$ such that $ Q_1(n) \\equal{} 0$ by interchanging at most one pair of coefficients of $ Q.$ Find all integers $ n$ for which $ n$-independent polynomials exist.", "Solution_1": "First we can remove the condition that $P(n)=0$ (it's superfluous).\n\nLet $d = 2000$. Clearly $0,1$-independent polynomials of degree $d$ exist, so suppose $n\\ne 0,1$ and an $n$-independent $P$of degree $d$ exists, with coefficients $a_00$, arbitrarily large positive integers $N$ with $\\| N a_i \\| \\le \\epsilon$ for all $i$. Then we can take $b_i = [Na_i]$ (where $[x] = x - \\|x\\|$) if $\\epsilon |1+n+\\cdots+n^d| < 1$, $N>\\frac{2}{\\min|a_i - a_j|\\setminus\\{0\\}}$, and $N>\\frac{1}{\\min|\\sum a_{\\pi(k)} n^k| \\setminus\\{0\\}}$. (The second guarantees the $b_i$ are pairwise distinct, while the first and third guarantee $\\sum b_{\\pi(k)} n^k$, which is an integer, vanishes only when $\\pi$ is good---this follows from the triangle inequality.)\n\n[b]Lemma.[/b] Suppose there are $r_i$ indices $j$ with $b_j \\equiv i \\pmod{n}$. Then $\\sum_{k=0}^{d} b_{\\pi(k)} n^k = C$ has at most $r_0!\\cdots r_{n-1}!$ solutions $\\pi$. Furthermore, if equality holds, then for each $j$ with $b_j \\equiv C \\pmod{n}$, $\\sum_{k=0}^{d-1} b_{\\sigma(k)} n^k = \\frac{C-b_j}{n}$ must have exactly $r_0!\\cdots (r_C-1)! \\cdots r_{n-1}!$ solutions $\\sigma$ (with $\\sigma$ restricted to the indices $k\\ne j$); in particular, there must be at least one index $k\\ne j$ with $b_k \\equiv \\frac{C-b_j}{n} \\pmod{n}$.\n\n[i]Proof.[/i] Induct on $r_0 + \\cdots + r_{n-1} = d+1 \\ge 1$ using the requirement $a_{\\pi(0)} \\equiv C \\pmod{n}$. $\\blacksquare$\n\nWLOG assume $\\gcd(b_0,b_1,\\ldots,b_d) = 1$, so $00$; WLOG $b_d \\equiv t \\pmod{n}$. Now focus on the $\\pi$ with $\\pi(0) = d$: they must all be bad. On the other hand, any good $\\pi$ \"saves\" (via a single transposition) exactly one bad $\\pi'$ with $\\pi'(0) = d$. But each bad $\\pi'$ (with $\\pi'(0) = d$) is saved by at least one good $\\pi$, so we must have at least $d!$ good $\\pi$. Yet by the lemma and the fact that $a! b! < (a+1)! (b-1)!$ whenever $a\\ge b\\ge 1$, we have at most $d! 1!$ good $\\pi$, with equality only when $\\{r_0,r_t\\} = \\{1,d\\}$.\n\nNow we will invoke the equality clause of the lemma to derive a contradiction. If $r_0 = 1$, then WLOG $b_0 \\equiv 0\\pmod{n}$ and $b_1,\\ldots,b_d \\equiv t\\pmod{n}$. It follows that $\\sum_{k=0}^{d-1} b_{\\pi(k)} n^k = \\frac{-b_0}{n}$ for all $\\pi$, from which we easily obtain $b_1 = \\cdots = b_d$ (e.g. consider the effect of a transposition), contradiction.\n\nOtherwise, $r_t = 1$, so $b_0,\\ldots,b_{d-1}\\equiv0\\pmod{n}$. Take a good $\\pi$ and suppose $\\pi(M) = d$. If $M\\le d-2$, let $\\{u,v\\} = \\{0,1,\\ldots,d\\}\\setminus\\{\\pi(0),\\ldots,\\pi(d-2)\\}$. Then by repeated application of the equality clause, $b_{\\sigma(d-1)}+nb_{\\sigma(d)} = -\\frac{\\sum_{k=0}^{d-2} b_{\\pi(k)} n^k}{n^{d-1}}$ has $2!$ solutions $\\sigma$ (where $\\{\\sigma(d-1),\\sigma(d)\\} = \\{u,v\\}$), since $b_u,b_v \\equiv 0\\pmod{n}$; hence $b_u = b_v$, which is impossible.\n\nThus $M\\ge d-1$ for any $\\pi$, whence the equality clause gives us $n^k \\mid b_0,\\ldots,b_{d-1}$ for $k=1,2,\\ldots,d-2$ (just induct on $k$). Now fix some $\\pi$; the equality clause tells us there exists $\\pi' \\ne \\pi$ with $\\pi'(i) = \\pi(i)$ for $i=0,1,\\ldots,d-3$. We cannot have $\\pi'(d-2) = d$ (by the previous paragraph) or $\\pi'(M) = \\pi(M) = d$ ($\\pi',\\pi$ can't agree in any more places), so WLOG $M = d-1\\implies \\pi(d-1) = d$ and $\\pi'(d) = d$. Thus\n\\begin{align*}\n0 &= \\sum (b_{\\pi(k)} - b_{\\pi'(k)}) n^k \\\\\n&= (b_{\\pi(d-2)} - b_{\\pi'(d-2)})n^{d-2} + (b_d - b_{\\pi'(d-1)})n^{d-1} + (b_{\\pi(d)} - b_d)n^d,\n\\end{align*}which is impossible mod $n^d$ (we have $n^2\\mid n^{d-2} \\mid b_{\\pi(d-2)},b_{\\pi'(d-2)},b_{\\pi'(d-1)}$, but $n\\nmid b_d$).\n\n[b]Comment:[/b] The official solution uses $|n|^d > |n|^{d-1}+\\cdots+1$ instead to show $a_d n^d + \\cdots \\ne a_0 n^d + \\cdots$, whence one finds an inductive bound of $|\\{\\sum a_{\\pi(k)} n^k\\}| \\ge 2^{d-1}$ (to finish, we have $\\{\\sum a_{\\pi(k)} n^k\\} \\subseteq \\{(a_{\\pi(j)} - a_{\\pi(i)})(n^j - n^i)\\}$, and the latter is too small). However, only the above method generalizes if we replace the \"polynomial basis\" $(n^0,\\ldots,n^d)$ with $(n_0,\\ldots,n_d)$, where for each $k$, $\\frac{n_k}{n^k}$ is an integer relatively prime to $n$.\n\n[b]Case 2: $n=-1$.[/b] Then for any indices $i_10$ we have the inequality:\r\n $\\frac{x_1}{\\sqrt{x_1+x_2}}+\\frac{x_2}{\\sqrt{x_2+x_3}}+...+\\frac{x_n}{\\sqrt{x_n+x_1}}\\leq\\sqrt{n-\\frac{23}{16}}$ if $x_1+x_2+...+x_n=1$.\r\n I have no idea whether we can find a constant $k<1$ which does not depend on $n$ such that $\\frac{x_1}{\\sqrt{x_1+x_2}}+\\frac{x_2}{\\sqrt{x_2+x_3}}+...+\\frac{x_n}{\\sqrt{x_n+x_1}}\\leq\\ k\\sqrt{n}$.", "Solution_1": "[quote=\"harazi\"]I don't even know I am posting it here (probably just to add it to the list of topics with no answer), but here it goes:\n For any $x_1,...,x_n>0$ we have the inequality:\n $\\frac{x_1}{\\sqrt{x_1+x_2}}+\\frac{x_2}{\\sqrt{x_2+x_3}}+...+\\frac{x_n}{\\sqrt{x_n+x_1}}\\leq\\sqrt{n-\\frac{23}{16}}$ if $x_1+x_2+...+x_n=1$.\n I have no idea whether we can find a constant $k<1$ which does not depend on $n$ such that $\\frac{x_1}{\\sqrt{x_1+x_2}}+\\frac{x_2}{\\sqrt{x_2+x_3}}+...+\\frac{x_n}{\\sqrt{x_n+x_1}}\\leq\\ k\\sqrt{n}$.[/quote]\r\n\r\n :)", "Solution_2": "I think I have a kind of solution.\r\n\r\nIt's clear that out sum does not exceed $\\sum_{i=1}^n \\sqrt{x_i}$. If one of the $x_i$ exceeds $\\frac 14$, $\\sum_{i=1}^n \\sqrt{x_i}$ will be at most $\\frac 12+\\sqrt{\\frac 34 (n-1)}<\\sqrt{n}-\\frac 12$ for $n>50$. By the methods below, replacing 3), we can prove that our sum doesn't exceed $\\sqrt n-\\frac 12$ for $n\\geq 7$. Now suppose $n>7$. Then we may assume none of the numbers exceed $\\frac 14$. \r\nWe need some preliminary results:\r\n1)Let $x_1,x_2,\\ldots,x_n$ be real numbers. Then $\\sum_{i,j}(x_i-x_j)^2\\geq \\frac{n+5}8 \\sum_{i=1}^n (x_i-x_{i+1})^2$ (subscripts are taken modulo $n$).\r\n\r\nProof: let $\\delta_k=\\sum_{i=1}^n (x_{i+k}-x_i)^2$. By using the inequality $x^2+y^2\\geq \\frac{(x+y)^2}2$ we get $(x_{i+k}-x_i)^2+(x_{i+k+1}-x_i)^2\\geq \\frac{(x_{i+k+1}-x_{i+k})^2}2$. Now summing over all $i$ we get $\\delta_k+\\delta_{k+1}\\geq \\frac{\\delta_1}2$. From here, we easily derive that our sum is $\\delta_1+\\frac{\\delta_2+\\delta_3+\\ldots+\\delta_{n-2}}2\\geq \\frac{n+5}8 \\delta_1$, as desired.\r\n\r\n2)Denote by $s=\\sum_{i=1}^n \\frac{(x_{i+1}-x_i)^2}{x_i+x_{i+1}}^2$. Then $\\sum_{i=1}^n \\sqrt{x_i} \\leq \\sqrt{n-\\frac{n+5}{16}s}$.\r\n\r\nProof: we have $n(\\sum_{i=1}^n x_i)-(\\sum_{i=1}^n \\sqrt{x_i})^2=\\sum_{i,j}(\\sqrt{x_i}-\\sqrt{x_j})^2\\geq \\frac{n+5}8 \\sum_{i=1}^n (\\sqrt{x_i}-\\sqrt{x_{i+1}})^2$, according to (1). Finally, $(\\sqrt{x_i}-\\sqrt{x_{i+1}})^2=\\frac{(x_i-x_{i+1})^2}{(\\sqrt{x_i}+\\sqrt{x_{i+1}})^2}\\geq \\frac{(x_i-x_{i+1})^2}{2(x_i+x_{i+1})}$. From here we deduce the conclusion.\r\n\r\n3)We have $\\sum_{i=1}^n \\frac{x_i}{\\sqrt{x_i+x_{i+1}}}\\leq \\sum_{i=1}^n \\sqrt{x_i}- \\frac{x_ix_{i+1}}{x_i+x_{i+1}}$. \r\n\r\nProof: We have $\\frac{x_i}{\\sqrt{x_i+x_{i+1}}}=\\sqrt{x_i}-\\frac{\\sqrt{x_i}(\\sqrt{x_i+x_{i+1}}-\\sqrt{x_i})}{\\sqrt{x_i+x_{i+1}}}=\\sqrt{x_i}-\\frac{x_{i+1}\\sqrt{x_i}}{\\sqrt{x_i+x_{i+1}} (\\sqrt{x_i+x_{i+1}}+\\sqrt{x_i})}$. We can deduce $\\frac{\\sqrt{x_i}}{(\\sqrt{x_i+x_{i+1}}+\\sqrt{x_i})(\\sqrt{x_i+x_{i+1}})}>\\frac{x_i}{x_i+x_{i+1}}$ for $x_i<\\frac 14$, so our term is at most $\\sqrt{x_i}-\\frac{x_ix_{i+1}}{(x_i+x_{i+1})}$. By summing over all (i), we get our conclusion.\r\n\r\nNow, let's finish the problem. Set $s$ as in 2). Then $\\sum_{i=1}^n \\frac{x_ix_{i+1}}{x_i+x_{i+1}}=\\sum_{i=1}^n \\frac{x_i+x_{i+1}}4-\\frac{(x_i-x_{i+1})^2}{4(x_i+x_{i+1})}=\\frac 12-\\frac s4$. \r\n\r\n$\\sum_{i=1}^n \\frac{x_i}{\\sqrt{x_i+x_{i+1}}}\\leq \\sum_{i=1}^n \\sqrt{x_i}- \\frac{x_ix_{i+1}}{x_i+x_{i+1}}\\geq \\sqrt{n-\\frac{n+5}8s}-\\frac 12+\\frac s4$. However, it's clear that $\\sqrt{n-\\frac{n+5}8s}<\\sqrt n-\\frac{n+5}{16\\sqrt n} s<\\sqrt n-\\frac s4$ for $n\\geq 7$. So for $n\\geq 7$ our sum is at most $\\sqrt n-\\frac 12$, which is more than we wanted.\r\n\r\nBy replacing $\\frac 14$ by $a=O(\\sqrt[3]{n})$, we can deduce that the sum is at most $\\sqrt n- \\lambda \\sqrt[6] n$, for some constant $\\lambda$." } { "Tag": [ "inequalities", "number theory unsolved", "number theory" ], "Problem": "Prove that n!m!(m+n)!|(2n)!(2m)! where n, m are natural numbers.", "Solution_1": "Posted before . \r\nUse the inequalities : \r\n$ [2x]\\plus{}[2y]\\geq [x]\\plus{}[y]\\plus{}[x\\plus{}y],\\forall x,y\\geq 0$", "Solution_2": "if $ p^a$ devides $ n!$ and $ a$ is the largest integer such that this holds, than $ a\\equal{}\\sum_{i\\equal{}1}^{\\infty}[\\frac{n}{p^i}]$\r\n\r\nwe have to prove that the exponent of every prime number $ p$ is smaler in integer factorization of $ m!n!(m\\plus{}n)!$ than in $ (2n)!(2m)!$.\r\nso we have to prove the following inequality:\r\n$ \\sum_{i\\equal{}1}^{\\infty}[\\frac{n}{p^i}] \\plus{} \\sum_{i\\equal{}1}^{\\infty}[\\frac{m}{p^i}] \\plus{} \\sum_{i\\equal{}1}^{\\infty}[\\frac{n\\plus{}m}{p^i}] \\leq \\sum_{i\\equal{}1}^{\\infty}[\\frac{2n}{p^i}]\\plus{}\\sum_{i\\equal{}1}^{\\infty}[\\frac{2m}{p^i}]$\r\n\r\nwhich is true because of the inequality TTsphn mentioned above, when used for every $ i$" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "I've got a question that I'm having some issues with if someone can give me some guidance, it would be appreciated.\r\n \r\nIf $ y$ is given by a composite function of $ x$, use the Chain Rule to show that:\r\n \r\n\\[ \\frac{d^{2}y}{dx^{2}}=\\frac{d^{2}y}{du^{2}}\\cdot\\left(\\frac{du}{dx}\\right)^{2}+\\frac{dy}{du}\\cdot\\frac{d^{2}u}{dx^{2}}\\]\r\n\r\nThanks in advance.", "Solution_1": "I think this works...\r\n\r\n$ \\frac{d^{2}y}{dx^{2}}= \\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)\\\\ = \\frac{d}{dx}\\left(\\frac{dy}{du}\\cdot\\frac{du}{dx}\\right)\\\\ = \\frac{d}{dx}\\left(\\frac{dy}{du}\\right)\\cdot\\frac{du}{dx}+\\frac{d^{2}u}{dx^{2}}\\cdot\\frac{dy}{du}\\\\ = \\frac{d^{2}y}{du^{2}}\\cdot\\left(\\frac{du}{dx}\\right)^{2}+\\frac{dy}{du}\\cdot\\frac{d^{2}u}{dx^{2}}$", "Solution_2": "...or\r\nLet $ y=f(u), u=g(x)$\r\n\r\n$ dy=f'(u)g'(x)\\,dx$\r\n\r\n$ d^{2}y=(f'(u)g'(x)\\,dx)'\\,dx$\r\n\r\n$ d^{2}y=(f''(u)u'g'(x)+g''(x)f'(u))dx^{2}$\r\n\r\n$ \\frac{d^{2}y}{dx^{2}}=f''(u)(g'(x))^{2}+g''(x)f'(u)$\r\n\r\n$ \\frac{d^{2}y}{dx^{2}}=\\frac{d^{2}y}{du^{2}}\\cdot\\left(\\frac{du}{dx}\\right)^{2}+\\frac{d^{2}u}{dx^{2}}\\cdot\\frac{dy}{du}$" } { "Tag": [ "ARML" ], "Problem": "ARML is putting together their next book of contests, and they need the full list of winners from 2004 and 2005. They have first place, but are looking for the rest. If you have that information, can you post it here?", "Solution_1": "don't they already have the results? After all, they are the ones that make the contest!\r\n\r\nI actually think the ARML winners can be found at the ARML website.", "Solution_2": "They have full team results, but not individuals; that was settled in the tiebreaker, and apparently their records got lost. The web site has the individual top ten for 2006 and later only.\r\n\r\nHere's what I can reconstruct from old threads here:\r\n\r\n2004:\r\n1. Aaron Pixton, Upstate New York\r\n2. Jongmin Baek, SFBA A\r\n3. Anders Kaseorg, North Carolina\r\nOther perfect scores:\r\nRicky Biggs, TJHSST A\r\nBrian Lawrence, Montgomery County A\r\nEric Price, TJHSST A\r\n\r\nSevens included:\r\nDarren Yin, SFBA A\r\nAdam Hesterberg, Washington A\r\nAlison Miller, Upstate NY\r\nDmitriy Karabash, NYC A\r\nYan Zhang (compucomp), Lehigh Valley?\r\n(rand%(0)), Missouri?\r\n\r\nThere were about 10-15 more sevens, and they may not have been in the tiebreaker at all. In fact, the individual results seem to have been announced differently at the three sites, with only the Penn State site hearing about all of the perfect scores.\r\n\r\nFor 2005, I can't find any individual results at all.", "Solution_3": "I have to agree with abcak; it strikes me as extremely strange that such a prominent national, well-organized competition would not have the results of their competition from just a few years ago...", "Solution_4": "The 7's from Iowa in 2005 were Zach Abel and one other student whose name I don't remember (he was on the Chicago team, and I seem to recall that he might not have been on Chicago A). I think Zach placed second or third; the other student did not place. I'm getting into pretty wild guesses here, but I think Ryan Ko won that year (or if not, he was second and Zach was third).\r\n\r\nThat was also the first year site awards were given out, so that other student would have been the Iowa site winner. If anybody out there thinks they might recognize him, I've got his picture and can send it to you.", "Solution_5": "The following is all from what I remember from actually being at ARML 4 years ago. There was only one 7 at the San Jose ARML site in 2005, and that was Curtis Liu from SFBA. I believe it was a 3-way tiebreaker over the different sites to determine first place.", "Solution_6": "The SoCal team in 2005 had three people who scored 6, and I think I remember them taking part in some form of tiebreaker - although I'm pretty sure that would have been for site honors, not national honors. What \"Myself\" is saying is consistent with what I remember, but I don't remember anything about the result of the national tiebreaker.\r\n\r\nWe had one person with a 6 in 2004; again, that didn't put him in the national tiebreaker.", "Solution_7": "According to Wikipedia, Ryan Ko won in 2005." } { "Tag": [], "Problem": "Perform the addition or subtraction or operation and write the result in standard form\r\n\r\n-(3/4 + 7/5i) - (5/6 -1/6i)\r\n\r\n\r\n(3 + sqrt of -5)(7 - sqrt of -10)", "Solution_1": "This is exersise for ... well, I'll better silent.\r\nIn any case it is not good for this section." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "assume that $ a\\sqrt{6} \\plus{} b\\sqrt{10} \\plus{} c\\sqrt{15}\\in\\mathbb{Q}$ for some $ a,b,c\\in\\mathbb{Z}$, \r\n\r\ndoes it imply that $ a\\equal{}b\\equal{}c\\equal{}0$ ?\r\n\r\n(and why, if so....)", "Solution_1": "A bit of theory: $ \\mathbb{Q}[ \\sqrt{2}, \\sqrt{3}, \\sqrt{5} ]$ has dimension $ 8$ over $ \\mathbb{Q}$ and can be given a basis $ \\{ 1, \\sqrt{2}, \\sqrt{3}, \\sqrt{5}, \\sqrt{6}, \\sqrt{10}, \\sqrt{15}, \\sqrt{30} \\}$. This can be proven by adjoining $ \\sqrt{2}, \\sqrt{3}, \\sqrt{5}$ separately, and generalizes to the square root of every prime." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "search" ], "Problem": "How can we find all eigenvalues of nxn all-ones matrix?", "Solution_1": "In other words, the matrix known as $ J.$\r\n\r\nThis is really pretty simple.\r\n\r\nDirect multiplication gives us that $ J^2\\equal{}nJ.$ That means that $ J$ satisfies the polynomial $ x^2\\minus{}nx$ (in fact, that's its minimal polynomial), which means that any eigenvalues of $ J$ must also satisfy $ x^2\\minus{}nx\\equal{}0.$ So the only possible eigenvalues are $ 0$ and $ n.$ \r\n\r\nI'll let you finish it: how many of each of those must we have? (That is, I'm asking about the algebraic multiplicity of each of those two eigenvalues.)", "Solution_2": "Thx, I have gotten the answer.\r\nThere are (n-1) 0s and 1 n\r\nActually, the full question is that\r\n[img]http://img131.imageshack.us/img131/9812/12842740.jpg[/img]\r\n\r\nI am afraid I get the wrong eigenvectors\r\nHere is my work\r\n[img]http://img131.imageshack.us/img131/949/ans1.jpg[/img]\r\n\r\nIt is for eigenvalue 0\r\n\r\nThere are (n-1) eigenvectors like that column vector.\r\n\r\nHowever, when I want to search for the answer form the internet.\r\nI find that the answer is one -1 and one 1 in the column vector.\r\n\r\nfor the eigenvalues of n, the answer is all 1. I can't find this answer. Is it right? I am afraid of the wrong eigenvectors will get the wrong matrix P.\r\n\r\nfor the following part. I think I can try to cope with it.\r\nIn a iii , P should be the component of corresponding eigenvectors.\r\n\r\na iv Form the \"trend\" of the eigenvectors value, i actually guess that P may be an orthogonal matrix.. Otherwise, I think it is quite difficult to obtain it.", "Solution_3": "It seems that I can think of where the problems lie on. I can solve the whole question. Thankyou." } { "Tag": [ "probability" ], "Problem": "Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.", "Solution_1": "[quote=\"pascal_1623\"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]\r\n\r\n[hide]I got $\\frac{19}{35}$...don't know if it's right[/hide]", "Solution_2": "[hide]We need to consider the number of partitions of the 4 chocolate chips. \n\nThere are 4 partitions if we put all 4 chips into one cookie.\nThere are 12 partitions if we put 3 chips in one cookie and 1 in another.\nThere are 6 partitions if we put 2 chips into one cookie and 2 into another.\nThere are 12 partitions if we put 2 chips into one cookie and put the other two in different cookies.\nAnd finally there is 1 partition if we put each chip into a different cookie.\n\nThere are a total of 35 partitions with 19 of them having no more than two chips in a cookie.\n\nThus our answer is $\\frac{19}{35}$[/hide]", "Solution_3": "[quote=\"pascal_1623\"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]\r\n\r\nTwo questions:\r\n\r\nAre the cookies distinct?\r\nAre the chips distinct?\r\n\r\nI remember the answer being something different but then again, I did it over a year ago.", "Solution_4": "[hide=\"Solution\"]We will find the probability that there are [i]more[/i] than two chips in some of the cookies and subtract that from 1. There are two cases:\n\n[i]Case 1:[/i] There are 3 chips in one cookie and 1 chip in another. There are a total of 4 x 3 ways to select the cookies and 4 ways to choose the chips. Thus there are a total of 48 ways for this case.\n\n[i]Case 2:[/i] There are 4 chips in one cookie. There are just 4 ways to select the cookie.\n\nNow we add up the total number of ways that satisfy either of these cases and divide by $4^4$, the total number of ways to put the chips in the cookies (each of 4 chips can go into one of 4 cookies), and the probability of either case 1 or 2 occurring is $\\frac{52}{256}=\\frac{13}{64}$, so our answer is $1-\\frac{13}{64}=\\boxed{\\frac{51}{64}}$.[/hide]", "Solution_5": "[quote=\"Max300\"][hide]We need to consider the number of partitions of the 4 chocolate chips. \n\nThere are 4 partitions if we put all 4 chips into one cookie.\nThere are 12 partitions if we put 3 chips in one cookie and 1 in another.\nThere are 6 partitions if we put 2 chips into one cookie and 2 into another.\nThere are 12 partitions if we put 2 chips into one cookie and put the other two in different cookies.\nAnd finally there is 1 partition if we put each chip into a different cookie.\n\nThere are a total of 35 partitions with 19 of them having no more than two chips in a cookie.\n\nThus our answer is $\\frac{19}{35}$[/hide][/quote]\n\n[hide=\"Comment\"]\nThe answer is incorrect since you are distinguishing among the cookies but not among the chips.\n\nYou have to distinguish among both.[/hide]", "Solution_6": "[quote=\"SplashD\"][quote=\"pascal_1623\"]Tamyra is making four cookies and has exactly four chocolate chips. If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.[/quote]\n\nTwo questions:\n\nAre the cookies distinct?\nAre the chips distinct?\n[/quote]\r\n\r\nAm I supposed to consider my questions answered?\r\n\r\nI think nebula's answer is right, I remember the answer being fifty-something over 64", "Solution_7": "[quote=\"SplashD\"][quote=\"SplashD\"]\nTwo questions:\n\nAre the cookies distinct?\nAre the chips distinct?\n[/quote]\n\nAm I supposed to consider my questions answered?\n\nI think nebula's answer is right, I remember the answer being fifty-something over 64[/quote]\r\n\r\nYes I think I sort of answered it in my previous post.\r\n\r\nThey both are distinct.", "Solution_8": "I got something along those lines, and basically did the same thing as [b]Max300[/b]...i think my answer is a little different\r\n\r\n[hide=\"something along the lines of Max300\"]\n$9/17$\n\nhere's the work...(\"o\"s are chips, \"x\"s are cookies, not that it matters)\n\n[b]o o o o \n\nx x x x[/b]\n\n4 chips per cookie\n\n4000\n0400\n0040\n0004\n\n3 chips per cookie + 1 chip per cookie\n\n1300\n1030\n1003-3\n3010\n3100\n3001-6\n0031\n0013\n0103-9\n0301\n0130\n0310-12\n\n2 chips per cookie + 2 chips per cookie\n\n2020\n2200\n2002-3\n0022\n0202\n0220-6\n\n2 chips per cookie + 2 chips in different cookies\n\n2110\n2011\n1021-3\n1012\n1102\n1120-6\n1210\n2011\n2101-9\n0112\n0121\n0211-12\n\n\n12+6+12+4=34\n6+12=18\n\n9/17[/hide]", "Solution_9": "[quote=\"rruszyk\"]I get the same answer.[/quote]\r\n\r\nYou're probably going to be banned for impersonation, but before you are, who are you?", "Solution_10": "[quote=\"Ignite168\"][quote=\"rruszyk\"]I get the same answer.[/quote]\n\nYou're probably going to be banned for impersonation, but before you are, who are you?[/quote]\r\nHey, we don't have to click on your sig to quote anymore.", "Solution_11": "[quote=\"bpms\"][quote=\"Ignite168\"][quote=\"rruszyk\"]I get the same answer.[/quote]\n\nYou're probably going to be banned for impersonation, but before you are, who are you?[/quote]\nHey, we don't have to click on your sig to quote anymore.[/quote]\r\n\r\nThat's because you need a double hide tag to hide the quote button.", "Solution_12": "51/64 is correct. I have seen the answer key.", "Solution_13": "Bruh what is this revive though" } { "Tag": [ "vector", "linear algebra", "matrix" ], "Problem": "1.) Is the set D = {(1,1,2),(5,-2,3)} a spanning set?\r\n\r\n2.) If the vector(1,0,0) is added to set D, will the enlarged set be a spanning set?", "Solution_1": "Spanning set of what? Unless you specify what space or subspace you are trying to span, the question cannot be interpreted.", "Solution_2": "[quote=\"Kent Merryfield\"]Spanning set of what? Unless you specify what space or subspace you are trying to span, the question cannot be interpreted.[/quote]\r\n\r\nI guess there is something wrong with the book. I suppose, it is 3-space.", "Solution_3": "All right, we'll assume you're talking about spanning $ \\mathbb{R}^3.$\r\n\r\nTwo vectors cannot possible span $ \\mathbb{R}^3.$ An spanning set for a three-dimensional space must have at least 3 elements.\r\n\r\nNow if we add that particular third vector, would the resulting set span $ \\mathbb{R}3?$ There's a way to find out. Write these vectors as columns, and stack them together as a matrix:\r\n\\[ \\begin{bmatrix}1&1&5\\\\0&1&\\minus{}2\\\\0&2&3\\end{bmatrix}\\]\r\nNow, perform Gaussian elimination. When you reach row echelon form, is there a pivot in every row? If there is if pivot in every row, then the set does span $ \\mathbb{R}^3.$ Otherwise, it doesn't.\r\n\r\nThat's also why the two-element set cannot span $ \\mathbb{R}^3.$ Take the $ 3\\times 2$ matrix you get and row eliminate it. There can be at most two pivots, and with three rows, there can't be a pivot in every row." } { "Tag": [ "trigonometry", "function", "algebra", "functional equation", "system of equations", "algebra unsolved" ], "Problem": "Determine all functionf $f : \\mathbb{R} \\mapsto \\mathbb{R}$ such that \r\n\r\n\\[ f(x+y) = f(x)f(y) - c \\sin{x} \\sin{y} \\] for all reals $x,y$ where $c> 1$ is a given constant.", "Solution_1": "First put in $y=0$. You get $f(x)=0 \\; \\forall x$ or $f(0)=0$. So $f(0)=0$, because the first function doesn't work.\r\nNow put $x=t,y=\\pi$ for any real $t$. You get $f(\\pi + t)=f(\\pi )f(t)$. For $x=\\frac{\\pi }{2}+t,y=\\frac{\\pi }{2}$ you get $f(\\pi +t)=f(\\frac{\\pi }{2})f(\\frac{\\pi }{2} + t) - c \\cos t$. Finally for $x=\\frac{\\pi }{2}, y=t$ you get $f(\\frac{\\pi }{2} + t)=f(\\frac{\\pi }{2})f(t)-c \\sin t$. Now you have a system of equations (if you put in the original functional equation $x=y=a$ ($a=f(\\frac{\\pi }{2})$) you get $f(\\pi )=a^2-c$) and from it you get $f(t)=a \\sin t + \\cos t$. To get $a$ you put $t=\\pi$ and from that $a_1= \\sqrt{c-1}$ and $a_2=- \\sqrt{c-1}$." } { "Tag": [], "Problem": "Using two distinct members of the set $ \\{0, 1, 2, 3, 4\\}$, how many different sums can be attained?", "Solution_1": "The least sum is $ 0\\plus{}1\\equal{}1$ and the greatest is $ 4\\plus{}3\\equal{}7$. Clearly all sums in between can be attained. Then we have $ 7\\minus{}1\\plus{}1\\equal{}\\boxed{7}$ different sums." } { "Tag": [], "Problem": "Provide a synthesis of acetophenone from ethylbenzene.", "Solution_1": "I am not sure if this works (just starting to learn about Organic Synthesis). Please guide me along where I am going wrong, thanks.\r\n\r\nI would start off by oxidising the ethylbenzene by a powerful oxidant (e.g. hot, concentrated, acidified $ KMnO_4$):\r\n\r\n$ C_6H_5C_2H_5 \\plus{} 6[O] \\rightarrow C_6H_5COOH \\plus{} 2H_2O \\plus{} CO_2$\r\n\r\nI am thinking along the lines of reducing the carboxyl group to a primary alcohol (e.g. $ LiALH_4$) and maybe oxidising that to an aldehyde.\r\n\r\nThe aldehyde can then be converted to a secondary alcohol using the Grignard Reagent ($ CH_3MgI$) and then oxidised to the ketone desired.\r\n\r\nAre the reductants that can convert carboxylic acids directly to aldehydes?", "Solution_2": "[quote=\"BanishedTraitor\"]Please guide me along where I am going wrong, thanks.[/quote]\n\nIn organic synthesis it is common practice to not show balanced chemical reactions: we are just to show the relevant information: starting material(s), product(s), and reaction conditions.\n\nYour synthesis is correct, good job. However, it uses too many steps: two are enough.\n\n[quote=\"BanishedTraitor\"]Are the reductants that can convert carboxylic acids directly to aldehydes?[/quote]\r\n\r\nYes there are (via the formation of some intermediate or by some indirect route), but the common practice (specially in organic chemistry books) is to first convert the acid into an acid chloride, esther, or cyanide, and then reduce one of these with a reducing agent not so powerfull as LiAlH4 (for example, DIBALH). Here's an indirect route that I particularly like:\r\n\r\n1) Treat the acid with two equivs. of butyl lithium;\r\n2) Treat with one equiv. of ethyl formate.\r\n3) Make an acidic workup and heat the mixture.\r\n\r\nThe explanation of these steps will be left as an additional exercise.", "Solution_3": "Hmmm...... I really need to learn more organic reactions.\r\n\r\nP.S. What are some useful chain lengthening and chain degradation process?", "Solution_4": "[quote=\"BanishedTraitor\"]What are some useful chain lengthening and chain degradation process?[/quote]\r\n\r\nIn this particular case you don't need that (just oxidation reactions). However:\r\n\r\n1) Chain lengthening reactions: aldol, SN with alkyne anions and organometalics, Fridel-Crafts, Wittig, Michael, Diels-Alder.\r\n\r\n2) Chain degradation reactions: oxidative cleavage of alkenes and alkynes (ozonolysis), oxidation of side chains in aromatic compounds.", "Solution_5": "Well I tried working backward but I always get stuck at some point when I need the branch to be halegen substitued but I only know using $ hf$ and $ X_2$ (radical reaction).", "Solution_6": "That's very close. Here are some of my approaches:\r\n\r\n[hide=\"Two step synthesis\"]1. $ C_6H_5CH_2CH_3 \\plus{} Br_2/light \\longrightarrow C_6H_5CHBrCH_3$\n\n2. $ C_6H_5CHBrCH_3 \\plus{} DMSO \\longrightarrow C_6H_5COCH_3$.[/hide]\n\n[hide=\"Three step synthesis\"]1. $ C_6H_5CH_2CH_3 \\plus{} KMnO_4/HO^\\minus{}/heat \\longrightarrow C_6H_5CO_2H$\n\n2. $ C_6H_5CO_2H \\plus{} 2\\,CH_3Li \\longrightarrow C_6H_5C(CH_3)(O^\\minus{})_2$\n\n3. $ C_6H_5C(CH_3)(O^\\minus{})_2 \\plus{} H_2O \\longrightarrow C_6H_5COCH_3$.[/hide]\n\n[hide=\"One step synthesis\"]$ C_6H_5CH_2CH_3 \\plus{} SeO_2 \\longrightarrow C_6H_5COCH_3$.\n\nOther oxidizing agents include CrO3/acetic acid, Jones reagent, and PCC.[/hide]" } { "Tag": [], "Problem": "I started to write mock SAT. For this summer, SAT is all I'm doing and I thought that I would score better if I can think as a ETS's view as well. There isn't official date yet but I wanted to see how many people will be interested on taking this mock SAT.\r\n\r\nBy the way, math section will be harder than the usual so you can even use this for AMC practice haha.\r\n\r\nAnyone? :)", "Solution_1": "I 'll do it but what are you gonna do with the essay section?", "Solution_2": "Sure I'll take it but I'll probably end up skipping the essay because its so boring and I hate writing.", "Solution_3": "[quote=\"SplashD\"]I 'll do it but what are you gonna do with the essay section?[/quote]\r\n\r\nFor essay section, I'll grade as follows:\r\n\r\nI'll take all answers and go through them once for 3 minute. This will be the first score. Then after I went through them once, I'll take about 5-6 minute detailed examination to give the second score. So, your score will be holistic, just like in reality (except I play two roles).\r\n\r\nI would recommend you to do essays because if you don't, your writing score will be relatively low.", "Solution_4": "i'll give it a try. no harm.", "Solution_5": "count me in, but I probably won't turn in the essay. I will try it of course, so it won't go to waste.", "Solution_6": "Are you certified to holistically grade an essay?\r\n\r\nCan we not include the math section? Everybody knows over 90% of AoPsers can get a perfect 800 on it, so it would be pointless.", "Solution_7": "I will try to see if I have time to do it.\r\n\r\nSo I guess I'm in. :)", "Solution_8": "[quote=\"SplashD\"]Are you certified to holistically grade an essay?\n\nCan we not include the math section? Everybody knows over 90% of AoPsers can get a perfect 800 on it, so it would be pointless.[/quote]\r\n\r\nI'm not certified. I'm just junior student who just wants to feel like a test maker. If you don't like my grading, fine. But I'm going to do my best to give a fair score.\r\n\r\nMath section has to be included. I know that 90% of AoPSers can get perfect score. However, that doesn't mean that all of them will get perfect scores. In math section, perfect score depends on not making careless mistakes, which are also important on the real one.\r\n\r\nPlus, I'm going to try to make it harder than usual SAT style.", "Solution_9": "[quote=\"Silverfalcon\"]\nPlus, I'm going to try to make it harder than usual SAT style.[/quote]\r\n\r\nThe math section or the whole test? \r\n\r\nJust out of curiosity what was your highest SAT score so far?", "Solution_10": "OK, I'll try, if you will be so kind as to remind me by PM before the test. :)", "Solution_11": "[quote=\"SplashD\"][quote=\"Silverfalcon\"]\nPlus, I'm going to try to make it harder than usual SAT style.[/quote]\n\nThe math section or the whole test? \n\nJust out of curiosity what was your highest SAT score so far?[/quote]\r\n\r\nNothing is certain yet.\r\n\r\nMy SAT score is pretty low (I rather not say the number but I'm still practicing). As I said, this is not anything that's goign to happen very soon. I'm going to do lots of practices to see the styles of questions and create (and borrow some problems from books) a test that will be very similar to real one.", "Solution_12": "Would you be able to finish it before school starts?", "Solution_13": "[quote=\"SplashD\"]Would you be able to finish it before school starts?[/quote]\r\n\r\nYup.\r\n\r\nMoreover, for all students who submit answers, they're going to receive full solution and for essay section, your feedback and sample essays.", "Solution_14": "[quote=\"Silverfalcon\"][quote=\"SplashD\"]Would you be able to finish it before school starts?[/quote]\n\nYup.\n\nMoreover, for all students who submit answers, they're going to receive full solution and for essay section, your feedback and sample essays.[/quote]\r\n\r\nAre you going to write the sample essay? By the way, are you going to make up the topic for the essay or take an existing one?", "Solution_15": "The topic should involve penguins.\r\n\r\nAnd count me in.", "Solution_16": "bump\r\n\r\nalso, is this going to happen seeing as how summer's almost over?", "Solution_17": "Is this not going to happen? :maybe:", "Solution_18": "if it is harder, then how is a mock sat i? it should be mock sat 2? and also where are teh problems", "Solution_19": "[quote=\"moogra\"]if it is harder, then how is a mock sat i? it should be mock sat 2? and also where are teh problems[/quote]\r\nProblems haven't been posted yet... :huh:", "Solution_20": "alright. i'll be waiting :lol:" } { "Tag": [ "function", "algebra", "polynomial", "ratio", "rational function", "functional equation", "number theory" ], "Problem": "Find all functions such that f (3x + 2y) = f (2x) + f (3y) + 6xy\r\nIf there does not exist such the function justify your answer.\r\nx,y are rationals and f (x) is a rational function\r\n\r\n[hide=\"Some thoughts/questions related to this problem and other functional equations in general\"]I was wondering is it possible to assume f(x) = (a (x^3)) + (b (x^2)) + (c x) + d and solve for f (x) ? Even if it does give a solution would it be possible to find any other functions? Also if it is the only function, how would you prove it?[/hide]", "Solution_1": "Regarding the functional equation\r\n[hide]\n$ f(3x \\plus{} 2y) \\equal{} f(2x) \\plus{} f(3y) \\plus{} 6xy$; $ x$, $ y$ rational; $ f$ rational-valued.\n\nAssume such a function $ f$ exists.\nPutting y=0 in the original equation,\n$ f(3x) \\equal{} f(2x) \\plus{} f(0)$.\nPutting x=0 in the original equation,\n$ f(2y) \\equal{} f(3y) \\plus{} f(0)$.\nPutting x=y and adding these two equations,\n$ f(0) \\equal{} 0$.\nThus\n$ f(3x \\plus{} 2y) \\equal{} f(2x) \\plus{} f (3y) \\plus{} 6xy \\equal{} f(3x) \\plus{} f(2y) \\plus{} 6xy \\equal{} f(2x \\plus{} 3y)$.\n\nNow put $ x \\equal{} \\minus{} \\frac {3}{2}y$ above and get\n\n$ f( \\minus{} \\frac {5}{2}y) \\equal{} f(0)$\nso $ f$ is constant, but that would mean $ 6xy$ is constant, which is false.\n\nThus such a function does not exist.\n[/hide]\nRegarding your question, [hide]\nyou may indeed do that to find a particular solution. I don't think you get marks \"for the answer.\"\nHowever, first you might try to prove that $ f$ takes a particular form. I do not see this so often; instead usually you just use the properties in the question to construct the answer (perhaps with some guessing in the scrap work).\nIf you don't show that f is a particular kind of function (for example, polynomials $ F$would arise under certain conditions; and linear functions arise in $ F(m \\plus{} n) \\equal{} F(m) \\plus{} F(n)$ for $ m$ and $ n$ rational; and multiplicative arithmetical functions arise in $ F(ab) \\equal{} F(a)F(b)$ for $ a$ and $ b$ relatively prime), but to show that the solution $ F$ to a functional equation is unique, perhaps you may\n\"Let $ F$ and $ G$ be solutions to --(some given functional equation)--\" and then show that $ F(t) \\minus{} G(t) \\equal{} 0$ or $ F(t)/G(t) \\equal{} 1$ everywhere, or even \"suppose there exists some $ t$ for which $ F(t)$ > $ G(t)$\" (and prove by contradiction) etc. I don't know what to say except that the conclusion you will be looking for will be similar to proving to real numbers are equal, or that two sequences are identical. \nInstead what we usually do is just ... find the function directly. I suppose if you use that method to find a particular solution, and if you are \"confident\" that there are no others, you may use facts about your particular solution as hints.\nIt becomes much different if $ F$ is known to be continuous! Then you can consider small changes. But then, when you know more facts about the function you would just be using those facts...[/hide]\r\n\r\nI thought \"rational function\" usually means \"ratio of polynomials with integer coefficients.\" Correct me." } { "Tag": [], "Problem": "Find all pairs of non-negative integers $ (m, n)$ such that $ m^{n}\\equal{}n^{m}$.", "Solution_1": "[quote=\"AndrewTom\"]Find all pairs of non-negative integers $ (m, n)$ such that $ m^{n} \\equal{} n^{m}$.[/quote]\r\nDid you mean to place the restriction that m doesn't equal n?", "Solution_2": "I guess so. Thanks.", "Solution_3": "(0,0) and (2,4)", "Solution_4": "(4,2) is too", "Solution_5": "[quote=\"varunrocks\"]>> (0,0) << and (2,4)[/quote]\r\n\r\nI say you can't use $ (0, 0)$ because $ 0^0$ is indeterminate\r\nto begin with.", "Solution_6": "Taking the ln of both sides, $ \\frac {ln (m)}{m} \\equal{} \\frac {ln (n)}{n}$. But $ f(x) \\equal{} \\frac {ln (x)}{x}$ has a maximum at $ x \\equal{} e$, so if we don't have $ m \\equal{} n$, we must have one of $ m$,$ n$ be either $ 1$ or $ 2$. There are clearly no solutions for $ m \\equal{} 1$ and as said before, we have the solutions $ (2,4)$ and $ (4,2)$." } { "Tag": [ "quadratics" ], "Problem": "The sum of a negative integer, $ N$, and its square is 6. What is\nthe value of $ N$?", "Solution_1": "$ N^2\\plus{}N\\minus{}6\\equal{}0$\r\nSolving quadratics, we get $ N\\equal{}\\minus{}3$, because $ N<0$" } { "Tag": [ "email" ], "Problem": "Hi all\r\n\r\nI have done 5 out of 6 given questions in INMO09. What are my chances???? :maybe: :huh: \r\n\r\nDo keep in mind i am in XII\r\n\r\n\r\n\r\n\r\n\r\n\r\nDo Reply to \r\n :arrow: http://www.mathlinks.ro/Forum/viewtopic.php?t=252391", "Solution_1": "You have the best chances provided u have written all of them very properly. I suppose u didnt solve the combi. :) {An INidan would be so often}.\r\n12thers here say they have got only 3 or 3 and a half. So obvioulsy u have great chances.\r\nCongrats. I fpossible pl tel, how others {11th and below} did in ur centre. And continue this in the INMO Discussion Thrad.", "Solution_2": "ya mostly in my center were saying they did 2 -3 questions. \r\n\r\nthe one quest i m not sure is 4th.\r\n\r\n& last question i did by geometry.", "Solution_3": "Very good Naveen.\r\nYeah a nice solution exists by geometry.\r\nAnd also, 4th sum is misleading wit its geometric progression.", "Solution_4": "Anyone getting 4 or 5 correct and is from class 12 :?:\r\n\r\nand 12thers what are ur scores in Fiitjee PT3?", "Solution_5": "well i cracked 4 out of 6 wat are my chances. and for naveen great if i get through we shall surely meet at the IMOTC. best of luck", "Solution_6": "Hello naveen and inabash. \r\n\r\nLast year the cut-offs were 40 for everyone not in 12th standard and 50 for those students in 12th standard.\r\nLiterally 50 translates to just getting three problems correct. :D However finally it just boils down to whether the solutions are \"rigorous\". Yeah last year a 12thie from my center got through with 4 questions right. The cutoffs never go above 50 or 55 for 12thies. So all the best and hope u make it", "Solution_7": "Did naveen/inabash qualify?", "Solution_8": "Well there is some1 in the name of Sanjit in the list .. hope he is naveenkr, coz his email address is Sanchitml@gmail.com . ch and j ... or maybe his name is sth else ...", "Solution_9": "[quote=\"Akashnil\"]Well there is some1 in the name of Sanjit in the list .. hope he is naveenkr, coz his email address is Sanchitml@gmail.com . ch and j ... or maybe his name is sth else ...[/quote]\r\nActually, I think Sanchit and Naveen are best friends and use the same ID? :rotfl:\r\nBut I don't think either qualified coz they are in 12th to the best of my knowledge. :huh:" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "$ \\lim_{p \\to \\infty} \\left( \\frac{ (px\\plus{}1)^e}{p^{e\\minus{}1} \\sqrt[p]{e^{px\\plus{}1}}} \\right) \\minus{} \\lim_{p \\to \\infty} \\left( \\frac{px^e}{e^x} \\right) \\equal{}$\r\n\r\n\r\nNB : $ x$ is a variable, and $ e \\approx 2.71828...$", "Solution_1": "Hmm, no one?\r\nBig hint : Let $ f(x) \\equal{} \\frac{x^e}{e^x}$", "Solution_2": "The correct statement should be : $ \\lim_{p \\to \\infty} \\left( \\frac{ (px\\plus{}1)^e}{p^{e\\minus{}1} \\sqrt[p]{e^{px\\plus{}1}}} \\minus{} \\frac{px^e}{e^x}\\right)$ \r\n\r\nFrom $ \\frac{ (px\\plus{}1)^e}{p^{e\\minus{}1} \\sqrt[p]{e^{px\\plus{}1}}} \\minus{} \\frac{px^e}{e^x} \\equal{} \\frac{ (x\\plus{}1/p)^ee^{\\minus{}1/p} \\minus{} x^e}{1/p}$ it's easy to see that the required limit is $ f'(0)$ where $ f(t)\\equal{}(x\\plus{}t)^ee^{\\minus{}t}$ .\r\n :cool:" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "geometric transformation", "reflection", "symmetry", "Miquel point" ], "Problem": "Triangle $ ABC$ is inscribed in circle $ \\omega$. The tangent lines to $ \\omega$ at $ B$ and $ C$ meet at $ T$. Point $ S$ lies on ray $ BC$ such that $ AS \\perp AT$. Points $ B_1$ and $ C_1$ lie on ray $ ST$ (with $ C_1$ in between $ B_1$ and $ S$) such that $ B_1T \\equal{} BT \\equal{} C_1T$. Prove that triangles $ ABC$ and $ AB_1C_1$ are similar to each other.", "Solution_1": "Nice problem! What number was it?\r\n\r\nUnfourtunately, I have a very long and ugly proof for this, but I hope it is worth mentioning. A shorter proof would definitely be more appropriate and appreciated:\r\n[hide=\"Solution\"]\nAllow us to take points $ B'$ and $ C'$ so that $ AT$ bisects $ \\angle BAC'$, $ AT$ bisects $ \\angle B'AC$, and $ TB \\equal{} TC \\equal{} TB' \\equal{} TC'$ (so that $ B'$ is closer to $ B_1$ than $ C_1$ and $ C'$ is closer to $ C_1$ than $ B_1$). Also, allow the intersection between $ B'C'$ and $ Bc$ to be $ S'$. Now, we call $ \\angle BAB' \\equal{} \\alpha$ and $ \\angle B'AT \\equal{} \\beta\\implies \\angle TAC \\equal{} \\beta$ and $ \\angle CAC' \\equal{} \\alpha$. \n\n[b]Lemma 1: $ B'$, $ T$, and $ C'$ are collinear[/b]\n[i]Proof:[/i] We notice that $ \\angle BAC \\equal{} \\angle TBC \\equal{} \\angle TCB$ and that $ \\angle BAC \\equal{} 2\\beta \\plus{} \\alpha$. Thus, $ \\angle BTC \\equal{} 180 \\minus{} 2\\angle TBC \\equal{} 180 \\minus{} 4\\beta \\minus{} 2\\alpha$. Now, we have that $ AT$ bisects $ \\angle B'AC$ and $ B'C \\equal{} TC$, so $ ACTB'$ is cyclic. Similarly, $ BAC'T$ is cyclic. Thus, $ \\angle CTB' \\equal{} 180 \\minus{} \\angle B'AC \\equal{} 180 \\minus{} 2\\beta$ and $ \\angle BTC' \\equal{} 180 \\minus{} \\angle BAC' \\equal{} 180 \\minus{} 2\\alpha \\minus{} 2\\beta$. So, $ \\angle BTB' \\equal{} \\angle B'TC \\minus{} \\angle BTC \\equal{} 180 \\minus{} 2\\beta \\minus{} (180 \\minus{} 4\\beta \\minus{} 2\\alpha) \\equal{} 2\\alpha \\plus{} 2\\beta$. We can also find that $ \\angle CTC' \\equal{} \\angle BTC' \\minus{} \\angle BTC \\equal{} 180 \\minus{} 2 \\minus{} \\alpha \\minus{} 2\\beta \\minus{} (180 \\minus{} 4\\beta \\minus{} 2\\alpha) \\equal{} 2\\beta$. Thus, we find that $ \\angle B'TC' \\equal{} \\angle BTB' \\plus{} \\angle BTC \\plus{} \\angle CTC' \\equal{} 2\\beta \\plus{} 180 \\minus{} 4\\beta \\minus{} 2\\alpha \\plus{} 2\\beta \\plus{} 2\\alpha \\equal{} 180$, and thus $ B'$, $ T$, and $ C'$ are collinear and Lemma 1 is proven.\n\n[b]Lemma 2: $ AS'\\perp AT$[/b]\n[i]Proof:[/i] We notice that $ C'T \\equal{} TB' \\equal{} TC \\equal{} TB$ and thus $ CBB'C$ is cyclic since it has center $ T$. Furthermore, $ C'$, $ T$, and $ B$ are collinear, and thus $ C'B'$ is the diameter of the circumcircle of $ CB'BC'$. Thus, we find that $ \\angle SCB' \\equal{} \\angle BCB' \\equal{} \\frac {\\angle BTB'}{2} \\equal{} \\beta$. Also, we have isoceles triangle $ CTB'$ and $ \\angle CTB' \\equal{} 180 \\minus{} 2\\alpha \\minus{} 2\\beta$. Thus, $ \\angle CB'T \\equal{} \\alpha \\plus{} \\beta$. Now, we notice that\n\\[ \\angle CSB' \\plus{} \\angle SCB' \\equal{} \\angle CB'T\\implies \\beta \\plus{} \\angle CSB' \\equal{} \\alpha \\plus{} \\beta\\implies \\angle CSB' \\equal{} \\alpha \\equal{} \\angle BAB'\n\\]\nand thus $ \\angle BSB' \\equal{} \\angle BAB'$ thus giving us cyclic quaderilateral $ ASBB'$. Now, this implies that\n\\[ \\angle B'AS \\equal{} \\angle B'BS \\equal{} 180 \\minus{} \\angle B'BC \\equal{} \\angle TC'C \\equal{} \\frac {180 \\minus{} \\angle CTC'}{2} \\equal{} \\frac {180 \\minus{} (2\\alpha \\plus{} 2\\beta)}{2} \\equal{} 90 \\minus{} \\alpha \\minus{} \\beta\n\\]\nThus, we see that $ \\angle TAS \\equal{} \\angle TAB \\plus{} \\angle BAB' \\plus{} \\angle B'AS \\equal{} \\beta \\plus{} \\alpha \\plus{} 90 \\minus{} \\alpha \\minus{} \\beta \\equal{} 90\\implies TA\\perp AS'$.\n\n[b]Lemma 3: $ B' \\equal{} B_1$, $ C' \\equal{} C_1$, and $ S' \\equal{} S$. [/b]\n[i]Proof: [/i] We know that $ S$ is on $ BC$ and $ SA\\perp AT$ and we know that $ S'$ is on $ BC$ and that $ S'A\\perp BC$. Thus, we have that the intersection between the perpindicular to $ TA$ at $ A$ and $ BC$ is both $ S$ and $ S'$. Two lines only meet at one point, so $ S \\equal{} S'$. Furthermore, $ B'$ and $ C'$ are located on $ ST$ so that $ B \\equal{} B'$ and $ C \\equal{} C'$. Now, $ B_1$ and $ C_1$ therefore follow the same conditions as $ B'$ and $ C'$ respectively, so $ B_1 \\equal{} B'$ and $ C_1 \\equal{} C'$. \n\n[b]Solution to Problem[/b]\nFrom here, the problem finishes quite quickly. We notice that $ \\angle CAS \\equal{} \\angle TAS \\plus{} \\angle CAC' \\plus{} \\angle C'AT \\equal{} 90 \\plus{} \\alpha \\plus{} \\beta$. Also, we have already seen that $ \\angle TC'C \\equal{} 90 \\minus{} \\alpha \\minus{} \\beta$. Thus, $ \\angle SC'C \\plus{} \\angle CAS \\equal{} 90 \\plus{} \\alpha \\plus{} \\beta \\plus{} 90 \\minus{} \\alpha \\minus{} \\beta \\equal{} 180$. So, we obtain that $ CASC'$ is cyclic. Thus, $ \\angle ACB \\equal{} \\angle ACS \\equal{} \\angle AC'S \\equal{} \\angle AC'B'$. Similarly, we can show that $ \\angle ABC \\equal{} \\angle AB'C'$ and thus $ \\triangle ABC\\sim \\triangle AB'C'$. [/hide]\r\nThis is my 500th post :D !", "Solution_2": "This was 5... so as long as you got the (apparently much easier) 1,2,4, you made the team by solving this one. I'm told the intended solution uses symmedians.", "Solution_3": "[quote=\"MellowMelon\"]This was 5... so as long as you got the (apparently much easier) 1,2,4, you made the team by solving this one. I'm told the intended solution uses symmedians.[/quote]\r\n\r\nPretty much...I remember how distressed I was to find two geometry problems among the six total.\r\n\r\nAh..here we go...here's their solution:\r\n\r\n[hide]\nSo $ AT$ is a symmedian of the triangle...this is well-known and can be easily proven (I can prove it if people want).\n\nNow let $ M$ be the midpoint of $ BC$. Given that, we have\n$ m\\angle{TBA}\\equal{}180^{\\circ}\\minus{}m\\angle{C}$\nSo, by the Law of Sines,\n\\[ \\frac{BT}{AT}\\equal{}\\frac{\\sin{\\angle{BAT}}}{\\sin{\\angle{TBA}}}\\equal{}\\frac{\\sin{\\angle{CAM}}}{\\sin{\\angle{BCA}}}\\]\n\\[ \\equal{}\\frac{MC}{AM}\\]\nBut $ BT\\equal{}TC_1$, so $ \\frac{TC_1}{TA}\\equal{}\\frac{MC}{MA}$.\nSo because $ m\\angle{TMS}\\equal{}m\\angle{TAS}\\equal{}90^{\\circ}$, quadrilateral $ TMAS$ is cyclic. So $ m\\angle{AMC}\\equal{}m\\angle{AMS}\\equal{}m\\angle{ATS}\\equal{}m\\angle{ATC_1}$. By SAS similarity, then, we have $ \\triangle{AMC} \\sim \\triangle{ATC_1}$. Then because $ \\frac{BC}{MC}\\equal{}\\frac{B_1C_1}{TC_1}\\equal{}2$, we get again by $ SAS$ similarity that $ \\triangle{ABC} \\sim \\triangle{AB_1C_1}$, as desired.\n[/hide]", "Solution_4": "A very nice problem. This solution isn't nearly as nice as the official solution, it's a completely different approach: \r\nLemma $ 1$: Consider a circle centre $ O$, radius $ r$, and a point $ A$ outside the circle. Let a line $ l$, parallel to $ AO$, intersect the circle at $ D, E$. If the perpendicular from $ A$ to $ l$ meets $ l$ at a point $ B$, choose $ C$ so that $ A$ is the midpoint of $ BC$. Let $ AD$ meet the circle again at $ G$, and $ AE$ meet the circle again at $ F$. Then $ G, F, C$ are collinear. \r\nProof: Let's use inversion about a circle centre $ A$, radius $ \\sqrt {AD.AG}$. Clearly, this inversion switches $ D,G$; $ F,E$, suppose $ C$ is taken to a point $ C'$. Proving that $ G, F,C$ are collinear reduces to proving that their images lie on a circle through $ A$, i.e. that $ DECC'$ is cyclic, or equivalantly $ BA*BC' \\equal{} BD*BE$. But we note \r\n$ BD*BE \\equal{} BO^2 \\minus{} r^2 \\equal{} BA^2 \\plus{} AO^2 \\minus{} r^2 \\equal{} BA^2 \\plus{} AC*AC' \\equal{} BA^2 \\plus{} BA*AC' \\equal{} BA*(BA \\plus{} AC') \\equal{} BA*BC'$, \r\nas required. Here we use the fact that $ AC*AC' \\equal{} AO^2 \\minus{} r^2$, since both expressions are equal to the power of $ A$ with respect to the circle. \r\nNow let's solve the problem:\r\nLet the circle centre $ T$, radius $ TB \\equal{} TC$ intersect $ AB$ again at $ B_{1}$ and $ AC$ again at $ C_{1}$. A quick angle chase tells us that $ B_{1}, C_{1}, T$ are collinear, and moreover, $ AB_{1}C_{1}$ is similar to triangle $ ACB$. Let $ B_{3}, C_{3}$ be the reflections of $ B_{1}, C_{1}$ about the line $ AT$ - by symmetry, we now have that $ AC_{3}B_{3}$ is similar to $ ABC$, $ T$ is the midpoint of $ B_{3}C_{3}$, and $ B_{1}C_{3}$, $ B_{3}C_{1}, AT$ are parallel. Let $ AB_{3}, AC_{3}$ intersect our circle (the circle centre $ T$, radius $ BT$), again at $ B'$ and $ C'$ respectively. Let the perpendicular from $ A$ to $ C_{1}B_{3}$ intersect $ C_{3}B_{1}$ at a point $ X$. From Lemma $ 1$ applied to this configuration, we have: $ B', C, X$ are collinear. But from Pascal applied to the $ 6$ points $ B_{3}, C_{3}, B, C, B_ {1}, B'$, we have that the $ 3$ points, $ BC \\cap B_{3}C_{3}$, $ BB_{1} \\cap B_{3}B'$, $ CB' \\cap C_{3}B_{1}$, are collinear - i.e. $ A, X$ and $ B_{3}C_{3} \\cap BC$ are collinear. If $ S'\\equal{}B_{3}C_{3} \\cap BC$ , we then have $ \\angle TAS' \\equal{} 90$. Since $ S'$ lies on $ BC$, going back to the problem statement, we have that, $ S' \\equal{} S$. We also, the $ B_{3}$ in this diagram is $ B_{1}$ in the problem statement (not the $ B_{1}$ in my figure, sorry for the confusion.), and $ C_{3}$ in this diagram, is $ C_{1}$ in the problem statement. Since $ AB_{3}C_{3}$ is similar to $ ABC$ (already proven), the conclusion follows.", "Solution_5": "Nice problem! :blush: Could anyone tell me the author of this problem?\r\nSome more properties:\r\nLet $ SA \\cap B_1C\\equal{}{K}$.Let K,A,B,C lie on a circle.\r\n\r\nAnd a similar one:\r\nLet $ BB_1 \\cap CC_1\\equal{}{L}$.Then L is lie on the symmedian wrt triangle ABC and on the circumcircle of triangle ABC.", "Solution_6": "Let $\\Gamma$ be the circle centered at $T$ with radius $TB$. Extend $AB$ and $AC$ to intersect $\\Gamma$ at $B_2$ and $C_2$, respectively. It is clear that $\\triangle ABC$ and $\\triangle AC_2B_2$ are directly similar. Since $BC$ and $B_2C_2$ are antiparallel with respect to $AB$ and $AC$ and $AT$ is a symmedian of $\\triangle ABC$, $AT$ is a median of $\\triangle AC_2B_2$, i.e., $T$ is the midpoint of $B_2 C_2$. \n\nReflect $B_2$ and $C_2$ across $AT$ to get $C_1'$ and $B_1'$, respectively. Since this reflection maps $\\Gamma$ to itself, $B_1'$ and $C_1'$ both lie on $\\Gamma$, and $B_1'$, $T$, and $C_1'$ are collinear. Since $\\triangle AC_2 B_2$ and $\\triangle AB_1' C_1'$ are directly similar, $\\triangle ABC$ and $\\triangle AB_1' C_1'$ are directly similar. \n\nLet $BC$ intersect $B_1' C_1'$ at $S'$ (which may be infinite.) Because $S' = BC \\cap B_1' C_1'$, $T$ is the center of $BCC_1' B_1'$, and $A$ is the Miquel point of $BCC_1'B_1'$, we have $TA \\perp AS'$. Hence, $S' = S$, whence $B_1 = B_1'$ and $C_1 = C_1$', so $\\triangle ABC \\sim \\triangle AB_1C_1$, as desired.", "Solution_7": "Let $D$ and $E$ be on $AB$ and $AC$ such that $BCED$ is cyclic and $DE$ passes through $T$. By tangent angle theorem, $\\angle{BCT}=\\angle{CBT}=\\angle{BAC}$. Hence $\\angle{BDT}=\\angle{ACB}=180^\\circ - \\angle{CBT} - \\angle{ABC}=\\angle{TBD}$. Hence $TD=TB=TC$. By the same argument, it follows that $TD=TB=TC=TE$. Let $M$ be the midpoint of $BC$. Since $BCED$ is cyclic, triangles $ABC$ and $AED$ are similar. Since $T$ is the midpoint of $DE$, it follows that $TB/TA=TD/TA=MC/MA$. Since $BTC$ is isosceles, $TM \\perp BC$ which implies that $TMAS$ is cyclic. Hence $\\angle{AMS}=\\angle{ATS}$ which since $C_1 T/TA=BT/TA=MC/MA$ implies that triangles $ATC_1$ and $AMC$ are similar. By the same argument, triangles $ATB_1$ and $AMB$ are similar. This implies that $ABC$ and $AB_1 C_1$ are similar.", "Solution_8": "Dear Mathlinkers,\nthe circle centered at T passing through B1, C1, B and C leads to the Boutin's theorem... then we can think to the Liquel situation... in order to have a synthetic proof.\nSincerely\nJean-Louis", "Solution_9": "Not so easy!", "Solution_10": "[quote=\"Zhero\"]Let $\\Gamma$ be the circle centered at $T$ with radius $TB$. Extend $AB$ and $AC$ to intersect $\\Gamma$ at $B_2$ and $C_2$, respectively. It is clear that $\\triangle ABC$ and $\\triangle AC_2B_2$ are directly similar. Since $BC$ and $B_2C_2$ are antiparallel with respect to $AB$ and $AC$ and $AT$ is a symmedian of $\\triangle ABC$, $AT$ is a median of $\\triangle AC_2B_2$, i.e., $T$ is the midpoint of $B_2 C_2$. \n\nReflect $B_2$ and $C_2$ across $AT$ to get $C_1'$ and $B_1'$, respectively. Since this reflection maps $\\Gamma$ to itself, $B_1'$ and $C_1'$ both lie on $\\Gamma$, and $B_1'$, $T$, and $C_1'$ are collinear. Since $\\triangle AC_2 B_2$ and $\\triangle AB_1' C_1'$ are directly similar, $\\triangle ABC$ and $\\triangle AB_1' C_1'$ are directly similar. \n\nLet $BC$ intersect $B_1' C_1'$ at $S'$ (which may be infinite.) Because $S' = BC \\cap B_1' C_1'$, $T$ is the center of $BCC_1' B_1'$, and $A$ is the Miquel point of $BCC_1'B_1'$, we have $TA \\perp AS'$. Hence, $S' = S$, whence $B_1 = B_1'$ and $C_1 = C_1$', so $\\triangle ABC \\sim \\triangle AB_1C_1$, as desired.[/quote]\n\n how you prove $A$ is Miquel of $BCC_1B_1$ very nice.", "Solution_11": "If $K = \\overline{BB_1} \\cap \\overline{CC_1}$ then $\\angle BKC = \\tfrac{\\widehat{B_1C_1} - \\widehat{BC}}{2} = \\tfrac{1}{2} \\left( 180 - \\angle BTC \\right) = \\angle BAC$, so $B,K,A,C$ are concyclic. So $A$ is the unique point with $AS \\perp AT$ and $A \\in (KBC)$; therefore $A$ is the Miquel point of $B_1BCC_1$, i.e. the center of the spiral similarity giving $\\triangle ABC \\sim \\triangle AB_1C_1$.\n\n[asy]/* DRAGON 0.0.9.6 \nHomemade Script by v_Enhance. */\n\nimport olympiad; \nimport cse5; \nsize(12cm); \nreal lsf=0.8000; \nreal lisf=2011.0; \ndefaultpen(fontsize(10pt));\n\n/* Initialize Objects */\npair B_1 = (3.0, -3.0);\npair C_1 = (11.0, -3.0);\npair T = midpoint(B_1--C_1);\npair B = (6.451128850753037, 0.9621636086265188);\npair C = (10.082123836968163, -0.45038970554465996);\npair S = IntersectionPoint(Line(B,C,lisf),Line(B_1,C_1,lisf));\npair K = IntersectionPoint(Line(B,B_1,lisf),Line(C,C_1,lisf));\npair A = foot(T,K,S);\n\n/* Draw objects */\ndraw(CirclebyPoint(T,B_1), rgb(0.0,0.6,0.6));\ndraw(circumcircle(A,B,C), rgb(0.0,0.6,0.8) + linetype(\"4 4\"));\ndraw(T--B, rgb(0.0,0.8,0.8));\ndraw(T--C, rgb(0.0,0.8,0.8));\ndraw(K--S, rgb(0.0,0.6,0.8));\ndraw(circumcircle(K,B_1,C_1), rgb(0.0,0.8,0.8) + linewidth(1.0) + dotted);\ndraw(T--A, rgb(0.0,0.8,0.8) + linetype(\"4 4\"));\ndraw((abs(dot(unit(T-A),unit(S-A))) < 1/2011) ? rightanglemark(T,A,S) : anglemark(T,A,S), rgb(0.0,0.8,0.8));\ndraw(B--B_1, rgb(1.0,0.6,0.0) + linewidth(1.0));\ndraw(B_1--C_1, rgb(1.0,0.6,0.0) + linewidth(1.0));\ndraw(C_1--C, rgb(1.0,0.6,0.0) + linewidth(1.0));\ndraw(C--B, rgb(1.0,0.6,0.0) + linewidth(1.0));\ndraw(K--B, rgb(0.0,0.8,0.8) + dashed);\ndraw(K--C, rgb(0.0,0.8,0.8) + dashed);\ndraw(C--S, rgb(0.0,0.8,0.8) + dashed);\ndraw(C_1--S, rgb(0.0,0.8,0.8) + dashed);\n\n/* Place dots on each point */\ndot(B_1);\ndot(C_1);\ndot(T);\ndot(B);\ndot(C);\ndot(S);\ndot(K);\ndot(A);\n\n/* Label points */\nlabel(\"$B_1$\", B_1, lsf * dir(225));\nlabel(\"$C_1$\", C_1, lsf * dir(-45));\nlabel(\"$T$\", T, lsf * dir(-90));\nlabel(\"$B$\", B, lsf * dir(140));\nlabel(\"$C$\", C, lsf * dir(70) * 1.414);\nlabel(\"$S$\", S, lsf * dir(-45));\nlabel(\"$K$\", K, lsf * dir(60));\nlabel(\"$A$\", A, lsf * dir(45));\n[/asy]", "Solution_12": "But how do you know $K, A$ and $S$ are collinear?", "Solution_13": "Sorry, $TA \\perp AS$ was meant rather than $TA \\perp KS$. We don't need this collinearity at all since Miquel points does this all for us. \n\nI will edit my post when I get home, seems I am unable to do so on my phone.", "Solution_14": "Let $X$ be the intersection of $\\omega$ and $AT$ and let $BX$ intersect $ST$ at $Y$ and $CX$ intersect $ST$ at $Z$ and let the foot of perpendicular from $X$ on $BC$ be $R$.Not hard to observe that $\\angle ABX =\\angle ARC,, .ARXS$ is cyclic $\\Rightarrow \\angle ABX=\\angle ATY \\Rightarrow ABTY$ is cyclic.$\\Rightarrow \\angle XBT=\\angle XAB=\\angle BYT \\Rightarrow Y \\equiv C_1$ .Similarly $Z \\equiv B_1.,, \\Rightarrow \\angle C_1BB_1=90 ,\\angle SBB_1=90+\\angle C_1BS=90+\\angle XB_1C_1=90+\\angle XAB_1=\\angle B_1AS \\Rightarrow ABB_1S$is cyclic.Similarly $ACC_1S$ is cyclic. $\\Rightarrow \\angle ABC=\\angle AB_1C_1$ and $\\angle ACB=\\angle AC_1B_1$ and we are done.", "Solution_15": "dame dame", "Solution_16": "Let $Q$ - Miquel point of $BCC_1B_1$. With $M,T$ being midpoints of $BC, B_1C_1$, by gliding principle we know that $Q\\in (SMT)$. Also, with $T$ being circumcenter of $BCC_1B_1$ we know that $\\angle SQT=90^\\circ$.\nClearly $A\\in (SMT)$ ($ST$ - diameter) and $\\angle SAT=90^\\circ$ by the statement. Hence $A\\equiv Q$ and that is all we need.\n\n[hide=fun stuff]That is what I originally wrote:\nWith $M$ - midpoint of $BC$, clearly $M\\in (ATS)$. Gliding principle + $\\angle TAS=90$ give us that $A$ - Miquel point of $BCC_1B_1$.\n\nBut then decided that some explanations won't hurt. Trying to find the right balance.[/hide]", "Solution_17": "[hide=an elementary proof]WLOG, $AB \\ge AC$, so will now not worry about configuration issues. Let $M$ be the mid-point of side $BC$. Then $TM \\bot SM$. So points $S,T,M,A$ lie on the circle with diameter $ST$. Note that it suffices to show $\\triangle ACC_1 \\sim \\triangle ABB_1$. Let $\\angle CTC_1 = 2 \\alpha,\\angle BTB_1 = 2 \\beta$. Now, an easy angle chase yields $\\angle ACC_1 = \\angle ABB_1$. So it suffices to show $\\frac{CC_1}{BB_1} = \\frac{AC}{AB}$. Now, $\\frac{CC_1}{BB_1} = \\frac{\\sin \\alpha}{\\sin \\beta}$. Also, we have that\n\\begin{align*}\n & 2 \\alpha = \\angle STC = \\angle STA - \\angle CTA = \\angle SMA - (\\angle B - \\angle CAT) = (\\angle SMA - \\angle B) + \\angle CAT = \\angle BAM + \\angle CAT = 2 \\angle BAM \\\\\n& \\implies 2 \\beta = 2 \\angle A - 2 \\alpha = 2 \\angle CAM\n\\end{align*}\nSo we have that\n$$\\frac{CC_1}{BB_1} = \\frac{\\sin \\alpha}{\\sin \\beta} = \\frac{\\sin \\angle BAM}{\\sin \\angle CAM} = \\frac{MB \\cdot \\frac{\\sin \\angle AMB}{AB}}{MC \\cdot \\frac{\\sin \\angle AMC}{AC}} = \\frac{AC}{AB} $$\nas desired. $\\blacksquare$[/hide]", "Solution_18": "It's easy to see $BCB_1C_1$ is cyclic with center $T$ and diameter $B_1C_1$. Now, let $BC_1 \\cap CB_1 = A_1$. It suffices to show $A \\in (A_1BC), (A_1B_1C_1) = \\Omega$, as the Spiral Center Lemma would imply $ABC \\sim AC_1B_1$.\n\nBecause $\\angle B_1BC_1 = \\angle C_1CB_1 = 90^{\\circ}$, we know $B_1B \\cap C_1C = H$ is the orthocenter of $A_1B_1C_1$.\n\n[b][color=#f00]Claim:[/color][/b] $H$ lies on $\\omega$.\n\n[i]Proof.[/i] We know $T$ is the midpoint of $B_1C_1$ and $TB, TC$ are tangent to $\\omega$. Because $B$ and $C$ are feet of altitudes of $A_1B_1C_1$, the Three Tangents Lemma applied to $A_1B_1C_1$ implies the desired result. $\\square$\n\nThus, we know $A_1$ also lies on $\\omega$.\n\nNow, let the second intersection of $\\omega$ and $\\Omega$ be $A_2$. It suffices to show $A_2 = A$.\n\n[b][color=#f00]Claim:[/color][/b] $A_1A_2$, $BC$, $B_1C_1$ concur at $S$.\n\n[i]Proof.[/i] Applying the Radical Axis Theorem to $\\omega$, $\\Omega$, $(BCB_1C_1)$ yields the desired result. $\\square$\n\nThus, $S, A_2, A_1$ are collinear. Because $AS \\perp AT$, it suffices to show $\\angle SA_2T = 90^{\\circ}$ (which is equivalent to $\\angle A_1A_2T = 90^{\\circ}$ due to the collinearity).\n\nObserve $\\angle A_1AH = 90^{\\circ}$, as $A \\in \\omega$ and $A_1H$ is a diameter by orthocenter configuration properties. Now, define $A_1'$ to be the antipode of $A_1$ wrt $\\Omega$. It follows that $\\angle A_1AA_1' = 90^{\\circ}$, so $A, H, A_1'$ are collinear. \n\nBecause $H$ is the orthocenter of $A_1B_1C_1$ inscribed in $\\Omega$, we know $T \\in HA_1'$ by the Orthocenter Reflection Lemma. Thus $\\angle A_1AT = \\angle A_1AH = 90^{\\circ}$ as desired. $\\blacksquare$\n\n\n[b]Disclaimer:[/b] My solution to this problem is horrendous. Do not attempt to replicate at home. \n\nUnfortunately, I constructed $S$ on the incorrect ray, so part of my diagram is essentially flipped. As a result, the similarity in my diagram is actually $ABC \\sim AC_1B_1$. \n\nAlso, my (incorrect) GGB can be found [url=https://www.geogebra.org/geometry/fbecrxas]here[/url]!", "Solution_19": "[asy] /* coordinates from geogebra */\nimport geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(3,6); B=(-5,-1); C=(4,-1); T=(-0.5,-7.9); S=(30.8,-1); B_1=(-8.6,-9.7); C_1=(7.6,-6.1); D=(-1.7,7.2); fill(A--B--C--cycle,purple+white+white+white); fill(B--C--C_1--B_1--cycle,blue+white+white+white+white); draw(circle(A,B,C),dashed); draw(circle(B_1,C_1)); draw(B--C--C_1--B_1--cycle,blue); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,pink+red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C); dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label(\"$A$\",A,NE); label(\"$B$\",B,2W); label(\"$C$\",C,NE); label(\"$T$\",T,SW); label(\"$S$\",S,2E); label(\"$B_1$\",B_1,SW); label(\"$C_1$\",C_1,SE); label(\"$D$\",D,2N); [/asy]\n\nLet $D=\\overline{BB_1}\\cap\\overline{CC_1}.$ Notice that $$\\measuredangle BDC=\\pi+\\measuredangle DB_1C_1+\\measuredangle DC_1B_1=\\frac{\\pi+\\measuredangle BTC}{2}=\\measuredangle BAC$$ since $T$ is the center of $(BB_1C_1C).$ Hence, $ACBD$ is cyclic. Also, $\\angle TAS=90,$ so $A$ it is the Miquel Point of $BB_1C_1C$ and therefore the center of a spiral similarity that maps $\\overline{BC}$ and $\\overline{B_1C_1}.$ $\\square$", "Solution_20": "\nNotice that $BT=B_1T=CT=C_1T$ which implies $T$ is the circumcentre of $BCC_1B_1$\nLet $BB_1 \\cap CC_1=X$ and we know that $BC \\cap B_1C_1=S$\nBy angle chasing,\nLet $\\angle BB_1T=x$ and $\\angle CC_1T=y$\n$$\\angle BTC=2x+2y-180 \\implies \\angle CBT=\\angle BCT=180-(x+y)=\\angle BAC=\\angle BXC$$\nwhere the last part comes from the cyclic quadrilateral.\nand therefore $X \\in \\odot(ABC)$\nClubbing all the information it follows that, $A$ is the Miquel point of cyclic quadrilateral $BCC_1B_1$ and now taking spiral symmetry we get that $\\triangle ABC \\sim \\triangle A_1B_1C_1$ as desired.\n", "Solution_21": "Easy one\nLet $BC_1$ and $CB_1$ meet at $K$. $\\angle BKC = \\angle 180 - \\angle BC_1T - \\angle CB_1T = \\angle TBC = \\angle BAC$ so $BAKC$ is cyclic so $A$ is Miquel point of $BCB_1C_1$. Now the problem is almost Done the rest is just angle chasing...", "Solution_22": "Let $F=SA\\cap(ABC)$, $L=AT\\cap(ABC)$, $Q=FL\\cap BC$ and $M=SL\\cap (ABC)$\n\\\\ Claim 1. M-Q-A\n\\\\ Pf. Notice that $$-1=(A,L;B,C)\\stackrel{F}{=}(S,Q;B,C)\\stackrel{L}{=}(M,F;B,C)\\stackrel{A}{=}(MA\\cap BC,S;B,C)$$ which implies that $MA\\cap BC=Q$\n\n\\\\ See how $(M,F;B,C)=-1$ also shows that $T-M-F$, and now by looking at cyclic quadrilateral $(FALM)$ by Brocard we get that $ST$ is the polar of $Q$. Thus, if we let $O$ be the circumcenter of this circumference, then $OQ\\perp ST$, but as $\\angle LAF= 90$ we get that $FL\\perp ST$. Deffine $N=FL\\cap ST$, $C_1*=FC\\cap ST$, and $B_1*=C_1*L\\cap ST$. By Brocard on $(BFCL)$ we conclude $B_1*-B-F$, and thus $L$ is the orthocenter of $FC_1*B_1*$, which implies $C_1*=C_1$ and $B_1*=B_1$. \n\\\\ Now we're done as $L$ and $A$ are inverses with respect to $(C_1BCB_1)$, and now it is well-known that $A$ is the Miquel point that sends $BC$ to $B_1C_1$ which implies the desired result.", "Solution_23": "A quick angle chasing gives that $M=CC_1\\cap BB_1$ lies on $(ABC)$. We will prove that if $N=CB_1\\cap BC_1$, then $A, T, N$ are collinear, implying that $A$ is a Miquel point of $C_1CBB_1$.\nLet $AB\\cap ST=P, AT\\cap (ABC)=N_1$ and $BN_1\\cap ST=C_0$. Then $-1=(B,C;A,N_1)\\stackrel{B}{=} (T,S;P,C_0)$ and $\\angle SAT=90^\\circ$,so $AT$ is a bisector of $\\angle BAC_0$. Then $\\angle C_0AT=\\angle TAB=\\angle N_1BT=\\angle C_0BT$, so $AC_0TB$ is cyclic, implying that $TC_0=TB$, so \n$C_0\\equiv C_1$.Similarly, $CB_1\\cap AT\\in (ABC)$, so $N=N_1$ and $N\\in AT, N\\in (ABC)$, so we are done.", "Solution_24": "Not hard at all.\n\nSuppose that [i]B[size=50]2[/size][/i] and [i]C[size=50]2[/size][/i] are the intersection points of [i]ST[/i] and [i](SAB)[/i] and [i](SAC)[/i]. Let [i]M[/i] be the midpoint of [i]BC[/i]. Since that [i]TM[/i] is perpendicular to [i]BC[/i], we have that [i]ASTM[/i] is cyclic.\n\nIt's well-known that [i]AT[/i] is a symmedian of triangle [i]ABC[/i]. After that it's easy using just anglechase prove that [i]TC[size=50]2[/size]=BT=TB[size=50]2[/size][/i].", "Solution_25": "[asy] /* coordinates from geogebra */\nimport geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(-0.89,1.11); B=(1.96,-1.44); C=(-1.44,-1.06); T=(-0.1,-4.5); S=(-9.76,-0.13); B_1=(3.26,-6.02); C_1=(-3.47,-2.98); D=(1.16,1.4); fill(A--B--C--cycle,white+white); fill(B--C--C_1--B_1--cycle,white+white+white+white); draw(circle(A,B,C)); draw(circle(B_1,C_1)); draw(circle(A,C_1,B_1));draw(C_1--D--B_1--cycle,green); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C);draw(S--C_1);dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label(\"$A$\",A,NE); label(\"$B$\",B,2W); label(\"$C$\",C,NE); label(\"$T$\",T,SW); label(\"$S$\",S,2E); label(\"$B_1$\",B_1,SW); label(\"$C_1$\",C_1,SE); label(\"$D$\",D,2N); [/asy]\n$B_1B \\cap C_1C= D$ . Chase, $\\angle BDC= 180^{\\circ}-(\\angle DB_1T +DC_1T) = \\frac{180^{\\circ}- \\angle BTC}{2}=\\angle BCT$.Thus, $ACBD$ is cyclic. As $T$ is the center$(B_1C_1CB)$ , and as $A,D,S$ are collinear by radical axes ,we use [url=https://artofproblemsolving.com/community/c6h60787p366594]IMO 1985/5 [/url], which proves $A$ is the miquel point of the complete quad $B_1C_1CBDS $. The property that the miquel point is the centre of the spiral sim. which sends $BC$ to $B_1C_1$ finishes the proof.", "Solution_26": "Thanks to [b]SatisfiedMagma[/b] for teaching me how to make good asy diagrams. :)\n\n[asy]\nimport graph; size(12cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -2, xmax = 4, ymin = -4.7, ymax = 2; /* image dimensions */\n\n\ndraw((0.33709926477356134,0.6527195501031977)--(0.2983270446038727,0.5051548519589105)--(0.44589174274816,0.4663826317892219)--(0.4846639629178486,0.6139473299335092)--cycle, linewidth(0.75) + blue); \n /* draw figures */\ndraw((0.4846639629178486,0.6139473299335092)--(-0.6435760050714912,-0.10350482168263458), linewidth(0.5)); \ndraw((-0.6435760050714912,-0.10350482168263458)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); \ndraw((0.5736726499363176,-0.10910136722290036)--(0.4846639629178486,0.6139473299335092), linewidth(0.5)); \ndraw(circle((-0.033620893008616766,0.18314254712320893), 0.673952485546976), linewidth(0.5)); \ndraw((-0.6435760050714912,-0.10350482168263458)--(-0.04083567317250101,-1.386072138521609), linewidth(0.5)); \ndraw((-0.04083567317250101,-1.386072138521609)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); \ndraw((-0.04083567317250101,-1.386072138521609)--(0.4846639629178486,0.6139473299335092), linewidth(0.5)); \ndraw((-0.2732001930556983,0.8130740606736085)--(3.283969388430581,-0.12156250165275902), linewidth(0.5)); \ndraw((3.283969388430581,-0.12156250165275902)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5)); \ndraw(circle((-0.04083567317250101,-1.386072138521609), 1.4171361366930244), linewidth(0.5)); \ndraw((-0.2732001930556983,0.8130740606736085)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5)); \ndraw((1.2837365045968172,-0.8823029693571232)--(-0.2732001930556983,0.8130740606736085), linewidth(0.5)); \ndraw((-1.3654078509418188,-1.8898413076860952)--(3.283969388430581,-0.12156250165275902), linewidth(0.5)); \ndraw((-0.2732001930556983,0.8130740606736085)--(0.48686186746905835,-1.1853750711357969), linewidth(0.5)); \ndraw((-0.6435760050714912,-0.10350482168263458)--(1.2837365045968172,-0.8823029693571232), linewidth(0.5) + linetype(\"4 4\")); \ndraw((0.5736726499363176,-0.10910136722290036)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5) + linetype(\"4 4\")); \ndraw(circle((1.9313054361793436,0.42502876216911906), 1.4589248703776139), linewidth(0.5) + blue); \ndraw(circle((1.3118282790787361,-1.9326634023434865), 2.677578575444292), linewidth(0.5) + blue); \n /* dots and labels */\ndot((0.4846639629178486,0.6139473299335092),linewidth(3pt) + dotstyle); \nlabel(\"$A$\", (0.5270237491555813,0.7436458119447124), NE * labelscalefactor); \ndot((-0.6435760050714912,-0.10350482168263458),linewidth(3pt) + dotstyle); \nlabel(\"$B$\", (-0.961798164673135,-0.08347747351568333), NE * labelscalefactor); \ndot((0.5736726499363176,-0.10910136722290036),linewidth(3pt) + dotstyle); \nlabel(\"$C$\", (0.6564865242711218,-0.04032321514383659), NE * labelscalefactor); \ndot((-0.04083567317250101,-1.386072138521609),linewidth(3pt) + dotstyle); \nlabel(\"$T$\", (-0.10590537363150583,-1.6298383985068579), NE * labelscalefactor); \ndot((-0.2732001930556983,0.8130740606736085),linewidth(3pt) + dotstyle); \nlabel(\"$Q$\", (-0.4511394406062806,0.9090704690367915), NE * labelscalefactor); \ndot((3.283969388430581,-0.12156250165275902),linewidth(3pt) + dotstyle); \nlabel(\"$S$\", (3.3680124253021653,-0.07628509712037554), NE * labelscalefactor); \ndot((-0.033620893008616766,0.18314254712320893),linewidth(3pt) + dotstyle); \nlabel(\"$O$\", (-0.005212104097196526,0.18264045311070484), NE * labelscalefactor); \ndot((0.20595840703846502,-0.4467889664271907),linewidth(3pt) + dotstyle); \nlabel(\"$P$\", (0.23213631694796114,-0.3639801529326871), NE * labelscalefactor); \ndot((-1.3654078509418188,-1.8898413076860952),linewidth(3pt) + dotstyle); \nlabel(\"$B_1$\", (-1.7529595681569938,-1.9822648418769395), NE * labelscalefactor); \ndot((1.2837365045968172,-0.8823029693571232),linewidth(3pt) + dotstyle); \nlabel(\"$C_1$\", (1.4044936693831338,-1.0616406632775426), NE * labelscalefactor); \ndot((0.48686186746905835,-1.1853750711357969),linewidth(3pt) + dotstyle); \nlabel(\"$S^*$\", (0.48386949078373437,-1.4068747302523164), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n[/asy]\n\nLet $Q$ denote the $P$ antipode in $\\odot(ABC)$.\n\nNow firstly we thus have that $\\overline{Q-A-S}$ are collinear. Now Pascal on $BBPCCQ$ gives that $\\overline{T-BP\\cap CQ-CP\\cap BQ}$ are collinear and then Pascal on $QAPBCC$ gives that $\\overline{S-T-BP\\cap QC}$ are collinear and thus we conclude that $BP\\cap QC$ and $CP\\cap BQ$ lie on $ST$.\n\nWe redefine $B_1=QB\\cap CP$ and $C_1=BP\\cap QC$.\n\nNow note that, $\\measuredangle B_1CC_1=\\measuredangle PCC_1=\\measuredangle PCQ=90^\\circ$ and similarly that $\\measuredangle B_1BC_1=90^\\circ$. So we have that the center of $\\odot(BCC_1B_1)$ is the midpoint of $B_1C_1$ and that it lies on the perpendicular bisector of $BC$ and thus $T$ becomes the center. This indeed matches the definitions of the points $B_1$ and $C_1$ as in the problem statement.\n\nNow, we have, $\\measuredangle CAS=\\measuredangle CAQ=\\measuredangle CBQ=\\measuredangle CBB_1=\\measuredangle CC_1B_1=\\measuredangle CC_1S\\implies ACC_1S$ is cyclic. Similarly, $ABB_1S$ is also cyclic.\n\nNow note that the circles $\\left\\{\\odot(BCB_1C_1),\\odot(ABC)\\right\\}$ are orthogonal to each other. So upon performing an Inversion $\\mathbf I(\\odot(BCC_1B_1))$, we get that $A\\leftrightarrow P$. Now $S^*$ denote the inverse of $S$ after the Inversion. Note that then as we have $ACC_1S$ is cyclic, we thus have $PCC_1S^*$ is also cyclic, and similarly $PBB_1S^*$ is also cyclic.\n\nNow a Radax on $\\left\\{\\odot(PBB_1S^*),\\odot(PCC_1S^*),\\odot(BCC_1B_1)\\right\\}$ gives that $\\overline{Q-P-S^*}$ are collinear. We finally have $SA\\cdot SQ=SC\\cdot SB=SC_1\\cdot SB_1\\implies AQB_1C_1$ is also cyclic.\n\nThus to finish, we have that as $Q=BB_1\\cap CC_1$, thus as $A=\\odot(QBC)\\cap\\odot(QB_1C_1)$, it becomes the miquel point mapping $\\overline{BC}\\mapsto\\overline{B_1C_1}$ and thus we are done.", "Solution_27": "Let $D = \\overline{BB_1} \\cap \\overline{CC_1}$ and $E = \\overline{B_1C} \\cap \\overline{BC_1}$. Note that $T$ is the center of $(BB_1C_1C)$ so $\\overline{B_1B} \\perp \\overline{BC_1}$ and $\\overline{B_1C} \\perp \\overline{CC_1}$. Hence $E$ is the orthocenter of $\\triangle KB_1C_1$, and $\\overline{TB}$ and $\\overline{TC}$ are tangent to $(DBEC)$ by three tangents lemma, which is enough to imply that $(DBEC)$ and $(ABC)$ coincide. \n\nThus, we now know that $A$ is one of the intersections of $(ST)$ with $(DBEC)$; denote these two intersections as $A_1$ and $A_2$. Then it's well-known that $A_1$ and $A_2$ are just the $K$-Humpty and Queue points of $\\triangle DB_1C_1$ (as $S$ is the $D$-ex point); WLOG assume that $A_1$ is the Humpty point and $A_2$ is the Queue point. If $A = A_1$, then $A$ lies on $(EB_1C_1)$ so $\\triangle ABC \\sim \\triangle AC_1B_1$. If $A = A_2$, then $A$ lies on $(DB_1C_1)$ so $\\triangle ABC \\sim \\triangle AB_1C_1$.", "Solution_28": "hello\n\nLet $G=BB_1\\cap CC_1$, notice that we have the good old orthocenter configuration hence $G\\in \\omega$. Let $A'$ be the foot of the altitude from $T$ to $SG$, it's well-known that $A'$ is the Miquel Point of $BCC_1B_1$. Hence $A'=\\omega\\cap (ST)$ which is also just $A$ so we're done??\n\nWait I think I fakesolved", "Solution_29": "Note that $BCB_1C_1$ is cyclic with circumcenter $T$. let $P$ be the intersection of $BC_1$ and $CB_1$.\\\\\n\nIt is well known that $TB$ and $TC$ are tangent to $(PBC)$ (this can be shown by angle chasing and showing that $\\angle BTC=2\\angle B_1+2\\angle C_1-180$). Hence, $P$ lies on $(ABC)$.\\\\\n\nThus, $A$ lies on the circle with diameter $ST$ as well as the $(PBC)$. Thus, it is the Miquel point, which establishes the spiral similarity, as desired." } { "Tag": [ "ratio", "complex analysis", "complex analysis unsolved" ], "Problem": "Prove that $ \\mathbb{C}\\setminus \\{ 0,1,2 \\}$ is not conformally isomorphic to $ \\mathbb{C}\\setminus \\{ 0,1,3 \\}$", "Solution_1": "What's a conformal isomorphism?", "Solution_2": "A conformal isomorphism is a 1-1 conformal map between sets; a conformal map is one that preserves angles between curves. This angle-preserving condition is equivalent to acting as multiplication by a nonzero complex number at a point (in $ \\mathbb{R}^{2}$), so conformal maps in the plane are complex analytic functions.", "Solution_3": "Assume to the contrary that there exists a conformal isomorphism $ f:{\\mathbb C}\\setminus \\{0,1,2\\}\\to{\\mathbb C}\\setminus \\{0,1,3\\}$. Note that each of the points $ 0, 1,2$ and $ \\infty$ is either a simple pole or a removable singularity of $ f$. Therefore, $ f$ can be extended to a conformal automorphism of $ {\\mathbb C}\\cup \\{\\infty\\}$. Hence $ f$ is a M\u00f6bius transformation. So it must preserve the cross ratio. Since $ f$ maps the set $ \\{0, 1, 2, \\infty\\}$ to $ \\{0, 1, 3, \\infty\\}$, the cross ratio of points in the former set equals the cross ratio of the points in the latter set (appropriately ordered).\r\nNow the cross ratio of $ 0, 1, 2, \\infty$ equals\r\n\\[ (0, 1; 2, \\infty) = \\frac{0-2}{1-2}= 2. \\]\r\nThe cross ratio of points $ 0$, $ 1$, $ 3$, $ \\infty$ equals (depending on the order):\r\n\\[ \\frac{0-1}{0-3}= \\frac13 \\text{ or }\\frac{0-3}{0-1}= 3 \\text{ or }\\frac{1-3}{1-0}=-2 \\text{ or }-\\frac{1}{2}\\text{ or }\\frac{3-0}{3-1}= \\frac{3}{2}\\text{ or }\\frac{2}{3}. \\]\r\nNone of these numbers equals $ 2$. We got a contradiction." } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO" ], "Problem": "When writing a proof for an inequality on the USAMO, is it necessary to state when equality occurs even if the problem doesn't specifically ask to do so?\r\n\r\nThanks.", "Solution_1": "I don't think it does, but that is just my opinion. I think they care more about how you arrive at the inequality itself.", "Solution_2": "You are not required to write equality cases unless the problem asks for them. (The problem-writers and graders do not assume that the test-takers are psychic.)", "Solution_3": "I agree with the responses. Figuring out the equality cases, however, can be useful in trying to prove an inequality. They can sometimes suggest promising approaches and rule out unpromising ones. That's often one of the first things I do when facing an inequality.", "Solution_4": "Yeah, sometimes it's helpful to find the equality cases. But sometimes it's harder to find the equality cases than it is to prove that inequality holds.", "Solution_5": "Ok, thanks for the clarification.", "Solution_6": "[quote=\"Ravi B\"]I agree with the responses. Figuring out the equality cases, however, can be useful in trying to prove an inequality. They can sometimes suggest promising approaches and rule out unpromising ones. That's often one of the first things I do when facing an inequality.[/quote]\r\n\r\nThat was one of the things that threw me off with zscool's last posted inequality - I could have sworn that there were no equality conditions.", "Solution_7": "It's probably reasonably stylish to toss in the equality conditions at the end, but I wouldn't do that unless you'd already proven them inside your work -- no point to doing additional work that count for anything." } { "Tag": [ "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "I am interested in knowing the degree of curvature of the earth's\r\nsurface. I am attempting to find a method which would allow me to \r\nconceptualize and calculate the making of such a depiction on paper.\r\n\r\nThat is to say, if the earth curves from point A to point B, points \r\nexactly 10 kilometers apart, if one were to draw a straight line \r\nbetwen the two points, then WHAT WOULD BE THE HEIGHT of the highest \r\npoint the curvature AB would be above the line AB? Moreover, if the \r\npoint AB were now to be 20 kilometers apart, would the height be 2 \r\ntimes the height calculated earlier or would it be something \r\naltogether different?", "Solution_1": "why is this here? do you want us to do it?", "Solution_2": "radius of the earth = 3960 mi. \r\nset up an isosceles triangle with two legs = 3960 mi. \r\nthe third leg is 10 km (in miles)\r\nuse law of cosines to find the angle the two 3960 legs make with each other\r\nFind 3960 - (height of the triangle)", "Solution_3": "Law of Cosines is trigonometry.\r\n\r\nI have not seen trig in over 25 years.\r\n\r\nCan you show me how to use trig to answer this question?", "Solution_4": "[quote=\"Interval\"]Law of Cosines is trigonometry.\n\nI have not seen trig in over 25 years.\n\nCan you show me how to use trig to answer this question?[/quote]\r\n\r\nyeah, trig generally does not belong in this forum but whatever.\r\n\r\nlaw of cosines: (i really hope you have gone to yahoo and google)\r\n\r\n$a^{2}=b^{2}+c^{2}-2bc\\cos A$ where $a,b,c$ are sides of a triangle and side $a$ is opposite of $\\angle A$.", "Solution_5": "Now that we got law of cosines here, can you show me how to use it to answer the given question?", "Solution_6": "[quote=\"Interval\"]Now that we got law of cosines here, can you show me how to use it to answer the given question?[/quote]\r\n\r\nno, you will have to figure that out urself. ask if you have any questions", "Solution_7": "jli:\r\n\r\nI appreciate your replies but make the hints a bit more easy for people who are not math-inclined.\r\n\r\nCan you do that?\r\n\r\nIf not, then please do not answer my posts.\r\n\r\nThanks,\r\nInterval", "Solution_8": "[quote=\"Interval\"]jli:\n\nI appreciate your replies but make the hints a bit more easy for people who are not math-inclined.\n\nCan you do that?\n\nIf not, then please do not answer my posts.\n\nThanks,\nInterval[/quote]\r\n\r\nall right.\r\n\r\nill do an example with my avatar an see if you can do the problem.\r\n\r\nwe are trying to find the value of $x$. let $a=x$ so $b=3$ and $c=3$. Also, the angle opposite of $a$ is $90$ degrees. \r\n\r\nuse the formula:\r\n\r\n$x^{2}=3^{2}+3^{2}-2\\cdot 3\\cdot 3\\cdot \\cos 90=18$.\r\n\r\nso $x=3\\sqrt{2}$\r\n\r\nget it?", "Solution_9": "[hide]\n\nUse pythagorean :D \n\n$3^{2}+3^{2}=\\_\\_$.\n\n$18=\\_\\_$\n\n$3\\sqrt{2}=18$!\n[/hide]" } { "Tag": [ "trigonometry", "geometry", "inequalities", "inequalities proposed" ], "Problem": "Prove that for x in (0, pi/2) we have sin(x)/x<3/(4-cos(x))<1", "Solution_1": "3<4-cosx easy to prove on )0,pi/2(\r\n\r\nf(x)=3x-sinx(4-3cosx) study the variation on )0;pi/2(\r\n\r\nWhy did you put the trig inq in geometry Lagrangia ?", "Solution_2": "Well, I have put them because I though that trigonometry and geometry go together... I may be wrong.. :D :D", "Solution_3": "Would be nice to get a geometrical proof of this inequality", "Solution_4": "for sin(x)/x<1 it's easy to see a geometrical proof. but for sin(x)/x<3/(4-cos(x)) .. it's harder.. (x in (0,pi/2))" } { "Tag": [ "algebra", "polynomial" ], "Problem": "If 1,2, and 3 are solutions to the equation $ x^4 \\plus{} ax^2 \\plus{} bx \\plus{} c \\equal{} 0,$ then $ a\\plus{}c$ equals\r\n\r\nA. -12\r\nB. 24\r\nC. 35\r\nD. -61\r\nE. -63", "Solution_1": "[hide]We know that $ (x \\minus{} 1)(x \\minus{} 2)(x \\minus{} 3)$ is a factor of our quartic because each individual factor also is. This means that $ (x^2 \\minus{} 3x \\plus{} 2)(x \\minus{} 3) \\equal{} x^3 \\minus{} 3x^2 \\minus{} 3x^2 \\plus{} 9x \\plus{} 2x \\minus{} 6 \\equal{} x^3 \\minus{} 6x^2 \\plus{} 11x \\minus{} 6$ is also a factor.\n\nLet the fourth root be $ q$, we have $ (x \\minus{} q)(x^3 \\minus{} 6x^2 \\plus{} 11x \\minus{} 6) \\equal{} x^4 \\plus{} ax^2 \\plus{} bx \\plus{} c$. Looking at the absent $ x^3$ term, we know that $ \\minus{} 6x^3 \\minus{} qx^3 \\equal{} 0$ by finding ways to get $ x^3$ from distribution. $ q \\equal{} \\minus{} 6$, so our polynomial in factored form is $ (x \\minus{} 1)(x \\minus{} 2)(x \\minus{} 3)(x \\plus{} 6)$.\n\nUsing Viete's formulae, we know that $ c \\equal{} ( \\minus{} 1)( \\minus{} 2)( \\minus{} 3)(6) \\equal{} \\minus{} 36$ and $ a \\equal{} ( \\minus{} 1)( \\minus{} 2) \\plus{} ( \\minus{} 1)( \\minus{} 3) \\plus{} ( \\minus{} 1)(6) \\plus{} ( \\minus{} 2)( \\minus{} 3) \\plus{} ( \\minus{} 2)(6) \\plus{} ( \\minus{} 3)(6) \\equal{} \\minus{} 25$\n\n$ a \\plus{} c \\equal{} \\minus{} 36 \\minus{} 25 \\equal{} \\minus{} 61 \\equal{} > D$[/hide]" } { "Tag": [], "Problem": "$35y^{2}+2y-24$\r\n\r\nfactor the above...is it prime? i thought it was but then again working on AIM is harder than you think it is", "Solution_1": "[quote=\"7h3.D3m0n.117\"]$35y^{2}+2y-24$\n\nfactor the above...is it prime? i thought it was but then again working on AIM is harder than you think it is[/quote]\r\n\r\n[hide]Here is the answer: (5y + 4)(7y - 6)[/hide]", "Solution_2": "OOPS :blush: .........forgot to cross multiply something....thanks anyway!", "Solution_3": "pn12345, you made a mistake you wrote:\r\n[quote]Here is the answer: (5y + 4)(7y - 6)[/quote]\r\n(5y + 4)(7y - 6) = 35y^2 - 30y + 28y - 24 which is wrong.\r\nShould be (5y - 4)(7y + 6)", "Solution_4": "Oh yeah... sorry. (Careless error!)" } { "Tag": [ "function", "calculus", "derivative", "vector", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "given acceleration $ a \\equal{} 1 \\plus{} ln x$\r\ni can find that $ \\Delta v^2 \\equal{} 2xlnx$ and since it is given that when $ t \\equal{} 0, x \\equal{} 1,v \\equal{} 0$\r\n$ \\therefore v^2 \\equal{} 2xlnx$\r\nhowever i have been asked to prove $ v > 0 \\; when \\;t > 0$ i have no idea how to explain it in mathematcial terms, can anyone please give any suggestion?\r\nalso howcome the symbols are not working", "Solution_1": "What is $ x$? What is $ a$? I'm going to take a guess and say that $ x$ is a function from $ \\mathbb{R}$ to $ \\mathbb{R}$, a twice-differentiable function in fact, and that $ a$ is the second derivative of $ x$. Is that at all correct, or am I way off?", "Solution_2": "yes you are correct a=f(x) where a is accel and x is bounded vector displacement,you can use d(v^2/2)/dx=a\r\nbut the resulting v has power two, and i am asked to eliminate the negative sign, how do you do that mathematically\r\n and how come my latex codes are not appearing right", "Solution_3": "What is $ v$? Also, I guessed that $ a \\equal{} x''$, and you said I was right but then you said $ a \\equal{} f(x)$, which seems to be different than what I said. What is $ f$?\r\n\r\nI don't know why your LaTeX wasn't showing up correctly. Were you enclosing it in dollar signs?", "Solution_4": "v is the velocity, a is accel and x is bounded vector displacement", "Solution_5": "Suppose that $ x$ is a twice-differentiable function from the non-negative reals to the reals and that $ x''(t) \\equal{} 1 \\plus{} \\ln(x(t))$ for all $ t \\ge 0$. Suppose also that $ x(0) \\equal{} 1$ and $ x'(0) \\equal{} 0$. I'll outline a proof that if $ t > 0$, then $ x'(t) > 0$.\r\n\r\nAssume the claim is false. Let $ t_0 \\equal{} min(\\{t > 0 | x'(t) \\equal{} 0 \\})$. (It takes a little work to prove that such a minimum exists.) By Rolle's theorem, $ x''(c) \\equal{} 0$ for some $ c$ between $ 0$ and $ t_0$. Thus $ 1 \\plus{} \\ln(x(c)) \\equal{} 0$, so $ x(c) \\equal{} e^{\\minus{}1} < 1$. By the mean value theorem, there exists $ d$ between $ 0$ and $ c$ such that $ x'(d) < 0$. By the intermediate value theorem, there exists $ w$ between $ 0$ and $ d$ such that $ x'(w) \\equal{} 0$. (It takes a little work to justify that step, but the idea is that $ x'$ is positive just to the right of $ 0$.) This contradicts the definition of $ t_0$.\r\n\r\nI wouldn't be surprised if there is some easier proof I haven't thought of." } { "Tag": [], "Problem": "My school finally did it this year. Are you doing it this year? Do you have any tips for beginners?", "Solution_1": "Hmm the spelling bee...I think I might actually do decently well (at least pwn everyone in my school), not like make it to nationals but possibly states. Well I don't have a whole lot of time to study, with tennis and piano and math. \r\n\r\nI would guess learn your roots and the languages, and if according to Akeelah and the Bee, if you have no clue, spell it the way it sounds.", "Solution_2": "did it in elementary school. and got 2nd in 3rd and 2nd in4th grade school spelling bee.. yeah i suck. \r\n\r\nthe school i went to in 5th didnt do it.6th grade- didn't do it\r\n\r\n7th- got 3rd in my school , missed some poetry word grr..... could have gotten 2nd easily\r\n\r\nI NEVER studied for any spelling stuff. The prepared ones always win. so I would say, PREPARE\r\n\r\nI don't care for spelling anyway, who cares if you know how to spell some odd word no one's ever going to use, and neither will you?", "Solution_3": "You need to get How to Spell Like a Champ, by Trinkle, Kimble, and Andrews, some stuff from http://hexco.com/nsb.htm, and the Paidiea or Spell-It, whatever they call the word list this year. After you've learned all the words in there get the official dictionary of the NSB (isn't that Webster's Unabridged?) and learn it all. But the important part is committing to study.", "Solution_4": "i highly prefer math", "Solution_5": "or vocabulary", "Solution_6": "Meh, vocabulary: I'm working on studying my first actual unabridged dictionary. At my present rate it'll take 32 more hours of reading and marking words I don't know, and I'll end up with 16,000 words I don't know yet. And I intend to learn every one of them, because I love vocabulary :D." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ 0 x_{0}^{3} \\minus{} x_{1}^{3} \\plus{} x_{1}^{3} \\minus{} ... \\minus{} x_{669}^{3} \\ge 3\\sum_{i \\equal{} 0}^{668}x_{i}x_{i \\plus{} 1}(x_{i} \\minus{} x_{i \\plus{} 1})$\r\n$ wlog, x_{0}\\ge x_{1}\\ge ...\\ge x_{669}$\r\nThen, $ x_{i}x_{i \\plus{} 1}(x_{i} \\minus{} x_{i \\plus{} 1}) \\le |x_{i}x_{j}(x_{i} \\minus{} x_{j})|$ for all j.\r\nSo, $ min\\{x_{i}x_{i \\plus{} 1}(x_{i} \\minus{} x_{i \\plus{} 1}) \\} \\equal{} k$, then $ 2007k < 1$ and $ k > 0$.\r\n\r\nwe can get conclusion" } { "Tag": [], "Problem": "can u ppl tell me your nsec scores.\r\ni'm expecting somewhere around 210. is that gud enough???\r\n\r\nplz reply.", "Solution_1": "Have you checked with the solns.\r\nThey are in the Indian forum so plz. check them.", "Solution_2": "Out of curiosity, just what is the NSEC?", "Solution_3": "NSEC - National Standard Examination in Chemistry.\r\n\r\nIts the first round in India leading to the ICHO. I think this thread belongs in the Indian section.", "Solution_4": "Actually, I don't mind it's being here. Since we have only one chemistry forum and no subforums, I feel that anything related to chemistry has a place here.\r\n\r\nNo reason to be US centric about these things. That being said, an explanation of topics probably unfamiliar to someone from the US would be nice!" } { "Tag": [ "integration", "calculus", "function", "calculus computations" ], "Problem": "If a>1 then evaluate $ \\int_{0}^{1}\\frac {ln\\frac {1 \\plus{} ax}{1 \\minus{} ax} dx}{x\\sqrt {1 \\minus{} x^{2}}}$", "Solution_1": "somebody please help me :(", "Solution_2": "[quote=\"vijaymenon\"]If a>1 then evaluate $ \\int_{0}^{1}\\frac {ln\\frac {1 \\plus{} ax}{1 \\minus{} ax} dx}{x\\sqrt {1 \\minus{} x^{2}}}$[/quote]\r\nIs numerator ok?\r\n(1 + ax/1 -ax)<0 when x belongs to interval (1/a,1),a>1 and we want to integrate at [0,1]", "Solution_3": "hello \r\n[hide=\"Solution\"]\nConsider \n $ f(a) \\equal{} \\int_{0}^{1}\\frac {ln\\frac {1 \\plus{} ax}{1 \\minus{} ax}dx}{x\\sqrt {1 \\minus{} x^{2}}}$\n$ f(a) \\equal{} \\int_{0}^{1}\\frac {ln(1 \\plus{} ax)dx}{x\\sqrt {1 \\minus{} x^{2}}} \\minus{} \\int_{0}^{1}\\frac {ln (1 \\minus{} ax)dx}{x\\sqrt {1 \\minus{} x^{2}}}$\nDifferentiating wrt $ a$ using $ Leibnitz Rule$\n$ f' (a) \\equal{} \\int^{1}_{0} \\frac {dx}{(1 \\plus{} ax)\\sqrt {1 \\minus{} x^{2}}} \\plus{} \\frac {dx}{(1 \\minus{} ax)\\sqrt {1 \\minus{} x^{2}}}$ \n$ f' (a) \\equal{} \\int ^{1}_{0} \\frac {2 dx}{(1 \\minus{} (ax)^2)\\sqrt {1 \\minus{} x^{2}}}$\n put $ x \\equal{} \\frac {1}{t}$ \nIt will become \n$ f'(a) \\equal{} \\int^{\\infty}_{1} \\frac {2t dt }{(t^2 \\minus{} a^2)(\\sqrt {t^{2} \\minus{} 1})}$ \nLet $ t^2 \\equal{} u$ and find the integral value .\nand then integrate wrt to a. \n\n[/hide]\r\nIf u find problem in solving further kindly tell me . I will solve it fully . \r\nThank u", "Solution_4": "Ya i almost forgot that it $ \\ a$ can be taken a a parameter. \r\n\r\nThanks KABI :)", "Solution_5": "the definite integral is not right. the argument of ln will be negative if x > 1/a. x varies from 0 to 1 and 1/a < 1. so, after x = 1/a, the function won't be defined.", "Solution_6": "I don see anything wrong there" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "I had nothing to do so I thought I could post a nice and easy problem :):\r\n\r\nIf $m,n\\in\\mathbb N^*$, show that we can find integers $a_1,a_2,\\ldots,a_n,b_1,b_2,\\ldots,b_m$ s.t. $a_ib_j$ are all distinct $\\pmod {mn}$.", "Solution_1": "[quote] I had nothing to do so I thought I could post a nice and easy problem \n\n[/quote]\r\n\r\nSo, why don't you solve all the remaining unsolved problems? ;) \r\n\r\nPierre." } { "Tag": [ "geometry", "rectangle", "analytic geometry", "geometric transformation", "homothety", "circumcircle", "ratio" ], "Problem": "Let $ ABCD$ be a rectangle with $ AB\\equal{}b, AD\\equal{}a$. A circle with center $ A$ and radius $ AC$ intersect line $ BD$ and $ E$ and $ F$. Bisector of angle $ ECF$ intersect chord $ EF$ at $ N$. Calculate $ \\dfrac{EN}{NF}$.", "Solution_1": "first apply the angle bisector theorem and thentry inversion through :) $ C$ then the solution follows easily", "Solution_2": "Please post your solution if you have already solved it.\r\n\r\nI have a really ugly solution using coordinates: :oops: \r\n\r\n[hide=\"My Solution\"]From the angle bisector theorem, $ \\frac {EN}{NF} = \\frac {CE}{CF}$. Consider the homothety that sends $ A$ to $ A'$ such that $ \\overrightarrow{CA'} = 2\\overrightarrow{CA}$. This sends rectangle $ ABCD$ to rectangle $ A'B'C'D'$ and we can restate our problem as follows: $ A'B'CD'$ is a rectangle and $ A,B,D$ are the midpoints of $ CA',CB'$ and $ CD'$ respectively. $ BD$ intersects the circumcircle of $ A'B'CD'$ at $ E$ and $ F$. Then we need to find $ \\frac {CE}{CF}$.\n\nConsider a coordinate system with $ A$ as the origin, and the axes along $ AD$ and $ AB$. Then $ D = (a,0),B = (0, - b), C = (a, - b)$. Hence the equation of $ BD$ (or $ EF$) is $ bx - ay = ab$ and the circumcircle of $ A'B'C'D'$ is $ x^2 + y^2 = a^2 + b^2$. From these we find that $ E = \\left(\\frac {a(b^2 + \\sqrt {D})}{a^2 + b^2},\\frac {b( - a^2 + \\sqrt {D})}{a^2 + b^2}\\right)$ and $ F = \\left(\\frac {a(b^2 - \\sqrt {D})}{a^2 + b^2},\\frac {b( - a^2 - \\sqrt {D})}{a^2 + b^2}\\right)$. Here $ D = a^4 + a^2b^2 + b^4$.\n\nThus (after simplification) $ CE = \\frac {\\sqrt {b^2(b^2 + \\sqrt D)^2 + a^2(a^2 - \\sqrt D)^2}}{a^2 + b^2}$ and ${ CF = \\frac {\\sqrt {a^2(a^2 + \\sqrt D)^2 + b^2(b^2 - \\sqrt D)^2}}{a^2 + b^2}}$. And finally $ \\frac {CE}{CF} = \\sqrt {\\frac {b^2(b^2 + \\sqrt D)^2 + a^2(a^2 - \\sqrt D)^2}{{a^2(a^2 + \\sqrt D)^2 + b^2(b^2 - \\sqrt D)^2}}}$.\n\nI haven't found any nice expressions. :blush: [/hide]", "Solution_3": "I don't think that my ratio is quite well. But at least it is a ratio with $ a$ and $ b$.\r\n\r\n[color=red]Here goes my solution,[/color]\r\nConsider a dilatation with center $ C$ and ratio $ 2$ which maps the rectangle $ ABCD$ to $ A'B'CD'$.\r\nNow using angle bisector theorem we get, ratio,$ k \\equal{} \\frac {EN}{NF} \\equal{} \\frac {EC}{CF} \\equal{} \\frac {\\sin \\angle {EFC}}{\\sin \\angle{CEF}} \\equal{} \\frac {\\sin \\angle {EA'C}}{\\sin \\angle{CA'F}}$\r\n\r\n$ B'EFD'$ is an isosceles trapezoid (as it is a cyclic one). So, $ B'E \\equal{} D'F$ and $ \\angle B'A'E \\equal{} \\angle FA'D'$\r\n\r\nLet $ \\angle BB'A \\equal{} B$ and $ \\angle ACD \\equal{} C$\r\n\r\nBy easy angle chasing we get that, $ \\angle EA'C \\equal{} C \\minus{} B/2$ and $ \\angle CA'F \\equal{} \\pi/2 \\minus{} C \\minus{} B/2$\r\nso,\r\n$ \\frac {\\sin \\angle {EA'C}}{\\sin \\angle{CA'F}} \\equal{} \\frac {\\sin (C \\minus{} B/2)}{\\cos (C \\plus{} B/2)} \\equal{} \\frac {\\sin C.\\cos{B/2} \\minus{} \\cos C.\\sin{B/2}}{\\cos C.\\cos{B/2} \\minus{} \\sin C.\\sin{B/2}}$\r\n\r\nLet $ AC \\equal{} r \\equal{} \\sqrt {a^2 \\plus{} b^2}$\r\nSo,\r\n$ \\sin {B/2} \\equal{} \\sqrt {\\frac {r \\minus{} a}{2r}}$ and $ \\cos B/2 \\equal{} \\sqrt {\\frac {r \\plus{} a}{2r}}$ and $ \\sin C \\equal{} \\frac {a}{r}$ and $ \\cos C \\equal{} \\frac {b}{r}$\r\nso,\r\n$ \\frac {\\sin C.\\cos{B/2} \\minus{} \\cos C.\\sin{B/2}}{\\cos C.\\cos{B/2} \\minus{} \\sin C.\\sin{B/2}} \\equal{} \\frac {{\\frac {a}{r}}\\sqrt {\\frac {r \\plus{} a}{2r}} \\minus{} {\\frac {b}{r}}\\sqrt {\\frac {r \\minus{} a}{2r}}}{{\\frac {b}{r}}\\sqrt {\\frac {r \\plus{} a}{2r}} \\minus{} {\\frac {a}{r}}\\sqrt {\\frac {r \\minus{} a}{2r}}}$\r\nAfter cancelling common terms we get,\r\n$ k \\equal{} \\frac {a\\sqrt {r \\plus{} a} \\minus{} b\\sqrt {r \\minus{} a}}{b\\sqrt {r \\plus{} a} \\minus{} a\\sqrt {r \\minus{} a}}$\r\n\r\n$ Q.E.D$\r\n\r\np.s. If the solution looks unclear please have a look at the diagram.\r\n\r\n[url=http://img146.imageshack.us/my.php?image=figure2dm8.png][img]http://img146.imageshack.us/img146/6676/figure2dm8.th.png[/img][/url]", "Solution_4": "I have another trignometric solution (wasnt as trivial as I thought first)\r\nlet the diagonals meet each other at $ K$\r\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac{CE}{CF}\\equal{} \\frac{ \\sin{x}}{ \\sin{y}}$\r\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\r\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$\r\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ __________________________$ eq^{n} 1$\r\nalso $ AC \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} \\frac {ab}{c}$\r\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$\r\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \r\nyou get a quadratic in the ratio required", "Solution_5": "i can't see anything...... :( please correct your latex code", "Solution_6": "fixed\r\n :oops:", "Solution_7": "[quote=\"gauravpatil\"]I have another trignometric solution (wasnt as trivial as I thought first)\nlet the diagonals meet each other at $ K$\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac {CE}{CF} \\equal{} \\frac {\\sin{x}}{\\sin{y}}$\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ __________________________$ eq^{n} 1$\nalso $ AC \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} \\frac {ab}{c}$\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \nyou get a quadratic in the ratio required[/quote]\r\n\r\nCould you please write a bit clearly. most probably I am missing something or you have missed something. :maybe:", "Solution_8": "I :) have modified it a bit\r\nlet the diagonals meet each other at $ K$ let $ c$ be third side\r\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac {CE}{CF} \\equal{} \\frac {\\sin{x}}{\\sin{y}}$ by sine rule\r\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\r\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$ substituting $ \\cos{x \\plus{} y} \\equal{} \\cos{x} \\cos{y} \\minus{} \\sin{x} \\sin{y}$\r\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ ______$ eq^{n} 1$\r\nnow we calculate distance between a and bd in 2 ways $ AE \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} b \\sin{ABD} \\equal{} \\frac {ab}{c}$\r\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\cos{x \\plus{} y} \\equal{} \\sin{x} \\sin{y}$\r\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \r\nyou get a quadratic in the ratio required or rather an equation of this form $ ratio \\plus{} \\frac {1}{ratio} \\equal{} r$ where $ r$ is a known constant in $ a,b$ and $ c$\r\natleast i see nothing missing", "Solution_9": "[size=117][color=darkblue][b]Very nice application of the [u]Generalized Pythagoras' Theorem[/u] (G.P.T.) ![/b][/color][/size]\r\n\r\n[quote=\"Ahiles\"] [color=darkred]Let $ ABCD$ be a rectangle with $ AD = a$ , $ AB = b$ . The circle $ w = C(A,r)$ , \n\nwhere $ r = \\sqrt {a^2 + b^2}$ , intersects $ BD$ in $ E$ and $ F$ so that $ B\\in (ED)$ . Calculate $ \\frac {CE}{CF}$ .[/color] [/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Denote $ \\|\\begin{array}{c} BE = x \\\\\n \\\\\nDF = y\\end{array}$ . Observe that $ \\|\\begin{array}{c} \\cos\\widehat {ABD} = \\frac br \\\\\n \\\\\n\\cos\\widehat {ADB} = \\frac ar\\end{array}$ . Apply [b]G.P.T.[/b] to :\n\n$ \\|\\begin{array}{cccc} \\triangle\\ ABE\\ : & r^2 = x^2 + b^2 + 2bx\\cdot \\frac br & \\implies & rx^2 + 2b^2x - ra^2 = 0 \\\\\n \\\\\n\\triangle\\ ADF\\ : & r^2 = y^2 + a^2 + 2ay\\cdot\\frac ar & \\implies & ry^2 + 2a^2y - rb^2 = 0\\end{array}\\|$ $ \\implies$ $ \\|\\begin{array}{c} rx = \\sqrt {\\Delta} - b^2 \\\\\n \\\\\nry = \\sqrt {\\Delta} - a^2\\end{array}\\|\\ \\ (1)$ \n\nwhere $ \\boxed {\\ \\Delta = a^4 + a^2b^2 + b^4\\ }$ . Observe that $ \\|\\begin{array}{c} \\cos\\widehat {CBD} = \\frac ar \\\\\n \\\\\n\\cos\\widehat {CDB} = \\frac br\\end{array}$ . Apply [b]G.P.T.[/b] to :\n\n$ \\|\\begin{array}{cc} \\triangle\\ BCE\\ : & CE^2 = x^2 + a^2 + 2ax\\cdot\\frac ar \\\\\n \\\\\n\\triangle\\ DCF\\ : & CF^2 = y^2 + b^2 + 2by\\cdot\\frac br\\end{array}\\|$ $ \\implies$ $ (\\frac {CE}{CF})^2 = \\frac {rx^2 + 2a^2x + ra^2}{ry^2 + 2b^2y + rb^2}\\stackrel {(*)}{\\ = \\ }$ $ \\frac {(ra^2 - 2b^2x) + 2a^2x + ra^2}{(rb^2 - 2a^2y) + 2b^2y + rb^2} =$\n\n$ \\frac {ra^2 + (a^2 - b^2)x}{rb^2 - (a^2 - b^2)y} =$ $ \\frac {r^2a^2 + (a^2 - b^2)\\cdot rx}{r^2b^2 - (a^2 - b^2)\\cdot ry} =$ $ \\frac {r^2a^2 + (a^2 - b^2)(\\sqrt {\\Delta } - b^2)}{r^2b^2 - (a^2 - b^2)(\\sqrt {\\Delta} - a^2)} =$ $ \\frac {a^4 + b^4 + (a^2 - b^2)\\sqrt {\\Delta}}{a^4 + b^4 - (a^2 - b^2)\\sqrt {\\Delta}} =$\n\n$ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)^2(a^4 + a^2b^2 + b^4)} =$ $ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)(a^6 - b^6)} =$ $ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{a^2b^2(a^2 + b^2)^2}$ .\n\nIn conclusion, $ \\boxed {\\ \\frac {CE}{CF} = \\frac {|a^4 + b^4 + (a^2 - b^2)\\sqrt {\\Delta}|}{ab(a^2 + b^2)}\\ }$ . Remark that $ x - y = \\frac {a^2 - b^2}{\\sqrt {a^2 + b^2}}$ .[/color]" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Five positive integers less than 100 have 12 positive integer factors. Which of these five integers is not divisible by 12?", "Solution_1": "90 is not.", "Solution_2": "how did you do that?", "Solution_3": "[hide=\"Like this\"]$12=2^2\\cdot 3$, and the number of positive divisors that a number $n=p_1^ap_2^b...$ is $(a+1)(b+1)...$, where $p_1,p_2...$ are distinct primes. So, we take all possible combinations of the prime factorization of $12$, and find numbers $k=p_1^ap_2^bp_3^c$ so that $(a+1)(b+1)(c+1)=12$ (the largest possible number of primes is $3$ so that we can make $12$) and $k<100$. One of these numbers will not be divisible by $12$. That number is $90=2\\cdot 3^2\\cdot 5$, since it does not have two $2$'s in it.[/hide]", "Solution_4": "If you didn't understand that (I couldn't get everything), I'll put it in hopefully simpler terms.\r\n\r\n\r\n[hide]Like Jesusfreak said, the number of positive factors a number has is when you take it into prime factorization, the powers of its prime factors + 1 all multiplied by each other. (72 is $2^3 * 3^2$, so it has (3+1)(2+1) factors, or 12 factors)\n\nSo for a number to have 12 factors, $(a+1)(b+1)$... must equal 12.\n\nAll the ways to make 12 $(2*2*3)$\n\n$2*6, 3*4, 2*2*3.$\n\nso our numbers can be either $a*b^5$ (only 96 works, which is divisible by 12),$a^2*b^3$ (only 72 works, which is divisible by 12) , or $a*b*c^2$ (must be).\n\nand this can't be divisible by $2^2*3.$ \n\ntry numbers, with c not being 2. \n\n$2*3*5^2$ is too high, so c has to be 3.\n\n$2*5*3^2$ works. Our answer is 90.\n\n\nSorry if this was a bit too long, but hopefully it was clear enough.[/hide]" } { "Tag": [ "inequalities", "geometry", "inequalities unsolved" ], "Problem": "IN triangle prove that\r\n(ma*mb)/(ra*mc)+(mb*mc)/(rc*ma)+(mc*ma)/(rb*mb)>=4/3\r\nma,mb,mc medium of triangle ABC\r\nra,rb,rc :radius of the circles inscribed in A,B,C and tangent to BC,CA,AB", "Solution_1": "I think you have many typos in your post, [b]blue allrise[/b]. First of all I suppose that you wanted to write symmetric sum. I think here is the right inequality:\r\n$\\sum_{cyclic}{\\frac{m_{a}\\cdot{m_{b}}}{r_{c}\\cdot{m_{c}}}}\\geq{3}$.\r\nI have a very nice proof for this one. We have to prove that:\r\n$\\frac{1}{2}\\sum_{cyclic}{(\\frac{m_{a}\\cdot{m_{b}}}{r_{c}\\cdot{m_{c}}}+\\frac{m_{b}\\cdot{m_{c}}}{r_{a}\\cdot{m_{a}}})}\\geq{3}.$\r\nNow: ${(\\frac{m_{a}\\cdot{m_{b}}}{r_{c}\\cdot{m_{c}}}+\\frac{m_{b}\\cdot{m_{c}}}{r_{a}\\cdot{m_{a}}})=\\frac{m_{b}}{S}(\\frac{m_{a}(p-a)}{m_{c}}+\\frac{m_{c}(p-c)}{m_{a}}})$\r\n${\\frac{m_{b}}{S}(\\frac{m_{a}(p-a)}{m_{c}}+\\frac{m_{c}(p-c)}{m_{a}}})\\geq{\\frac{2m_{b}}{S}\\cdot{\\sqrt{(p-a)\\cdot{(p-c)}}}}=\\frac{2m_{b}}{\\sqrt{p(p-b)}}=\\frac{\\sqrt{2a^{2}+2c^{2}-b^{2}}}{\\sqrt{p(p-b)}}\\geq{\\frac{\\sqrt{(a+c-b)(a+c+b)}}{\\sqrt{p(p-b)}}}=2.$\r\nWe can do the others in similar way and then add them up to get the desired result.\r\nHereby $m_{a}$ is the median from vertice $A$, $r_{a}$ is the exradius, $S$ is the area of the triangle, $a,b,c$ are the sides of the triangle. :)", "Solution_2": "Well i also think that blue allrise's inequality needs 3 as a RHS\r\nHere is my solution :\r\nUse [tex] m_a\\geq\\sqrt{s(s-a)} [/tex] to find [tex]m_am_bm_c\\geq s^2r =r_ar_br_c[/tex] . The rest is AM-GM to the initial inequality .\r\n\r\n[tex] s [/tex] is semiperimeter , and the other symbols are the well know ones." } { "Tag": [], "Problem": "Between $ 1$ and $ 8000$ inclusive, find the number of integers which are divisible by neither $ 14$ or $ 21$ but divisible by either $ 4$ or $ 6$.", "Solution_1": "can't we some how use the fact that every 7th of 4's multiple is divisible by 14 and every 21st one divisible by 21 and every 7th multiple of 6 is divisible by 14 and 21.then to exclude double counting eve", "Solution_2": "Yes, but it's awfully hard to keep track of all the details. My answer is [hide]2287[/hide] but I'm not sure if that's correct.", "Solution_3": "yep the answer is 2287.there r 2667 nums that r multiples of either 4 or 6.there are 285 multiples of 4 that are also divisible by 14 and 95 that r divisible by 21 & 95 that are div by 42(l.c.m. of 14&21).thus 285 multiples of 4 that r divisible by either 14 or 21.doing the same we find 190 nums for 6.there r 95 multiples of 12(l.c.m of 4&6) that r also divisible by 14,95 for 21 & 95 for 42.\r\nso the answer is 2667-285-95+95-190-190+190+95=2287." } { "Tag": [ "geometric series" ], "Problem": "Hi,\r\n\r\nThis problem is from the book AoPS Vol. 1. It states:\r\n\r\nTwo dogs, each traveling $ 10ft/sec$, run toward each each other from $ 500$ feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at $ 25ft/sec$. The flea flies between the dogs in this manner until it is crushed when dogs collide. How far did the flea fly?\r\n\r\nI first tried:\r\n\r\nLet $ x \\equal{}$ the number of seconds until the dogs collide\r\n\r\nTherefore, we have $ 10x \\plus{} 10x \\equal{} 500 \\Rightarrow 20x\\equal{}500 \\Rightarrow x\\equal{}25$\r\n\r\nSo there are $ 25$ seconds until the dogs collide. In that $ 25$ seconds, the flea flies: $ 25(25) \\equal{} 625ft$\r\n\r\nSo the flea flies $ 625ft$ until it is crushed. And according to the Solution manual $ 625ft$ is the right answer. But how does that make sense? Cant the flea's distance not exceed $ 500ft$ because the collision happens in that range? \r\n\r\nYour help will be appreciated!\r\n\r\nRegards,\r\nlimac", "Solution_1": "[quote=\"limac\"]Hi,\n\nThis problem is from the book AoPS Vol. 1. It states:\n\nTwo dogs, each traveling $ 10ft/sec$, run toward each each other from $ 500$ feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at $ 25ft/sec$. The flea flies between the dogs in this manner until it is crushed when dogs collide. How far did the flea fly?\n\nI first tried:\n\nLet $ x \\equal{}$ the number of seconds until the dogs collide\n\nTherefore, we have $ 10x \\plus{} 10x \\equal{} 500 \\Rightarrow 20x \\equal{} 500 \\Rightarrow x \\equal{} 25$\n\nSo there are $ 25$ seconds until the dogs collide. In that $ 25$ seconds, the flea flies: $ 25(25) \\equal{} 625ft$\n\nSo the flea flies $ 625ft$ until it is crushed. And according to the Solution manual $ 625ft$ is the right answer. But how does that make sense? Cant the flea's distance not exceed $ 500ft$ because the collision happens in that range? \n\nYour help will be appreciated!\n\nRegards,\nlimac[/quote]\r\n\r\nNot nessacarily.\r\nx----------------------x\r\nIf I were to move back and forth between these two x's, I could move 500,1000,1500, etc.", "Solution_2": "Thank you for the reply, BOGTRO! :)\r\n\r\nAh, so we are considering that the fly will move back and forth to gain a distance a distance of 625? But the problem says that the fly wants to go from the nose of the first dog to the nose of the other dog. So did that mean that the fly will be traveling back and forth across the two dogs' noses?\r\n\r\nRegads,\r\nlimac", "Solution_3": "[quote=\"limac\"]Thank you for the reply, BOGTRO! :)\n\nAh, so we are considering that the fly will move back and forth to gain a distance a distance of 625? But the problem says that the fly wants to go from the nose of the first dog to the nose of the other dog. So did that mean that the fly will be traveling back and forth across the two dogs' noses?\n\nRegads,\nlimac[/quote]\r\n\r\nYes. If you consider infinite geometric series this should beocme more clear to you.", "Solution_4": "Gotcha! Thank you very much for your time BOGTRO :D" } { "Tag": [ "Support" ], "Problem": "How many positive integers less than 1000 are multiples of 7 and have a units digit of 8?\r\n\r\n\r\nThe explanation was very confusing and I don't understand how to do it. Can you please help?\r\n\r\nThanks!", "Solution_1": "Hint:\r\nThe first number that matches those terms is 28, and it repeats in cycles of 10.\r\ni.e first is 28, then 98, and there are 10 multiples of 7 between 28 and 98, so next number would be 168.\r\n :D", "Solution_2": "Well, first you find the number of multiples of seven less than a thousand.\r\n1000/7=142.(some random digits) so the largest multiple of 7 under a thousand is 142*7=994, and smallest is 7, so there are 994/7=142 multiples.\r\n\r\nnow we have to find the ones that end in eight, and if you multiply 7x1 7x2... a bit, you'll find that the units digit is only affected by the other numbers units digit. So for it to have a units digit of 8, it needs 7 times a number that ends in 4. (4, 14, 24, etc.) because 4x7=2[8]. you can count these out: 4, 14, 24, 34, 44, 54,...134.\r\nso there are fourteen.\r\n\r\nI'm not sure if I explained well enough, sorry if thats the cause.", "Solution_3": "Please post your questions in the questions selection of the Forum, please." } { "Tag": [ "quadratics", "algebra", "difference of squares", "special factorizations" ], "Problem": "Do You Have Math Anxiety?\r\nA Self Test\r\n\r\nRate your answers from 1 to 5; add them up and check your score below. (1) = Disagree, (5) = Agree.\r\n\r\n*I cringe when I have to go to math class. 1 2 3 4 5 \r\n*I am uneasy about going to the board in a math class. 1 2 3 4 5 \r\n*I am afraid to ask questions in math class. 1 2 3 4 5 \r\n*I am always worried about being called on in math class. 1 2 3 4 5 \r\n*I understand math now, but I worry that it's going to get really difficult soon.1 2 3 4 5 \r\n*I tend to zone out in math class. 1 2 3 4 5 \r\n*I fear math tests more than any other kind. 1 2 3 4 5 \r\n*I don't know how to study for math tests. 1 2 3 4 5 \r\n*It's clear to me in math class, but when I go home it's like I was never there. 1 2 3 4 5 \r\n*I'm afraid I won't be able to keep up with the rest of the class. 1 2 3 4 5 \r\n\r\nCHECK YOUR SCORE:\r\n \r\n40-50 Sure thing, you have math anxiety. \r\n30-39 No doubt! You're still fearful about math.\r\n20-29 On the fence!.\r\n10-19 Wow! \r\n \r\nWhat is math anxiety?\r\n \r\nMath anxiety is an emotional reaction to mathematics based on a past unpleasant experience which harms future learning. A good experience learning mathematics can overcome these past feelings and success and future achievement in math can be attained.\r\n \r\nDo you agree or disagree with this statement?\r\n\r\nWhy? Why not?", "Solution_1": "10 :D \r\n\r\nhave to do my homework now, so I won't expand on this", "Solution_2": "I got a 24, but I think i answered \"5\" for different reasons than intended; for example on the question \"I don't know how to study for math tests.\" I said 5 because I can never find any way that studying would do anything since everything all makes sense and seems natural without memorization needed or studying required. It seems that 5's are suppose to be negative, yet that is clearly a positive thing... similar reasons for my other scores above 1.", "Solution_3": "[quote=\"tjhance\"]I said 5 because I can never find any way that studying would do anything since everything all makes sense and seems natural without memorization needed or studying required.[/quote]\r\n\r\nHaha I was thinking the exact same thing; none of the math classes at my school pose any challenge...", "Solution_4": "2\r\n1\r\n1\r\n1\r\n1\r\n5\r\n1\r\n5\r\n1\r\n1\r\n=19", "Solution_5": "18....\r\n\r\n1\r\n1\r\n1\r\n1\r\n1\r\n5...Uhhh i know it already\r\n1\r\n5...I don't need to study\r\n1\r\n1", "Solution_6": "[quote=\"sharkman\"]Do You Have Math Anxiety?\nA Self Test\n\nRate your answers from 1 to 5; add them up and check your score below. (1) = Disagree, (5) = Agree.\n\n*I cringe when I have to go to math class. 1 2 3 4 5 \n\nhmm sometimes 2\n\n*I am uneasy about going to the board in a math class. 1 2 3 4 5\n\n hmm I don't go to the board\n\n*I am afraid to ask questions in math class. 1 2 3 4 5\n\n uhh kinda 3\n\n*I am always worried about being called on in math class. 1 2 3 4 5 \n\nwtf you don't get called on\n\n*I understand math now, but I worry that it's going to get really difficult soon.1 2 3 4 5 \n\nuh I only somewhat understand math now and it is definitely going to get very difficult soon 5\n\n*I tend to zone out in math class. 1 2 3 4 5 \n\nyeah sometimes 2\n\n*I fear math tests more than any other kind. 1 2 3 4 5 \n\nno I fear them about the same as any other kind 3\n\n*I don't know how to study for math tests. 1 2 3 4 5 \n\nno, I can study for them. I'm not very good at it, but I can 2\n\n*It's clear to me in math class, but when I go home it's like I was never there. 1 2 3 4 5 \n\nno this is stupid 1\n\n*I'm afraid I won't be able to keep up with the rest of the class. 1 2 3 4 5 \n\nhaha no 1\n\nCHECK YOUR SCORE:\n \n40-50 Sure thing, you have math anxiety. \n30-39 No doubt! You're still fearful about math.\n20-29 On the fence!.\n10-19 Wow! \n \nWhat is math anxiety?\n \nMath anxiety is an emotional reaction to mathematics based on a past unpleasant experience which harms future learning. A good experience learning mathematics can overcome these past feelings and success and future achievement in math can be attained.\n \nDo you agree or disagree with this statement?\n\nWhy? Why not?[/quote]\r\n\r\nso uhh like 21 or so maybe", "Solution_7": "1\r\n3\r\n3\r\n3\r\n2\r\n2\r\n1\r\n1\r\n1\r\n1\r\n\r\nI answered 3's on #2-4 because I usually never really concentrate on the questions in class much", "Solution_8": "5 I cringe not from the actual class but from the teacher\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1 but I never study.\r\n1\r\n1\r\n\r\n=14\r\n\r\nI don't have math anxiety.", "Solution_9": "38?\r\n5\r\n5\r\n5\r\n5\r\n1\r\n5\r\n5\r\n5\r\n1\r\n1", "Solution_10": "All those who are scoring in the math anxiety range are getting it for reasons other than the test intended (if you get a 5 for cringing when you go to math class, it's not supposed to be because the math class will bore you so much that you don't want to go :P). I had math anxiety when I was a young'n, I can remember shaking with nervousness if the teacher asked me a question, or walking into exams (...yeah, exams in the 5th grade) on the brink of tears. It's not so much that I was bad at math, but I was in constant fear of making a mistake, or not being able to solve a problem the first time I saw it, or having kids in the class understand before me. None of you guys have math anxiety, you wouldn't be here if you did.", "Solution_11": "1\r\n1\r\n1\r\n1\r\n1\r\n5\r\n1\r\n1\r\n1\r\n1\r\n\r\nhmm I should stop reading algebra books in math class...", "Solution_12": "[quote=\"max_tm\"]All those who are scoring in the math anxiety range are getting it for reasons other than the test intended (if you get a 5 for cringing when you go to math class, it's not supposed to be because the math class will bore you so much that you don't want to go :P). I had math anxiety when I was a young'n, I can remember shaking with nervousness if the teacher asked me a question, or walking into exams (...yeah, exams in the 5th grade) on the brink of tears. It's not so much that I was bad at math, but I was in constant fear of making a mistake, or not being able to solve a problem the first time I saw it, or having kids in the class understand before me. None of you guys have math anxiety, you wouldn't be here if you did.[/quote]\r\nWell for example, I always feel uneasy when going to the board because sometimes the method in the textbook isn't the best way to solve the problem, and I have to force myself to show the generic solution.\r\nLike in solving $(x+25)(x+27)=8$ the textbook solution is to expand, combine, and plug into the quadratic, I can't bear to go through all those ugly steps when the simple substitution x=n-26 trivializes the problem.", "Solution_13": "Don't you just love those problems that are like, solve for x in $(x-7)(x+3)=0$ by expanding and using the quadratic formula.", "Solution_14": "[quote=\"pianoforte\"]Don't you just love those problems that are like, solve for x in $(x-7)(x+3)=0$ by expanding and using the quadratic formula.[/quote]\r\n :rotfl: \r\n\r\n\r\nin school, I'd have said 14 (I zone out in any class whatsoever, me not zoning out in a class is a rare thing, one of the rarest in the history of classes I should say)\r\n\r\nin the past year (especially in the last sem), everything's changed. :( It's all maths now (next year we have a physic course though), and I'm doing miserably. I carried the idea that 'nobody studies for [i]maths[/i] tests, pooh, it's [i]maths[/i] after all' into college and have been shocked. I must review my policies for the coming year. :(", "Solution_15": "That is a large part of the problem with high schools. College courses are so much more rigorous than high-school courses that many students, even those in \"college prep\" curricula, are shocked by the sudden increase in difficulty between high school and college.", "Solution_16": "[hide=\"Breakdown\"]I cringe when I have to go to math class.\n2 We keep it entertaining\n\nI am uneasy about going to the board in a math class.\n1 Meh\n\nI am afraid to ask questions in math class.\n2 I usually don't get answers anyway... unless it doesn't have anything to do with math.\n\nI am always worried about being called on in math class.\n2 It's distracting\n\nI understand math now, but I worry that it's going to get really difficult soon.\n1 I hope it does though.\n\nI tend to zone out in math class.\n4 Every now and then\n\nI fear math tests more than any other kind.\n5 It means I can't do homework that class\n\nI don't know how to study for math tests.\n5 I don't know how I would\n\nIt's clear to me in math class, but when I go home it's like I was never there.\n5 Both parts of the sentence are true\n\nI'm afraid I won't be able to keep up with the rest of the class.\n1 No...[/hide]\n\n[hide=\"Total\"]28[/hide]\r\n\r\nBut yeah, I agree in part with bpms...\r\nI remember in 8th grade, we had the problem $x^{2}=256$\r\nSo I did $x=\\pm \\sqrt{256}=\\pm 16$ and the teacher yelled at me.\r\n\r\nApparently he wanted:\r\n$x^{2}=256$ $\\implies x^{2}-256=0$ $\\implies (x+16)(x-16)=0$ $\\implies x+16 = 0 \\text{ or }x-16=0$ $\\implies x=-16 \\text{ or }x=16$\r\n\r\nHow was I supposed to know we were on the difference of squares chapter? Is it even a reasonable expectation for students to know what chapter their class is on? Not in that class.\r\n\r\nIt just shows how mathematics is taught as a set of procedures to be followed, rather than a set of tools to be used and applied in efficient and creative ways. I don't think that promotes clean thinking.", "Solution_17": "$x^{2}=256 \\implies x^{2}-256=0 \\implies (x+16)(x-16)=0 \\implies x+16 = 0 \\text{ or }x-16=0 \\implies x=-16 \\text{ or }x=16$\r\n :o How did you do that?!? :o \r\nSorry, just spazzing, don't mind me...\r\n\r\nI got an 18, don't know how to study, can't pay attention in class, too boring, play (my own version!!!) of the arithmetic game on my calculator (or read math books.)", "Solution_18": "[hide=\"Breakdown\"][b]I cringe when I have to go to math class. [/b]\n1. I love math class.\n\n[b]I am uneasy about going to the board in a math class.[/b]\n1. I don't mind it.\n\n[b]I am afraid to ask questions in math class.[/b]\n2. Afraid that I'll ask a stupid question.\n\n[b]I am always worried about being called on in math class.[/b]\n3. I don't know the reason...\n\n[b]I understand math now, but I worry that it's going to get really difficult soon.[/b]\n1. It's never going to get difficult.\n\n[b]I tend to zone out in math class.[/b]\n1. No.\n\n[b]I fear math tests more than any other kind.[/b]\n1. I don't fear math tests at all.\n\n[b]I don't know how to study for math tests.[/b]\n1. I know how to study, but I never do... :D \n\n[b]It's clear to me in math class, but when I go home it's like I was never there.[/b]\n1. It's always clear to me.\n \n[b]I'm afraid I won't be able to keep up with the rest of the class.[/b]\n1. I go ahead of the class.[/hide]\n\n[hide=\"Total\"]13[/hide]", "Solution_19": "Here's my breakdown:\r\n\r\nI cringe when going to math class.\r\n(1) I enjoy it. My math teachers have been very nice. I can't say anything about next year, but I'm sure it'll still hold.\r\n\r\nI am uneasy about going to the board in math class.\r\n(1) I feel comfortable with going to the board. I've done so many times.\r\n\r\nI am afraid to ask questions in math class.\r\n(1) I don't usually ask questions, but if there's something I'm confused about, I'm not afraid to ask.\r\n\r\nI'm always worried about being called on in math class.\r\n(1) I don't mind being called on.\r\n\r\nI understand math now, but I worry that it's going to get really difficult soon.\r\n(2) I feel a little nervous about the level of difficulty of math in years to come, but I'm sure that I'll be able to catch on very quickly.\r\n\r\nI tend to zone out in math class.\r\n(5) I know most of the stuff.\r\n\r\nI fear math tests more than any other kind.\r\n(1) Nope. They're easy to me.\r\n\r\nI don't know how to study for math tests.\r\n(3) I've never found the need to study for a math test, so I can't say I absolutely know how. But I have a pretty good guess.\r\n\r\nIt's clear to me in math class, but when I go home it's like I was never there.\r\n(1) It's always clear to me.\r\n\r\nI'm afraid I won't be able to keep up with the rest of the class.\r\n\r\n\r\nMy score: 17\r\n(1) I feel that I'm ahead of most people in my class." } { "Tag": [ "integration", "calculus", "derivative", "calculus computations" ], "Problem": "Given the voltage drops for a resistor and capacitor are $ E_R \\equal{} iR$ and $ E_c \\equal{} \\frac {1}{c}\\int{i.dt}$ respectively, show that an RC-circuit with resistance R, capacitance C, voltage E and current i has equation $ R.\\frac {di}{dt} \\plus{} \\frac {1}{c}i \\equal{} \\frac {dE}{dt}$\r\n\r\nhow to proof this formula?", "Solution_1": "The sum across the voltage is 0 (\"The Loop Law\" I believe its called, that's not important really).\r\n\r\nThus $ E\\minus{}E_R\\minus{}E_C\\equal{}0$ or $ E\\minus{}iR\\minus{}\\frac{1}{c}\\int i dt\\equal{}0$. \r\n\r\nTake the derivative with respect to time to get $ \\frac{dE}{dt}\\minus{}\\frac{di}{dt}R\\minus{}\\frac{i}{c}\\equal{}0$. Now rearrange terms." } { "Tag": [ "modular arithmetic" ], "Problem": "True or False?\r\n\r\na) 15 =35 (mod 20)\r\n\r\nTRUE. 35/20 has a remainder of 15. Thus 15 is congruent to 35 (mod 20). ( this is from [url=http://www.cs.usask.ca/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp03TFsoln.html]this website[/url])\r\n\r\nIs 15=35 (mod 20) the same as 35=15(mod 20?\r\n\r\nThanks", "Solution_1": "[quote=\"myc\"]True or False?\n\na) 15 =35 (mod 20)\n\nTRUE. 35/20 has a remainder of 15. Thus 15 is congruent to 35 (mod 20). ( this is from [url=http://www.cs.usask.ca/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp03TFsoln.html]this website[/url])\n\nIs 15=35 (mod 20) the same as 35=15(mod 20?\n\nThanks[/quote]\r\n\r\n(15-35)/20 is an integrer, so it's true I think...", "Solution_2": "What do mean by \"are they the same thing\"? They're both true, if you want to know...", "Solution_3": "how is 15 equal to 35 (mod 20)? doesn't 15=5(mod 20)? how did they get 35. 35= 15(mod 20), because 35 divided by 20 =1 R 15. or am I wrong? \r\n\r\nThanks", "Solution_4": "For any integer $n$, we have $15+20n\\equiv 15\\pmod{20}$.\r\nIt is because $20n$ is divisible by $20$ no matter what $n$ is.\r\n\r\nThis is true even if $n$ is a negative integer; for example, if $n=-1$, then $15+20n=-5\\equiv 15\\pmod{20}$.", "Solution_5": "Properly, you shouldn't use the equals sign (you should use 'congruency,' or $\\equiv$), and you shouldn't think of it in terms of remainders (which is what it looks like you're thinking).\r\n\r\n$x \\equiv y \\bmod z$ is true if and only if $x - y$ is divisible by $z$ (or, actually, if you still want to think about it in terms of remainders, if $x$ and $y$ [i]both give the same remainder upon division by $z$[/i]). For example, $15 \\equiv 35 \\bmod 20$ because $15 - 35 = -20$, which is divisible by $20$. \r\n\r\nBut you can also say, for example, that $15 \\equiv 35 \\equiv 55 \\equiv 75 \\equiv -5 \\equiv -25 \\bmod 20$ - each of those statements is true because each of those numbers differ from each other by some multiple of $20$. \r\n\r\nAn equivalent way of thinking about it: $x \\equiv y \\bmod z$ if $x$ and $y$ end in the same digit in base $z$ (applies only if they all have the same sign). For example, $3 \\equiv 13 \\equiv 23 \\equiv 33 \\equiv 43 \\bmod 10$ because they differ from each other by $10$. \r\n\r\n\r\nAlso, when you say $15 \\equiv 5 \\bmod 20$, this is incorrect because $15 - 5 = 10$.", "Solution_6": "Yeah, I'm sure you meant $15 \\equiv -5 \\pmod 20$" } { "Tag": [], "Problem": "X, Y and Z are pairwise disjoint sets of people. The average ages of people in the sets X, Y, Z, X $ \\cup$ Y, X $ \\cup$ Z and Y $ \\cup$ Z, respectively, are as follows: 37, 23, 41, 29, 39.5, 33\r\n\r\nFind the average age of the people in set X $ \\cup$ Y $ \\cup$ Z", "Solution_1": "[hide=\"Answer (for reference)\"]$ A\\equal{}34$[/hide]" } { "Tag": [ "analytic geometry", "Mandelbrot Competition", "probability", "expected value" ], "Problem": "pkZKMbyW4UPdkRsgJ5xP+j8i+qckkW5ZznguEmtF5pAz8FyMwxsvrm2qBbia", "Solution_1": "ok.... here's the answer to your question to regional level for round three:\r\nThe question writers probably found 88, and rinted it; but afterwards they changed the answer to 6... both answers would have counted (they should have).... the test never stated if a single digit integer is a palindrome", "Solution_2": "i think both answers to the palindrome question are accepted.\r\n\r\nas for the lightbulb problem, the answer should indeed be 13/9. You interpreted the problem wrong (Or at least the wording is too vague). Our team actually thought the same way you did, at first, and also got an answer of 1. But here's why it's incorrect: looking at part iii (i think it was), they phrased the problem something like: \"prove that [some result] is true for all k >= 2\" And the thing is, with YOUR assumption that ALL lights go off, this result (i forgot what it was)would NOT be true, making that entire question faulty. \r\nSince that is obviously not the intent of the question writers, we assumed the other possible interpretation: that once a lightbulb has been touched twice, that particular light goes off and that's it. It turns out that with this interpretation part iii became solvable.\r\n\r\nbut yea, i agree, the wording of the setup was pretty vague. Our team argued about this very thing for a long time, before coming to our [ultimately correct] interpretation", "Solution_3": "pkZKMbyW4UPdkRsgJ5xP+j8i+qckkW5ZznguEmtF5pAz8FyMwxsvrm2qBbia", "Solution_4": "[quote=\"Mandelbrot Contest\"]A row of n lights is situated above the stage of the school auditorium, each controlled by a separate switch on the lightning panel offstage. It is late in the afternoon after math team practice, and all the lights are currently off. Using a calculator, an idle team member decides to choose a number from 1 to n at random, then flip the corresponding switch on the panel. He then repeats this process; if he happens to choose the same switch as before then all the lights will again be off, otherwise there will be two lights on. He continues picking numbers at random and flipping switches until an irate custodian puts an end to the activity after a total of k turns. On average, how many lights are on at this point?[/quote]The question seems pretty clear to me. The first sentence says that each light is controlled by a separate switch. The fourth sentence is referring only to the second turn -- otherwise why would there have to be zero or two lights on? If the switches are separate, then how can one switch turn many lights off? These are just ordinary switches.", "Solution_5": "I agree with Ravi -- that wording isn't ambiguous at all, you just decided it was more complicated than it really was. I doubt you'll get any credit for it. Sorry.", "Solution_6": "Our team had no problem with the interpretation as far as I know. Make sure to read problem statements carefully." } { "Tag": [ "geometry", "rectangle" ], "Problem": "A window frame consisting of 6 equal rectangles is shown\r\n\r\n[url=http://img162.imageshack.us/my.php?image=40617920db0.png][img]http://img162.imageshack.us/img162/1614/40617920db0.th.png[/img][/url]\r\n\r\na) Let the entire frame have height $ h$m and width $ y$m. If 12m of frame is available, show that y=$ \\frac{1}{4}(12-3h)$", "Solution_1": "[hide]So basically, you get the equation $ 3h+4y=12$. Next, you plug in $ 3h+4y=12$ into the equation $ y=\\frac{1}{4}(12-3h)$ and you get $ \\frac{1}{4}(4y)$, which obviously equals $ y$. [/hide]", "Solution_2": "[hide]I'm guessing the \"frame\" means all the lines in the picture..\nSince there are three vertical lines and four horizontal lines, $ 3h+4y=12$.\n$ 4y=12-3h$\n$ y=\\frac{1}{4}(12-3h)$\nIs there a b)?[/hide]" } { "Tag": [], "Problem": "In the pulley system attached to this post, shown are two masses of 10kg each. The pulley on the left is fixed to the ceiling while the pulley on the right is attached to the right mass of 10kg. What will be the accelaration of each mass? (assume the pulleys and rope are frictionless and the rope has negligible mass)", "Solution_1": "[hide=\"Answer\"]If I am in no mistake, $ a_{left}\\equal{}\\frac{2g}{3}$ and $ a_{right}\\equal{}\\frac{g}{3}$.[/hide]", "Solution_2": "I'm not sure if your answers are correct, but you did you arrive with those answers?", "Solution_3": "Let T be the tension along the rope. The left mass is labeled 1 and that on the right labeled 2. Then I used\r\n\r\n$ P\\minus{}T \\equal{} ma_{1}$, $ P\\minus{}2T \\equal{} ma_{2}$, and $ a_{1}\\equal{} 2a_{2}$.", "Solution_4": "What is $ P$? By the way, my teacher says the answer is $ 6g$", "Solution_5": "P is the weight of each mass. The answer is 6g? Then I have done something quite wrong. Can you provide your reasoning? By the way, how can the acceleration be greater than g if each body is restricted in its motion due to another body?", "Solution_6": "Well, the bodies have a mass of 10kg each, so it's possible for the acceleration to be 6g? Because gravity exerts a force of 9.81 Newtons per kilogram.\r\n\r\nI will ask my teacher for the working", "Solution_7": "remember gravity is not the only force acting on the block...." } { "Tag": [ "\\/closed" ], "Problem": "I think it would be an extremely convenient feature if when a post was made in one of someone's \"Favorites\" threads then the button would light up (similar to the private message icon).\r\n\r\nThanks,\r\n-diophantient", "Solution_1": "It used to be that the favorites were arranged in the order of the time of the last post (like it is in all fora) and just a quick look was enough to see if someone posted in one of your favorite threads but now thay are arranged according to when you added them. There is some logic in this arrangement, of course, but I would prefer to have it the old way or, even better, to be able to sort the favorites in the same way as the usual forum threads :).", "Solution_2": "I will try to add both options (although the one about lighting up the button is not that easy to implement, while the second is not that complicated) at some point in the future." } { "Tag": [ "geometry", "parallelogram", "geometry solved" ], "Problem": "Consider parallelogram $ABCD$. Point $E$ is chosen such that $BE=BC$ and $B$ lies on a segment $AE$. Line passing through $C$ and perpendicular to $BD$ intersects line through $E$ perpendicular to $AB$ in $F$. Prove that $\\angle BAF =\\angle DAF$.", "Solution_1": "This is very simolar problem. I hope it will help you.\r\n\r\n\r\n http://www.mathlinks.ro/Forum/viewtopic.php?highlight=parallelogram+AND+perpendicular&t=56498\r\n\r\n[color=red][Moderator edit: This is actually the same problem, stated in a different way.][/color]\r\n\r\n [u]Babis[/u]" } { "Tag": [ "vector", "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "Hi, I really need help for this question\r\n\r\nLet G be the group $\\mathbb{Z}^n_2$, the set of binary vectors ($a_1$,$a_2$,...,$a_n$) in which $a_i$=0 or 1. The binary operation is vector addition modulo 2, in which 1+1=0. \r\n\r\n(i) Let H be the subset of G consisting of all binary vectors with an even number of 1's. Show that H is a subgroup of G. Determine the order of G in terms of n.\r\n\r\n(ii) Construct the group table for H for n=4.\r\n\r\nThanks before", "Solution_1": "For (i)\r\nConsider $v_1,v_2\\in H$. $v_1$ has $2a$ 1's and $v_2$ has $2b$ number of 1's. Let $u$ be the number of 1's which are on the same position in $v_1$ and $v_2$.\r\nThen $v_1+v_2$ will have $1$ on a position $i$ iff $v_1$ and $v_2$ have different values on position $i$ and 0 iff $v_1$ and $v_2$ have the same value on position $i$.\r\nSo, the number of positions where $v_1$ has $1$ and $v_2$ has $0$ is $2a-u$ and the number of positions where $v_1$ has $0$ and $v_2$ has $1$ is $2b-u$, hence the number of positions where $v_1+v_2$ has $1$ is $2a+2b-2u$ which is even. \r\nSo, if $v_1,v_2\\in H\\Rightarrow v_1+v_2\\in H$, so $H$ is a stable part(i'm not sure how to call it in english) of $\\mathbb{Z}_2^n\\Rightarrow$ H is a subgroup of $\\mathbb{Z}_2^n$\r\nThe order of $G$ is $2^n$.", "Solution_2": "H closed under addition Frozen" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $P$ be a convex polygon with no two parallel sides. For each side of $P$ we consider the angle under which this side is seen from the farthest vertex in $P$. Prove that the sum of all these angles (for each side of $P$) is $180^\\circ$.", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=40281 .\r\n\r\n darij" } { "Tag": [], "Problem": "A gram of fat contains 9 calories, and a gram of carbohydrates and protein contains 4 calories. 10 pieces of McDonald's Chicken selects contains 66 grams of fat, 92 grams of carbohydrates, and 77 grams of protein. What percentage of the total amount of calories come from protein? Express your answer to the nearest tenth.", "Solution_1": "[hide]66 x 9 + 92 x 4 + 77 x 4 = 1270\n(77 x 4)/1270 = 308/1270 = [b]24.3%[/b][/hide]", "Solution_2": "[hide=\"answer\"]There are a total of $9\\times66+4\\times92+4\\times77$ calories. Of those, $4\\times77$ are from protein. That makes a total of 1270 calories, with 308 from protein. Therefore, $\\frac{1270}{308}\\approx0.2425$, which is $24.3\\%$.[/hide]", "Solution_3": "That is some sieriously unhealthy mcdonalds food.\r\nThe chicken selects are really filling.", "Solution_4": "Click Mr. Green for the answer!\r\n[hide=\" :D \"]66x9+92x4 +77x4=1270 \n(77x4)/1270\n=308/1270\n=0.2425\n=24.3%[/hide]", "Solution_5": "[hide]\n66*9=594\n92*4=368\n77*4=308\n594+368+308=1270\n$\\frac{308}{1270}\\cdot100\\approx24.25$\n\nThe answer is 24.3%[/hide]" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Find all functions $ f : \\mathbb{R} \\to \\mathbb{R}$ such that \\[ f(x\\plus{}y)f(f(x)\\minus{}y)\\equal{}xf(x)\\minus{}yf(y)\\] for all $ \\ x,y \\in \\mathbb{R}.$", "Solution_1": "[hide=\"First solution :D \"]\n$ f(x)\\equal{}x$\n$ f(x)\\equal{}0$\n[/hide]", "Solution_2": "$ f(0) \\equal{} c$, $ f(x) \\equal{} 0$ for all real number $ x$ except 0 could be solution too :P", "Solution_3": "I'll rewrite my solution in another way:\r\nSolution:\r\n$ x \\equal{} 0$ in the main equation gives us:\r\n$ f(y)f(f(0) \\minus{} y) \\equal{} \\minus{} yf(y)$,\r\nso it is either $ f(y) \\equal{} 0$,or $ f(c \\minus{} y) \\equal{} \\minus{} y$,where $ c \\equal{} f(0)$.\r\nHence $ f(c \\minus{} y) \\equal{} \\minus{} y$,for all $ y\\in\\mathbb{R}$,such that $ f(y)\\neq 0$.\r\n$ y\\rightarrow c \\minus{} y$,so $ f(y) \\equal{} y \\minus{} c$,for all $ y$,such that $ f(c \\minus{} y)\\neq 0$.\r\nIf $ f(x) \\equal{} x \\minus{} c$,for some $ x\\neq c$ then we make substitution\r\n$ y \\equal{} 0$:\r\n$ (x \\minus{} c)f(x \\minus{} c) \\equal{} x(x \\minus{} c)$.\r\nIf $ f(x \\minus{} c) \\equal{} x$ and $ x\\neq 0$,then $ c \\equal{} 0$.\r\nIf $ x \\equal{} 0$,then $ c \\equal{} 0$.\r\nSo $ f(x) \\equal{} x$,if $ f( \\minus{} x)\\neq 0$.\r\nSuppose that there exist $ x_0,y_0\\in\\mathbb{R}$,such that $ f(x_0) \\equal{} 0,f(y_0) \\equal{} y_0,y_0\\neq 0$.\r\nIf we put $ x \\equal{} x_0,y \\equal{} y_0$:\r\n$ f(x_0 \\plus{} y_0)f( \\minus{} y_0) \\equal{} \\minus{} y_0^2$\r\nIt is obvious that $ f( \\minus{} y_0),f(x_0 \\plus{} y_0)\\neq 0$,thus $ f(x_0 \\plus{} y_0) \\equal{} x_0 \\plus{} y_0$,hence $ x_0 \\equal{} 0$.\r\nSo if $ f(x) \\equal{} x$,for some $ x\\in\\mathbb{R}$,then it is true for all $ x\\in\\mathbb{R}$.\r\nIf $ f(y) \\equal{} 0$,then it is true for all $ y\\in\\mathbb{R}$,except $ y \\equal{} 0$.\r\nSo the answer is $ f(y) \\equal{} y$ and $ f(y) \\equal{} 0,f(0) \\equal{} c,y\\neq 0$.", "Solution_4": "[quote=\"Erken\"]Very strange,that an easy problem like this was at Japan MOF.[/quote]\r\n\r\nI agree with you.", "Solution_5": "[quote=\"kunny\"][quote=\"Erken\"]Very strange,that an easy problem like this was at Japan MOF.[/quote]\n\nI agree with you.[/quote]\r\nWhere is the other problems?\r\nThank you.", "Solution_6": "[quote=\"Erken\"]\nnow since $ f$ is injective and $ f(2x)\\neq 0,$,then\n$ f(f(x) \\minus{} x) \\equal{} 0 \\equal{} f(0)$.\n[/quote]\r\n\r\nWhy is $ f$ injective?\r\n\r\nIn fact, there are strange solutions:\r\n\r\n$ f(x) \\equal{} 0 (x \\neq 0), f(0) \\equal{} c$", "Solution_7": "[quote=\"ringos\"][quote=\"Erken\"]\nnow since $ f$ is injective and $ f(2x)\\neq 0,$,then\n$ f(f(x) \\minus{} x) \\equal{} 0 \\equal{} f(0)$.\n[/quote]\n\nWhy is $ f$ injective?\n\nIn fact, there are strange solutions:\n\n$ f(x) \\equal{} 0 (x \\neq 0), f(0) \\equal{} c$[/quote]\r\nIt is injective,because $ f(f(x)) \\equal{} x$.\r\nOh yes,i forgot to consider $ f(0)$,when $ f(x)\\equiv 0$,for all $ x\\in\\mathbb{R},x\\neq 0$.So $ f(0)\\equal{}c,f(x)\\equal{}0,\\forall x\\in\\mathbb{R},x\\neq 0$ is a solution too.Thank you.", "Solution_8": "[quote=\"Erken\"][quote=\"ringos\"][quote=\"Erken\"]\nnow since $ f$ is injective and $ f(2x)\\neq 0,$,then\n$ f(f(x) \\minus{} x) \\equal{} 0 \\equal{} f(0)$.\n[/quote]\n\nWhy is $ f$ injective?\n\nIn fact, there are strange solutions:\n\n$ f(x) \\equal{} 0 (x \\neq 0), f(0) \\equal{} c$[/quote]\nIt is injective,because $ f(f(x)) \\equal{} x$.\nOh yes,i forgot to consider $ f(0)$,when $ f(x)\\equiv 0$,for all $ x\\in\\mathbb{R},x\\neq 0$.So $ f(0) \\equal{} c,f(x) \\equal{} 0,\\forall x\\in\\mathbb{R},x\\neq 0$ is a solution too.Thank you.[/quote]\r\n\r\n$ f(f(x)) \\equal{} x$ for some $ x$, not for all $ x$.", "Solution_9": "[quote=\"Erken\"][quote=\"kunny\"][quote=\"Erken\"]Very strange,that an easy problem like this was at Japan MOF.[/quote]\n\nI agree with you.[/quote]\nWhere is the other problems?\nThank you.[/quote]\r\n\r\nHere are.\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=188109]Problem 1[/url]\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=188111]Problem 3[/url]\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=188110]Problem 5[/url]", "Solution_10": "I've posted another solution,i hope it is clear now.See above.", "Solution_11": "This problem isnt so easy as it looks like, because just like ringos wrote f(f(x))=x just for some x. I will post my solution as soon as i have time. I just hope i didnt make a mistake.", "Solution_12": "[quote=\"primoz2\"]This problem isnt so easy as it looks like, because just like ringos wrote f(f(x))=x just for some x. I will post my solution as soon as i have time. I just hope i didnt make a mistake.[/quote]\r\nI've already changed my solution,i didn't use $ f(f(x))\\equal{}x$,i replaced the old proof,by new,see above.", "Solution_13": "[quote=\"Erken\"]I've posted another solution,i hope it is clear now.See above.[/quote]\r\n\r\n[color=green]Where is the first one, Erken?[/color]", "Solution_14": "I replaced it by a new proof few minutes ago.", "Solution_15": "[color=green]Uhm, and what about it ? Did you have a mistake with this solution? Show it again please, friend.[/color]", "Solution_16": "[quote=\"Erken\"]\nIf $ x \\equal{} 0$,then $ c \\equal{} 0$.\n[/quote]\r\n\r\nCould you explain this part?\r\n\r\nI couldn't solve this problem in the competition.", "Solution_17": "[quote=\"primoz2\"]This problem isnt so easy as it looks like, because just like ringos wrote f(f(x))=x just for some x. I will post my solution as soon as i have time. I just hope i didnt make a mistake.[/quote]\r\n\r\nAt first sight, I thought that this problem is not difficult, but when I began to solve, it seemed to be unable to solve so easily.", "Solution_18": "[quote=\"ringos\"][quote=\"Erken\"]\nIf $ x \\equal{} 0$,then $ c \\equal{} 0$.\n[/quote]\n\nCould you explain this part?\n\nI couldn't solve this problem in the competition.[/quote]\r\nWe have that $ f(x)\\equal{}x\\minus{}c$,so if $ x\\equal{}0$,then $ f(0)\\equal{}\\minus{}c\\equal{}\\minus{}f(0)$,hence $ c\\equal{}0$.So in any case $ c\\equal{}0$.", "Solution_19": "[quote=\"Erken\"]We have that $ f(x) \\equal{} x \\minus{} c$,so if $ x \\equal{} 0$,then $ f(0) \\equal{} \\minus{} c \\equal{} \\minus{} f(0)$,hence $ c \\equal{} 0$.So in any case $ c \\equal{} 0$.[/quote]\r\n\r\nThank you, I understood. Nice solution.", "Solution_20": "[quote=\"quangpbc\"][color=green]Uhm, and what about it ? Did you have a mistake with this solution? Show it again please, friend.[/color][/quote]\r\nNo,i think this solution is correct,friend. :)\r\nP.S:I've got a nice number of posts:$ 666$. :lol:", "Solution_21": "[quote=\"Erken\"]\nIf $ f(x) \\equal{} x \\minus{} c$,for some $ x\\neq c$ then we make substitution\n$ y \\equal{} 0$:\n$ (x \\minus{} c)f(x \\minus{} c) \\equal{} x(x \\minus{} c)$.\n[/quote]\r\n\r\nWhere do you substitute $ y\\equal{}0$?", "Solution_22": "[quote=\"Jumbler\"][quote=\"Erken\"]\nIf $ f(x) \\equal{} x \\minus{} c$,for some $ x\\neq c$ then we make substitution\n$ y \\equal{} 0$:\n$ (x \\minus{} c)f(x \\minus{} c) \\equal{} x(x \\minus{} c)$.\n[/quote]\n\nWhere do you substitute $ y \\equal{} 0$?[/quote]\r\nIn the main equation", "Solution_23": "Thank for your post. Thank you very much.", "Solution_24": "[hide=\"Answer\"]$ f(x)\\equal{}x$ and $ f(x)\\equal{}\\left\\{\n\\begin{array}{ll}\nc\\ (x\\equal{}0)\\ c: arbiterary\\ real\\ numbers &\\quad \\\\\n0\\ (x\\neq 0) &\\quad\n\\end{array}\n\\right.$[/hide]", "Solution_25": "[quote=\"kunny\"][hide=\"Answer\"]$ f(x) \\equal{} x$ and $ f(x) \\equal{} \\left\\{ \\begin{array}{ll} c\\ (x \\equal{} 0)\\ c: arbiterary\\ real\\ numbers & \\quad \\\\\n0\\ (x\\neq 0) & \\quad \\end{array} \\right.$[/hide][/quote]\r\nI've already solved this problem,see the previous list :wink:", "Solution_26": "[quote=\"Erken\"]I'll rewrite my solution in another way:\nSolution:\n$ x \\equal{} 0$ in the main equation gives us:\n$ f(y)f(f(0) \\minus{} y) \\equal{} \\minus{} yf(y)$,\nso it is either $ f(y) \\equal{} 0$,or $ f(c \\minus{} y) \\equal{} \\minus{} y$,where $ c \\equal{} f(0)$................\n......................................\n[/quote]\r\n\r\n I'm afraid that this conclution is not valid. Generally, if $ f(x)g(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ , we can not take $ f(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ or $ g(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ .Do you have something more in mind ?\r\n\r\n Babis", "Solution_27": "See here (at the last post ) for a nice straightforward solution \r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=194064", "Solution_28": "[quote=\"silouan\"]See here (at the last post ) for a nice straightforward solution \n\nhttp://www.mathlinks.ro/viewtopic.php?t=194064[/quote]\r\n\r\n Thank you Silouan!\r\n Good luck by the IMO 2008 !!!\r\n\r\n [b]NOTE[/b]\r\n\r\n The solution at the last post(in the greek forum) is in english , so everybody can read it.\r\n\r\n Babis", "Solution_29": "[quote=\"kunny\"]Find all functions $ f : \\mathbb{R} \\mapsto \\mathbb{R}$ such that $ f(x \\plus{} y)f(f(x) \\minus{} y) \\equal{} xf(x) \\minus{} yf(y)$ for all $ \\ x,y \\in \\mathbb{R}.$[/quote]\r\n\r\n+ $ f \\equal{} 0$\r\n\r\n+ $ f \\neq 0$\r\n\r\n$ y \\equal{} 0\\ \\equal{} > \\ f(x).f(f(x)) \\equal{} x.f(x) \\ \\equal{} > \\ f(f(x)) \\equal{} x \\ \\ \\ (1)$\r\n\r\n$ y: \\equal{} f(x)\\ \\equal{} > \\ f(x \\plus{} f(x)).f(0) \\equal{} x.f(x) \\minus{} f(x).f(f(x))\\ \\ \\ (2)$\r\n\r\n$ (1)\\ \\ vs\\ \\ (2) \\ \\ f(0) \\equal{} 0$\r\n\r\n$ x \\equal{} 0\\ \\equal{} > \\ f(y).f( \\minus{} y) \\equal{} \\minus{} y.f(y) \\ \\equal{} > \\ f( \\minus{} y) \\equal{} \\minus{} y\\ \\equal{} > \\ f(x) \\equal{} x$\r\n\r\nTwo solutions : $ f(x) \\equal{} 0 \\ \\ ; \\ \\ f(x) \\equal{} x$", "Solution_30": "[quote=\"hophinhan\"] \nTwo solutions : $ f(x) \\equal{} 0 \\ \\ ; \\ \\ f(x) \\equal{} x$[/quote]\r\n\r\nYou should read the entire thread.\r\nThere exist other solutions :\r\n\r\n$ f(0)\\equal{}a\\neq 0$ and $ f(x)\\equal{}0$ $ \\forall x\\neq 0$", "Solution_31": "[quote=\"stergiu\"][quote=\"Erken\"]I'll rewrite my solution in another way:\nSolution:\n$ x \\equal{} 0$ in the main equation gives us:\n$ f(y)f(f(0) \\minus{} y) \\equal{} \\minus{} yf(y)$,\nso it is either $ f(y) \\equal{} 0$,or $ f(c \\minus{} y) \\equal{} \\minus{} y$,where $ c \\equal{} f(0)$................\n......................................\n[/quote]\n\n I'm afraid that this conclution is not valid. Generally, if $ f(x)g(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ , we can not take $ f(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ or $ g(x) \\equal{} 0 , \\forall x \\in \\mathbb R$ .Do you have something more in mind ?\n\n Babis[/quote]\r\n\r\nPlease read it through more carefully, I didn't make such a conlusion.", "Solution_32": "we have 2 case first is f(x)=0 second is not.\r\nfor second;\r\nlet y=x so we have $ f(2x).f(f(x)\\minus{}x)\\equal{}0$ and plugging y=0 we have $ f(f(x))\\equal{}x$\r\nso taking f of two sides in first equation we have $ f(x)\\equal{}f(0)\\plus{}x$ writing it to first function we get $ f(x)\\equal{}x$ or from first case $ f(x)\\equal{}0$", "Solution_33": "I didn't know that this problem is from Japan.\r\n\r\nThank you for posting the problem that I wanted.", "Solution_34": "Oh!!I must mention that there are too many useless posts of this problem.", "Solution_35": "[quote=\"kunny\"]Find all functions $ f : \\mathbb{R} \\mapsto \\mathbb{R}$ such that $ f(x\\plus{}y)f(f(x)\\minus{}y)\\equal{}xf(x)\\minus{}yf(y)\\quad (1)$ for all $ \\ x,y \\in \\mathbb{R}.$[/quote]\nLet $a=f(0).$ Replacing $y=0,$ we get \\[f(x)\\cdot f\\big(f(x)\\big)=x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (2)\\] Now, replacing $y=f(x)$ and using $(2),$ we get \\[a\\cdot f\\big(x+f(x)\\big)=0.\\quad (3)\\] There are two cases to consider:\n\n[b]Case 1: [/b]$\\mathbf{a=0}.$ Replacing $x=0$ in $(1),$ we get \\[f(y)\\cdot f({-y})=-y\\cdot f(y),\\quad \\forall y \\in \\mathbb R. \\quad (4)\\] Replacing $y$ by $-y$ in $(4),$ we get \\[-y\\cdot f(y)=f(y)\\cdot f({-y})=y\\cdot f({-y}),\\] so by combining with the fact $f(0)=0,$ we get \\[f({-y})=-f(y),\\quad \\forall y \\in \\mathbb R. \\quad (5)\\] Using this result in $(4),$ we obtain \\[f(y)=0\\vee f(y)=y,\\quad \\forall y \\in \\mathbb R. \\quad (6)\\] Assume that, there exist $u,\\,v \\ne 0$ such that $f(u)=u$ and $f(v)=0.$ Replacing $x=u,\\,y=v$ in $(1),$ we get \\[f(u+v)\\cdot f(u-v)=u^2.\\] Since $u^2 \\ne 0,$ we must have $f(u+v) \\ne 0$ and $f(u-v)\\ne 0.$ Therefore, $f(u+v)=u+v,\\, f(u-v)=u-v.$ It follows that \\[u^2=(u+v)(u-v)=u^2-v^2,\\] and hence $v=0,$ a contradiction. So, we must have $f(x)=0,\\, \\forall x \\in \\mathbb R$ or $f(x)=x,\\, \\forall x \\in \\mathbb R.$ It is easy to check that these functions satisfy our condition.\n\n[b]Case 2: $\\mathbf{ a\\ne 0}.$[/b] From $(3),$ we get \\[f\\big( x+f(x)\\big) =0,\\quad \\forall x \\in \\mathbb R. \\quad (7)\\] Now, by replacing $x$ by $x+f(x),$ $y$ by $x$ in $(1),$ we get \\[f\\big(2x+f(x)\\big)\\cdot f({-x})=-x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (8)\\] Replacing $y$ by $x+ f(x),$ we also have \\[f\\big( 2x+f(x)\\big)\\cdot f({-x})=x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (9)\\] Combining $(8)$ and $(9),$ we get \\[f(x)=0,\\quad \\forall x \\ne 0.\\] Therefore, \\[f(x)=\\begin{cases} 0 &\\text{if } x \\ne 0 \\\\ a &\\text{if } x =0 \\end{cases}\\] It is easy to check that this function satisfies the given condition.", "Solution_36": "Let us denote the given statement by $P(x,y)$. Then,\n$P(x,f(x))\\Rightarrow f(x+f(x))f(0)=xf(x)-f(x)f(f(x))$.\n$P(x,0)\\Rightarrow f(x)f(f(x))=xf(x)$.\nSo, $f(x+f(x))f(0)=0 \\forall x\\in \\mathbb{R}$.\nSo we have three cases here.\n\n[b]Case 1:[/b] When $f(x+f(x))=0 \\forall x\\in \\mathbb{R}$.\n$P(x+f(x),-x)\\Rightarrow f(f(x))f(x)=xf(-x)$. But $f(f(x))f(x)=xf(x)$. So, $f(x)=f(-x)\\forall x\\in \\mathbb{R}$.\n$P(x,0)\\Rightarrow f(0)f(x+f(x))=0=2xf(x)$. This yields us the solution:\n$f(x)=0\\forall x\\neq 0$ and $f(0)=c\\in \\mathbb{R}$. In particular, if $c=0$, then it is the constant solution to the given equation.\n\n[b]Case 2:[/b] When $f(x+f(x))\\neq 0\\forall x\\neq 0$.\nThen $f(0)=0$. The only constant solution belongs to the family of solution in case $1$. So let $f$ be non-constant.\n$P(x,-x)\\Rightarrow f(x)=-f(-x)\\forall x\\in \\mathbb{R}$.\n$P(0,y)\\Rightarrow f(y)(f(y)-y)=0 \\forall y\\in \\mathbb{R}.....(*)$\nNow since $f$ is non-constant, there exists $d\\neq 0$ such that $f(d)\\neq 0$. Then from equation$(*)$, $f(d)=d$.\nIf there is a $k\\neq 0$ such that $f(k)=0$, then,\n$P(k,d-k)\\Rightarrow df(k-d)=(d-k)f(k-d)\\Rightarrow kf(k-d)=0\\Rightarrow f(k-d)=0$. So,\n$P(k-d,d)\\Rightarrow 0=-df(d)\\Rightarrow d^{2}=0\\Rightarrow d=0$ which is a contradiction.\nSo, $r=0$ if and only if $f(r)=0$. Thus from $(*)$ and from $f(0)=0$, we have that $f(y)=y\\forall y\\in \\mathbb{R}$.\n\n[b]Case 3:[/b] When $\\exists a,b$ such that $f(a+f(a))=0$ and $f(b+f(b))\\neq 0$, where both $a,b\\neq 0$.\nAccording to our previous results, $f(b+f(b))f(0)=0$. This means that $f(0)=0$. So,\n$P(0,y)\\Rightarrow f(y)[f(y)-y]=0$. Set $y=b+f(b)$, then we have $f(b+f(b))=b+f(b)$.\nLet $d=b+f(b)$ and $k=a+f(a)$. Now unless $k=0$, we will have $d=0$, i.e. $f(b+f(b))=b+f(b)=0$ which contradicts our assumption. So, $k=f(a)+a=0$. Thus,\n$P(a+f(a),-a)\\Rightarrow f(a)=f(-a)$. Then $f(-a)=f(a)=-a$. So, \n$P(-a,a)\\Rightarrow 0=-af(-a)-af(a)=2a^{2}\\Rightarrow a=0$ which is also a contradiction. So no such $a,b$ exist.\nThus either $f(x+f(x))=0\\forall x\\in \\mathbb{R}$, or $f(x+f(x))$ is non-zero for all non-zero $x$.\n\nSo finally the solutions to the given equations are:\n$f(x)=0\\forall x\\neq 0$ and $f(0)=c\\in \\mathbb{R}$\nor,\n$f(y)=y\\forall y\\in \\mathbb{R}$.\n\nIt is easy to check that these are indeed the solutions.", "Solution_37": "y=0 => f(f(x))=x (1)\ny=f(x) => f(0)=0 (2)\ny=-x => f(x)= -f(-x) (3)\nIf you change x and y =>\nf(x+y)f(f(y)-x)=yf(y)_xf(x)\nAnd with compare to the general equation and (3)\n=> f(x-f(y))=f(f(x)-y)\nAnd if you put y=0\n=> f(x-f(0))=f(f(x)) => (2) f(x)=f(f(x)) => (1) f(x)=x", "Solution_38": "[quote=Erken]I'll rewrite my solution in another way:\nSolution:\n$ x \\equal{} 0$ in the main equation gives us:\n$ f(y)f(f(0) \\minus{} y) \\equal{} \\minus{} yf(y)$,\nso it is either $ f(y) \\equal{} 0$,or $ f(c \\minus{} y) \\equal{} \\minus{} y$,where $ c \\equal{} f(0)$.\nHence $ f(c \\minus{} y) \\equal{} \\minus{} y$,for all $ y\\in\\mathbb{R}$,such that $ f(y)\\neq 0$.\n$ y\\rightarrow c \\minus{} y$,so $ f(y) \\equal{} y \\minus{} c$,for all $ y$,such that $ f(c \\minus{} y)\\neq 0$.\nIf $ f(x) \\equal{} x \\minus{} c$,for some $ x\\neq c$ then we make substitution\n$ y \\equal{} 0$:\n$ (x \\minus{} c)f(x \\minus{} c) \\equal{} x(x \\minus{} c)$.\nIf $ f(x \\minus{} c) \\equal{} x$ and $ x\\neq 0$,then $ c \\equal{} 0$.\n\nsorry but I dont understand why here you get c=0\nexplain it please :oops: ", "Solution_39": "Nice problem!\n---------------\n$ P(x,0)\\implies f(x)f(f(x))=xf(x) $\n-------------\n[b][color=#f00]Case 1:[/color][/b]\n$ f(x)=0 $\n---------------\n[b][color=#f00]Case 2:[/color][/b]$ x\\neq 0\\implies f(x)\\neq 0$\n$ f(f(x))=x $\n--------------\n$ f(f(x))=x\\implies bijective $\n$ P(x,x)\\implies f(2x)f(f(x)-x)=0 $\n$ f(2x)\\neq 0\\implies f(f(x)-x)=0 $\n$$ f(f(x)-x)=0\\implies f(f(f(x)-x))=f(0)=0=f(x)-x\\implies f(x)=x $$\n$ Then $\n\n$ f(x)=x,f(x)=0 $", "Solution_40": "[quote=silouan]See here (at the last post ) for a nice straightforward solution \n\nhttp://www.mathlinks.ro/viewtopic.php?t=194064[/quote]\nin order not to miss the solution mentioned by [b]silouan [/b] as link went dead, here is the correct [url=https://artofproblemsolving.com/community/c267h194064p1066876]one[/url] \nand the solution given by [b]savvas_michael[/b] [hide=follows]\n[quote=savvas_michael]Set $ y \\equal{} 0$ in the given equation and get $ f(x) \\cdot f(f(x)) \\equal{} x \\cdot f(x) \\ (*)$.\n\nTherefore there exists at least one $ c \\in R$ so that $ f(c) \\equal{} 0$.\n \nNow plug in $ y \\equal{} f(x)$ to get according to (*) that $ f(0) \\cdot f(f(x) \\plus{} x) \\equal{} 0$\n \nand then $ y \\equal{} \\minus{} x$ to get $ f(0)f(f(x) \\plus{} x) \\equal{} x(f(x) \\plus{} f( \\minus{} x))$ and thus\n\n$ f( \\minus{} x) \\equal{} \\minus{} f(x)$, for every $ x \\neq 0$.\n\nPlugging in now $ x \\equal{} c$ and $ y \\neq 0$ we get $ f(y) \\cdot (f(y \\plus{} c) \\minus{} y) \\equal{} 0$. \n\nAs a result for every $ x \\neq 0$ we have either $ f(x) \\equal{} 0$ or $ f(x \\plus{} c) \\equal{} x$. \n\nLet those $ x \\neq 0$ satisfying the first equation form set $ A$ and the remaining $ x \\neq 0$ form set $ B$. \n\nWe distinguish cases:\n\n1. $ A$ is empty. Then we must have $ c \\equal{} 0$ and therefore $ f(x) \\equal{} x,\\ \\forall x \\in R$.\n\n2. $ A$ is non-empty. If $ B$ is empty then $ f$ is identically zero everywhere except for its value at $ 0$ which can be arbitrary. \n\nSuppose now that $ B$ is non-empty. We will reach a contradiction. \n\nPick $ a \\in A$ and $ b \\in B$ and plug in $ x \\equal{} a \\plus{} b$ in (*) to get\n\n$ f(b) \\equal{} b\\plus{}a$. Therefore it is obvious that $ A$ has a single element $ a$. Then $ f(x)\\equal{}x\\plus{}a, \\forall x \\neq 0,a$\n\nand hence as $ f$ is odd in $ R*$ we obtain $ a\\equal{}0$ which is a contradiction.[/quote]\n\n\n\n[/hide]", "Solution_41": "[quote=CanVQ][quote=\"kunny\"]Find all functions $ f : \\mathbb{R} \\mapsto \\mathbb{R}$ such that $ f(x\\plus{}y)f(f(x)\\minus{}y)\\equal{}xf(x)\\minus{}yf(y)\\quad (1)$ for all $ \\ x,y \\in \\mathbb{R}.$[/quote]\nLet $a=f(0).$ Replacing $y=0,$ we get \\[f(x)\\cdot f\\big(f(x)\\big)=x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (2)\\] Now, replacing $y=f(x)$ and using $(2),$ we get \\[a\\cdot f\\big(x+f(x)\\big)=0.\\quad (3)\\] There are two cases to consider:\n\n[b]Case 1: [/b]$\\mathbf{a=0}.$ Replacing $x=0$ in $(1),$ we get \\[f(y)\\cdot f({-y})=-y\\cdot f(y),\\quad \\forall y \\in \\mathbb R. \\quad (4)\\] Replacing $y$ by $-y$ in $(4),$ we get \\[-y\\cdot f(y)=f(y)\\cdot f({-y})=y\\cdot f({-y}),\\] so by combining with the fact $f(0)=0,$ we get \\[f({-y})=-f(y),\\quad \\forall y \\in \\mathbb R. \\quad (5)\\] Using this result in $(4),$ we obtain \\[f(y)=0\\vee f(y)=y,\\quad \\forall y \\in \\mathbb R. \\quad (6)\\] Assume that, there exist $u,\\,v \\ne 0$ such that $f(u)=u$ and $f(v)=0.$ Replacing $x=u,\\,y=v$ in $(1),$ we get \\[f(u+v)\\cdot f(u-v)=u^2.\\] Since $u^2 \\ne 0,$ we must have $f(u+v) \\ne 0$ and $f(u-v)\\ne 0.$ Therefore, $f(u+v)=u+v,\\, f(u-v)=u-v.$ It follows that \\[u^2=(u+v)(u-v)=u^2-v^2,\\] and hence $v=0,$ a contradiction. So, we must have $f(x)=0,\\, \\forall x \\in \\mathbb R$ or $f(x)=x,\\, \\forall x \\in \\mathbb R.$ It is easy to check that these functions satisfy our condition.\n\n[b]Case 2: $\\mathbf{ a\\ne 0}.$[/b] From $(3),$ we get \\[f\\big( x+f(x)\\big) =0,\\quad \\forall x \\in \\mathbb R. \\quad (7)\\] Now, by replacing $x$ by $x+f(x),$ $y$ by $x$ in $(1),$ we get \\[f\\big(2x+f(x)\\big)\\cdot f({-x})=-x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (8)\\] Replacing $y$ by $x+ f(x),$ we also have \\[f\\big( 2x+f(x)\\big)\\cdot f({-x})=x\\cdot f(x),\\quad \\forall x \\in \\mathbb R. \\quad (9)\\] Combining $(8)$ and $(9),$ we get \\[f(x)=0,\\quad \\forall x \\ne 0.\\] Therefore, \\[f(x)=\\begin{cases} 0 &\\text{if } x \\ne 0 \\\\ a &\\text{if } x =0 \\end{cases}\\] It is easy to check that this function satisfies the given condition.[/quote]\n\nCanVQ, Perfect !\n\n" } { "Tag": [ "logarithms", "trigonometry", "function", "calculus", "derivative" ], "Problem": "$0 \\leq x \\leq \\pi$ find $x$ such that,\r\n$\\log(\\sin(x))=\\sin(\\log(x))$", "Solution_1": "[quote=\"mathpk\"]$0 \\leq x \\leq \\pi$ find $x$ such that,\n\\[ \\log \\sin x = \\sin \\log x \\][/quote]\r\nI'm sure that the proof uses function, but I cannot find the function's derivative now, just because I have no paper and ball point pen!", "Solution_2": "[quote=\"mathpk\"]$0 \\leq x \\leq \\pi$ find $x$ such that,\n$\\log(\\sin(x))=\\sin(\\log(x))$[/quote]\r\nIf you graph these functions, you'll see that this is never true. Is that what you want us to prove?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a$, $ b$, and $ c$ be positive real numbers such that $ ab + bc + ca = 3$. Prove the inequality\r\n\\[ 3 + \\sum_{\\mathrm{\\cyc}} (a - b)^2 \\ge \\frac {a + b^2c^2}{b + c} + \\frac {b + c^2a^2}{c + a} + \\frac {c + a^2b^2}{a + b} \\ge 3.\r\n\\]", "Solution_1": "[quote=\"Raja Oktovin\"]Let $ a$, $ b$, and $ c$ be positive real numbers such that $ ab + bc + ca = 3$. Prove the inequality\n\\[ 3 + \\sum_{\\mathrm{\\cyc}} (a - b)^2 \\ge \\frac {a + b^2c^2}{b + c} + \\frac {b + c^2a^2}{c + a} + \\frac {c + a^2b^2}{a + b} \\ge 3.\\]\n[/quote]", "Solution_2": "I kinda remember that the left side could be changed into the following stronger inequality:\r\n\\[ 3\\plus{}\\frac12\\sum_{cyc}(a\\minus{}b)^2\\geq\\sum_{cyc}\\frac{a\\plus{}b^2c^2}{b\\plus{}c}\\]\r\nBut then I can't prove it. Anybody?", "Solution_3": "[quote=\"wangsacl\"]I kinda remember that the left side could be changed into the following stronger inequality:\n\\[ 3 \\plus{} \\frac12\\sum_{cyc}(a \\minus{} b)^2\\geq\\sum_{cyc}\\frac {a \\plus{} b^2c^2}{b \\plus{} c}\\]\nBut then I can't prove it. Anybody?[/quote]\r\nThis ineq is equivalent to :$ {a^2} \\plus{} {b^2} \\plus{} {c^2} \\ge \\sum {\\frac {a}{{b \\plus{} c}} \\plus{} } \\sum {\\frac {{{b^2}{c^2}}}{{b \\plus{} c}}}$\r\nEasy to see that : $ a \\plus{} b \\plus{} c \\ge 3$\r\nNotice :\r\n$ \\frac {a}{{b \\plus{} c}} \\equal{} \\frac {{a(ab \\plus{} bc \\plus{} ca)}}{{3(b \\plus{} c)}} \\equal{} \\frac {{{a^2}}}{3} \\plus{} \\frac {{abc}}{{3(b \\plus{} c)}} \\le \\frac {{{a^2}}}{3} \\plus{} \\frac {{a(b \\plus{} c)}}{{12}}$\r\nAnd $ \\frac {{{b^2}{c^2}}}{{b \\plus{} c}} \\le \\frac {{bc(b \\plus{} c)}}{4}$\r\nAnd we need to prove that:\r\n$ \\frac {{2({a^2} \\plus{} {b^2} \\plus{} {c^2})}}{3} \\ge \\frac {1}{2} \\plus{} \\frac {{ab(a \\plus{} b) \\plus{} bc(b \\plus{} c) \\plus{} ca(c \\plus{} a)}}{4}$ \r\n $ \\Leftrightarrow \\frac {{2({a^2} \\plus{} {b^2} \\plus{} {c^2})}}{3} \\plus{} \\frac {{3abc}}{4} \\ge \\frac {1}{2} \\plus{} \\frac {{(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca)}}{4} \\equal{} \\frac {1}{2} \\plus{} \\frac {{3(a \\plus{} b \\plus{} c)}}{4}$ \r\nBy Cchur ineq, we have : $ \\frac {1}{4}({a^2} \\plus{} {b^2} \\plus{} {c^2} \\plus{} 3abc) \\ge \\frac {1}{4}({a^2} \\plus{} {b^2} \\plus{} {c^2} \\plus{} \\frac {{9abc}}{{a \\plus{} b \\plus{} c}}) \\ge \\frac {3}{2}$\r\nThe proof'll complete if we can show that\r\n$ \\frac {{5({a^2} \\plus{} {b^2} \\plus{} {c^2})}}{{12}} \\plus{} 1 \\ge \\frac {{3(a \\plus{} b \\plus{} c)}}{4}$\r\nEasy to see that $ LHS \\ge \\frac {{5{{(a \\plus{} b \\plus{} c)}^2}}}{{36}}$\r\nAnd from the ineq $ \\frac {{5{{(a \\plus{} b \\plus{} c)}^2}}}{{36}} \\plus{} 1 \\ge \\frac {{3(a \\plus{} b \\plus{} c)}}{4} \\Leftrightarrow (a \\plus{} b \\plus{} c \\minus{} 3)(a \\plus{} b \\plus{} c \\minus{} \\frac {{12}}{5}) \\ge 0$ ( which is true )\r\nShow that the ineq is true. So we are done !" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all positive integer solutions to $a^{2}=b^{3}+1$.", "Solution_1": "someone...", "Solution_2": "$a^{2}= b^{3}+1 = (b+1)(b^{2}-b+1)$\r\n$(b+1,b^{2}-b+1) | (b^{2}-b+1-(b+1)(b-2),b+1) = (3,b+1) \\in \\{1,3 \\}$\r\n\r\nDivide into cases :\r\na) $(b+1,b^{2}-b+1) = 1$, then $b^{2}-b+1$ is a square.\r\nBut $(b-1)^{2}< b^{2}-b+1 < b^{2}$, otherwise $b = 1 \\Rightarrow a^{2}= 2$ which gives no integer solution\r\n\r\nb) $(b+1,b^{2}-b+1) = 3$, then $b+1 = 3p^{2}$ and $b^{2}-b+1 = 3q^{2}$\r\nWe have $3p^{4}-3p^{2}+1 = q^{2}$\r\n\r\ni) $q = 3k-1 \\Rightarrow p^{2}(p^{2}-1) = k(3k-2) \\Rightarrow (2p^{2}-1)^{2}= s(3s+2)$ with $k \\in \\mathbb{N}, s = 2k-1$\r\nSince $(s,3s+2) = 1$, then $s = m^{2}, 3s+2 = n^{2}$ and we have contradiction in $mod 3$\r\n\r\nii) $q = 3k+1 \\Rightarrow p^{2}(p^{2}-1)=k(3k+2) \\Rightarrow (2p^{2}-1)^{2}= s(3s-2)$ with $k \\in \\mathbb{N}, s = 2k+1$\r\nSince $(s,3s-2) = 1$, then $s = m^{2}, 3s-2 = n^{2}\\Rightarrow 3m^{2}-n^{2}= 2$\r\nI think we use the similar way Andrei Negut used in proving the previous problem... One of the solution is $(3,2)$", "Solution_3": "AFAIK it's definitely NOT easier then the previous one. The previous one is an excercise for these people who are familiar with $Z[i]$ but for this one probably all solutions are based on the elliptic curves (if I remember correctly one can also make it elementary but the idea will be the same and it won't be easy).", "Solution_4": "you're right. my solution wasn't correct. :blush:", "Solution_5": "[quote=\"RDeepMath91\"]\ni) $q = 3k-1 \\Rightarrow p^{2}(p^{2}-1) = k(3k-2) \\Rightarrow (2p^{2}-1)^{2}= s(3s+2)$ with $k \\in \\mathbb{N}, s = 2k-1$\nSince $(s,3s+1) = 1$, [/quote]\r\n\r\nyou wrote $3s+1$ instead of $3s+2$ in the last line above. so your solution has some gaps to be filled in.", "Solution_6": "Yup, thx for correcting.. \r\nBut the answer is $1$ since $s = 2k-1$ and $(s,2) = 1$, so $(s,3s+2) = 1$ :roll:", "Solution_7": "[quote=\"RDeepMath91\"]I think we use the similar way Andrei Negut used in proving the previous problem... One of the solution is $(3,2)$[/quote]\r\nSorry, I can't understand this step,Could you explain more ,RDeepMath91?", "Solution_8": "I mean if you read the link given by someone (I forgot) in the previous topic : 'diophantine equation'. There are 2 solutions, but Andrei Negut gave a solution with a lemma concerning Pell's equation.. So, I think we can use a similar way he used to find the Pell's form since in my solution, it hasn't been in the correct form of Pell's equation. :)" } { "Tag": [ "ratio" ], "Problem": "two triangles are similar one with hypotenuse x and leg y+1, the other one has a hypotenuse x+1 and leg y. is this possible? if not, then why not?", "Solution_1": "My solution, not sure if its correct:\n\n[hide]\n\nNot possible.\n\n\n\nx / y+1 =x+1 / y\n\n\n\n(x+1)(y+1) = (xy)\n\nxy + y +x + 1 = xy\n\nx+y = -1\n\n\n\nIf their sum is -1 then one of them has to be negative, and negative lengths don't exist in euclidian geometry.[/hide][/hide]", "Solution_2": "That looks ok so far, but I think you've only looked at one of two possible cases.", "Solution_3": "Good point, but I think the problem is a little ambiguous.\n\n\n\nHowever, it seems like :sqrt: (y+1) :^2: - x :^2: and :sqrt: y :^2: - (x+1) :^2: are really hard expressions to work w/. Maybe it works for a few cases?[/hide][/quote]", "Solution_4": "Well you have to consider the case where y and y+1 are not corresponding sides. Then, work out the missing two legs by similar triangle proportions, and set up ratios using the side lengths u found, checking to see if it all makes sense. I did find that this case works." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$\\text{Let}\\; a,b,c,d\\; \\text{ be positive real numbers, prove that}$\r\n\\[ \\frac1{a^2+b^2+ab+cd}+\\frac1{b^2+c^2+bc+da}+\\frac1{c^2+d^2+cd+ab}+\\frac1{d^2+a^2+da+bc}\\leq\\frac1{\\sqrt{abcd}}. \\]", "Solution_1": "please edit it.It is wrong when $a=b=c=d=1$ :)" } { "Tag": [ "videos", "Support" ], "Problem": "How are the stats calculated?\r\n\r\nFor ex: (Karma) Dojo 2060 under daily top 10. My total well... 2295.\r\n\r\nWhats going on?", "Solution_1": "The stats represents karma earned, not total karma. You earn karma by rating problems, videos, and submitting comments to videos. That being said, looking at the karma right now (11 pm pacific, 2 am eastern), it shows you with the most karma earned today. \"Today\" has only been 2 hours according to the server and it shows you with over 2000 karma earned? That seems like an awful lot and I'm wondering if we got a bug in the system.", "Solution_2": "Probably because now BOGTRO has 4,330 karma and it's only 7:30", "Solution_3": "Yeah, and, on the high scores, he has 4330 total...\r\n\r\nOn some days, Jongao has like 11K karma. His percentage would be MUCH lower if he really answered that many problems and rated, and only got however many points he has.", "Solution_4": "[quote=\"levans\"] That being said, looking at the karma right now (11 pm pacific, 2 am eastern), it shows you with the most karma earned today. \"Today\" has only been 2 hours according to the server and it shows you with over 2000 karma earned? That seems like an awful lot and I'm wondering if we got a bug in the system.[/quote]\r\n\r\nYeah when the stats refreshed at the new day, it started me off with 2000 some karma.\r\n\r\nIf I may, I would suggest just make the rankings for daily the amount earned that day, monthly be amount earned per month and overall be actual karma.", "Solution_5": "Yes, and Dojo now has 4755 karma for today, 6 minutes into the day :P\r\n\r\nAnd he has 2 points.", "Solution_6": "Yeah I've realized that my Monthly and Daily Karma earned are the always the same and I'm always on the leaderboard as like 4495 Karma earned before I do any quesiton (or rate anything)" } { "Tag": [ "vector", "linear algebra", "matrix" ], "Problem": "Hi ,\r\nI have a problem about system of equation:\r\n\r\nBelow are partially reduced augmented matrices for systems of equations. In each case, give the number of solutions to the system and when consistent express the soution as linear combination of n*1 vectors:\r\n\r\n[b][1 1 2 1 | 6 ] This is a matrice . I just dont know how to present it\n[0 1 6 2 | 17]\n[0 0 2 4 | 30][/b]\r\n\r\n \"[b]|[/b]\" in the matrix, marks the right side of the system.\r\n\r\nI need help to find the number of solution.\r\nCan we say that because the last line is not all 0 0 0 0 0, the system is inconsistent ( no solutions)??\r\n\r\nBertrand", "Solution_1": "Absolutely not; you can only say that if all of the entries on the coefficient side of such a row are zero. \r\nIn this case, there are infinitely many solutions (a translated 1-dimensional solution space)." } { "Tag": [ "logarithms", "ceiling function" ], "Problem": "Five coins look the same, but one is a counterfeit coin with a different weight than each of the four genuine coins. Using a balance scale, what is the least number of weighings needed to ensure that, in every case, the counterfeit coin is found and is shown to be heavier or lighter?", "Solution_1": "take four of the coins, and put two on each side\r\nif they balance, then the one you didn't weigh is the counterfeit\r\n\r\nhowever, if one side is heavier, take the lighter side and divide it into two parts again\r\nthe lighter side is the counterfeit\r\n\r\ntherefore, the least number of weighings is 2", "Solution_2": "[quote=\"vallon22\"]take four of the coins, and put two on each side\nif they balance, then the one you didn't weigh is the counterfeit\n\nhowever, if one side is heavier, take the lighter side and divide it into two parts again\nthe lighter side is the counterfeit\n\ntherefore, the least number of weighings is 2[/quote]\r\n\r\nHowever, we do not know if it is heavier or lighter. And, the second weighing might have both sides equal, so you would have to weigh again. By the third weighing, we would know which pile of 2 coins have different weights, so you randomly take 1, and weigh with one of the other ones. \r\n\r\nI can't find a solution under 4...", "Solution_3": "????????\r\n\r\nlike, the first weighing, just randomly choose 4 coins, and put two on each side\r\n\r\nif one side is heavier than the other, take the lighter side and split it\r\nthe lighter side after that weighing will have the lighter coin?\r\n\r\nhow does that not work?", "Solution_4": "We do not know if the counterfeit coin is heavier or lighter than real coins.", "Solution_5": "OH\r\ni see\r\n\r\nthen we take the lighter side and measure\r\nif it's equal, we take the heavier side and measure\r\n\r\nthen we would have to compare it to another coin and see....\r\nhmmm...\r\n\r\ni think the answer's 4... but i'm not sure", "Solution_6": "I think I got 3...\r\n\r\nLet A,B,C,D,E be the coins\r\n\r\nfirst weighing:\r\nA B | C D\r\nIf the two are equal, then E is the counterfeit. If the two are not equal, assume that A B is higher than C D\r\n\r\nsecond weighing:\r\nA C | E D\r\nIf A C is higher than E D, then either A is lighter than a regular coin or D is heavier than a regular coin. This case would then take 3 weighings (third weighing could be A | E). If A matches E, then D is heavier, and if A is lighter than E, then we're done.\r\nIf A C is the same as E D, then B is obviously the counterfeit. The first weighing tells you that it's lighter.\r\nIf A C is lower than E D, then C must be the counterfeit coin (think it through a bit). It also follows that C is heavier.\r\n\r\nsorry if that was unclear...\r\n\r\ndid anyone find a way to get it in two?", "Solution_7": "I believe that I asked this question a few days ago and recall that the correct answer was 3.\r\n\r\nThanks...yea...I thought 2 as well, but I forgot that I did not know whether the mystery coin was heavier or light than the others", "Solution_8": "[quote=\"5849206328x\"]I think I got 3...\n\nLet A,B,C,D,E be the coins\n\nfirst weighing:\nA B | C D\nIf the two are equal, then E is the counterfeit. If the two are not equal, assume that A B is higher than C D\n\nsecond weighing:\nA C | E D\nIf A C is higher than E D, then either A is lighter than a regular coin or D is heavier than a regular coin. This case would then take 3 weighings (third weighing could be A | E). If A matches E, then D is heavier, and if A is lighter than E, then we're done.\nIf A C is the same as E D, then B is obviously the counterfeit. The first weighing tells you that it's lighter.\nIf A C is lower than E D, then C must be the counterfeit coin (think it through a bit). It also follows that C is heavier.\n\nsorry if that was unclear...\n\ndid anyone find a way to get it in two?[/quote]\r\n\r\nNo. That is correct. I have done this problem before, and there is a formula for any amount of coins, including 999999999999999999999999999999999999. :)", "Solution_9": "You have piqued our curiosity. What is the formula?", "Solution_10": "In general, $ n$ weighings are sufficient to \ffind a bad coin among $ \\frac{3^n\\minus{}1}2$ coins (this reminds us of dividing into $ 3$ groups, which always is the minimum for numbers that can be obtained from this formula). Note that this formula only finds the MAXIMUM number of coins that these weighings can find, and many values can't be obtained.\r\n\r\nThus, if we let $ \\frac{3^n\\minus{}1}2$, then we need to solve for $ n$ in terms of $ x$.\r\n\r\n$ 2x\\plus{}1\\equal{}3^n\\implies n\\equal{}\\log_3{2x\\plus{}1}$.\r\n\r\nBut this is not always an integer, so the formula is $ \\boxed{n\\equal{}\\lceil\\log_3{2x\\plus{}1}\\rceil}$.\r\n\r\nIn this particular problem, we have $ n\\equal{}\\lceil\\log_3{11}\\rceil\\equal{}\\boxed3$." } { "Tag": [], "Problem": "I need help on this problem. My book doesn't show the solution except its answer. :(", "Solution_1": "please post a clearer diagram", "Solution_2": "Sorry but it is just impossible to view the image.", "Solution_3": "From Next Time Please Make a Neatly Labelled Figure.\r\n\r\nI guess I May be right . However ,What Answer Does The Book Show???" } { "Tag": [ "geometry", "number theory", "national olympiad" ], "Problem": "[i]1st Balkan Student Mathematical Competition was organized by Young talented mathematicians \"Marin Getaldic\", a group of high-school students and university students (which act as mentors to high-school students and try to improve their knowledge of mathematics and their result in various mathematical competitions) based in Zagreb. It was held in November 2008, with help from various people from 5 countries (Bosnia and Herzegovina, Croatia, Macedonia, Montenegro and Serbia), who acted as local organizers. 103 students from 11 schools participated altogether.[/i]\r\n\r\n[b]Problems[/b]\r\nGrade 9\r\n1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339430#1339430\r\n2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339433#1339433\r\n3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339436#1339436\r\n4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339440#1339440\r\n\r\nGrade 10\r\n1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339444#1339444\r\n2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339456#1339456\r\n3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339459#1339459\r\n4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339446#1339446\r\n\r\nGrades 11 and 12\r\n1. (Algebra) http://www.mathlinks.ro/viewtopic.php?p=1339448#1339448\r\n2. (Combinatorics) http://www.mathlinks.ro/viewtopic.php?p=1339451#1339451\r\n3. (Geometry) http://www.mathlinks.ro/viewtopic.php?p=1339462#1339462\r\n4. (Number Theory) http://www.mathlinks.ro/viewtopic.php?p=1339453#1339453\r\n\r\nAll the problems, official solutions, results and statistics can be seen (English version is included, of course) in files contained in the following zip file: http://www.2shared.com/file/4401926/280ec27f/BSMC.html (click on \"Save file to your PC: click here\" near the end of the page)\r\n\r\n[i]Writing time was 240 minutes and each problem was worth 10 points. We are especially proud of the fact that 9 of 12 problems were original.\nIf there are any questions or suggestions about the competition or if there are any problems with the zip file above, please, PM me and we'll see how I can help you.[/i]", "Solution_1": "Moderators, how about putting this competition into \"Resources\" page? (there are 9 original problems out of 12 :) )", "Solution_2": "Who can participate in htis contest?", "Solution_3": "I asked it in FY forum, and the answer was I think: your school must participate and then everybody who is interested can participate in their own school.", "Solution_4": "Yes, Dimitris, Bugi is correct (mostly :lol: ), as I've answered in the (Former) Yugoslavia community.\r\n\r\nThe original idea was to make a fun, interesting and simple competition. As the competition was completely unofficial (i.e. lacking any funding whatsoever :D ) and the organizers (both locally and globally) were usually high-school teachers and university students (like myself), it was obvious that it could not be held at a single place. Therefore, we have decided to ask all the people we knew who hold preparations for competitions in their schools if they'd like to organize the competition at their school.\r\n\r\nThere were no formal prerequisites for competing (although it was obvious the problems were tough and ment for more advanced competitors), so anybody who wanted to compete, could.\r\n\r\nIn the end, 9 schools held the competition. However, 2 more students (from two other schools) came, on their own initiative, to nearby schools which were holding the competition and competed there. Thus, students from 11 schools participated in the contest.\r\n\r\nI hope this explains it. I know, it's kind of confusing, but, in the beginning (and partly until the end), BSMC was not ment to be a really serious, formal competition - it was more of a way to motivate students and allow them to see what areas of mathematics they're good at, and what areas they need to work on more.\r\n\r\nIf you've got more questions, please, do that (you can also PM me). I'm here to answer (and I also like to talk about this competition way too much :D ).", "Solution_5": "Sorry, it seems the link in the opening post has timed out. This is the new link: http://www.2shared.com/file/5357102/45bfdefc/BSMC.html ." } { "Tag": [ "MATHCOUNTS" ], "Problem": "This is an old problem we actually put a skit about it when I was in elementary school (that was many decades ago)\r\n\r\nIt was lunch time and there were only three young men in the car of the train. When the luch boxes were opened, Ashok had $3$ rotis (packed by his mom), Bob had $5$ roties, but no one packed a lunch for Chou. But in any case, all three divided their roties equally among all three and had a nice luch.\r\n\r\nNext stop, Chou got down from the train, thanked A & B for sharing their lunch with him and handed Rs $8$ to them.\r\n\r\nBob said something like \" Well $8$ Ruppes for $8$ rotis so I will keep $5$ and you keep $3$\"\r\n\r\nAshok on the ohter hand wanted to equally divide the money and wanted $4$ Rupees.\r\n(In the skit we put, here brief intermission at this stage after a big argument erupted among the two young men - and audience was asked by the narrator \"how the money should be divided - assuming it is to be done in a mathematically correct way\")\r\n\r\nSo how the money should be divided?\r\n\r\n(Funny - After so many decades I saw the same problem -without a skit and the nice story above - in an old mathcounts test)", "Solution_1": "[quote=\"Gyan\"]This is an old problem we actually put a skit about it when I was in elementary school (that was many decades ago)\n\nIt was lunch time and there were only three young men in the car of the train. When the luch boxes were opened, Ashok had $3$ rotis (packed by his mom), Bob had $5$ roties, but no one packed a lunch for Chou. But in any case, all three divided their roties equally among all three and had a nice luch.\n\nNext stop, Chou got down from the train, thanked A & B for sharing their lunch with him and handed Rs $8$ to them.\n\nBob said something like \" Well $8$ Ruppes for $8$ rotis so I will keep $5$ and you keep $3$\"\n\nAshok on the ohter hand wanted to equally divide the money and wanted $4$ Rupees.\n(In the skit we put, here brief intermission at this stage after a big argument erupted among the two young men - and audience was asked by the narrator \"how the money should be divided - assuming it is to be done in a mathematically correct way\")\n\nSo how the money should be divided?\n\n(Funny - After so many decades I saw the same problem -without a skit and the nice story above - in an old mathcounts test)[/quote]\r\n\r\n[hide=\"answer\"]\nWell, if they shared it equally, all of them got 8/3 rotis. So Ashok gave 1/3 of a roti and Bob gave 7/3 of a roti. So 1 rupee goes to Ashok and 7 rupees goes to Bob.\n[/hide]", "Solution_2": "Above answer is corret.", "Solution_3": "What are rotis anyway?", "Solution_4": "Its a kind of bread, very popular in India, (and other places):\r\n[url=http://www.saveurs.sympatico.ca/ency_6/triperie/foie-roti.htm]Check it out , for example...[/url]\r\n\r\nOr [url=http://www.lasan.co.uk/photos/press/Roti%20sag.jpg]For a close up view ...[/url]", "Solution_5": "With that filling it looks really yummy :) .\r\n\r\nExcept... how do you split that into thirds :P ?" } { "Tag": [], "Problem": "Points $ A$, $ B$, $ C$, $ D$, $ E$, $ F$, and $ G$ are all consecutive vertices on an inscribed 7-gon. $ \\angle A \\equal{} 120^\\circ$, $ \\angle B \\equal{} 160^\\circ$, and $ \\angle E \\equal{} 154^\\circ$. Find the measure of minor arc $ FG$.", "Solution_1": "Is a measure something other than distance? because I think we need more info \r\n\r\n[geogebra]b2d1e7ebeac1b9d7350d92ae3f83a655b7a6820e[/geogebra] \r\n\r\nThose sides of the septagon can just be slid round the circle, therefore the minor arc $ FG$ can take a range of values. Also, depending on how big you scale the circle can the minor arc take values $ [0,\\infty)$?\r\n\r\nI think i must be missing something :?:", "Solution_2": "The problem was asking for [i]angle[/i] measure. (I assume this because they usually ask for the length when referring to the length of the arc and measure when referring to the angle measure.) You also have to consider that the interior angles of septagon always sum up to one value. Maybe you should also take a look at the exterior angles too.\r\n\r\nBy the way, the problem is right, it's exactly as it appeared on the paper. (But of course, there might be chance that there was a mistake in writing the problem itself.)", "Solution_3": "Well, it doesn't make a difference. We can slide $ \\widehat{DEF}$, giving different answers." } { "Tag": [ "algebra", "polynomial", "search" ], "Problem": "Given \r\n\r\n$a+b+c=4$\r\n$a^{2}+b^{2}+c^{2}=10$\r\n$a^{3}+b^{3}+c^{3}=22$\r\n\r\nwhat is the value of $a^{4}+b^{4}+c^{4}$.", "Solution_1": "[quote=\"Franz Joseph\"]Given \n\n$a+b+c=4$\n$a^{2}+b^{2}+c^{2}=10$\n$a^{3}+b^{3}+c^{3}=22$\n\nwhat is the value of $a^{4}+b^{4}+c^{4}$.[/quote]\r\nDon't you love brute force?[hide]I don't. $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+4(a^{3}b+a^{3}c+b^{3}c+b^{3}a+c^{3}a+c^{3}b)+6(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})=256$ Ouch. Let's see how that can be simplified. $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=16$ which gives $ab+bc+ac=3$. Ugh. I think I am on the right track... [/hide]", "Solution_2": "I think this has something to with the Theory of Symmetric Polynomials. :huh:", "Solution_3": "Ew...\r\n\r\n[hide]$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=16$\n$2(ab+bc+ac)=6 (ab+bc+ac)=3$\n\n$(ab+bc+ac)(a+b+c)=a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+3abc=12$\n\n$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc)=64$\n$3(a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc)=42$\n$a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc=14$\n$abc=-2$\n\nLet $a^{2}bc+ab^{2}c+abc^{2}=x$\n$abc(a+b+c)=a^{2}bc+ab^{2}c+abc^{2}=-8$\n$x=-8$\n\nLet $a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}=y$\n$(a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}=100$\n$y=100$\n\nLet $a^{4}+b^{4}+c^{4}+a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3}=z$\n$(a+b+c)(a^{3}+b^{3}+c^{3})=a^{4}+b^{4}+c^{4}+a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3}=88$\n$z=88$\n\n$(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+4(a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3})+3(2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2})+12(a^{2}bc+ab^{2}c+abc^{2})=256$\n\n$(a+b+c)^{4}-4z-3y-12x=-6(a^{4}+b^{4}+c^{4})=-300$\n$\\boxed{a^{4}+b^{4}+c^{4}=50}$\n\n?\n[/hide]", "Solution_4": "Yea basically, just brute force this problem.", "Solution_5": "In addition, the use of the Theory of Symmetric Polynomials and an understanding of elementary symmetric polynomials are also used when solving this problem.", "Solution_6": "[quote=\"Imperial Effect\"]In addition, the use of the Theory of Symmetric Polynomials and an understanding of elementary symmetric polynomials are also used when solving this problem.[/quote]\r\n\r\nSymettric polynomials? Um....\r\n\r\nPlease provide a solution to back up your claim.", "Solution_7": "[quote=\"neelnal\"][quote=\"Imperial Effect\"]In addition, the use of the Theory of Symmetric Polynomials and an understanding of elementary symmetric polynomials are also used when solving this problem.[/quote]\n\nSymettric polynomials? Um....\n\nPlease provide a solution to back up your claim.[/quote]\r\nAn example has already been provided. Look at the way besttate manipulates the equations. \r\n\u201cIn mathematics, a symmetric polynomial is a polynomial P(X1,X2,...,Xn) in n variables, such that if some of the variables are interchanged, the polynomial stays the same.\u201d \r\n\r\nHere are some links to help understand symmetric polynomials:\r\nhttp://en.wikipedia.org/wiki/Symmetric_polynomials\r\nhttp://planetmath.org/encyclopedia/ElementarySymmetricPolynomial.html\r\nhttp://everything2.com/index.pl?node=symmetric%20polynomial\r\n\r\n\r\nI\u2019m sure there are a lot more just search on [url]http://google.com[/url] and you will find more.\r\n\r\nP.S Do not Google \"[b]Symettric[/b] polynomials\" instead try \"[b]symmetric [/b]polynomials\"", "Solution_8": "Now use it to solve the problem :D", "Solution_9": "[quote=\"JavaMan\"]Now use it to solve the problem :D[/quote]\n\nIt already been solved :rotfl: !!\n\n[quote=\"besttate\"]\n[hide]$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=16$\n$2(ab+bc+ac)=6 (ab+bc+ac)=3$\n\n$(ab+bc+ac)(a+b+c)=a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+3abc=12$\n\n$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc)=64$\n$3(a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc)=42$\n$a^{2}b+b^{2}c+a^{2}c+ab^{2}+bc^{2}+ac^{2}+2abc=14$\n$abc=-2$\n\nLet $a^{2}bc+ab^{2}c+abc^{2}=x$\n$abc(a+b+c)=a^{2}bc+ab^{2}c+abc^{2}=-8$\n$x=-8$\n\nLet $a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}=y$\n$(a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}=100$\n$y=100$\n\nLet $a^{4}+b^{4}+c^{4}+a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3}=z$\n$(a+b+c)(a^{3}+b^{3}+c^{3})=a^{4}+b^{4}+c^{4}+a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3}=88$\n$z=88$\n\n$(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+4(a^{3}b+b^{3}c+a^{3}c+ab^{3}+bc^{3}+ac^{3})+3(2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2})+12(a^{2}bc+ab^{2}c+abc^{2})=256$\n\n$(a+b+c)^{4}-4z-3y-12x=-6(a^{4}+b^{4}+c^{4})=-300$\n$\\boxed{a^{4}+b^{4}+c^{4}=50}$\n\n?\n[/hide][/quote]", "Solution_10": "Ok, I just thought you had a less tedious way :)", "Solution_11": "Yea. The way you wrote, I thought there was a less brute-force solution waiting to be revealed.", "Solution_12": "It would be pretty cool if this could be done in an easier way.", "Solution_13": "http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums", "Solution_14": "I was gonna do that, but then i thought it'd take too long to figure out the coefficients of the polynomial.\r\n\r\nEDIT: Come to think of it, you can work out the coefficients pretty easily. Then work out S4 from that.", "Solution_15": "[quote=\"t0rajir0u\"]http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums[/quote]\r\n\r\nThat's an awesome way of doing it! Thanks [b]t0rajir0u[/b].", "Solution_16": "[quote=\"Franz Joseph\"][quote=\"t0rajir0u\"]http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums[/quote]\n\nThat's an awesome way of doing it! Thanks [b]t0rajir0u[/b].[/quote]\r\n\r\nDefinetely. Thanks t0rajir0u." } { "Tag": [ "search", "combinatorics proposed", "combinatorics" ], "Problem": "In a circle,let $ 35$ points$ A_1,A_2....A_n$ satisfying $ A_1A_2 \\equal{} A_2A_3 \\equal{} ...A_{34}A_{35} \\equal{} A_{35}A_1$ (it mean here is a regual $ 35$-gon,and thay were colored by $ 2$ colors:red or green.Prove that there are $ 5$ point (in them )were colored by a similar color,and they make two isosceles triangles(such that two one have a common point)\r\nIt is hard.who can help me", "Solution_1": "[quote=\"Friendly1\"]In a circle,let $ 35$ points$ A_1,A_2....A_n$ satisfying $ A_1A_2 \\equal{} A_2A_3 \\equal{} ...A_{34}A_{35} \\equal{} A_{35}A_1$ (it mean here is a regual $ 35$-gon,and thay were colored by $ 2$ colors:red or green.Prove that there are $ 5$ point (in them )were colored by a similar color,and they make two isosceles triangles(such that two one have a common point)\nIt is hard.who can help me[/quote]\r\nHa ha ha ha :rotfl: :rotfl: \r\nI think it is too hard for all Mathlinks member ,I am sure no one can solve it :lol: :lol:", "Solution_2": "Van der Waerden number $ V(2,4) \\equal{} 35$...\r\n\r\nPierre.", "Solution_3": "[quote=\"Friendly1\"][quote=\"Friendly1\"]In a circle,let $ 35$ points$ A_1,A_2....A_n$ satisfying $ A_1A_2 \\equal{} A_2A_3 \\equal{} ...A_{34}A_{35} \\equal{} A_{35}A_1$ (it mean here is a regual $ 35$-gon,and thay were colored by $ 2$ colors:red or green.Prove that there are $ 5$ point (in them )were colored by a similar color,and they make two isosceles triangles(such that two one have a common point)\nIt is hard.who can help me[/quote]\nHa ha ha ha :rotfl: :rotfl: \nI think it is too hard for all Mathlinks member ,I am sure no one can solve it :lol: :lol:[/quote]\r\ndon't spam all names!\r\n@pbornsztein:Can you post your full solution?", "Solution_4": "Label the vertices $ 1,2,\\cdots,35$ in clockwise sense.\r\nAny two colouring of the vertices induces a two colouring of the integers from $ 1$ to $ 35$. But, since Van der Vaerden's number $ V(2,4)\\equal{}35$ it means that for any two colouring of the integers $ 1,2,\\cdots,35$ there exists a monochromatic sequence of length $ 4$, say $ a,b,c,d$ in that order.\r\nTherefore the vertices corresponding to these labels give two isoscele triangles, namely $ abc$ and $ bcd$, as desired.\r\n\r\nPierre.", "Solution_5": "[quote=\"Math pro\"]\n\ndon't spam all names!\n@pbornsztein:Can you post your full solution?[/quote]\r\nall names,who is he?,and who are you,mathpro :rotfl: :rotfl: \r\nAny way I am quite :mad: ,with your reply,a post without any information and I think it is exactly spam post(Although I dont know all names I guess he is not a spamer as you)\r\nAnd I think you should think more before post any post because our site is Mathlinks where is not the \"land\" for a spamer(as you again)\r\n @Add :rotfl: :rotfl: :I have just checked your profile and found \"I'm not a spamer!\" in your signature\r\nMorever,I dont see no post of you http://www.mathlinks.ro/search.php?search_author=Math+pro\r\narent spams post,\r\nBut I am very thank to pbornsztein,the first who care my hard problem,thank you,\r\nBut can you show me what is Van der Vaerden's number ,and who have a conceivable proof?,I really want to know all your help" } { "Tag": [ "conics", "ellipse", "geometry", "geometric transformation", "homothety", "circumcircle", "inradius" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14596[/img]\r\n\r\n :(", "Solution_1": "call $ P: AT_1 \\cap BT_2$ and $ T_3: CP \\cap AB$. Then the conic must pass through $ T_3$ (Carnot + Ceva theorem). Also it's know that the center of an inscribed conic with perspector P it's the complement of the isotomic conjugate of P and so we get other 3 point.", "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14607[/img]\r\n\r\n :(", "Solution_3": "the point P* that you draw in you figure is the perspector, not the center, for obtain the center call Q the isotomic conjugate of P* and take an homothety with radio $ \\minus{}\\frac{1}{2}$ with center the centroid and Q go in P that is the center of the ellipse.", "Solution_4": "[quote=\"\u00ac[\u0192(Gabriel)\u00b3\u00b2\u00b9\u00ba]\u00bc\"]call $ P: AT_1 \\cap BT_2$ and $ T_3: CP \\cap AB$. Then the conic must pass through $ T_3$ (Carnot + Ceva theorem). Also it's know that the center of an inscribed conic with perspector P it's the complement of the isotomic conjugate of P and so we get other 3 point.[/quote]\r\n\r\nHello, what carnot's theorem are you referencing? I am only familiar with the carnot's theorem that relates the lengths of the projections of the circumcenter to the inradius and circumradius. Surely there is something else that you are refering to. Could you please explain?", "Solution_5": "[quote=\"Altheman\"]Hello, what carnot's theorem are you referencing? I am only familiar with the carnot's theorem that relates the lengths of the projections of the circumcenter to the inradius and circumradius. Surely there is something else that you are refering to. Could you please explain?[/quote]\r\n[b]Carnot's Theorem[/b]\r\n\r\n[i]if a conic intersect the sides BC at X, X', CA at Y, Y' and AB at Z, Z' then[/i]\r\n\r\n$ \\frac {BX}{XC}\\cdot\\frac {BX'}{X'C}\\cdot\\frac {CY}{YA}\\cdot\\frac {CY'}{Y'A}\\cdot\\frac {AZ}{ZB}\\cdot\\frac {AZ'}{Z'B} \\equal{} 1$\r\n\r\n[i]and if the relation is true then the 6 points are on a conic.[/i]", "Solution_6": "[b]\"for obtain the center call Q the isotomic conjugate of P* and take an homothety with radio - 1/2 with center the centroid and Q go in P that is the center of the ellipse.\"[/b]\r\n\r\nyou mean that center of ellipse is midpoint of GP* ? but it's not true, so [u][b]where's the center of the ellipse[/b][/u]?\r\n\r\n :(", "Solution_7": "...see at this figure:\r\n\r\n[img]http://img100.imageshack.us/img100/8445/geometria36eg6.png[/img]", "Solution_8": "[b]GT = 1/2*QG ?[/b]" } { "Tag": [ "algorithm", "algebra proposed", "algebra" ], "Problem": "Noone posted a solution for this in the intermediate section so I'm reposting it here:\r\n\r\nShow that if the root of the algebraic equation [tex]f(x)=0[/tex] is of the form [tex]a+bi[/tex] and [tex]a,b\\in\\mathbb{R}[/tex] \r\n [tex]f(x)\\in\\mathbb{R}[x][/tex], then the number [tex]a-bi[/tex] is another root of the equation.", "Solution_1": "???\r\n\r\nWe can write $0=\\overline{f(a+bi)}=f(a-bi)$. That's all.", "Solution_2": "Let be $z$ any root of the equation $\\sum_{k=0}^n a_k x^k = f(z)=0$. For the conjugate of $z \\in \\mathbb{C}$ I will write $z'$.\r\nThen $0=f(z)'=(\\sum_{k=0}^n a_k z^k)'=\\sum_{k=0}^n (a_k z^k)'=\\sum_{k=0}^n a_k' (z^k)'=\\sum_{k=0}^n a_k z'^k=f(z')$.", "Solution_3": "Maybe a proof which does not use the properties of the conjugate is what Xixas wants:\r\nLet $P(x)=(x-a-bi)(x-a+bi)=x^2-2ax+a^2+b^2\\in \\mathbb {R}[X]$.\r\nThe long division algorithm gives the existence of $c,d\\in \\mathbb {R}$ and $Q\\in \\mathbb {R}[X]$ such that $f(x)=P(x)Q(x)+cx+d$. Plugging $x=a+bi$ yields $c(a+bi)+d=0$, hence $ac+d=bc=0$. \r\nWe have $f(a-bi)=c(a-bi)+d=(ac+d)-bci$ hence $f(a-bi)=0$, as well.", "Solution_4": "Thank you for your replies." } { "Tag": [], "Problem": "Solve:\r\n\r\n$ x(16 \\minus{} y) \\equal{} 78$\r\n$ y(16 \\minus{} x) \\equal{} 14$\r\n\r\nFIND $ (x,y)$", "Solution_1": "[hide]The first one can be rewritten as $ 16x \\minus{} xy \\equal{} 78$ whereas the second one as $ 16y \\minus{} xy \\equal{} 14$\nsubtracting the second from the first we get that $ 16x \\minus{} 16y \\equal{} 64 \\equal{} > x \\minus{} y \\equal{} 4 \\equal{} > y \\equal{} x \\minus{} 4$\nSo substituting y in the first one we get $ x(16 \\minus{} x \\plus{} 4) \\equal{} 78 \\equal{} > 20x \\minus{} x^2 \\equal{} 78 \\equal{} > x^2 \\minus{} 20x \\plus{} 78 \\equal{} 0$ which has two solutions: $ x \\equal{} 10 \\minus{} \\sqrt22 and x \\equal{} 10 \\plus{} \\sqrt22$\nfor the first one we get that $ y \\equal{} 6 \\minus{} \\sqrt22$ and for the second $ y \\equal{} 6 \\plus{} \\sqrt22$\nso $ (x,y) \\equal{} (10 \\minus{} \\sqrt22,6 \\minus{} \\sqrt22) or (x,y) \\equal{} (10 \\plus{} \\sqrt22,6 \\plus{} \\sqrt22)$[/hide]", "Solution_2": "It helps that they are symmetric :)" } { "Tag": [], "Problem": "Suppose a runner moving at $ 0.75c$ carries a horizontal pole 15m long toward a barn that is 10m long. The barn has front and rear doors. An observer on the ground can instantly and simultaneously open and close the two doors by remote control. When the runner and pole are inside the barn, the ground observer closes and then opens both doors so that the runner and the pole are momentarily captured inside the barn and then proceed to exit the barn from the back door. Do both the runner and the ground observer agree that the runner makes it safely through the barn?", "Solution_1": "This is a very famous problem and the answer is that in the pole frame, the observer seems to open the doors in the wrong order. \r\n\r\nA variation: The back door cannot be opened. The observer ('farmer') closes the front door when the runner ('his son') is inside. What will each observe when the runner comes to rest? There are two ways to go about this. \r\n(Of course, technically, this cannot be done in special relativity but give a qualitative answer.)", "Solution_2": "The solution resolves it by saying that by relativity of simultaneity, the doors do not open and close and open at the same time from the pole carrier's point of view. The pole carrier gets through safely but he and the observer see different things happening =P\r\n\r\nThat variation is tricky.. will the length contraction stop and the pole expand or something? Lol..", "Solution_3": "You are on the right track. \r\nThe farmer will indeed see the length contraction stop and the rod expands till it is held compressed (under tension) by the door.\r\nWhat will the son see?", "Solution_4": "He will see the door get farther away? Because in his frame, the length contraction of the barn stops. Counterintuitive!", "Solution_5": "Yes. He will also see the rod in his hand contract due to it's own inertia, so that it finally fits in the barn. \r\nNote the use of [u]psuedo force[/u] in a non onertial frame (that of the son).", "Solution_6": "What's the pseudo force here? By the way, in the twin paradox, they argue that the travelling twin is in a non-inertial frame and there's the pseudo forces acting on him (he speeds up and slows down while turning his spacecraft around). But could it not also be that the twin on earth is accelerating while he is not? It's like the inertial and non-inertial frames could be swapped. Correct me if i'm wrong :P", "Solution_7": "When the son reaches the end, he accelerates backwards; hence in his frame, there is a psuedo force acting forwards (on the pole for instance).\r\n\r\nThe travelling twin experiences a [i]real[/i] force, that of the rocket engines. This makes his frame non inertial." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Consider two circles in a plane $ S_1$ and $ S_2$ that intersect at the points $ A$ and $ B$. Let $ Q$ be a point on $ S_1$ and outside $ S_2$, and let $ D$- the intersection of $ QA$ with $ S_2$ and let $ C$- the intersection of $ QB$ with $ S_2$. Suppose that the tangents to $ S_1$ at $ A$ and $ B$ intersect at $ P$. Prove that $ QP$ halves $ CD$.", "Solution_1": "Denote L is the midpoint of AB\r\nKAQB is a harmonic quadrilateral => \u2220LQB=\u2220KQA =>\u22bfLQB\u223d\u22bfMQD \r\nM is the midpoint of CD", "Solution_2": "Dear Mathlinkers,\r\n1. according to Reim's theorem, the tangent Tq at Q is parallel to CD\r\n2. according to Boutin's theorem, P is the midpoint of the segment C'D' determined by the parallel to Tq passing through P with endpoints on QB and QA\r\n3. Consider the trapez C'D'DC....\r\nfor more see \r\nhttp://perso.orange.fr/jl.ayme vol. 1 A propos du th\u00e9or\u00e8me de Boutin\r\nSincerely\r\nJean-Louis", "Solution_3": "It\u2019s a well known fact that PQ is simmedian in triangle ABQ, hence median in $ \\triangle CDQ$.\r\n\r\nBest regards,\r\nsunken rock", "Solution_4": "I solved it the same way sunken rock did, but I would like to post the details on why that fact is true.\r\nDraw the parallel through P to CD, let it meet DQ at R and CQ at S. Let K be any point on PB, extended past B. Then $ \\angle PBS\\equal{} \\angle QBK\\equal{} \\angle QAB$. Also, since ABCD is cyclic, $ \\angle PSB\\equal{}\\angle DCB\\equal{}\\angle QAB$. So $ \\triangle PBS$ is isoceles. Similarly $ \\triangle PAR$ is isoceles. So $ PS\\equal{}PB\\equal{}PA\\equal{}PR$. So $ QP$ bisects $ RS$, but $ CD \\parallel RS$, so $ QP$ bisects $ CD$." } { "Tag": [ "vector", "real analysis", "real analysis theorems" ], "Problem": "In a normed vector space $ (V,\\left\\|\\cdot\\right\\|)$, show that $ B_{r}(x)\\equal{}x\\plus{}B_{r}(0)\\equal{}\\{x\\plus{}y\\mid \\left\\|y\\right\\| f(y)$ or $ f(x) < f(y), \\forall y \\in (a, b), y \\neq x$?", "Solution_3": "First, a side remark. When I see ???????? at the end of a question (as well as other completely unnecessary excessive punctuation, capitalization, usage of large fat fonts in bright colors, etc.), I am strongly tempted to dismiss the entire post. As far as I know, I'm not the only one here with this attitude. ;)\r\n\r\nNow, closer to the subject. When $ y\\equal{}x$, your strict inequalities obviously cannot hold, so you must have meant something else.", "Solution_4": "Thanks, i haved edited" } { "Tag": [], "Problem": "find m.n is integer such that with p be a prime number and 2^p+3^p=m^n :D :?", "Solution_1": "I think this one has already been discussed. The idea is that $ 2^p+3^p$ is divisible by 5, but not by 25 for odd prime p." } { "Tag": [ "LaTeX" ], "Problem": "Hi,\r\n\r\nHow to extract data from a .tex file to SQL Server and then display it on ASP page with all the printable/mathematical symbols, what is available on that .tex file ?????? Extraction is not the big deal but how to display the symbols on ASP pages ???\r\n\r\nThanks,\r\nAlok.", "Solution_1": "A couple of possibilities come to mind:\r\n1. Convert the TeX file to HTML first using tools like [url=http://hutchinson.belmont.ma.us/tth/]TtH[/url] or [url=http://www.cis.ohio-state.edu/~gurari/TeX4ht/mn.html]Tex4ht[/url]. If this is to be done dynamically you will have to use TtH; Tex4ht is slower but uses images to give better results.\r\n\r\n2. Assuming you are happy with delving into PHP and ASP code, you could convert [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender]LatexRender[/url], which powers the on-the-fly [tex]\\LaTeX[/tex] conversion on this board, from PHP to ASP. [url=http://www.php-asp-convertor.com/index.php]The PHP to ASP converter[/url] may help. You will need [tex]\\LaTeX[/tex] to be installed on the server.\r\n\r\nI would be very interested if you do manage to convert LatexRender to ASP :)" } { "Tag": [ "function" ], "Problem": "Find $ f : R \\to R$ satisfying\r\n$ xf(y)\\minus{}yf(x)\\equal{}f(\\frac{y}{x})$ with all $ x;y \\in R;x \\neq 0$", "Solution_1": "Is this also cheating?\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259079", "Solution_2": "Probably:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=259082" } { "Tag": [ "MATHCOUNTS", "email" ], "Problem": "im just wondering if anyone has some national tests from mathcounts that are free to download. if so, can some please pm them to me.\r\n\r\nnote that they need to be ones that can be downloaded on to acrobatic reader or word. \r\n\r\np.s. sorry about the first post it was a mistake.", "Solution_1": "PM me too if you have any.", "Solution_2": "Pm or email them to me as well please.", "Solution_3": "You know those weeks when Mathcounts posts sample nationals problems on the Problem of the Week? I compiled all of those and saved them in a file. Anyone?", "Solution_4": "[quote=\"mr. math\"]but do you have any regular ones? :ninja: :roll:[/quote]\r\n\r\nWhat are \"regular ones?\"", "Solution_5": "mathnerd314, can you pm me those problems :ninja: \r\n :rotfl:", "Solution_6": "They've been posted in my blog. Please, no more PM-ing--I'm too lazy to copy and paste a hundred million times.\r\n\r\nWhy the :ninja: ?" } { "Tag": [ "MATHCOUNTS", "Support", "\\/closed" ], "Problem": "How do i change my username. Suppose i tell an admin to change it for me, will the online classes, currently MATHCOUNTS, be terminated? Currently, i don't want to change it though.", "Solution_1": "You cannot change your username.", "Solution_2": "I thought you could, that's what happened to another user in the support forum.", "Solution_3": "*I* (or any other admin) can change a username, but *you* cannot.\r\n\r\nSince it's a pain for us to change them, we strongly discourage it (i.e. you had better have a good reason for changing your username)." } { "Tag": [], "Problem": "\u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03af \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03bf\u03c5\u03c2 :\r\n\r\n$ x^3 \\minus{} y^3 \\equal{} xy \\plus{} 61$", "Solution_1": "[quote=\"meinstein\"]\u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03af \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03bf\u03c5\u03c2 :\n\n$ x^3 \\minus{} y^3 \\equal{} xy \\plus{} 61$[/quote]\r\n\r\n[hide]Pollaplasiazeis epi 27 kai meta euler.[/hide]", "Solution_2": "Sostos o Nick! :!:", "Solution_3": "See here http://www.mathlinks.ro/viewtopic.php?t=181090", "Solution_4": "hahahhahaha :lol: \r\n\r\npou to thimithikes????ego entaksei molis tin eida, katalava amesos oti einai gnosti mou alla den thymomoun to post klp klp...pantos exei plaka ee kai thn prohgoumenh fora me ton idio tropo eixa apanthsei...", "Solution_5": "[quote=\"Nick Rapanos\"]pou to thimithikes????[/quote]\r\n\r\n\u039e\u03ad\u03c1\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03ba\u03b1\u03bb\u03ae \u03bc\u03bd\u03ae\u03bc\u03b7 . :D \r\n\u038c\u03bd\u03c4\u03c9\u03c2 \u03b5\u03af\u03c7\u03b5\u03c2 \u03b1\u03c0\u03b1\u03bd\u03c4\u03ae\u03c3\u03b5\u03b9 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf :P" } { "Tag": [ "geometry", "perimeter", "rectangle", "angle bisector", "geometry unsolved" ], "Problem": "Let $ ABCDE$ be a pentagon whose perimeter is $ 4$, $ AB\\equal{}DE\\equal{}1$ and $ \\widehat{EAB}\\equal{}\\widehat{BCD}\\equal{}\\widehat{DEA}\\equal{}90^o$. Prove that the angle bisector of $ \\widehat{BCD}$ bisects the area of the pentagon.", "Solution_1": "Extend CB & CD to intersect AE @ P & Q resp. Let CR be angle bisector of C, R on AE.\r\nLet CD=[i]x[/i], CB=[i]y[/i] & AE=[i]z[/i].\r\nThen it is sufficient to prove that ar(CRP)-ar(CRQ)=ar(ABP)-ar(DEQ).\r\nThis reduces to proving [i]x[/i] + [i]y[/i] = [i]xy[/i] + [i]z[/i], which can be proved using [i]x+y+z[/i]=2 & [i]x^2+y^2=z^2[/i].", "Solution_2": "Extend CB & CD to intersect AE @ P & Q resp. Let CR be angle bisector of C, R on AE.\r\nLet CD=[i]x[/i], CB=[i]y[/i] & AE=[i]z[/i].\r\nThen it is sufficient to prove that ar(CRP)-ar(CRQ)=ar(ABP)-ar(DEQ).\r\nThis reduces to proving [i]x[/i] + [i]y[/i] = [i]xy[/i] + [i]z[/i], which can be proved using [i]x+y+z[/i]=2 & [i]x^2+y^2=z^2[/i].", "Solution_3": "[quote=\"Ajinkya Dahale\"]Then it is sufficient to prove that ar(CRP)-ar(CRQ)=ar(ABP)-ar(DEQ).[/quote]\n\nWell, to say this you need to prove that the bisector of $ C$ intersects the segment $ AE$, right?\n\n[quote=\"Ajinkya Dahale\"]This reduces to proving x + y = xy + z[/quote]\r\n\r\nCan you give more detais? In my picture, it is not true that $ x^2 \\plus{} y^2 \\equal{} z^2$... But my picture is not very good.", "Solution_4": "Feliz, first, your pictures are always not good :P\r\n\r\nI will explain x^2+y^2=z^2, it's easy.\r\n\r\nNote that BA//DE and BA=DE=1, by =0 ,so ab>=a+b ,then lna+lnb>=ln(a+b) ===>\r\nlnc/(lna+lnb)<=lnc/ln(a+b)\r\nso its easy now by adding this 3 inequality and using Nesbit inequality" } { "Tag": [ "modular arithmetic", "number theory proposed", "number theory" ], "Problem": "Does there exist a square-free positive integer $ n$ such that\r\n\r\n$ n\\plus{}1,n\\plus{}2,...,n\\plus{}2007$ are not square-free, and $ n\\plus{}2008$ is not a power of some positive integer?", "Solution_1": "Call $ p_1,..,p_{2007}$ is 2007 distant prime number . \r\nConsider the system of congruence equation : \r\n$ x\\equiv \\minus{}i (\\mod p_i^2)$\r\nFrom chinese remainder theorem this system have a root . Call smallest natural root is a . \r\n$ M\\equal{}\\prod _{i\\equal{}1}^{2007}p_i^2$\r\nThen $ n\\equal{}a\\plus{}mM$ is satisfy for all m .\r\nEasy prove that exist m for $ a\\plus{}mM\\plus{}2008$ is not a power of an interger .", "Solution_2": "[quote=\"TTsphn\"]\nThen $ n \\equal{} a \\plus{} mM$ is satisfy for all m .\n[/quote]\r\n\r\nHow do you know $ n$ is square-free?", "Solution_3": "[quote=\"nguyentrang\"][quote=\"TTsphn\"]\nThen $ n \\equal{} a \\plus{} mM$ is satisfy for all m .\n[/quote]\n\nHow do you know $ n$ is square-free?[/quote]\r\nFrom your problem it is not square free. Because $ p_i^2|x\\plus{}i$", "Solution_4": "I said $ n$ is square-free, but $ n\\plus{}1,n\\plus{}2,...,n\\plus{}2007$ is not :)", "Solution_5": "[Edited to fix some mistakes.]\r\n\r\nI will rely on the following lemma.\r\n\r\n[b]Lemma:[/b] Let $ m$ be a positive integer and $ a$ be an integer such that $ \\gcd(a, m)$ is square-free. Then there is a square-free positive integer that is $ a \\pmod{m}$.\r\n\r\nThe proof of the lemma is by adapting the usual proof of the density of square-free integers.\r\n\r\nNow, let $ p_1$, $ p_2$, ..., $ p_{2007}$ be distinct odd primes such that $ p_i^2$ is not a divisor of $ i$ (for all $ i$ from 1 to 2007). Consider the equations\r\n\\begin{align*} n & \\equiv - i \\pmod{p_i^2} \\\\\r\nn & \\equiv 2\\pmod{4}. \\end{align*}\r\nLet $ m = 4 p_1^2 p_2^2 \\dots p_{2007}^2$. By the Chinese Remainder Theorem, there is an integer $ a$ such that the solution to the above system of congruences is $ n \\equiv a \\pmod{m}$. Then $ n + i$ is a multiple of $ p_i^2$, and so is not square-free. Also $ n + 2008$ is $ 2 \\pmod{4}$, and so is not a power of an integer.\r\n\r\n\r\nFinally, we can check that $ \\gcd(a, m)$ is square-free, because $ p_i^2$ is not a divisor of $ i$. So the Lemma says we can choose $ n \\equiv a \\pmod{m}$ such that $ n$ is square-free.", "Solution_6": "[quote=\"Ravi B\"]\n[b]Lemma:[/b] Let $ m$ be a positive integer and $ a$ be an integer such that $ \\gcd(a, m)$ is square-free. Then there is a square-free positive integer that is $ a \\pmod{m}$.\n\nThe proof of the lemma is by adapting the usual proof of the density of square-free integers.\n\n[/quote]\r\n\r\nI think it's a very nice lemma. Please show its proof.", "Solution_7": "Sorry, I don't really have time now. One high-powered way is to use Dirichlet's theorem on primes in arithmetic progressions. That theorem gives the lemma almost immediately.\r\n\r\nIn fact, we can modify our construction to make $ \\gcd(a, m) \\equal{} 1$, not merely square-free. Back to the original problem, that means we can make $ n$ prime, not merely square-free.\r\n\r\nTo modify the construction, we may have to change $ n \\equiv 2 \\pmod{4}$ to $ n \\plus{} 2008 \\equiv 3 \\pmod{9}$.", "Solution_8": "[quote=\"Ravi B\"]\nIn fact, we can modify our construction to make $ \\gcd(a, m) \\equal{} 1$, not merely square-free. Back to the original problem, that means we can make $ n$ prime, not merely square-free.\n\n[/quote]\r\n\r\nWhat do you mean? How to modify it?", "Solution_9": "First, change $ n \\equiv 2 \\pmod{4}$ to $ n \\plus{} 2008 \\equiv 3 \\pmod{9}$. Let $ p_i$ be distinct primes different from 3 such that $ p_i$ (not just $ p_i^2$) is not a divisor of $ i$. Finally, change $ m$ to $ 9 p_1^2 p_2^2 \\dots p_{2007}^2$. With these modifications, we can check that $ \\gcd(a, m) \\equal{} 1$." } { "Tag": [ "Columbia", "geometry", "trigonometry", "geometric transformation", "dilation", "calculus", "reflection" ], "Problem": "I've taken core classes such as History, English, Math, and Foreign language in school because it was mandatory. It's also occured to me many of these core classes are mandatory in college as well.\r\n\r\nOut of pure curiosity, why is it necessary for everyone to take these classes?\r\nI suppose I take English because it is important to learn to communicate with others, and I take math cause it's fun, but it seems so weird to me that future writers, history professors, lawyers, etc. has to learn DeMoivre's theorem (this is included in Alg 2 which is mandatory to graduate from highschool), and that future doctors and engineers have to learn what year Jackson was president and the major philosophers of the traditionalist and revisionist views in America.\r\n\r\nI realize that the world is rapidly changing and that lawyers and customer service secretaries may indeed need to learn math etc far beyond what might be necessary today. But if schools were trying to teach us every possible thing that we might ever need in our career and lifetime, all mathletes might as well learn calligraphy as well, because it might help to have nice handwriting when they write a letter to trying to impress their boss, business partner, etc.", "Solution_1": "It depends on what college you attend. Some schools like Columbia and Chicago have rigorous core curriculums while others like Hopkins and Brown have none at all.", "Solution_2": "Maybe subject irrelevant to the major are crammed into college curricula because colleges do not believe that they were learned to any depth in high school. Maybe it is necessary to do that in college because K-12 schools cannot be trusted to give students a well-rounded education. Maybe requiring freshmen to take a broad range of courses keeps a number of professors employed who otherwise would not have a job. It seems to be the usual situation in the USA, but it does not need to be so. Since we have always done it that way, we will probably continue to do it that way. Because of our insular mentality, we do not care to find out if it is done differently anywhere else. It probably is not done the same way in many other countries.\r\n\r\nLong time ago, in a planet far, far away, I finished 10th grade with a lot of education under my belt, including geometry, algebra, trigonometry, physics, chemistry (general, inorganic, and organic), biology, music, art, Spanish, English, French, world history, world literature, philosophy, civics, geography, space science. At that point I was asked to choose a major and I chose one of the sciences, resultinh in my being required to take a couple of more years of world literature (why?), philosphy (under which they included psychology and logic too), chemistry, physics, biology, and math (including analytical geometry and calculus). After that I was free to start college as a science major and was never again required to study any humanities again. Do I remember Napoleon, Descartes, Sartre, Baudelaire, Smetana? Yes, if dimly. Do I think I would remember any more of Dante's description of the circles of hell if I had had to study it again in college? Not really. Getting rid of the humanities gave me a lot more room in my college schedule for math, physics and chemistry courses.", "Solution_3": "In my opinion, I think it is great that people who major in sciences have to take world literature and philosophy. Science may be your future profession, but in a sense, world literature and philosophy are life. They raise thought provoking questions about the nature of life and the universe. They raise complicated ethical and moral questions (and the really good literature leaves it to you to ponder and doesn't preach a certain set of morals). \r\n\r\nI happen to get rather annoyed at my fellow grad students when I try to talk about literature or make a literature reference that no one gets. Being well-versed in the arts makes communication easier. That's just my opinion, though. \r\n\r\nThe way my undergrad worked was for there to be certain \"categories\" and then we could pick electives that fit those categories. That way they didn't force any \"core\" courses we didn't want to do. We got a broader range of classes that way and we could pick what seemed interesting or useful to us at the same time.", "Solution_4": "My undergraduate school works similar to what HilbertThm90 described in that there are 5 or 6 categories and we have to take a certain number of classes in each category. In general, I think this is a bad idea. I think college students should be mature and responsible enough to select the courses that will prepare them most for the challenges and experiences that they will face. Forcing college students to take a pre-decided number of course in various areas outside of their concentration just treats them like na\u00efve children who need administrators to tell them what they should learn. \r\n\r\nI have no idea what the college administrators are thinking when they sit down and decide that every student who graduates from here should take this many history courses, this many math courses, this many science courses, this many foreign language courses, etc as if anyone who takes less than this will have some sort of a gap in their education. I think many of the administrators are using this archaic notion of a \u201cRenaissance man\u201d who has a complete and well-rounded education as a standard in the guidelines they set for current university students. Maybe if I lived in 16th century academia in Italy, this standard would be okay. But here in 21st century, there is a completely different set of skills needed to succeed. \r\n\r\nWith that said, I do think being able to discuss a variety ideas from different academic disciplines and being reasonably fluent at this is good. But, I think the best place to achieve this is outside of the classroom. Many colleges are moving to a residential college system where students, at least for their first one or two years, live together with a variety of faculty members and interact with each other at meals, social events, guest lectures, etc. This fosters intelligent discussion and learning in many different areas. I think this is a much more effective way to make students well-rounded than to force them to sit through many hours of listening to a professor go into incredible depth on a single topic.\r\n\r\nFour years of college is certainly more than enough to allow a student to specialize and so there is some room to accommodate these extra liberal arts classes. But for students who are very focused and want to graduate in two or three years, I think it is quite unreasonable to tack on an extra semester or an extra year to their college education, possibly costing them tens of thousands of dollars, just to try to make them \"well-rounded\".\r\n\r\nSo, I guess for students who attend college for all four years, it might make sense for them to take at least some courses outside of their concentration. I think college students should basically know for example what an English course or a math course is like, just so that it is not a mystery to them. But in general, I think that universities are way too draconian about this. I doubt that graduation requirements are the best way to achieve this. I think it would be enough to give students a general recommendation about what classes to take and let them read it and discuss it with their advisers and decide whether some of the recommendations don\u2019t apply to them for special reasons.", "Solution_5": "I've often thought about this. I've sat in English class learning compound-complex sentences and wondering WHY I MUST DO THIS! Although I'm not sure if it is true, I have heard of some places testing children from early ages and determining what they would be most successful at, and placing them in a program to help them do that. I think that could work in some cases, although some students may have a change of heart on what they desire to become. \r\n\r\nMy sister attends a university known for it's science and medical programs (she's planning to be a doctor). The university offers an Honors program for advanced students in which they learn all of their \"core\" classes. This only takes up one day a week of their schedules, and the rest of the week can be spent with classes for their major. My sister finds the Honors program to be a great thing. She roomed in her freshman year with a non-honor student, who found it excrutiating to study for finals in subjects she would never use. \r\n\r\nMy guess is that the administrators of schools want students to be \"well-rounded\".", "Solution_6": "How do you feel when so much of the world has a \"Math? Who needs that?\" reaction?\r\n\r\nIf my university did not have a requirement that every student take at least some mathematics, then my department would be a whole lot smaller - so much smaller that I might never have been hired in the first place.", "Solution_7": "No one's given a good reason why students should have to be \"well-rounded\" and meet distribution requirements and whatnot.\r\n\r\n[quote=\"HilbertThm90\"]In my opinion, I think it is great that people who major in sciences have to take world literature and philosophy. Science may be your future profession, but in a sense, world literature and philosophy are life. They raise thought provoking questions about the nature of life and the universe. They raise complicated ethical and moral questions (and the really good literature leaves it to you to ponder and doesn't preach a certain set of morals).[/quote]\n\nYes, almost all subjects are interesting (to me at least; that's why I'm not good at anything). And it is wonderful that science majors have the opportunity to study a variety of subjects in the humanities. But to be forced against their will to? That's a waste of time - not good. They'll probably forget everything they learn if they learn it to meet some requirement...\n\nSome argue that student's eyes need to be opened, that subjects really start to get fascinating when one delves deeper. But if a student shows no interest in a subject by the time they get to [i]college[/i], I don't think it's worth it to keep pushing them. Students should have the choice to specialize much earlier.\n\n[quote=\"Kent Merryfield\"]How do you feel when so much of the world has a \"Math? Who needs that?\" reaction? [/quote]\n\nI feel sorry for people like that; they're missing out on something awesome. But I don't feel that they should be forced to learn math.\n\n[quote]If my university did not have a requirement that every student take at least some mathematics, then my department would be a whole lot smaller - so much smaller that I might never have been hired in the first place.[/quote]\r\n\r\nI don't want you to lose your job or anything, but that isn't a reason to require students to take mathematics courses.\r\n\r\nI'm sure you would agree that trying to teach math to people who simply aren't interested is not nearly as satisfying as teaching it to those who [i]chose[/i] the course.", "Solution_8": "[quote=\"Kent Merryfield\"]How do you feel when so much of the world has a \"Math? Who needs that?\" reaction?[/quote]\n\nWell, however much I think they're wrong, people have a right to believe what they think is valuable. I see math as something essential to everyday life (especially living in a NASA and extremely-high-on-engineers town), who else would program our microwaves and high-def TV's, and what else would I do on Tuesdays after school? While others see it as something petty, I don't. You have to respect others opinions. \n\n\n[quote]If my university did not have a requirement that every student take at least some mathematics, then my department would be a whole lot smaller - so much smaller that I might never have been hired in the first place.[/quote]\r\n\r\nAnd you wouldn't be able to find a job anywhere else? Like, leoxlin, I don't want you to lose your job, but....", "Solution_9": "[quote=\"leoxnlin\"]I feel sorry for people like that; they're missing out on something awesome. But I don't feel that they should be forced to learn math. [/quote]\r\nI'm not a big advocate of the whole \"core\" thing and if my university had a core, I wouldn't be attending. Sure people should be exposed to new things but if they don't like it, don't push them-as it's just a waste of the student and instructor's time.\r\n\r\nI completely disagree with the quoted statement. Math, along with the ability to read and write, is essentially for functioning in the world. The world is driven by math in some form, whether it be the stock market or prices at the supermarket. You need to have some level of skill at math or you won't survive. I'm not saying that a history major should be exposed to real analysis, but at least calculus.", "Solution_10": "I agree that people will probably forget the \"details,\" but I think the main concepts will stay. I also don't think very many schools force students to take extremely specific classes with no choice. There is usually a choice, and the choices should be broad enough that you can find one that will be useful to you.\r\n\r\nThis is especially true of say \"English\" or \"writing\" classes. Can you think of a single profession that requires a college education, but doesn't involve a great deal of writing? Learning to be clear communicators is arguably part of what the university has to teach students. People will stop hiring people from your university if it is clear that students consistently graduate and don't acquire this skill, which in turn will lead to less enrollment.\r\n\r\nThe same is true of many other subjects. It isn't necessarily to be \"well-rounded.\" You just never know when you might need to know about those things. If the time comes and you need to and don't, then that will look bad for your university. We then are back to the lower enrollment if people are walking around saying they graduated from University X, but are unable to do basic things outside of their major (technology breaks and you can't do some math by hand, you write a business letter full of errors, you respond poorly in an ethical dilemma, etc). \r\n\r\nJust because you don't like or enjoy something doesn't mean you will never have to do it. I don't like combinatorics (some may say that I hate it, but my harsh feelings have weakened in the past year), but I still learned the basics since I have no idea if it will ever come up in what I do. It would probably look bad on my university if a basic combinatorial problem came up, and I said, oh, this isn't my area I don't know how to do it. That excuse only goes so far.\r\n\r\n(Note that two posts went up in the time it took me to write this).", "Solution_11": "[quote=\"HilbertThm90\"]I also don't think very many schools force students to take extremely specific classes with no choice. There is usually a choice, and the choices should be broad enough that you can find one that will be useful to you.[/quote]\r\nI think for the most part this is true, but the notable exception is Columbia, which does require that all students take the same classes as part of the core.", "Solution_12": "[quote=\"math_Llama\"]I've taken core classes such as History, English, Math, and Foreign language in school because it was mandatory. It's also occured to me many of these core classes are mandatory in college as well.\n\nOut of pure curiosity, why is it necessary for everyone to take these classes?\nI suppose I take English because it is important to learn to communicate with others, and I take math cause it's fun, but it seems so weird to me that future writers, history professors, lawyers, etc. has to learn DeMoivre's theorem (this is included in Alg 2 which is mandatory to graduate from highschool), and that future doctors and engineers have to learn what year Jackson was president and the major philosophers of the traditionalist and revisionist views in America.\n\nI realize that the world is rapidly changing and that lawyers and customer service secretaries may indeed need to learn math etc far beyond what might be necessary today. But if schools were trying to teach us every possible thing that we might ever need in our career and lifetime, all mathletes might as well learn calligraphy as well, because it might help to have nice handwriting when they write a letter to trying to impress their boss, business partner, etc.[/quote]\r\n\r\nlucky...the only thing our school does with complex #s is like $ i^ 2\\equal{}\\minus{}1$, and addition, subtraction, and stuff like that.", "Solution_13": "Some people learn what they need without any teacher. Others sit through classes and in the long run end up being incompetent anyway. Basic math, science, reading and writing should have been learned long before high school graduation. In an ideal world, a certain competency should be demanded from the earliest grades; those who do not develop it should be told to try again. Knowledge should be retained for the long term rather than learned for a unit test and then forgotten.\r\nA high school diploma should mean that a \"well-rounded education\" has been already attained. Colleges could assign core classes based on results of placement tests, or they could allow students to \"challenge\" core courses and test out of them. That would convince me that they care about the educational attainment of the students more than they care for the tuition dollars and the continued employment of old tenured professors. In its turn, a college degree should guarantee competence in the major listed. Reality seems to be otherwise. Maybe it is because grade points are given piecemeal for sundry assignments rather than on the basis of cumulative exams. Maybe too many grade points are awarded for effort and participation. Maybe grading is too often done \"on a curve.\" Maybe \"learners\" are not motivated to learn for the long term.[quote=\"HilbertThm90\"]I agree that people will probably forget the \"details,\" but I think the main concepts will stay.[/quote]I am not so sure. I work with people who seem to have forgotten most of what they should have learned in their major. For example, a PhD pharmacologist and pharmacist, with degrees from prestigious US universities and teaching pharmacology and toxicology in a US medical school could not tell apart (the names or chemical formulas of) sodium cyanate and sodium cyanide; confused gamma rays with beta radiation; needed help spelling simple words or interpreting simple chemical formulas. A pharmacist working on a doctoral research project argued about how to make a dilution with an undergraduate student. The undergraduate was right. I could list examples for hours. \n[quote=\"HilbertThm90\"]This is especially true of say \"English\" or \"writing\" classes. Can you think of a single profession that requires a college education, but doesn't involve a great deal of writing? Learning to be clear communicators is arguably part of what the university has to teach students. People will stop hiring people from your university if it is clear that students consistently graduate and don't acquire this skill, which in turn will lead to less enrollment.[/quote]Regarding writing and communicating, I have worked with US-born and educated people with advance degrees who cannot spell, but they rely on their administrative assistants who are very good writers. I know that some people do not learn to write in high school, but many of those do not learn in college either. I was once shown samples of work from advanced undergraduate students in a creative writing class. Not only were the plots and timelines full of holes, but also there were atrocious \"spellchecked\" misspellings in many of the stories. One writer even used \"I\" and \"eye\" interchangeably.[quote=\"HilbertThm90\"]The same is true of many other subjects. It isn't necessarily to be \"well-rounded.\" You just never know when you might need to know about those things. If the time comes and you need to and don't, then that will look bad for your university. We then are back to the lower enrollment if people are walking around saying they graduated from University X, but are unable to do basic things outside of their major ...[/quote]Again with the enrollment! If I were to enroll in a college, I would look at the professors, their work, their attitudes, and references from former students. A school may have fame for something unrelated, and be really lousy for the major you have in mind. If people choose to pay big money for big college names, I say let them. Sadly, you cannot judge the quality of the education by the reputation of the school that issued the diploma. After working with some utterly incompetent graduates from big name universities, and seeing other people do excellent work out of their field or with a degree from a little known small college, or without any degree at all, I do not judge people based on the prestige of their alma mater. I listen to what they (and their references) can tell me.[quote=\"HilbertThm90\"]You just never know when you might need to know about those things. ... if people are walking around saying they graduated from University X, but are unable to do basic things outside of their major (technology breaks and you can't do some math by hand, you write a business letter full of errors, you respond poorly in an ethical dilemma, etc).[/quote]So, does anyone know what degree Bernard Madoff had and from what university? I wonder if he took an Ethics course in college.", "Solution_14": "[b]Cores in College:[/b]\r\n\r\n\"Cores\" can exist in colleges, programs within colleges (ie, liberal arts, engineering, etc.) and within a major. A core is just a set of classes that you have to take. There is nothing wrong with that in principle -- in order to get a particular degree, there are certain courses you must have taken. Similarly, in order to graduate with a certain major, the major requirements must be met.\r\n\r\nWhat you probably are objecting to are the liberal arts distribution requirements that exist in almost every college and university. The strictness of the requirements depend on the institution and the program within the institution. Though, keep in mind they're there for a good reason. See the below section on high schools for a list of what I think are some of the important skills developed through humanities courses.\r\n\r\n[b]Cores in High School:[/b] [i]The below is largely adapted from a post I made in another Round Table thread.[/i]\r\n\r\nI think the following skills should be...\r\n\r\n[b]Mandatory:[/b]\r\n\r\n1. Literacy and communication - [i]English, foreign language, general learning environment[/i]\r\n2. Basic interpersonal skills - [i]General learning environment[/i]\r\n3. A (very) basic outline of US history and government, which is necessary to maintain a minimum level of voter competency - [i]History[/i]\r\n4. Knowledge of basic quantitative tools and scientific concepts that will streamline and improve performance in nearly every profession and in daily tasks - [i]Math, science[/i]\r\n5. Awareness of basic principles of health, such as the hazards of smoking and drinking, and how not to get a venereal disease\r\n\r\n[b]Encouraged:[/b]\r\n\r\n1. An experience in the arts, because the arts provide a medium of communication and propagate culture\r\n2. Leadership through student-led initiative opportunities, such as student government, student clubs and organizations\r\n3. Understanding and interest in mathematics and science to drive academic and technological innovation\r\n4. Athletics, because they promote good health and healthy competition\r\n5. Community service, because if taken seriously, it is an eyeopening experience that exposes students to other environments, and also because it promotes good character\r\n\r\nWhile I don't think someone should be forced to know DeMoivre's Theorem to graduate from high school, I certainly think that four years of English, a few years of history (at least one of which is American), some foreign language, and some art are [b]very [/b]reasonable as graduation requirements. There is [i]no[/i] reason why anyone should be certified as a high school graduate without this basic, important coursework.", "Solution_15": "@TZF, I strongly agree with the things you identified as important in a high school education and I think you outlined those things quite well (that could easily be the mission statement for a real high school). However, I think you should differentiate your reasons for high school and college.\r\n\r\nI think having a comprehensive set of liberal arts requirements for college makes college too similar to high school. I think the education system in the United Stated is in general quite slow in moving students up through the system toward a position where they can solve real-world problems and contribute to some profession or some field of research. Obviously, we live in a very technologically and intellectually advanced era where there is a lot of preparation necessary to achieve mastery of some sort of field. But still, I don't think college should just basically be a repetition of high school with more just more advanced classes. \r\n\r\nI think the reasons for going to college should be fundamentally different than the reasons for going to high school and I think the college curriculum should be structured accordingly. Let me try to make an analog of TZF's list for college. This is only a partial list:\r\n\r\n[b]Mandatory (for college):[/b]\r\n\r\n1) Develop communication skills, both written and verbal, to the point where you feel comfortable writing or talking at length about subject matter at a high but non-technical level.\r\n\r\n2) Concentrate in an area that you intend to pursue. Obtain the skills necessary to find work or enter graduate school after college.\r\n\r\n3) You should have significant interaction both with other students and adults. You should probably live in dorm and learn how to co-exist with a variety of other people outside your immediate family. Interaction with professors and graduate students and other people outside your age group should be required regularly.\r\n\r\n[b]Encouraged (for college):[/b]\r\n\r\n1) Maintain a healthy exercise routine, healthy eating habits, healthy relationships. These things should have been learned in high school and they should just be continued.\r\n\r\n2) Participate in clubs and other extracurricular activities that involve competitions but which are more oriented to solving real-world problems or making real-world change than at the high school level. \r\n\r\n3) Continue volunteering and community service activities, like in high school.", "Solution_16": "@KMST: I'm pretty certain that every single argument you presented is flawed or changed my argument into something it is not (and hence was a flaw).\r\n\r\nFor the first thing, I will just repeat my statement \"I agree that people will probably forget the \"details,\" but I think the main concepts will stay,\" that you seem to be refuting, but really are just reinforcing. Every example you have is of someone forgetting details. I agree with this.\r\n\r\nYour general argument seems to be that since it doesn't work the way I say it should, then that is reason to get rid of it. This is a poor attitude. The US healthcare system is not working, thus we should get rid of it. No need for it anymore. The arguments are about the necessity of it, not on whether it works or not.\r\n\r\nNot to mention that you keep using very specific examples. Just because a few people push their way through college and learn nothing, doesn't mean that everyone does this. It is possible a majority are becoming better communicators, learning how to behave ethically, etc. That is called the fallacy of generalization.\r\n\r\n\"Sadly, you cannot judge the quality of the education by the reputation of the school that issued the diploma. After working with some utterly incompetent graduates from big name universities, and seeing other people do excellent work out of their field or with a degree from a little known small college, or without any degree at all, I do not judge people based on the prestige of their alma mater.\" \r\n\r\nThis is irrelevant to what I was saying. I agree with this. It is good that you don't judge based on degrees, but rather on the person. The sad truth is that a majority of people are not like this (note that I named that fallacy above so that I could mention that this just means a majority of people do not make logical decisions, most people generalize).", "Solution_17": "I think that yes, people should be reasonably well-versed in all subjects, especially communication in English. In an ideal world, this education should be attained in high school. College should be where you study [i]whatever you want to[/i], whether you choose to be highly specialized or continue studying a variety of things. But this isn't an ideal world. Darn it.\r\n[size=59]\n[hide]Escape: But math is ideal. We can study that instead. YAY![/hide]\n[/size]\r\n\r\nWe seem to just be debating back and forth over nothing in this thread. One person says, \"we should have to study more stuff\" and another says \"we shouldn't have to study more stuff\", when in reality, everyone agrees quite well on what should and shouldn't be learned.\r\n\r\nOr perhaps I like communication skills. I hope not.", "Solution_18": "[b]KSMT:[/b] Examples of undergraduates being more competent than graduates are really irrelevant here. The point is that there are certain basic skills that should be developed during a college education.\r\n\r\nBy the way, I am curious to know what perspective you argue from\r\n\r\n\r\n[b]To Everyone:[/b] Three things:\r\n\r\n[b]#1:[/b] Different colleges have different approaches toward trying to make this happen, and [b]it is the responsibility of the student to get into a program that suits his/her particular philosophy of learning[/b]. Examples:\r\n\r\n- Students that want a rigorous, complete program in some study that has been laid out by the gods of the field can go to a place with such a program, understanding full well that a lot of specific coursework will be expected of them.\r\n- Students that want to be able to try their hand at many fields can go to a college with fewer major requirements. Students who want the wisdom of established professors and who would benefit from taking the same classes as their peers can go to a college with an established core curriculum.\r\n- Students who love to self-study and don't want to be told what classes they are qualified to take can go to a college that doesn't enforce pre-requisites.\r\n- Students who want to forge their own path through education can go to a place where faculty are encouraging of students designing their own majors.\r\n\r\nAgain, it is the student's responsibility to do the research, apply to, and get themselves into a program that they want to be in -- if they fail to do so, it is nobody else's fault!\r\n\r\n[b]#2:[/b] Having established that it is the student's problem if they are in a program with an unwanted set of mandatory classes, we move to distribution requirements. They're not as bad as you make them out to be.\r\n\r\nDistribution requirements are there because it is commonly believed (by people more qualified than you) that everyone can benefit from them.\r\n- There are some places where AP credits or a proficiency exam will get you credit for a particular class, and thus fulfill the requirement. No place is going to say that you [i]have to[/i] take single-variable calculus [i]no matter what[/i].\r\n- There are other classes that you can't jump out of. For example, suppose some schools require a writing seminar for first-year students, and offer a distribution of topics on which they are taught. The supposition is that no matter how good of a writer you are, and no matter what your interests are, you should be able to find a writing seminar that will aid you in your personal growth as a writer. It is highly arrogant for any student to think otherwise.\r\n- Then, in the true spirit of distribution requirements, many places have guidelines like, \"you must take classes from at least four of the following eight groups of classes\", and then list groups like \"literature\", \"foreign language\", \"social behavior and analysis\", \"mathematics\", \"natural science\", ... or whatever. This doesn't force anyone to take, say, history, if they abhor it. I think these are quite reasonable -- obviously nobody goes to college to take five math courses every semester for eight semesters.\r\n\r\n[b]#3:[/b] I also want to note that everyone can benefit from some degree of distribution.\r\n- If a student knows the material cold in a class that they have to take, then they can surely find other ways to benefit from the class. For example, someone who knows how to write a good essay might use a writing class as an opportunity to experiment a bit with their writing and try out new styles and techniques.\r\n- Students can meet other students with different backgrounds and interests and gain from their perspectives\r\n- A student might uncover a new interest in a topic they did not know much about or that they thought would be boring as a result of a preconceived notion or a bad previous teacher. This is actually quite big -- universities with many courses of study know that most students don't know enough about the fields of their interest to commit to them from day one, and that most students havent seen enough of other interesting fields. They want students to experiment with their interests a bit before asserting what they want to do. This is an example of the institutional philosophies I was talking about in point #1.\r\n\r\n[b]Final thoughts:[/b]\r\n\r\nFor any situation, it is one's responsibility to do one or more of the following:\r\n1. Avoid an undesirable situation (see point #1)\r\n2. If in the undesirable situation, find something to gain from it (see point #2)\r\n3. If the person can do none of the above, then suck it up, reflect and see what you could've done differently, and don't get into a similar situation in the future.\r\n\r\nThis applies to a student's academic career, too. A student should try to choose an academic program that fits their learning needs and philosophy of education. If they can't, they should try to make the best of what they have, knowing that there are likely people that would be more than happy to help them figure out how to do this. And if none of that works, they better give a good thought to where their life is going.\r\n\r\nCore classes and distribution requirements are a property of a student's particular academic career. There are good reasons why a diverse education is encouraged in many places. And if that approach doesn't suit a particular student's needs, they are welcome to find a place with a different approach (because there probably is such a place).", "Solution_19": "[quote=\"TZF\"][b]it is the responsibility of the student to get into a program that suits his/her particular philosophy of learning[/b]. Examples:\n[/quote]\r\n\r\nI second that and I think it is very awful how many students don't do that at all. In my opinion this should be one of the most important criteria in choosing an undergraduate university. I think every prospective student should thoroughly read through the distribution requirements of their prospective university. And admissions department should make these requirements readily accessible.\r\n\r\n I hate how prospective students focus on things like how beautiful the campus is or other peripheral things and don't pay much attention to what coursework is actually required of them of graduate. Maybe high school students just don't realize there is a such a variety of different requirements at different universities since almost all high schools have the same general requirements.", "Solution_20": "\"I hate how prospective students focus on things like how beautiful the campus is or other peripheral things and don't pay much attention to what coursework is actually required of them of graduate. Maybe high school students just don't realize there is a such a variety of different requirements at different universities since almost all high schools have the same general requirements.\"\r\n\r\nThanks for the warning. :)", "Solution_21": "[quote=\"TZF\"][b]KSMT:[/b] Examples of undergraduates being more competent than graduates are really irrelevant here. The point is that there are certain basic skills that should be developed during a college education.\n\nBy the way, I am curious to know what perspective you argue from.[/quote]My points obviously did not get across. It must be because I never took a college writing class in any language, and English is not my native language. If I really cannot communicate, maybe I should go and take a college writing course. Maybe those courses are as necessary as advertised.\r\n\r\nI fully agree that there are basic skills that should be developed. They should also be maintained. My previous post stated that many of such skills, if ever developed, are not maintained. My use of examples distracted from the fact that I was also making general statements, and led to the idea that the examples I was citing were isolated case of \"a few people\" \r\n\r\nI fully agree with TZF's requirements for high school, as long as the requirements include learning and remembering at least half of the material. However, I do not think that the requirement to take a class, or to get a passing grade in a class means much. If a student takes American Government in 11th grade, and gets an A in that class, my congratulations on knowing how to work the system. If after graduation that student can tell me who represents him in Congress (House and Senate), for how long they were elected, and when their term expires, then we have a competent citizen. Unfortunately, I believe that such competent citizens are only a small minority of the voters. Likewise, the students who did not flunk English, should be competent enough to communicate in writing in an ideal world, and ideally each diploma should guarantee a certain competency. I believe the system is focused on process and paperwork, but does not focus on results and results are never guaranteed.\r\n\r\nAfter the competency in communications and other subjects deemed essential for living in a society but not related to a student's major has been established, there is no need for a college to impose the requirement to pay for and sit in a class in those subjects. That was my main point. Does anyone have a list of the US colleges that do that? \r\n\r\nMy secondary point is that people forget (maybe as soon as they walk out of each test/exam) what they were supposed to have learned for the long term. Skill/knowledge that is not interesting or continuously used is soon forgotten. That goes to reinforce my argument about the futility of forcing mature students to take classes they do not want to take. I'll agree that you would want to force small children to learn to read and write. College graduates will develop the ability to communicate in written form if and only if they value and need that ability. How much they practiced writing in a college writing class, or how much they forgot after they obtained their passing grade is not that important. If communicating in writing was necessary in college (e.g. essay questions in exams) and is needed in their life as graduates, the skill attained from K-12 school writing will be maintained and further developed.\r\n\r\nI was trying to make the point that people seem to forget not only the details, but also much of the essence of what they supposedly learned in college, even knowledge related to their major, which should be most likely to be retained. I gave examples because I do not know of any study that has examined how much of the supposedly learned material is retained. Would the average chemist out of college for over 5 years pass the AP chemistry exam? I am not sure. The chemists around me are competent in the specialized knowledge that they use daily. A few are interested in and know about some other areas. Does the average pharmacist out of college for over 5 years remember the properties of drugs, the appropriate doses, and how to dilute them to get to the desired concentration? I would not bank on it. The examples I gave were not of people forgetting details, or of particular undergraduates being more competent than graduates. I gave examples of people being incompetent in the most essential parts of the knowledge that their various diplomas implied. A pharmacist should know about drugs and how to dilute them, as much as a mathematician should know about the order of operations. A person confusing sodium cyanate with sodium cyanide should not be a high school graduate, much less a professor of toxicology and pharmacology in a medical school. They both should know about chemical formulas taught in high school chemistry, and should be able to spell the words that elementary school students are taught." } { "Tag": [], "Problem": "How many permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 have: \r\n1 appearing somewhere to the left of 2, \r\n3 somewhere to the left of 4, and\r\n5 somewhere to the left of 6? \r\nFor example, 8 1 5 7 2 3 9 4 6 would be such a permutation. \r\n\r\n(a) $9\\cdot7!$\r\n(b) $8!$\r\n(c) $5!4!$\r\n(d) $8!4!$\r\n(e) $8!+6!+4!$", "Solution_1": "[hide]I believe that the answer would be[b] [i]E[/i][/b] because you have 8! places to place the 1, 6! places to place the 3, and 4! places to place the 5.[/hide]", "Solution_2": "You have that many places to put each of those numbers? O_O\r\n\r\nI'm really not sure how to solve it, as none of the ideas that I've come up with are answer choices. :?", "Solution_3": "[quote=\"mathgeek2006\"][hide]I believe that the answer would be[b] [i]E[/i][/b] because you have 8! places to place the 1, 6! places to place the 3, and 4! places to place the 5.[/hide][/quote]\r\n\r\nHow do you have 8! places to put the 1? :?", "Solution_4": "I got\r\n\r\n[hide]\n\n ( 9 choose 2 ) * ( 7 choose 2 ) * ( 5 choose 2) * 3! = 45360 \n\n This happens to be equal to the expression in A. [/hide]" } { "Tag": [], "Problem": "What is the greatest three-digit integer in which the product of the digits is 6?", "Solution_1": "$ \\boxed{611}$. Self-Explanatory..." } { "Tag": [ "function" ], "Problem": "Hello!\r\n\r\nI'd like to know if anyone could help me on this problem.\r\n\r\nLet F be a function from Q to Z where Q is the set of rational numbers and Z the set of integers. Show that there are two distinct rational numbers r and s such that [f(r)+f(s)]/2 <= f((r+s)/2).\r\n\r\nThanks!", "Solution_1": "[hide=\"Hint\"] Assume the opposite; namely, that $ f(r)+f(s) > 2f \\left( \\frac{r+s}{2}\\right)$ for any choice of $ r, s$. [/hide]" } { "Tag": [ "geometry", "3D geometry", "sphere", "advanced fields", "advanced fields theorems" ], "Problem": "If a $\\Delta$-complex has the same simplicial homology as the 3-sphere, does that imply it is homotopic to the 3-sphere?\r\n\r\nIn general, can simplicial homology be used to show homotopic equivalence or is singular homology absolutely required?", "Solution_1": "[url=http://en.wikipedia.org/wiki/Homology_sphere]Homology spheres[/url] do exist... The simplicial and singular homology groups of a $\\Delta$-complex are isomorphic: see [url=http://www.math.cornell.edu/~hatcher/AT/ATpage.html]Hatcher's book[/url] for a proof." } { "Tag": [], "Problem": "Join Ducky Tourney [size=200]NOW![/size] [size=150]NOW![/size] [size=100]NOW! [/size][size=75]NOW![/size] [size=59]NOW![/size] [size=42]NOW![/size] [size=25]NOW![/size]\r\n\r\nPlease sign up to the coolest tournament ever!!\r\n\r\n(this is a serious tournament btw)\r\n\r\nThis will be a team tournament, pick your teammates, and if a team is way too strong or too weak, i will have to switch players to make it more even. \r\n\r\nthe number of people per team will be decided later\r\n\r\nHeres how it goes:\r\n\r\nEveryone from team $ X$ plays everyone from team $ Y$\r\n\r\nPerson $ 1$ from Team $ X$ goes against Person $ 1$ from Team $ Y$.\r\n\r\nThe team with the most wins wins!\r\n\r\nThe winning team moves on to the next round. I will decide when each game occurs.\r\n\r\nPlease play responsibly. This may be dangerous for some minds.\r\n\r\n[size=50]MYYELLOWDUCKY82 will not be liable for any injury that will most likely result from the tournament. The tournament is not available in all states or minds. Thank you [/size]", "Solution_1": "OK!!! THIS IS A COMPLETE COPY OF the happy face tourney by (^_^)!!!! THATS ILLEGAL!!!!\r\n\r\n\r\nBY THE WAY I\"M JOINING!", "Solution_2": "Sign me up! :)", "Solution_3": "sorry guys i guess some other things i put earlier wasn't added. \r\n\r\nJoin Ducky Tourney [size=200]NOW![/size] [size=150]NOW![/size] [size=100]NOW! [/size][size=75]NOW![/size] [size=59]NOW![/size] [size=42]NOW![/size] [size=25]NOW![/size]\r\n\r\nPlease sign up to the coolest tournament ever!!\r\n\r\nThis will be a team tournament, pick your teammates, and if a team is way too strong or too weak, i will have to switch players to make it more even. \r\n\r\nthe number of people per team will be decided later\r\n\r\nHeres how it goes:\r\n\r\nEveryone from team $ X$ plays everyone from team $ Y$ and $ Z$\r\n\r\nPerson $ 1$ from Team $ X$ goes against Person $ 1$ from Team $ Y$. and person $ 1$ from team $ Z$\r\n\r\nThe team with the most wins wins!\r\n\r\nThe winning team moves on to the next round. I will decide when each game occurs.\r\n\r\nPlease play responsibly. This may be dangerous for some minds.\r\n\r\n[size=50]MYYELLOWDUCKY82 will not be liable for any injury that will most likely result from the tournament. The tournament is not available in all states or minds. Thank you [/size]", "Solution_4": "OK!!! THIS IS A COMPLETE COPY OF the happy face tourney by (^_^)!!!! THATS ILLEGAL!!!!\r\n\r\n\r\nBY THE WAY I\"M JOINING!", "Solution_5": "uh i changed it.", "Solution_6": "[quote=\"myyellowducky82\"][size=59]MYYELLOWDUCKY82 will not be liable for any injury that will most likely result from the tournament. The tournament is not available in all states or minds. Thank you [/size][/quote]\r\n\r\nThis is really:\r\n\r\nMYYELLOWDUCKY82 will not be liable for any injury that will most likely result from the tournament. The tournament is not available in all states or minds. Thank you", "Solution_7": "ITS STILL INFRINGEMENT!", "Solution_8": "sorry guys i guess some other things i put earlier wasn't added....AGAINNNNNNNN!\r\n\r\nJoin Ducky Tourney [size=200]NOW![/size] [size=150]NOW![/size] [size=100]NOW! [/size][size=75]NOW![/size] [size=59]NOW![/size] [size=42]NOW![/size] [size=25]NOW![/size]\r\n\r\nPlease sign up to the coolest tournament ever!!\r\n\r\nThis will be a team tournament.\r\n\r\nHeres how it goes:\r\n\r\nEveryone from team $ X$ plays everyone from team $ Y$ and $ Z$\r\n\r\nPerson $ 1$ from Team $ X$ goes against Person $ 1$ from Team $ Y$. and person $ 1$ from team $ Z$\r\n\r\nThe team with the most wins wins!\r\n\r\nThe winning team moves on to the next round. I will decide when each game occurs.\r\n\r\nPlease play responsibly. This may be dangerous for some minds.\r\n\r\n[size=50]MYYELLOWDUCKY82 will not be liable for any injury that will result from the tournament. The tournament is not available in all states or minds. Thank you [/size]", "Solution_9": "wth...\r\n\r\ni'm gonna sue you...", "Solution_10": "[quote=\"myyellowducky82\"]sorry guys i guess some other things i put earlier wasn't added....AGAINNNNNNNN!\n\nJoin Ducky Tourney [size=200]NOW![/size] [size=150]NOW![/size] [size=100]NOW! [/size][size=75]NOW![/size] [size=59]NOW![/size] [size=42]NOW![/size] [size=25]NOW![/size]\n\nPlease sign up to the coolest tournament ever!!\n\nThis will be a team tournament.\n\nHeres how it goes:\n\nEveryone from team $ X$ plays everyone from team $ Y$ and $ Z$\n\nPerson $ 1$ from Team $ X$ goes against Person $ 1$ from Team $ Y$. and person $ 1$ from team $ Z$\n\nThe team with the most wins wins!\n\nThe winning team moves on to the next round. I will decide when each game occurs.\n\nPlease play responsibly. This may be dangerous for some minds.\n\n[size=50]MYYELLOWDUCKY82 will not be liable for any injury that will result from the tournament. The tournament is not available in all states or minds. Thank you [/size][/quote]\r\n\r\n\r\nSO if these are the official rules, then does that mean you wont change teans around?", "Solution_11": "i'll change the teams as a last resort.\r\nsorry guyz for changing the rules a lot.", "Solution_12": "u change the rules a lot...\r\n\r\njust like dinger (neat name, huh?)", "Solution_13": "computer was weird.\r\nthats why.", "Solution_14": "meh. maybe dinger's comp was weird too...", "Solution_15": "[quote=\"ernie\"]meh. maybe dinger's comp was weird too...[/quote]\r\n\r\nwhatever....\r\n\r\nur just mad that i took u out of my tournament and ur trying to get back at me...", "Solution_16": "wait...r u talking 2 me happy face?\r\nif u r im not mad.\r\nreally.", "Solution_17": "[quote=\"(^_^)\"][quote=\"ernie\"]meh. maybe dinger's comp was weird too...[/quote]\n\nwhatever....\n\nur just mad that i took u out of my tournament and ur trying to get back at me...[/quote]\r\n\r\ndude... im rooting for u, and u put me down like that?\r\n\r\nharsh.", "Solution_18": "very harsh...\r\n\r\n\r\nno i wasn't talking to you ducky, sorry if it seemed like it.", "Solution_19": "[quote=\"(^_^)\"]very harsh...\n\n\nno i wasn't talking to you ducky, sorry if it seemed like it.[/quote]\r\noh haha i thought u were yelling at me for no reason.\r\ndo u want 2 join the tourney? i need to make a list of the participants.", "Solution_20": "NOTE: if some more people do not join, the tourney will be CANCELED!", "Solution_21": "sooo, is anyone actually joining\r\nor is this just a post for chatting?", "Solution_22": "Okay guys, \r\nthe tourney is officially CANCELLED!!!!!! sorry!\r\nI might make a new one soon, so watch for it!\r\n**someone please lock this one, thanks!" } { "Tag": [ "calculus", "integration", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "can you solve :D \r\n\r\n$ \\int_{0}^{\\infty}\\;\\frac{\\textbf dx}{\\sqrt{\\;x^{4}\\plus{}x^{2}\\plus{}1\\;}}$", "Solution_1": "\\begin{eqnarray*}\\int_{0}^{\\infty}\\frac{dx}{\\sqrt{1+x^{2}+x^{4}}}& = &\\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sec^{2}\\theta\\, d\\theta}{\\sqrt{\\sec^{4}\\theta-\\tan^{2}\\theta}}\\\\ & = &\\int_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta}{\\sqrt{1-\\sin^{2}\\theta\\cos^{2}\\theta}}\\\\ & = &\\int_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta}{\\sqrt{1-\\tfrac{1}{4}\\sin^{2}2\\theta}}\\\\ & = &\\int_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta}{\\sqrt{1-\\tfrac{1}{4}\\sin^{2}\\phi}}\\quad\\quad (\\phi = 2\\theta)\\\\ & = & K\\left(\\tfrac{1}{2}\\right)\\end{eqnarray*}\r\n\r\nwhere $ K(k)\\ =\\ \\int_{0}^{\\frac{\\pi}{2}}\\frac{d\\theta}{\\sqrt{1-k^{2}\\sin^{2}\\theta}}$ is the complete elliptic integral of the first kind." } { "Tag": [ "induction", "algebra proposed", "algebra" ], "Problem": "1. Let $a_1,..., a_n,...$ be positive real numbers such that $\\frac{a_n+a_{n+2}}{2} < a_{n+1}$ for every $n \\geq 1$. Show that $a_n < a_{n+1}$ for every $n \\geq 1$.", "Solution_1": "This is rather evident... :?\r\n\r\nWe can rewrite the condition as $a_{n+2}-a_{n+1} \\frac{a_{n+2}}{r}+2$ we have $a_{n+k}<0$, contradiction.", "Solution_2": "Yar, I dropped a mark or two because I negated the statement wrongly, I didn't make it greater [b]or equal[/b] :blush: :blush: :blush:", "Solution_3": "Andre, you know your mark?\r\nJack", "Solution_4": "nope, but adrian admonished me for the slip, if that's the right word (=" } { "Tag": [ "videos", "ARML", "rate problems" ], "Problem": "In this game, we carry on a conversation with various segues. In other words, we try to create an entire monologue like in this video:\r\n\r\n[youtube]OratYKWszU8[/youtube]\r\n\r\nI'll start.\r\n\r\n\r\nHello. My name's Francis Pumphandle, but everyone calls me Pip. I see that there are a lot of people on this forum that enjoy math. I enjoy math, as well. There are so many math contests to participate in these days. One contest in the United States is ARML, which occurs in three places at once - Nevada, Iowa, and Pennsylvania.\r\n\r\n[i]And then the next poster could talk about Pennsylvania.[/i]", "Solution_1": "Now there's a great state, Pennsylvania. Great schools, historical sites, and food. My favorite food is the yam, which is grown in Africa, Asia, Latin America, and Oceania. They also have great music. I love being serenaded with Japanese music. Japan also is the origin of the well known food sushi. It's rice, vegetables, and sometimes seafood wrapped in seaweed.\r\n\r\n@Next poster:Seaweed.\r\n\r\n\r\n\r\n\r\nI'm not really good at making monologues, though.", "Solution_2": "seaweed can be found in the sea...\r\nin the sea lives fish...\r\nexamples of fish are clown fish..\r\nsuch as NEMO!!\r\n\r\n\r\n@next poster: nemo", "Solution_3": "the name nemo of course brings back memories of the legendary captain of the nautilus.\r\nfew characters from the world of literature have been as awe inspiring.\r\namong those that come close one undoubtedly would be albus dumbledore.\r\n\r\n(note: the last statement is not my personal view, it's just to keep the game going)", "Solution_4": "Dumbledore, eh? Have you seen those Harry Potter movies? Great stuff! If you haven't you should make a family night out. It has something for everyone. Though the Harry kid does get on some peoples' nerves of course besides the fact that he is British. Well everyone from the movie is British, and it is kind of filmed in Britian. And to think of it the author of the books is British also. What do people have agianst the Brits? Not like they bother us they live across the Atlantic, well, of course, they do not call it the \"Atlantic\" they call it the \"Pond\", so I guess you could say that we live across the \"Pond\" from them though no one in the U.S. says \"Pond\" unless they mean a small body of water. Have you every gone swimming in a pond? What about a lake? It id kind of gross, you know, with the water being all brown and stuff and you can not see where you are swimming. Not to get off topic though. Back to those Brits. They have weird phrases and words that people from the U.S. do not use even thoguh they are technically the predecssors to Americans. For instance telly, I mean really the is one instance where American English sounds more sophisticated than the British English. Hm. Another! Watercloset, that can mean a bunch of things in the U.S., if an Brit came to the U.S. and asked where the watercloset was in a restuarent the waiter would give them a weird look and, after they continue to insist, they would eventually get shown the plumbing. The Brits have a bunch of differences with Americans. ...\r\n\r\n :P", "Solution_5": "American food is kind of gross, and the only real American foods (pizza and pasta: Italian) are mac & cheese (not bad, but unhealthy. Yeah, I guess you could work out and stuff, but that would be too tiring. Best just to not eat it) and hamburgers. Hamburgers are so disgusting nowadays. Just about the only hamburger fastfood restaurant worth eating from is In-n-Out, which assured you eat only the most fresh and delicous burgers. Companies like Burger King on the other hand.... BLEH! Especially McDonalds, the food is tasteless, dry, and... It just tastes so bad you can't possibly describe it. I don't see why so many people actually buy food from McDonalds....\r\n\r\nHave fun...", "Solution_6": "Woohoo! Someone with my viewpoint about McDonalds!\r\n\r\nMcDonalds food is basically tasteless white filth served with the lowest form of wheat: the bun. I doubt buns are even the real thing; they are too squishy. The drinks they serve there? Just \"\"pleasure\"\"*does finger thing* drinks that kill brain cells. Brain cells are the most important cells in the body. They don't grow back like your other cells. When they are dead, they are dead. You don't get a second chance. Just like \"Curiosity killed the cat.\".", "Solution_7": "the average rate of cats being killed by a car is increasing every year. however, the average rate of squirrels being killed by cars are skyrocketing...\r\nhowever, the lack of knowledge and eating McDonald's fries have left squirrels hopeless like this:", "Solution_8": "ahh...roadkill. did you know in West Virginia, you can take roadkill home for dinner? and WV is a great place. a great source of unrenewable resources and home to part of the Appalachian Mountains. Is that how you spell Appalachian? thats a funny word. i like funny words. Have you ever noticed the viewpoint of funny in diferent perspectives? Different things sound different to other people. and sound! what is sound? what is the sound of 1 hand clapping? i think thats part of a deep question used in Buddhism" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ x; y \\geq 0$with x + y = 2. Prove that:\r\n \\[ x^2y^2(x^2 \\plus{} y^2) \\leq 2\\].", "Solution_1": "Well, by applying twice Cauchy- Schwartz, we get \r\n$ \\frac{xy}{2}.2xy.(x^2 \\plus{} y^2) \\leq \\left(\\frac{x\\plus{}y}{2}\\right)^2\\frac{1}{2}\\left(\\frac {(x \\plus{} y)^2}{2}\\right)^2 \\equal{} 2$\r\nThe equality holds if and only if $ x \\equal{} y \\equal{} 1$", "Solution_2": "ineq$ \\Leftrightarrow x^2y^2(4 \\minus{} 2xy)\\leq 2\\Leftrightarrow (xy)^3 \\minus{} 2(xy)^2 \\plus{} 1\\geq 0\\Leftrightarrow(xy \\minus{} 1)(x^2y^2 \\minus{} xy \\minus{} 1)\\geq 0,Right ( xy\\leq 1)$", "Solution_3": "x^2 + y^2 = (x + y)^2 - 2xy = 4 - 2xy.\r\n\r\nNow,\r\n\r\n(xy)^2 * (4 - 2xy) =< 2 \r\n\r\n<=> 1/[(xy)^2] >= 2 - xy \r\n\r\n<=> 1/[(xy)^2] + xy >= 2 which follows from am gm as follows:\r\n\r\n1/[(xy)^2] + xy >= 2/[(xy)^(1/2)] >= 2 since xy <= 1", "Solution_4": "Well, since $ x\\plus{}y\\equal{}2$ do $ x\\equal{}1\\plus{}a$ and $ y\\equal{}1\\minus{}a$, with $ 0 \\le a \\le 1$.\r\nThen we get: $ ((1\\minus{}a)(1\\plus{}a))^2((1\\plus{}a)^2\\plus{}(1\\minus{}a)^2) \\le 2$.\r\nBut this is the same as $ (1\\minus{}a^2)^2(1\\plus{}a^2) \\le 1$, or, $ (1\\minus{}a^2)(1\\minus{}a^4) \\le 1$, which is obviously true.", "Solution_5": "$t=xy,f(t)=x^2y^2(x^2+y^2)=8t-6t^2(00,f(t)\\le f(1)=2,$\n\n$x^2y^2(x^2+y^2)\\le 2.$\n\nIrish Mathematical Olympiad 2000 \nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=196400&hilit=Ireland\uff1a\n$ \\iff$ $ x^2y^2(4\\minus{}2xy)\\,\\leq\\,2$\n$ \\iff$ $ x^3y^3\\minus{}2x^2y^2\\plus{}1\\,\\geq\\,0$\n$\\iff (xy\\minus{}1)^{2}\\plus{}(\\frac{1}{xy}\\minus{}1)\\geq 0$ $( xy\\in (0,1))$\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=523349\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=523362&p=2955008#p2955008\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=521368&p=2939566\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=344620#p344620]India 2002:[/url]\nIf $x$, $y$ are positive reals such that $x + y = 2$ show that $x^3y^3(x^3+ y^3) \\leq 2$", "Solution_6": "[b]Generalization of Ireland 2000[/b] \uff1a\nLet $x$, $y$ are positive reals such that $x + y = 2$ , for $n\\ge 1$ , we have\\[x^ny^n(x^2+ y^2) \\leq 2 .\\]\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=542249", "Solution_7": "[quote=\"stvs_f\"]Let $ x; y \\geq 0$with x + y = 2. Prove that:\n \\[ x^2y^2(x^2 \\plus{} y^2) \\leq 2\\].[/quote]\n\n$ x^2y^2(x^2 \\plus{} y^2)\\le xy(x^2 \\plus{} y^2)= 2xy(2-xy)\\le 2(\\frac{xy+2-xy}{2})^2= 2.$\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=360304&hilit=Bulgaria+2010]Bulgaria 2010:[/url]\nLet $a,b,c>0$and$a+b+c=3$,Prove that\\[abc(a^{2}+b^{2}+c^{2}) \\le 3\\].\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3129961#p3129961]Inspired by Bulgaria 2010:[/url]\nLet $a,b,c,d>0$ and $a+b+c+d=4$ , Prove that\\[abcd(a^{2}+b^{2}+c^{2}+d^{2}) \\le 4.\\]", "Solution_8": "http://www.artofproblemsolving.com/community/c6h1246458p6396061:\nIt's obvious that $xy\\le 1$, then \n$x^2y^2(x^2+y^2)=2x^2y^2(2-xy)\\le 2xy(2-xy)\\le 2$ \nby AM-GM.", "Solution_9": "[quote=stvs_f]Let $ x; y \\geq 0$with x + y = 2. Prove that:\n \\[ x^2y^2(x^2 \\plus{} y^2) \\leq 2\\].[/quote]\n[b][color=#f00]Irish 2000[/color][/b]\n[url=https://artofproblemsolving.com/community/c1068820h2028088p14282030] Estonia 2001[/url]\nLet $x$, $y$ are positive reals such that $x + y = 2$ , [url=https://artofproblemsolving.com/community/c6h542249p3128662]show that[/url]\n$$xy(x^2 \\plus{} y^2) \\leq 2$$" } { "Tag": [ "vector", "linear algebra" ], "Problem": "The given definition of a quotient vector space $V/W$ over a subspace $W$ of vector space $V$ is:\r\n$V/W=\\{v_1+w_1|v_1\\in V, w_1\\in W\\}$\r\n\r\nSince $w_1\\in W\\in V$, $w_1+v_1\\in V$ by closure of vector spaces. On the other hand, for any $v_1\\not\\in W$, choose the zero vector for $w_1$ to get $v_1+w_1=v_1$. That said, how is $V/W$ different from $V$?\r\nAlso, can someone please give an example of a quotient vector space? Thanks in advance.", "Solution_1": "I don't think that's the correct definition of a qoutient vector space. The one you proposed describes the [i]sum[/i] V + W of two subspaces of the same vector space. The correct one should be:\r\n\r\n$ V / W = \\{ W + v | v \\in V \\} $, where $ W $ is a subspace of $ V $\r\n\r\nand where the set $ W + v_0 $ is the set $ \\{ w + v_0 | w \\in W \\} $. Note that $ W / V $ is a set and not a family; the definition may produce many pairs $ v_1, v_2 $ for which $ W + v_1 = W + v_2 $, so we disregard repeated sets. Then $ V / W $ is a partition of $ V $ into sets of the form $ W + v $, called [i]cosets[/i], and in order to turn $ V / W $ into a vector space, addition and scalar multiplication are defined on it by $ (W + v_1 ) + (W + v_2) = W + (v_1 + v_2) $ and $ c(W + v_1) = W + cv_1 $ (you can check that these are well-defined operations that satisfy the vector space axioms).", "Solution_2": "I think there was an example about a quotient space in $\\mathbb{R}^2$.\r\nThe subspace was defined as the set of vectors $\\{(x,y)^T|y=3x\\}$.\r\nThen the quotient space is the set of all lines in $\\mathbb{R}^2$ parallel to $y=3x$.\r\nI don't really understand why that is the case.\r\n\r\nWhat's the advantage of having a quotient vector space then? Also, can I have another elementary example?" } { "Tag": [ "induction" ], "Problem": "thanx alot for the help", "Solution_1": "[hide=\"a\"]\nLet $a_{n}$ be the number of such good sequences. Given an $n+1$-digit sequence, remove the last digit. If the last $n$ digits form a good sequence, then any of the five options for the last digit work. Otherwise, a 0 occurs before any of the 1's in the remaining digits (that can't happen) or there are no more 1's (in this case, the removed digit must have been 1 and there can't be any 0's, but any combination of 2's, 3's, and 4's is fair game). So we get\n\\[a_{n+1}=5a_{n}+3^{n}\\]\n[/hide]\n\n[hide=\"b\"]\n$a_{1}=1$ because the 1 has to appear as a digit, and then... oh, of course! there's no more room!\n[/hide]\n\n[hide=\"c\"]\nWe find $\\boxed{a_{n}=\\frac{5^{n}-3^{n}}{2}}$. This may be proven via induction, or it may be [i]counted directly[/i] as follows:\n\nThere are $5^{n}$ total strings, and $3^{n}$ do not contain any 0's or 1's. Therefore $5^{n}-3^{n}$ contain 0 or 1 (or both). Now we can split these $5^{n}-3^{n}$ into pairs such that the second string in each pair looks like the first one, except that 0's and 1's have been switched. (i.e. [b]20342113[/b] gets paired with [b]21342003[/b]). In each of these pairs, exactly one string is good (because if a 1 comes before a 0 (if it appears) in one string, the other string will have a 0 before a 1 (if it appears), and vice versa.) Since there are $\\frac{5^{n}-3^{n}}{2}$ pairs, there are $\\frac{5^{n}-3^{n}}{2}$ good strings.\n[/hide]", "Solution_2": "I haven't found a recurrence relation, but I found an explicit formula:\r\n\r\n[hide=\"blah\"]\nConsider where the first $0$ is found. It could be found in the $2nd, 3rd,...,nth$ position or it could not exist at all.\n\nConsider the latter case. Since we could use only the numbers $1,2,3,4$ in the $n$ available spaces, there are $4^{n}$ combinations. However, because at least one $1$ must exist, we have to subtract the number of combinations with only $2,3,4$. There are $3^{n}$ combinations, and thus $4^{n}-3^{n}$ cases.\n\nNow the case where the first zero is placed at the $kth$ place. To the right of the zero, any of the five numbers could exist; therefore, there are $5^{n-k}$ choices there. To the left of the zero, we recall the previous information and conclude that there are $4^{k-1}-3^{k-1}$ combinations there. Thus, we have $5^{n-k}(4^{k-1}-3^{k-1})$.\n\nLet $f(n)$ denote the number of valid n-sequences.\nSumming, we have\n$f(n)=4^{n}-3^{n}+\\sum_{k=2}^{n}5^{n-k}(4^{k-1}-3^{k-1})$\n\nWe could split up the summation into components of $3^{k-1}$ and $4^{k-1}$.\n\nLet $k=\\sum_{k=2}^{n}5^{n-k}3^{k-1}$\n\nWe have\n$k=5^{n-2}3^{1}+5^{n-3}3^{2}+\\cdots+3^{n-1}$\n$\\frac{3}{5}k=5^{n-3}3^{2}+\\cdots+3^{n-1}+\\frac{1}{5}3^{n}$.\n\n$(1-\\frac{3}{5})k=5^{n-2}3^{1}-\\frac{1}{5}3^{n}$.\n\nWe can do the same with the $4^{k-1}$ component and then substitute in to the equation to get\n\n$f(n)=4^{n}-3^{n}+(4*5^{n-1}-4^{n})-\\frac{1}{2}(3*5^{n-1}-3^{n})$.\n$f(n)=\\frac{5^{n}-3^{n}}{2}$[/hide]", "Solution_3": "[quote=\"scorpius119\"][hide=\"c\"]\nWe find $\\boxed{a_{n}=\\frac{5^{n}-3^{n}}{2}}$. This may be proven via induction, or it may be [i]counted directly[/i] as follows:\n\nThere are $5^{n}$ total strings, and $3^{n}$ do not contain any 0's or 1's. Therefore $5^{n}-3^{n}$ contain 0 or 1 (or both). Now we can split these $5^{n}-3^{n}$ into pairs such that the second string in each pair looks like the first one, except that 0's and 1's have been switched. (i.e. [b]20342113[/b] gets paired with [b]21342003[/b]). In each of these pairs, exactly one string is good (because if a 1 comes before a 0 (if it appears) in one string, the other string will have a 0 before a 1 (if it appears), and vice versa.) Since there are $\\frac{5^{n}-3^{n}}{2}$ pairs, there are $\\frac{5^{n}-3^{n}}{2}$ good strings.\n[/hide][/quote]\r\n\r\nVery tricky direct counting :ninja:", "Solution_4": "thanx to all.... is there any easy way the above question could be solved...... any how i appreciate everybody help\r\n\r\nthanx much" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "x,y are natural number.find all x,y that:\r\n3^x =2^x *y +1", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?p=198299#198299" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "Hello,\r\nIs there an easy way for finding all of the cyclic subgroups of $A_{4}$. It is easy when the group is cyclic, then one can find a generator and then use it to find all the other subgroups, but what do you do in the case when this is not the case.", "Solution_1": "Try all possible generators. owk" } { "Tag": [ "probability" ], "Problem": "If I watch the sky for one hour, the probability that I see a shooting star is exactly 1%. Suppose that I watch the sky for [i]n[/i] hours. What is the probability that I see an odd number of shooting stars?", "Solution_1": "[hide=\"solution\"]The probability that she does not see a shooting star in one hour is $ \\frac{99}{100}$. The probability that she does not see a shooting star in 6 hours is $ \\frac{99^6}{100^6}$. The probability that she does see a shooting star in 6 hours is therefore $ \\frac{100^6\\minus{}99^6}{100^6}\\equal{}\\boxed{\\frac{58519850599}{1000000000000}}$[/hide]", "Solution_2": "@1=2\r\n\r\nYou didn't read the question too carefully.", "Solution_3": "This is $ \\dbinom n1\\cdot0.01^1\\cdot0.99^{n \\minus{} 1} \\plus{} \\dbinom n3\\cdot0.01^3\\cdot0.99^{n \\minus{} 3} \\plus{} \\cdots$...not sure how to simplify this.", "Solution_4": "replace all 6s with ns", "Solution_5": "@1=2: Read the problem again (if you haven't already). All of it.", "Solution_6": "Oh wow odd. I can't read. :wallbash:", "Solution_7": "I don't think I understand the question.\r\nIs it possible that two shooting stars are seen in one hour?", "Solution_8": "I assumed that you can only either see one or none within one hour, which gives the partial solution above.", "Solution_9": "Yes, you can only see one or none shooting stars per hour." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "x is integers\r\n\r\n$(4-x)^{4-x}+(5-x)^{5-x}+10 =4^{x}+5^{x}$", "Solution_1": "$2^{2}+3^{3}+10 =4^{2}+5^{2}$", "Solution_2": "And the solution is unique because $4^{x}+5^{x}$ is strictly increasing and $(4-x)^{4-x}+(5-x)^{5-x}$ is strictly decreasing (at least for $x\\le 3$)." } { "Tag": [ "limit", "trigonometry" ], "Problem": "Evaluate the following limit:\r\n\r\n$\\lim_{h \\to 0}\\frac{\\sqrt{h+2}-\\sqrt{2}}{\\sin(h)}$", "Solution_1": "[hide=\"Solution\"]\\begin{eqnarray*}\\lim_{h\\to 0}\\frac{\\sqrt{h+2}-\\sqrt{2}}{\\sin h}&=& \\lim_{h\\to 0}\\frac{\\sqrt{h+2}-\\sqrt{2}}{h}\\cdot\\frac{h}{\\sin h}\\\\ &=& \\lim_{h\\to 0}\\frac{h+2-2}{h(\\sqrt{h+2}+\\sqrt{2})}\\cdot\\frac{h}{\\sin h}\\\\ &=& \\lim_{h\\to 0}\\frac{1}{\\sqrt{h+2}+\\sqrt{2}}\\cdot\\frac{h}{\\sin h}\\\\ &=& \\frac{1}{2\\sqrt{2}}\\cdot 1\\\\ &=& \\frac{\\sqrt{2}}{4}\\end{eqnarray*}[/hide]", "Solution_2": "[hide]or just use L'Hopital's rule;\nthis is an indeterminate form of 0/0;\nso the limit is really:\nlim (h-->0) of (1/(2*sqrt(h+2) -0) /cos(h)\n =1/(2*(sqrt(2))/1\n=sqrt(2)/4\n\nsorry about the bad syntax.[/hide]", "Solution_3": "[hide]Also if $f\\left(x\\right)=\\sqrt{x}$, then you have \\begin{eqnarray*}\\lim_{h\\rightarrow 0}\\frac{\\sqrt{h+2}-\\sqrt{2}}{\\sin h}&=&f'\\left(2\\right)\\lim_{h\\rightarrow 0}\\frac{h}{\\sin h}\\\\ &=& \\boxed{\\frac{1}{2\\sqrt{2}}}\\end{eqnarray*} [/hide]" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "By considering each bracket separately, find the value of:\r\n\r\n${(2+x^2)^3}-{(2-x^2)^3}$", "Solution_1": "[hide=\"answer\"]$2x^6+24x^2$[/hide]", "Solution_2": "[hide=\"hint\"]use difference of cubes[/hide]", "Solution_3": "[hide]\nUsing difference of cubes, we get $2x^6+24x^2$[/hide]", "Solution_4": "[hide]Since $a^3+b^3=(a-b)(a^2+ab+b^2)$, \n\n$(2+x^2-2+x^2)((4+4x^2+x^4)+(4-x^4)+(4-4x^2+x^4))=2x^2(12+x^4)=24x^2+2x^6$. [/hide]" } { "Tag": [], "Problem": "Show that there are infinitely many $ (x,y) \\in \\mathbb{N}^2$ satisfying $ x^{x\\minus{}y} \\equal{} y^{x\\plus{}y}$.\r\n\r\n[b]Extra:[/b] How complete a solution set can you find?", "Solution_1": "[quote=\"The Zuton Force\"]Show that there are infinitely many $ (x,y) \\in \\mathbb{N}^2$ satisfying $ x^{x \\minus{} y} \\equal{} y^{x \\plus{} y}$.\n\n[b]Extra:[/b] How complete a solution set can you find?[/quote]\r\n\r\n[hide]Set $ x \\equal{} ky$ and you get $ (ky)^{y(k \\minus{} 1)} \\equal{} (y)^{y(k \\plus{} 1)}$, or $ k^{k \\minus{} 1}\\cdot y^{k \\minus{} 1} \\equal{} y^{k \\plus{} 1}$, or $ k^{k \\minus{} 1} \\equal{} y^2$ so we can take an odd integer $ k$ and get an integer $ y$, and an $ (x,y) \\in \\mathbb{N}^2$.[/hide]" } { "Tag": [ "blogs", "\\/closed" ], "Problem": "Why are there two Temperal's?\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=viewprofile&u=30944]Temperal[/url]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/usercp.php?u=28419&mode=viewprofile]Temperal[/url]\r\n\r\nAnd they have two different blogs?", "Solution_1": "The fake Temperal has the \"T\" in his name different from a usual Latin \"T\". Maybe it's a cyrillic \"T\" - this is the way some trolls have created fake accounts on the Wikipedia.\r\n\r\n Darij", "Solution_2": "Ok. But how did you figure that out?\r\n\r\nAnd, is there a way to stop this? Like check through usernames for these \"fake\" letters before continuing.", "Solution_3": "Any good ASCII converter will tell you. \r\n\r\ne.g. http://www.mikezilla.com/exp0012.html", "Solution_4": ":rotfl: \r\nSo uh, what type of 'T' is it?", "Solution_5": "Its a cryllic T as darij said.", "Solution_6": "Nope.\r\nGuess again.\r\n :|" } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "We want to select as many subsets of $ \\{1,2,3,...,n\\}$ as possible so that any two selected subsets have at least one element in common. What is the largest number of subsets we can select?", "Solution_1": "$ 2^{n\\minus{}1}$.\r\n\r\n- If you select $ 2^{n\\minus{}1}\\plus{}1$ (or more) subsets then, from the pigeon-hole principle, you will have select at least one subset and its complementary, so that the desired condition will not be satisfied.\r\n\r\n- By induction, it is easy to prove that we may select at least $ 2^{n\\minus{}1}$ subsets as desired : if it ok for $ n$, then select $ 2^{n\\minus{}1}$ such subsets $ A$ in $ \\{1, \\cdots, n \\}$ and add the $ 2^{n\\minus{}1}$ subsets $ A \\cup \\{n\\plus{}1\\}$ to have the result for $ n\\plus{}1$.\r\n\r\nPierre.", "Solution_2": "(or just take those sets that include the element $ 1$.)\r\n\r\nNonmathematically, there is no \"s\" in Mikl\u00f3s B\u00f3na's family name (and note also that he has two accents).", "Solution_3": "[quote=\"JBL\"](or just take those sets that include the element $ 1$.)\n[/quote]\r\n\r\nWell, since the induction initializes with $ \\{1\\}$ for $ n\\equal{}1$, this leads to the same sets. But, my way to say it is clearly more complicated, thus it may be thought that it has more depth :D \r\n\r\nPierre." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c>0$, then\r\n\r\n$ \\prod[(a^2\\plus{}b^2)^2\\minus{}a^2b^2]\\geq(\\sum a^3b^3)(\\sum a^2b)(\\sum ab^2)$", "Solution_1": "[quote=\"filomen\"]If $ a,b,c > 0$, then\n\n$ \\prod[(a^2 \\plus{} b^2)^2 \\minus{} a^2b^2]\\geq(\\sum a^3b^3)(\\sum a^2b)(\\sum ab^2)$[/quote]\r\n\r\n\r\n$ \\Longleftrightarrow \\sum_{sym} (2a^8b^4 \\plus{} a^8b^2c^2 \\minus{} 2a^7b^4c \\plus{} 4a^6b^4c^2 \\minus{} 2a^6b^3c^3 \\minus{} 3a^5b^5c)\\geq 0$.\r\n\r\n true due to Muirhead theorem.\r\n\r\n :lol:" } { "Tag": [ "advanced fields", "advanced fields theorems" ], "Problem": "i have been trying to generate pythagorean triplets. i could generate it using the recursive way starting from 3,4,5 using ,\r\n\r\n gentrip (a - 2 * b + 2 * c, 2 * a - b + 2 * c, 2 * a - 2 * b + 3 * c);\r\n gentrip (a + 2 * b + 2 * c, 2 * a + b + 2 * c, 2 * a + 2 * b + 3 * c);\r\n gentrip (-a + 2 * b + 2 * c, -2 * a + b + 2 * c, -2 * a + 2 * b + 3 * c);\r\n\r\nBut how do we find the nth one with sorted by c (hypotenuse). I mean scaling it would slow it considerably because we have to sort it. Is there any way to generate the primitive along with non primitive triplets . Can we generate the next term if we know the current term. \r\n\r\nWe can consider both (a,b,c) and (b,a,c) to be different triplets.\r\n \r\nI'm trying to generate these on a computer , the problem is scaling and huge memory and sorting them.\r\nn<10^6\r\n\r\nAny suggestions", "Solution_1": "You haven't defined a total order. After you sort by hypotenuse, how do you sort different triplets with the same hypotenuse? (You should be aware that there are different triplets with the same hypotenuse.)\r\n\r\nThat having been said, the set of all Pythagorean triplets is generated (uniquely) by\r\n\r\n$ (2kmn, k(m^2 \\minus{} n^2), k(m^2 \\plus{} n^2))$\r\n\r\nwhere $ \\gcd(m, n) \\equal{} 1$ and $ m \\not \\equiv n \\bmod 2$. Generating them in any kind of order would be much more annoying than generating a bunch of them and then sorting them afterwards (although I'm not sure why you would want to sort them).", "Solution_2": "[quote=\"t0rajir0u\"]You haven't defined a total order. After you sort by hypotenuse, how do you sort different triplets with the same hypotenuse? (You should be aware that there are different triplets with the same hypotenuse.)\n\nThat having been said, the set of all Pythagorean triplets is generated (uniquely) by\n\n$ (2kmn, k(m^2 \\minus{} n^2), k(m^2 \\plus{} n^2)$\n\nwhere $ \\gcd(m, n) \\equal{} 1$ and $ m \\not \\equiv n \\bmod 2$. Generating them in any kind of order would be much more annoying than generating a bunch of them and then sorting them afterwards (although I'm not sure why you would want to sort them).[/quote]\r\n\r\nThe triplets are sorted in the order (c,a,b) where c is the hypotenuse and (a < b). \r\nI want sort because if we generate triplets the nth triplet would generated from equation above needn't be the sorted nth triplet i want . Now generating the primitives (a,b,c), (b,a,c) and scaling them would generate in some order and we to sort to get the right order. We have 9,12,15 (scale of 3,4,5) after 5,12,13 . Now i want nth triplet with this order defined.Now nth triplet is n/2th triplet if we remove all (a,b,c) but that itself is a problem.\r\n\r\n\r\n> $ (2kmn, k(m^2 \\minus{} n^2), k(m^2 \\plus{} n^2)$\r\n Yes this generates all the triplets but it generates duplicates also. Where do we stop k for the nth one, Say we generate for m < 1000 and n_<", "Solution_1": "Shout out to Saskatchewan! I lived for 5 years there in Saskatoon. :) \r\n\r\nI'm in the same situation, because I'm living in Nova Scotia, which has an even smaller population than Sask. Frankly, I don't think schools anywhere do the stuff on the harder math contests (like proving inequalities...I looked at that, and I don't think any high school teaches that stuff). However, there's a reason many of the people here are ahead of you (and me): almost all the best math students post on this site. You have Euclid winners, CMO winners, Fermat winners, and all that jazz. Don't be discouraged! Try to get invited to UWaterloo Math Contest Seminar next year (if you weren't invited this year). It won't be too hard from Saskatchewan; you just need to win the Fermat there. That's how I got invited. :D", "Solution_2": "There all slow. Its not just your area of the world. Believe it or not, but most class rooms in the western world are HORRIBLE places to learn things. You learn alot more by self discovery, and with gentel guidance. Hang around here, order some books that really make you think your stupid, and then you'll be learning the math you should be.", "Solution_3": "Bah :( It's too bad that the math programs everywhere are so slow... we have enough kids in our school to run a special class or something, and a teacher that would do it, but I don't think the school board would approve. Our school got 7th in Canada on the Euclid with a team of all grade 11s, including 7th in Canada individually. I'll try to win the Fermat, I havn't received any kind of invitation anywhere as of yet, but I'd love to get one. I know that all the competition for the Fermat is inside my school, so as long as I stay above the 5 people biting at my heels I should be able to win it. I did all the Fermats on the CEMC site last year and I got around the 120-134 range on all of them, and since then I've completed 4 math credits, my grade 10, 11 and 2 grade 12s, so I might be able to do well :)\r\n\r\nI'll try to score well in the COMC and Euclid, I hear those are the most important for scholarships and camps. Actually, the CEMC site has decent workshops that outline the Euclid, so for self discovery it might not be that hard to find.\r\n\r\nAt least I might be able to take some university classes once I finish grade 11, in grade 11 I'm taking my last 2 high school math courses, one of which is Calculus, so I'm going to ask my friend to talk to his math professor father and see if I can get something going.\r\n\r\n:o Surprising to see someone else from Saskatoon :P", "Solution_4": "well same thing happens in BC too.\r\nEveryone thinks BC has a comparatively large population, thus the math education would be sufficient. BC hasn't had an IMO member for 5 years already(except for this year, Farzin made it). I'm taking math 10 now, I don't see any point to take such a dumb math course, but the school board always says 'you have the provincial exam this year, you have to take the course to write it.' Also, I can't get any help from the teachers with math problems beyond AMC. \r\nYesterday I talked about it with the math department head in our school. He suggested that I could go to SFU(which is only 3 miles from my school) for further math (calculus year II and linear algebra and stuff). Would that be a good idea? I would have much more schoolwork when I enter higher grades. Also, I'm not planning to take math as my major in university. If anyone has tried this way to make progress in math, please help me out.", "Solution_5": "Yeah, I'm surprised that you say that. I'd say go for it with SFU, but at the same time mattering what kind of person you are it could backfire. It'd be more efficient then just working on your own I'd say, because then you at least receive some recognition for it.\r\n\r\nPersonally I don't really intend on majoring in math in university, but after looking at some IMO stuff, the competitive part of me really desires to learn more... I regret not pushing forward more in elementary school, but partially I wasn't really aloud to, and then the other part was that I didn't know how to. Plus, primary school kids shouldn't be expected to have to be ambitious to excellerate their pace, I'd say that's more on the shoulders of the teacher. I had all my multiplication tables down and could divide and do long division when I was in kindergarten, so I mean if I didn't have to formally go through grade 1-5 here doing all that stuff I could have been way ahead, not to mention I could do algebra in grade 1 and stuff...\r\n\r\nBut I read about people outside of Canada that finish university when they're 20 ish, or I hear about these phenoms and it's frustrating because I'd say the school system here doesn't really allow for kids like that to attain what they could. I mean, they wouldn't let my friend take calculus this year, so he isn't taking a math because he's in grade 11 and done all his except calculus. He scored 7th in Canada on the Euclid. He's beyond all the math we're doing in school, and actually I think a surprisingly large amount of students are. Definately I'd say the curriculum doesn't reflect students ability. My friend was looking through some math textbooks from China, and some of the grade 3 textbooks from China teach our grade 8 curriculum. It's ridiculous. I wish they'd push us more here.\r\n\r\nWhat I'm trying to get at, I guess, is that if you can do it, you should. Don't let little technical things hold you back. Personally I wish I started looking into this stuff earlier :( I feel so behind now that I'm in grade 10, I doubt I could even do the first IMO question :blush: But this summer I hope I can change some of that :)\r\n\r\nIf you think that you won't have enough time to do your high school coursework plus the SFU coursework as well, you could always look into doing a program over the summer. But bleh! *Pokes math B30 textbook* This is supposed to be grade 12 curriculum? Give me a break x_x", "Solution_6": "actually do not over-expect education in other countries which seems spendid for you now. I've been in a so-called high-school-gifted-kid program for two years before, but the education I received can't guarantee me a place in IMO at all! the stuff I learnt was surely ahead of most people of my age, however I felt that it's somehow futile. \r\n\r\nThe goal of education is not to stuff as much knowledge in your brain as possible. Heurism is proven to be a more successful way. Math is not just knowing how to do calculations, it requires more logic and creativity(especially when it comes to olympiads questions), so a better ideation would be more likely to lead you to success in math. And that's also why I hate school math so much----everything with a calculator, then who's thinking?the dumb machine or my brain?\r\nFor contest math, I think the best way to practise is not to do tons of problems, but to have a general idea about each topic it would be testing. I try to analyze every question I do, and that does help me a lot do similiar questions", "Solution_7": "I got so much things to say about education.\r\n\r\nWhen I found out that our family is going to immigrate to Canada, I was actually very happy(opposed to my little brother, who hated losing friends). The main reason was that the education system in Korea was really disappointing. Students were pressured too much, always had so much to work on, yet the gifted students had no way to learn more in the system.\r\n\r\nWhen I came here, it first looked good. Now I actually had free time to do other things. However, after a while I found out that this is even worse. Every course was going too slow(I had hard time in English though, because of the obvious reason). I graduate in a month, and I feel that I gained very little from the school. I think the whole high school curriculum could be learned within a year.\r\n\r\nI really hope that the system is modified so that there is an option of going subjects over faster, for some people who can learn certain subject faster. Students shouldn't suffer through the course and learn less stuff, just because other students learn slower. :(", "Solution_8": "Yeah, the school system here is very... well, it tailors to the average, or slower students. I think in Asia the schools are more made so that the brightness and genius rise up, almost like sifting out the bad ones? But I've never been in Asian schools, that's just what I've heard. My dad skipped 2 grades in China when he was growing up though (I hear that's impossible? :P) but they barely let me skip 1 even though my test scores and comprehension proved that well... I could finish all of elementary school in probably a year (so instead, in grade 1 I just always played sick and then went in the last week and did all my work), and I really got discouraged academically :/\r\n\r\nIt's almost like they chain weights to your legs that you must drag behind yourself, and then you eventually just get tired and bored and collapse. If it wasn't for math contests, I'd probably have demanded that my parents send me to school in Ontario at least, preferrably some private school in the US that would let me be smart instead of dulling me to blend with the norm :/", "Solution_9": "i heard system in asia is changing to accomodate weaker students too... the truth is, no matter how easy the system is, there will always be ppl struggling, so no point making it easier. with the ontario government wanting to cut out gr 12 calc, which is :mad:. in all my courses, i've never had one that didn't use calculus!\r\n\r\nwhen i moved to canada in gr 3, i was learning the same stuff here that i was learning in gr 1 back in asia. in gr 5, we started learning stuff that i learnt in gr 3. basically gr 3,4,5 was learning pretty much the same stuff. if they want to cut the high school curriculum because it's too hard, why not change the curriculum at the elementry level?\r\n\r\ncanadian kids are not dumber, if u take psych courses, u'll see that there is virtually no difference in intelligence between races. just that the curriculum needs to be changed if canadians wish to remian competitive internationally. <---- quoted from a prof\r\n\r\nalso, avg canadians r not competitive in academics. i think the reason why is, there is really no top canadian university, plus there r so many schools, why worry which one u go into? even the top schools in the country are not that hard to get in at all, so why try?", "Solution_10": "Well I think a lot of the people here get lazy since their life is fine. In say China, there's WAY more people, WAY more competition, and the average person struggles a lot growing up. This stress and pressure pushes them to perform in my opinion. :/ I think that elementary school curriculum could be covered in maybe a bit more than 2 years for most \"gifted\" children. Kindergarten and Preschool are really just like daycare.", "Solution_11": "education is daycare.", "Solution_12": "Rant alert:\r\n[rant]\r\nI wouldn't know fully, because I've never been to Korea or China, but I know that in Canada we spend most of our elementary education reading children's books. Since the English language has such an abundance of literature, it's mandatory for multiple years in high school too. Frankly, studying literature is a waste of time. Writing it, or reading it, is different, but studying literature does not increase appreciation of literature nor does it teach us life skills. \r\nAnother problem I see is that much of the \"social studies\" we learn are redundant from year to year. Mostly, they focus on local or Canadian history up to Grade 10, which is an absolute waste of time. Canada has been around for 139 years; the world has been around for billions. I don't know if Asian schools waste so much time on teaching this crap.\r\nWhen I think back, I realize I learned about 5 things in math up to grade 5:\r\na) addition;\r\nb subtraction;\r\nc) multiplication;\r\nd) division; and\r\ne) that \"guess and check\" is an appropriate way to do math. \r\nTell me, what is the difference between adding a two-digit number to a three-digit number and adding a two-digit number to a four digit number? The textbooks I had in those grades were full of crap like this. In the Nova Scotia public system, there is absolutely NO streaming of students who find the work too easy. The homework was mostly art projects, and since my art skills are terrible, my grades were never very good. I don't know if my being in French Immersion slowed me down.\r\nI think what it comes down to is a culture of entitlement. All of the people at my school are going to unversity (it is a private school, but there are still dummies). I know people who have never read a book in their life who plan on going to university People say they want to major in English or Psychology, which seem to be a waste of time. Also, regardless of what you hear students whine about, post-secondary education is not overwhelmingly expensive. It [b]is[/b] a worthwhile investment if you are not going to party all day, skip most of your courses, or major in Comparitive Literature.\r\n[/rant] \r\n:diablo: 's day today!", "Solution_13": "I'm honestly happy I made it out of highschool with ambition and curriosity. There were several points where it actualy did KILL them. They grew back in new forms, but I was never again interested in some of the things I was when I went into highschool. \r\n\r\nI noticed something recently: If I hadn't found math in the last 12 months. I would have ended up as one of them.....\r\n\r\nNot that math is the 'only' path to 'salvation' :D , but I honestly had no interest in anything besieds getting out of highschool.\r\n\r\nI don't think I'm being dramatic when I say \"Highschool destroys children\"/", "Solution_14": "I liked high school. :) Made friends.", "Solution_15": "I agree that kids should challenge themselves, hell, my life became [b]fun[/b] when I relized I can a) choose my own constructive activities I can do, and b) work to better myself, reguardless of whats going on around me. \r\n\r\nBut I didn't figure this out untill begginning of grade 11. We all think the advice is obvious, but no one tells kids to 'work hard at something important to them'. Thats why things are so messed up, kids have no idea how to do better, or be constructive.\r\n\r\nTo be fair, its pretty easy to complain about the ontario cirriculum, becuase it resists [i]everything[/i] you do. It doesn't let kids move forward 'within' the cirriculum. If you want to advance faster, you HAVE to have your parents pay for private schools ( at least that was the only option left for my family).\r\n\r\nIts not that people aren't trying to fallow all of this easy, good advice, its the fact that the system in place is just so hard to work with. \r\n\r\nexample: when my brother got his gifted designation, my parents spent almost a month going to meetings with his grade 8 principle. They [/i] refused[i] to give him an IEP ( the Individual education plan ), that he would need to be let into the gifted program in grade 9. Why? becuase they had never done it before. They make IEPs all the time to get kids shoved into the sp-ed programs, but when a kid came in, trying to get an IEP, to help him excel, they actualy thought the system shouldn't be used this way, and resisted for as long as possible.\n\nexample 2: I love ultimate frisbee. since there was no team running for my school on my senior year, I went on a mission to start it up. They made this impossible. They told me I couldn't even play on school grounds witouth being a 'team' and having a 'teacher supervisor'. I asked every teacher in the school to be an advisor, but the apathy was too thick. after a few weeks of all this running in circles, trying to get some help, we just left, and play in a local shitty park on weekend. =(\n\n\nin conclusion: I agree kids should complain less, do more. But remember: there are some elaborate shut down systems in place, some people feel fine complaining after so much crap.[/i]", "Solution_16": "i think the issue of defining \"gifted\" is a big problem itself. one of my friends, he was invited to participate in 3 olympiads, would he be \"gifted\" by the standards discussed?\r\nhe isn't particularly smart, but he works really hard, he's got bronze in ioi and silver in ibo (only participated in 2 olympiads because third one conflicted)\r\nhe wasn't accepted to the typical gifted program we have here in ontario because his iq score wasn't high enough.\r\nthings didn't always come easy to him, but he definitly tried hard. would he be denied to such program?\r\nhow do u measure one's \"giftness\"?\r\nthe current iq test is quite flawed actually, so do we do it by academic achievements? if so, wouldn't we run into the same problems about gifted kids being not sociable?\r\nif a school will take kids who r simply \"gifted\" not by academic achievements, then why isn't university acceptance based on \"giftedness?\"\r\nu don't need to be gifted to excel in something, all u need is a certain standard of intelligence (not too high, i dunno what the level of measurement is), perserverance, interest, and hard work. for ex. the person who discovered the double helix of dna has iq of 104, he won nobel prize. thomas edision once said \"99% hard work, 1% intelligence\"\r\n\r\ni'm always interested about subjects of measurement of \"smarts\", let me know if u have any comments on that", "Solution_17": "To make J-I-M-B-O's point of view even more punching to young mathematicians, I'll add that most (if not all) of widely known mathematicians were \"stakhanovist's\". From Poisson, to Erdos, all of them worked much and hard ! Being gifted is not revelant. If you are \"gifted\" (whatever that means) you have a better starting point, but if you work harder than the \"gifted\", you learn more, hence you surpass him unevitably. \r\n\r\nAs to the IQ discussion it is to be noted that it heavily depends on your standard of living. Given the same kid, psychologist's measured an IQ of 100 in poor living conditions, and 115 in considerably better conditions. Having this in mind, don't you find IQ measurement in schools an awful perspective ? More generally, has the notion of \"gifted child\" any sense if it does oscillate in the limits of 15 IQ points (since with 105 you are normal, and with 120 that you possibly might have you are \"gifted\") ?", "Solution_18": "I think the whole IQ thing is complete nonsense.", "Solution_19": "[quote=\"lightrhee\"]I think the whole IQ thing is complete nonsense.[/quote]\r\n\r\nit is... it was proved to be flawed...\r\n\r\nthe point of iq test is to test a person's potential\r\nwhile the point of academic test is to test a person's knowledge\r\n\r\nhowever, one's iq score changes from year to year. it was shown that a person's iq drop over the summer when not in school. also, an interesting note was that avg person's iq increased from 70s to 100 since 1930s, which coincidently was when education was made widely available to the public. either human beings evolved intellectually over the past 70 years or iq is very muchly influenced by education or knowledge. \r\n\r\nso iq doesn't accurately test a person's potential but rather than test one's current knowledge.\r\n\r\nso the question is... how do we measure intelligence? current iq test gives us an IDEA, but not accurate but it does give us an idea of how we stand to the avg joe.", "Solution_20": "I think a comparison I first heard from singular is relavent:\r\n\r\n\" the discussion of nature vs nurture is interesting, but not incredibly important. Nature sets the stage and determines the small game, but nurture seems to win out every time in the big game\"\r\n\r\nSome of us have distinct advantages in the womb, but most people here would agree, its what you do with it, not what your born with.", "Solution_21": "Well, it's hard to say what effect schooling actually has on IQ. There could be the fact you learn things in school that make oneself aware of certain things. For instance, I remember on my IQ test in grade 3 I never could figure out how to get 4 litres of water given a 5 litre and 3 litre bucket. Of course, after seeing the solution to one of these questions, I was able to do them all. At the same time, even if you do not \"learn\" anything new from the schooling, it keeps your mind active. Over the summer, lots of people get lazy and their minds relax more, the synapses don't fire as quickly and what not, so getting a lower IQ would be very permissable due to inactivity.\r\n\r\nOf course, I think one can learn to score well on an IQ test, but I think that an IQ test is pretty accurate when someone is a kid and hasn't been exposed to questions that would be on an IQ test. I think it's useful to test a child's logical potential, but of course, that's just potential. But really, I don't think it can be that heavily weighted.\r\n\r\nEven tests such as the SATs are underfire for not being an accurate assessment. I think one of the best ways to reach someone's intellegence is to talk to them and observe them. I think most reasonably educated or smart people can differentiate between someone that is very smart, and someone that just says a bunch of BS.\r\n\r\nI also find it odd that the school system is more willing to cater towards children that have difficulty learning due to a disease, disability, or just the fact they aren't all that \"smart\" compared to say a gifted child that has difficulty learning because the curriculum doesn't teach all that much. I mean, I'm taking grade 11 chemistry this year and for me it's a joke. It's just basic math all over again. Stoiciometry, moles, gas laws, it's all pretty basic once you wrap your head around it. But I know kids that have a LOT of trouble wrapping their heads around it, and of course the teacher is more willing to give them extra work to practice, but none for me to further my knowledge base.\r\n\r\nIt's really hard to title someone gifted though, and I'll agree to that. But I think that some people at least have a sense for who is gifted and who isn't.\r\n\r\nAnd also, hard work I feel can only get you so far. I think that everyone needs a certain level of \"intellegence\" of giftedness to meld with their hard work. I mean, I know guys that try so hard in classes but still get 60s or 70s because they just don't have the head for that class. And I mean, that's normal. I think the whole thing is having a balance. I watched a show once... W-Five or the Fifth Estate or something that was about people with very high IQs that ended up being like.. bouncers and stuff, or just sit around and do nothing. It's a certain balance between intellegence and how far you want to take it. So I mean, someone with a 110 IQ compared to someone with a 150 IQ (assuming IQ is accurate, just for a hypothetical example) could score way better on tests, and get further academically and what not just because they push their intellegence to it's limit, and the 150 IQ person is just content doing whatever.\r\n\r\nI just think though all in all that the school system should cater to gifted children to some extent. I mean... in grade 5-8 my teachers just got me to mark other kids work and help other kids with their assignments when I finished early... and I was in a gifted program that is supposed to have a 130 IQ minimum... It was ridiculous.", "Solution_22": "i'd like to add that from the textbook i have, for students who succeed in post-secondary (i dunno what they mean by success) has 0.44 correlation with their iq tested in gr 3, while SAT score has correlation of 0.78.\r\nit's very true we can tell if the person is smart by talking to them, but there r a lot of exceptions, like einstein, who was thought to be slightly slow in his early teens.\r\nagain, some of the top students have trouble communicating ideas, thus sound/look stupid. \r\ni think it's much better to stick with the system we have right now and just tighten it up slightly. truly gifted students can simply study on their own.\r\n\r\nno institution will take anyone based on their intelligence score, they at most take combination of academic and intelligence. \r\n\r\nbtw, since u need certain intelligence to excel in something and especially when mentioned that ppl getting 60s and 70s because they're not smart enough to get the subject, wouldn't we just pick ppl based on their accomplishment??? just like that we r doing right now?? since ppl who accomplish big r ppl who ACTUALLY cares so they try and is smart as well. contests r also another good indicator!! thats what waterloo does. if u excel in contests and have sufficient marks, they'll take u.\r\n\r\nthis world is not just about nature, it's nature AND nurture.\r\nwhy take ppl who won't work hard but ridiculously smart, just because they r born lucky and put them in class with other smart ppl?\r\nhow is this feasible?\r\nit's like saying, there's a football star in my school. instead, since playing locally is impeding his progress and us normal ppl is slowing his progress, so he gets go to training camp with the pros. so far that hasn't happened yet, not to my knowledge anyways. even if it did happen, it's private company that pays for it. why would the government pay for that?\r\ni am sure that we're not the first ones to think of this, if it was a feasible plan, any country would of done it by now.", "Solution_23": "[quote=\"J-I-M-B-O\"]i'd like to add that from the textbook i have, for students who succeed in post-secondary (i dunno what they mean by success) has 0.44 correlation with their iq tested in gr 3, while SAT score has correlation of 0.78.\nit's very true we can tell if the person is smart by talking to them, but there r a lot of exceptions, like einstein, who was thought to be slightly slow in his early teens.\nagain, some of the top students have trouble communicating ideas, thus sound/look stupid. \ni think it's much better to stick with the system we have right now and just tighten it up slightly. truly gifted students can simply study on their own.\n\nno institution will take anyone based on their intelligence score, they at most take combination of academic and intelligence. \n\nbtw, since u need certain intelligence to excel in something and especially when mentioned that ppl getting 60s and 70s because they're not smart enough to get the subject, wouldn't we just pick ppl based on their accomplishment??? just like that we r doing right now?? since ppl who accomplish big r ppl who ACTUALLY cares so they try and is smart as well. contests r also another good indicator!! thats what waterloo does. if u excel in contests and have sufficient marks, they'll take u.\n\nthis world is not just about nature, it's nature AND nurture.\nwhy take ppl who won't work hard but ridiculously smart, just because they r born lucky and put them in class with other smart ppl?\nhow is this feasible?\nit's like saying, there's a football star in my school. instead, since playing locally is impeding his progress and us normal ppl is slowing his progress, so he gets go to training camp with the pros. so far that hasn't happened yet, not to my knowledge anyways. even if it did happen, it's private company that pays for it. why would the government pay for that?\ni am sure that we're not the first ones to think of this, if it was a feasible plan, any country would of done it by now.[/quote]\r\n\r\nActually I think some parts of China do it :P *pokes their IMO score* :| \r\n\r\nAnd I think that sure, gifted kids can study on their own but they shouldn't have to. The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain). They do this, but they aren't willing to title a child GIFTED and give him the same funding? I think that is skewed. I really do believe the school system is made for the lower hinge of kids, and I don't think that's fair... I'm just saying it would seem more logical to put all that power where the potential is.\r\n\r\nOh, and anyone know what kind of test scores and marks in high school classes one would need to go somewhere like Stanford or Harvard? I hear that it's mostly about extra curricular activities.", "Solution_24": "Chris, are you going to quit Advanced?\r\nI don't think that's a good idea--I personally still think it is a great experience. However--your crop of Advancies seems a bit stranger and more arrogant than ours...maybe that's just me.\r\n\r\nI think Advanced will do you good, especially the IRP (I recommend it despite the stress; you learn a lot more about how to learn and yourself from the project than regular will ever give you).\r\n\r\n*sigh* I wish we had IB.", "Solution_25": "[quote=\"{x}\"]To make J-I-M-B-O's point of view even more punching to young mathematicians, I'll add that most (if not all) of widely known mathematicians were \"stakhanovist's\". From Poisson, to Erdos, all of them worked much and hard ! Being gifted is not revelant. If you are \"gifted\" (whatever that means) you have a better starting point, but if you work harder than the \"gifted\", you learn more, hence you surpass him unevitably. [/quote]\r\n\r\nOf course being gifted is relevant. Probably some 'famous' mathematicians are more innately intelligent than others, but I'm pretty sure they would all rank in the 99.9th percentile of any standardized intelligence test.", "Solution_26": "[quote=\"CaucasianAsian\"] And I think that sure, gifted kids can study on their own but they shouldn't have to. The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain). They do this, but they aren't willing to title a child GIFTED and give him the same funding? I think that is skewed. I really do believe the school system is made for the lower hinge of kids, and I don't think that's fair... I'm just saying it would seem more logical to put all that power where the potential is.\n[/quote]\r\nyo man always remember that 144 or w/e on Cayley is not the best in Canada and someone else should get more funding than you bcz of the contest. is that fair?\r\nand also, potential is not only in math contests. the contests in high school are only as piddling in your life as kindergarten games in your education.", "Solution_27": "[quote=\"CaucasianAsian\"]\nAnd I think that sure, gifted kids can study on their own but they shouldn't have to. The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain). They do this, but they aren't willing to title a child GIFTED and give him the same funding? [/quote]\r\n\r\nLittle bit off topic. Kids with Aspergers AREN'T stupid, they tend to especially excel at one topic. And they actually need that TA, cuz they just don't function with large groups of people. I've worked with a couple kids like that, its crazy what they do. But agreed, the government and individual schools need to have some way of encouraging talent in other students. I suppose they just can't afford it in general, not now at least. Besides, those kids falling behind need the help more than kids who can potentially help themselves.", "Solution_28": "[quote=\"Sunny\"][quote=\"CaucasianAsian\"]\nAnd I think that sure, gifted kids can study on their own but they shouldn't have to. The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain). They do this, but they aren't willing to title a child GIFTED and give him the same funding? [/quote]\n\nLittle bit off topic. Kids with Aspergers AREN'T stupid, they tend to especially excel at one topic. And they actually need that TA, cuz they just don't function with large groups of people. I've worked with a couple kids like that, its crazy what they do. But agreed, the government and individual schools need to have some way of encouraging talent in other students. I suppose they just can't afford it in general, not now at least. Besides, those kids falling behind need the help more than kids who can potentially help themselves.[/quote]\r\n\r\nYeah, my point for that is just that it is a perfect example of everyone's tax money going towards one person's [well, I guess two people's] child.", "Solution_29": "[quote=\"nneonneo\"]Chris, are you going to quit Advanced?\nI don't think that's a good idea--I personally still think it is a great experience. However--your crop of Advancies seems a bit stranger and more arrogant than ours...maybe that's just me.\n\nI think Advanced will do you good, especially the IRP (I recommend it despite the stress; you learn a lot more about how to learn and yourself from the project than regular will ever give you).\n\n*sigh* I wish we had IB.[/quote]\r\n\r\nYeah I am Bobby, sorry to disappoint, but I hate the group of children that I am with for it. Nothing personal with them, but I'm not a big fan of comparing marks for the most part, especially in a class like english. IRP isn't going to be a big loss for me, I had to teach myself math 10 and 20 over last summer, so I have a decent idea of how I learn, and I mean... I know I won't be able to do the same with IRP, but I finished my MIRP essentially in one night over the course of 3 hours, or at least I did all the research, and all the writing except maybe a couple proof reading things. Not just that, but I ended up with one of the best marks in our class.... apparently though, our classes marks were bad :P I got a 90 I think. I also prepared for my presentation over the course of 5 minutes, before that I had no idea what I wanted to do for it, and I had no notes for reference, and Mr Davis said it was one of the best ones.\r\n\r\nA perfect example of why I'm leaving advanced though, is after he said that a couple kids of course tried to say how bad my presentation was, etc etc because of how competitive the advanced class is :blush: not to say I can't compete, but I'd rather enjoy my time in school and be relaxed than be all stressed over marks and comparing myself to others. We both know I'm a pretty....... confident person. :P" } { "Tag": [], "Problem": "here are a few exam archives from australia:\r\n\r\n4 unit maths:\r\n\r\nhttp://www.angelfire.com/ab7/fourunit\r\n\r\nhttp://www.geocities.com/fourunitmaths\r\n\r\nhttp://www.geocities.com/ext2papers\r\n\r\nmaybe some other australians can post up their papers here on mathlinks. ok?", "Solution_1": "That isn't really problem solving is it? :?\r\n\r\nIt's just the typical schoolish calculation...", "Solution_2": "Yeah, these above links are just \"4 unit maths\" papers which is just the end-of-school level maths here in Australia --- no problem solving. In terms of olympiad problem solving here we have a two-day contest, the Australian MO (run by the Australian Maths Trust --- http://www.amt.edu.au), in february each year. All the papers used to be on http://www.kalva.demon.co.uk until John Scholes suddely took them all off. I think he thought they were too easy, and they certainly are easy compared to APMO and IMO which we of course have in common with everyone else.", "Solution_3": "[quote=\"Dionysius\"]All the papers used to be on http://www.kalva.demon.co.uk until John Scholes suddely took them all off. I think he thought they were too easy, and they certainly are easy compared to APMO and IMO which we of course have in common with everyone else.[/quote]\r\n\r\nNo, he was forced to take them off. The AMT left him te choice: taking it off voluntarily, or ask a court what they think of it.\r\n\r\nThey did the same thing with the Chinese olympiad. I really wonder what the AMT has to say on the CMO but anyway, it's happened: http://www.mathlinks.ro/Forum/viewtopic.php?t=5304" } { "Tag": [ "probability" ], "Problem": "A fair coin is tossed 1997 times. Find the probability that at no point during the\r\ntossing are two heads flipped consecutively.", "Solution_1": "Uh..how is the answer expressed?\r\n\r\nAs a sum of combinations..?\r\n\r\nbeucase its too big..", "Solution_2": "[hide=\"Solution\"]\nI will consider 1 to be heads, and 0 to be tails. \n\n[b]Count number of binary strings that do not contain two consecutive 1s:[/b]\n\nLet $ f_{a}(n)$ denote the number of binary strings of length n, with last digit $ a$ (0 or 1).\n\n$ f_{0}(1) = 1\\; f_{1}(1) = 1$ (There's one binary string that ends in 0, and one that ends in 1).\n\nThe number of binary strings of length n ending with 0, is the sum of the number of binary strings of length n-1 ending in 0 or 1.\n$ f_{0}(n) = f_{0}(n-1)+f_{1}(n-1)$\nAlso the number of binary strings of length n ending with 1, is the sum of the number of binary strings of length n-1 ending in 0.\n$ f_{1}(n) = f_{0}(n-1)$\n\nAdding the previous two equations together; \n$ f_{0}(n)+f_{1}(n) = 2f_{0}(n-1)+f_{1}(n-1)$\n$ = [f_{0}(n-1)+f_{1}(n-1)]+[f_{0}(n-2)+f_{1}(n-2)]$\n\nThe LHS represents the total number of binary strings of length n without two consecutive heads. This is the fibonacci sequence since $ F_{n}= F_{n-1}+F_{n-2}$\n\nSo the answer is $ \\frac{F_{1997}}{2^{1997}}$\n[/hide]", "Solution_3": "[quote=\"JavaMan\"][hide=\"Solution\"]\nI will consider 1 to be heads, and 0 to be tails. \n\n[b]Count number of binary strings that do not contain two consecutive 1s:[/b]\n\nLet $ f_{a}(n)$ denote the number of binary strings of length n, with last digit $ a$ (0 or 1).\n\n$ f_{0}(1) = 1\\; f_{1}(1) = 1$ (There's one binary string that ends in 0, and one that ends in 1).\n\nThe number of binary strings of length n ending with 0, is the sum of the number of binary strings of length n-1 ending in 0 or 1.\n$ f_{0}(n) = f_{0}(n-1)+f_{1}(n-1)$\nAlso the number of binary strings of length n ending with 1, is the sum of the number of binary strings of length n-1 ending in 0.\n$ f_{1}(n) = f_{0}(n-1)$\n\nAdding the previous two equations together; \n$ f_{0}(n)+f_{1}(n) = 2f_{0}(n-1)+f_{1}(n-1)$\n$ = [f_{0}(n-1)+f_{1}(n-1)]+[f_{0}(n-2)+f_{1}(n-2)]$\n\nThe LHS represents the total number of binary strings of length n without two consecutive heads. This is the fibonacci sequence since $ F_{n}= F_{n-1}+F_{n-2}$\n\nSo the answer is $ \\frac{F_{1997}}{2^{1997}}$\n[/hide][/quote]\r\n\r\nUm if its $ \\frac{F_{1997}}{2^{1997}}$, then you have to note that $ F_{1}=2$ (then fibionacci), not $ 1$.Otherwise if $ F_{1}=1$and$ F_{2}=1$(then 2,3..so on), then its $ F_{1999}$ over $ 2^{1997}$", "Solution_4": "I should've been more explicit there, though I did mention $ f_{0}(1) = 1,\\; f_{1}(1) = 1 \\implies F(1) = 2$", "Solution_5": "[quote=\"JavaMan\"]I should've been more explicit there, though I did mention $ f_{0}(1) = 1,\\; f_{1}(1) = 1 \\implies F(1) = 2$[/quote]\r\nand i shouldve read your full solution..I just went straight to your answer and compared it to mine..because I was thinking about putting the answer as a summation of combinations, which results in that fibionacci pattern." } { "Tag": [ "function", "calculus", "derivative", "integration", "real analysis", "inequalities", "real analysis unsolved" ], "Problem": "i cant seem to pick the test functions right? the question asks if $ \\exists p \\in[1,\\infty)$ \r\n\r\ns.t. $ u \\in W^{1,p}([0,1])$ then show that the weak derivative $ u'$ exists and is in $ L^p(0,1)$ and that u is equal a.e. to an absolutely continous function.\r\n\r\ni think by definition of the sobolev space we get the u' exists and is in L^p, but im having trouble isolating something to look like\r\n\r\n$ \\sum_k |u(x_k) \\minus{} u(y_k)|$ without being able to use all the normal calculus tools..\r\n\r\npicking $ \\phi_m \\to \\sum_k 1_{[x_k,y_k]}(x)$ made sense to me, but then i use the FTC which im not sure is legal on a weak derivative.. help!\r\n\r\npart 2: asks us that if $ p\\in (1,\\infty)$ then show\r\n\r\n$ |u(x)\\minus{}u(y)| \\le |x\\minus{}y|^{1\\minus{}\\frac{1}{p}} \\left( \\int_0^1 |u'(t)|^pdt\\right)^{\\frac{1}{p}}$", "Solution_1": "Let $ v(x)\\equal{}u(0)\\plus{}\\int_0^x u'(t)\\,dt$. This is an absolutely continuous function because $ u'$ is Lebesgue integrable. It is also in $ W^{1,p}$, since $ u'$ is its weak derivative. Finally, prove that $ u\\minus{}v\\equal{}0$ a.e. Hint: if $ \\int (u\\minus{}v)\\phi \\equal{}0$ for any test function, then $ u\\minus{}v\\equal{}0$ a.e..\r\n\r\nThe second part is Holder's inequality applied to $ \\int_x^y 1\\cdot u'(t)\\,dt$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Find $ \\boxed{S\\equal{}\\left( \\frac{C_n^0}{1}\\right)^2\\plus{}\\left( \\frac{C_n^1}{2}\\right)^2\\plus{}\\left( \\frac{C_n^2}{3}\\right)^2\\plus{}...\\plus{}\\left( \\frac{C_n^n}{n\\plus{}1}\\right)^2}$", "Solution_1": "\\begin{align*}\r\n\\sum_{i = 0}^n \\left(\\frac{1}{i + 1} \\binom{n}{i}\\right)^2 & = \\sum_{i = 0}^n \\left(\\frac{n!}{(i + 1)!(n - i)!}\\right)^2 \\\\\r\n& = \\frac{1}{(n + 1)^2} \\sum_{i = 0}^n \\binom{n + 1}{i + 1}^2 \\\\\r\n& = \\frac{1}{(n + 1)^2} \\left(\\binom{2n + 2}{n + 1} - 1\\right).\r\n\\end{align*}", "Solution_2": "Yes. It's true and it's a nice solution. Thank you very much." } { "Tag": [ "Harvard", "college", "MATHCOUNTS", "Gauss", "AMC", "AIME", "USA(J)MO" ], "Problem": "Congrats to Alex Zhai on his perfect score on the IMO! :lol:", "Solution_1": "wow yeah. good job alex.", "Solution_2": "Whose Alex Zhai... is he from Illinois?", "Solution_3": "yes. he's a senior from downstate who (as you can see) is pretty beastly.", "Solution_4": "Oh this guy...\r\nhttp://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2008-ia/2008imoteamannounce.shtml\r\n\r\nhe's amazing", "Solution_5": "[quote=\"RussianRocket\"]Oh this guy...\nhttp://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2008-ia/2008imoteamannounce.shtml\n\nhe's amazing[/quote]\r\nYep. And he's going to Harvard this fall.", "Solution_6": "So did Alex win the whole thing? :o", "Solution_7": "He tied for 1st along with two members of the China team. :first:", "Solution_8": "Is china usually the best or are there other countries that are usually in the top?\r\nI wouldn't usually expect for a USA person to win.", "Solution_9": "China and Russia are always very good; US is usually in the top 5.\r\nI'm pretty sure the records for teams are online, but I can't find them right now.", "Solution_10": "Yeah my dad tied for first at the Soviet Union Olympiad and made the IMO in 1985 but he did not do as good as he wanted to there.\r\nThat year the Union of the Soviet socialist republics aka Russia got 6th. I am not sure what my dad got but I think he was second with a score of 28.\r\n\r\nHere is the website\r\nhttp://www.imo-official.org/year_country_r.aspx?year=1985", "Solution_11": "Wow...your dad is so much better than my dad in math...\r\n\r\n :( :( :(", "Solution_12": "Yah... :rotfl: right now im not living up to his expectations :wallbash_red: \r\nBut im getting better :football:", "Solution_13": "I challenged my dad to a MATHCOUNTS countdown and nuked him :D :D :D", "Solution_14": "The one game i can beat my dad at is 24. (you know when you take 4 numbers and you have to make 24 out of it).\r\nWe had 9 rounds to see woould figure out the answers faster and i beat him\r\n5-4. LOL! :D \r\nHe can destroy me at any other game. :rotfl:", "Solution_15": "dude, who is this stephen character? why would he ever sing the teletubbies song? geez. what a moron he must be. :roll:", "Solution_16": "i think he's stupidityismygam.", "Solution_17": "[quote=\"Aryth\"]dude, who is this stephen character? why would he ever sing the teletubbies song? geez. what a moron he must be. :roll:[/quote]\n\ni don't know, but if I ever meet him I am turning and running in the opposite direction. He sounds really really really really scary/silly.\n\n[quote=\"gauss1181\"]i think he's stupidityismygam.[/quote]\r\n\r\noh what a lame screen name, this furthers my utter fright and disrespect for this stephen character. I hope I never meet him at a math tournament or something :oops:", "Solution_18": "[quote=\"gauss1181\"]i think he's stupidityismygam.[/quote]\r\n\r\nuh, i was being sarcastic (again). i lived with him for 3 weeks once.", "Solution_19": "[quote=\"gauss1181\"][quote=\"avec_une_h\"]Do I seem like a guy? Because I'm definitely not.[/quote]\nI'm not saying you...they're saying ME :D[/quote]\r\n\r\nI thought you were a guy for a minute. Then I realize who you were.", "Solution_20": ":rotfl: :rotfl: :rotfl: \r\n\r\nby the way, avec_une_h are you gonna take AMCs for 2009?\r\n\r\nI AM!", "Solution_21": "The AMC's are so easy...\r\n\r\nIt doesn't take much to get a perfect on those...", "Solution_22": "how can you easily get a perfect AMC 12? :huh:", "Solution_23": "simple. be Alex Zhai.", "Solution_24": "[quote=\"gauss1181\"]:rotfl: :rotfl: :rotfl: \n\nby the way, avec_une_h are you gonna take AMCs for 2009?\n\nI AM![/quote]\r\n\r\nDefinitely. There's not much else in high school. (I'm still trying to work out how existent my high school's math team is.", "Solution_25": "Have you ever taken Amcs before? what scores do u get? (and on aime too?)", "Solution_26": "[quote=\"avec_une_h\"][quote=\"gauss1181\"]:rotfl: :rotfl: :rotfl: \n\nby the way, avec_une_h are you gonna take AMCs for 2009?\n\nI AM![/quote]\n\nDefinitely. There's not much else in high school. (I'm still trying to work out how existent my high school's math team is.[/quote]\r\n\r\nyeah. go amc. is there anything else in high school to do 4 math? actually i'm not sure if my school has a math team...", "Solution_27": "um\r\n\r\nARML, local contests (ICTM, NSML, IML, ...), Mandelbrot...\r\njust try to make it, and then do well, on the usamo. that's enough of a field to keep you busy for a while (forever probably)", "Solution_28": "the HMMT and USAMTS are good ones too", "Solution_29": "china is usually #1 or #2, and this year there was an american person in the top ;)" } { "Tag": [ "vector", "linear algebra", "matrix" ], "Problem": "Eight lamps, each with an on-off switch, are arranged in a circle. A lamplighter can flip (in\r\nother words, change the state of) the switches, but he is not allowed to flip just one at a time.\r\nWhen he switches lamps on or off, he is required to flip the switches of three consecutive\r\nlamps simultaneously. (For example, he can flip the switches of lamps 7, 8 and 1 at the same\r\ntime.) Prove that no matter what set of lamps was turned on at the start, the lamplighter\r\ncan turn all the lamps on.", "Solution_1": "did u get these from pomona?\r\n\r\nhttp://www.math.pomona.edu/ps1_06.pdf", "Solution_2": "[hide]It suffices to show that there is a combination that, overall, will switch one light and keep the others the same.\n\nSuppose you want to change light number 1 alone. You can switch {2,3,4}, {3,4,5}, {5,6,7}, {6,7,8}, {8,1,2}, which in effect switches all switches but 1 twice, and 1 once.[/hide]\r\n\r\n[b]Generalization:[/b] If there are $ n$ lights, and you can switch $ m$ consecutive lights at a time, for which ordered pairs $ (m,n)$ is this doable?", "Solution_3": "[hide=\"Solution\"] This is doable precisely when $ (m, 2n) \\equal{} 1$. \n\nFirst, we rephrase the problem in terms of linear algebra. We are given the $ n$ vectors in $ \\mathbb{F}_2^n$ consisting of $ m$ consecutive entries of $ 1$s with the rest zero and we wish to prove it spans $ \\mathbb{F}_2^n$. A rather general result in linear algebra tells us this is possible if and only if the $ n$ vectors are linearly independent. (This is actually easier to prove over finite fields than infinite ones.)\n\n- If $ (m, n) > 1$, then there are $ \\frac {2n}{(m,n)}$ vectors that sum to the zero vector. For example, $ (1, 1, 1, 0, 0, 0) \\plus{} (0, 0, 0, 1, 1, 1) \\plus{} (0, 1, 1, 1, 0, 0) \\plus{} (1, 0, 0, 0, 1, 1) \\equal{} \\mathbf{0}$.\n\n- If $ (m, 2) \\equal{} 2$, then all $ n$ vectors sum to the zero vector.\n\n- Finally, if $ m$ is odd and $ (m, n) \\equal{} 1$, the sum of $ (1, 1, 1, ...) \\plus{} (0, 1, 1, ...)$ is a vector that switches $ 1$ and $ m \\plus{} 1$. Generally, we can switch $ 1$ and $ am \\plus{} 1 \\bmod n$ simultaneously. Since $ am$ goes through all residue classes $ \\bmod n$, we can switch any two lamps simultaneously. Then starting from $ (1, 1, 1, ... )$ we can remove every $ 1$ except the first in pairs to obtain $ (1, 0, 0, ...)$, and since the same logic holds cyclically we can obtain every element of the standard basis of $ \\mathbb{F}_2^n$. (There is in all likelihood a much slicker way to handle this case. See [url=http://en.wikipedia.org/wiki/Circulant_matrix]circulant matrix[/url].) [/hide]" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Hi, i have a problem understanding a thing in the joined text.\r\nall is clear until the author introduces this \"t\" (see the picture attached).\r\n\r\nCould someone explain me what it means ? where does this t com from ?\r\nis it a known notation i don't know ?\r\n\r\nThank you very much", "Solution_1": "[quote=\"Klasz\"]Hi, i have a problem understanding a thing in the joined text.\nall is clear until the author introduces this \"t\" (see the picture attached).\n\nCould someone explain me what it means ? where does this t com from ?\nis it a known notation i don't know ?\n\nThank you very much[/quote]\r\n\r\nIt means [url=http://mathworld.wolfram.com/Transpose.html]Transpose[/url], go there, they have a nice definition and explanation.\r\n\r\nBest,", "Solution_2": "Great Thanks :blush: \r\n\r\nin france we put this at the upper left of the matrix name....", "Solution_3": "[quote=\"Klasz\"]Great Thanks :blush: \n\nin france we put this at the upper left of the matrix name....[/quote]\r\n\r\nYeah, not long ago Valentin mentioned that in Romania that was the standard notation too. But such is life. Here people use to write the t on the upper right of the matrix name, and most of the time they write a capital T instead of a lower case t.\r\n\r\nActually, there are tons of things that we use to assume they are standard notation, and they are not. For example, when you have an interval that is half open you write a round parenthesis towards the number, and in most of Latin America you see a bracket toward the other side. For example, you see $[0,1[$ instead of $[0,1)$, and $]1,2]$ instead of $(1,2]$ :? .\r\n\r\nBest," } { "Tag": [ "calculus" ], "Problem": "Does anybody have any idea how to get a solution manual for Spivak's second edition Calculus book??", "Solution_1": "I doubt they still make the solution book for the second edition, since there is now a 4th edition out. Some university bookstores may have old copies of the third edition but finding the second edition would be difficult." } { "Tag": [], "Problem": "What does this sign mean?\r\n\r\n U (pretend it's upside-down)\r\n\r\nAnd is there a formula for A U(upside-down) B?", "Solution_1": "You might be interested in [url=http://www.artofproblemsolving.com/Wiki/index.php/Set]this.[/url] It explains everything about sets and the notation you were wondering about.", "Solution_2": "That notation means \"intersection\", which is the set of numbers that two or more sets have in common. For example, if you have sets $ (1,2,3,4,5)$, $ (2,3,4,5,6)$ and $ (3,4,5,6,7)$ The intersection of these three sets are $ (3,4,5)$" } { "Tag": [ "linear algebra" ], "Problem": "Let $A = \\left({\\begin{array}{*{20}c}3 & 2 &{-5}\\\\ 2 & 6 &{-10}\\\\ 1 & 2 &{-3}\\\\ \\end{array}}\\right)$ and $B = \\left({\\begin{array}{*{20}c}6 &{20}&{-34}\\\\ 6 &{32}&{-51}\\\\ 4 &{20}&{-32}\\\\ \\end{array}}\\right)$ and let $A,B\\in Mat_{3}(\\mathbb{Q}).$ Prove that $A$ and $B$ are similar in $\\mathbb{Q}$ and find invertible $P\\in Mat_{3}(\\mathbb{Q})$ such that $P^{-1}BP=A.$", "Solution_1": "They're not similar, since their traces are different.", "Solution_2": "[quote=\"jmerry\"]They're not similar, since their traces are different.[/quote]\r\n\r\nI'm sorry. I corrected the problem.\r\nThank you.", "Solution_3": "traces are still different.." } { "Tag": [ "geometry", "perimeter", "ratio", "3D geometry" ], "Problem": "[color=red]Do not post the answers.(If you find the answer,pm me.)[/color]\r\n#1.\r\n[color=green]The students in Mr.Beaty\u2019s[/color] [color=red]Math[/color]Counts [color=green]designed the decorations that are hung on the city\u2019s street-light poles. Each decoration is either in the shape of an equilateral triangle or a regular hexagon. The perimeter of each decoration is 24 feet. What is the ratio of the area of the triangle to the area of the hexagon? Express your answer as a common fraction.[/color]\r\n--------------------------------------------------------------------------------\r\n#2.\r\n[color=red]In December, 1990, a holiday tree in the shape of a right circular cone, had a height of 8 feet and a radius of 2 feet. By December 1991, the tree\u2019s height and radius each increased by 10%, and the tree\u2019s measurements continued to increase each year at a rate of 10% of its previous year\u2019s size. In December, 2006, the tree will be decorated with lights. The lights will be placed along the curved lateral surface of the tree at the rate of one light per square foot. No lights will be placed within the branches or on the underside of the tree. The formula for the lateral area of a cone is LA = \u03c0 \u00d7 r \u00d7 l where r is the radius and l is the slant height of the cone. How many lights are needed to decorate the tree in December, 2007? Express your answer to the nearest ten. [/color]\r\n--------------------------------------------------------------------------------\r\n#3.\r\n[color=green]Each year[/color] [color=red]Mario :lol: [/color] [color=green]gives [/color][color=olive]Luigi [/color]:lol: [color=green]a gift (mushroom) that is wrapped in five boxes that are five similar rectangular prisms. After he unwraps the first box, there is another box inside to be unwrapped, and then another and another and another. He does not get to his gift until he has unwrapped all five boxes. The length, width, and height of the smallest box are 4 inches, 2 inches, and 1 inch respectively. The dimensions of each larger box are 20% greater than the dimensions of the previous smaller box. By what factor does the volume of the smallest box need to be multiplied to obtain the volume of the largest box? Express your answer as a decimal to the nearest tenth.[/color]", "Solution_1": "im guessing these r the \"holiday\" problems that u talked about.....", "Solution_2": "I'm going to post the answers to thesewhen 5 people send me a pm with the right answer.", "Solution_3": "Or I'll just post the answers when somebody asks me to.", "Solution_4": "Okay I'll post the answers now.\r\n1. [hide]$ \\frac{2}{3}$[/hide]\n\n\n\n2.[hide]$ 1090$[/hide]\n\n\n\n3. [hide]$ 5.9$[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "Can anyone give me a code (what to write in $\\LaTeX$) for 4x4 square in Problem 1?\r\n\r\nI can't reveal my reason for it but I just want it. Thanks in advance for anyone who can give me. :D", "Solution_1": "Anyone?\r\n\r\n :?", "Solution_2": "You could just make the square in Excel, copy it and save it as an image (in Paint or whatever), and implement it into $\\LaTeX$ using the \\includegraphics command.", "Solution_3": "\\[ \\begin{tabular}{|c|c|c|c|}\\hline x & y & z & w\\\\ \\hline &&&\\\\ \\hline &&&\\\\ \\hline &&&\\\\ \\hline \\end{tabular} \\]\r\n\r\n[code]\\begin{tabular}{|c|c|c|c|}\\hline\nx & y & z & w\\\\ \\hline\n&&&\\\\ \\hline\n&&&\\\\ \\hline\n&&&\\\\ \\hline\n\\end{tabular}[/code]", "Solution_4": "Anyone know how to make a thick border?", "Solution_5": "[quote=\"joml88\"]\\[ \\begin{tabular}{|c|c|c|c|}\\hline x & y & z & w\\\\ \\hline &&&\\\\ \\hline &&&\\\\ \\hline &&&\\\\ \\hline \\end{tabular} \\]\n\n[code]\\begin{tabular}{|c|c|c|c|}\\hline\nx & y & z & w\\\\ \\hline\n&&&\\\\ \\hline\n&&&\\\\ \\hline\n&&&\\\\ \\hline\n\\end{tabular}[/code][/quote]\r\nThat's exactly what I did - also, if you want to put several tables across one row, use the align command with something like \\quad between tables", "Solution_6": "[quote=\"chess64\"]Anyone know how to make a thick border?[/quote]\r\nsame question\r\nwhy do they not give out the latex code and images for the questions in each round?\r\nthat would help so much... we could just edit the images and copy-paste it into our solutions", "Solution_7": "Unfortunately, there's not much to glean from the LaTeX files. This year, we did our graphics externally (even the square) with METAPOST, so looking at the LaTeX files won't really show you anything.", "Solution_8": "Oh, I was going to use metapost but I wanted to know if it could be done in LaTeX first. (I can't find any good tutorials for Metapost :()", "Solution_9": "[quote=\"chess64\"]Oh, I was going to use metapost but I wanted to know if it could be done in LaTeX first. (I can't find any good tutorials for Metapost :()[/quote]\r\n\r\nOh, there's a number of them:\r\n[url=http://remote.science.uva.nl/~heck/Courses/mptut.pdf]remote.science.uva.nl/~heck/Courses/mptut.pdf[/url]\r\n[url=http://cm.bell-labs.com/who/hobby/MetaPost.html]cm.bell-labs.com/who/hobby/MetaPost.html[/url]\r\n[url=http://www.tug.org/metapost.html]www.tug.org/metapost.html[/url]", "Solution_10": "Thanks! :)", "Solution_11": "Thanks.\r\n\r\nI didn't really need the thick border line anyway. I just needed a way to make tables and this is one thing I should learn in $\\LaTeX$ (along with eqnarray..).", "Solution_12": "[quote=\"Silverfalcon\"]Thanks.\n\nI didn't really need the thick border line anyway. I just needed a way to make tables and this is one thing I should learn in $\\LaTeX$ (along with eqnarray..).[/quote]\r\nlol eqnarray is my favorite command. :lol:", "Solution_13": "[quote=\"ZennyK\"]lol eqnarray is my favorite command. :lol:[/quote]\r\n\r\nActually, if you're using $\\LaTeXe$ (you can tell by whether the document begins with documentclass, instead of documentstyle), I recommend using the align environment, or the split environment, instead of eqnarray. I think it looks nicer.", "Solution_14": "[quote=\"nsato\"][quote=\"ZennyK\"]lol eqnarray is my favorite command. :lol:[/quote]\n\nActually, if you're using $\\LaTeXe$ (you can tell by whether the document begins with documentclass, instead of documentstyle), I recommend using the align environment, or the split environment, instead of eqnarray. I think it looks nicer.[/quote]\r\nI've switched over to almost exclusively using the align environment - it's just as good as eqnarray but is more powerful.", "Solution_15": "I don't mean to change the topic or anything, but how do you an align environment?", "Solution_16": "Here's an example:\r\n[code]\n\\begin{align*}\na^2 + b^2 &= c^2, \\\\\n\nx + y &= z.\n\\end{align*}\n[/code]", "Solution_17": "Okay, but I don't see any difference.", "Solution_18": "[quote=\"nsato\"][quote=\"ZennyK\"]lol eqnarray is my favorite command. :lol:[/quote]\n\nActually, if you're using $\\LaTeXe$ (you can tell by whether the document begins with documentclass, instead of documentstyle), I recommend using the align environment, or the split environment, instead of eqnarray. I think it looks nicer.[/quote]\r\nThanks! I'll try using that in my solutions for round 2.", "Solution_19": "[quote=\"chess64\"]Okay, but I don't see any difference.[/quote]\r\n\r\nYou might have to put two examples next to each other. In the align/split environments, there's not as much space around the equal signs, or whatever you're aligning around.", "Solution_20": "[quote=\"nsato\"][quote=\"chess64\"]Okay, but I don't see any difference.[/quote]\n\nYou might have to put two examples next to each other. In the align/split environments, there's not as much space around the equal signs, or whatever you're aligning around.[/quote]\r\nOh, good. That's one thing I never liked about the eqnarray command; it always seemed kind of weird that there was so much space. :?", "Solution_21": "Oh, I see it. Thanks! (It's also less to type... :D)" } { "Tag": [ "MATHCOUNTS", "number theory", "Diophantine equation" ], "Problem": "I had this question during mathcounts, the answer is 10, but I can't think of a way to get to that besides guess and check.\r\n\r\nA number, if divided by 2, has a remainder of 0\r\n\r\nif divided by 3, remainder of 1\r\n\r\ndivided by 4, remainder of 2", "Solution_1": "I assume the question asks for the smallest positive integer? If, so\r\n\r\n[hide=\"Solution\"] Call the integer $ y$. Then from the second equation we get that $ y \\equal{} 3k \\plus{} 1$, for some positive integer $ k$. From the third condition, we get that $ y \\equal{} 4x \\plus{} 2$, for some positive integer $ x$ (we can ignore the first condition). Then, we have\n\n$ y \\equal{} 3k \\plus{} 1$\n\n$ y \\equal{} 4x \\plus{} 2$\n\nThis implies that \n\n$ 3k \\plus{} 1 \\equal{} 4x \\plus{} 2 \\implies 4x \\plus{} 1 \\equal{} 3k$\n\nThis is a Diophantine Equation with smallest solution at $ x \\equal{} 2$, $ k \\equal{} 3$. Thus, the smallest positive integer is\n\n$ y \\equal{} 4x \\plus{} 2 \\implies y \\equal{} 4(2) \\plus{} 2 \\implies y \\equal{} \\boxed{10}$.[/hide]", "Solution_2": "Another way to think of the question...\r\n\r\n[hide=\"Another Solution\"]Let's consider this: If we add 2 to the number, and then divide it by 2, 3, and 4, we would get a remainder of 0 for each answer. \n\nTherefore, what we are now looking for is the $ LCM(2,3,4)$ subtract 2. \n\n$ LCM(2,3,4) \\equal{} 12$ \n\n$ 12 \\minus{} 2 \\equal{} \\boxed{10}$[/hide]", "Solution_3": "I will have to remember that in the future,\r\n\r\nand oh, yes, it was for the smallest integer", "Solution_4": "you will see a lot of these problems in mathcounts where the difference is constant.\r\n\r\nFor example State Sprint a couple of years ago was:\r\n\r\nA rectangular band has a certain number of members between 100 and 200. When the band is divided into 8 rows there are 2 spots left unoccupied. When it is divided into 9 rows 3 spots are left unoccupied..how many people does the band have?", "Solution_5": "Well,what can i do for you? How to get the answer?" } { "Tag": [ "algorithm", "number theory", "prime numbers", "number theory unsolved" ], "Problem": "problem : $d,n \\in N$ so that $d|n$ and d prime.proove that $d\\leq \\sqrt{n}$.\r\n\r\n(i think i just wasted eight pages on this and got nowhere,i just\r\nthought i got two solutions,but then looking back i always found\r\nan error in my proof. I remember i used this to make allot of\r\nalgorithms,but i never had to proove it,it seemed natural,\r\nin Knuth they say that one should consider the fact\r\nthat $n=d\\cdot \\frac{n}{d}$,by this i thought about a solution\r\nwith a friend ,we made it but i feel that its...too complicated)", "Solution_1": "The way it is stated, it's wrong. Counterexample: let $n=100$, $d=20$. $20|100$ but $20>\\sqrt{100}=10$. I think you meant there exists a $d$ that divides $n$ such that $d\\le\\sqrt{n}$\r\n\r\n[EDIT]Well, my wording is too obvious by letting $d=1$. Hmm, maybe this: let $a,b\\in\\mathbb{Z}^+$ so that $n=ab$. Then prove that $\\min\\{a,b\\}\\le\\sqrt{n}$. :)", "Solution_2": "sorry about the mistake,i corrected it", "Solution_3": "The corrected version doesn't make sense either. Let $d=5$ and $n=10$. $d=5$ is prime but $5>\\sqrt{10}$.\r\n\r\n[EDIT]If you read the statement I put in my previous post under EDIT, you can realize that's a true statement.\r\n\r\nLet $a,b\\in\\mathbb{Z}^+$ so that $ab=n$. Prove that $\\min\\{a,b\\}\\le\\sqrt{n}$.\r\n\r\nProof: Without loss of generality (WLOG) let $a\\le b$. Now, we have to prove that $a\\le\\sqrt{n}$. We use a simple proof by contradiction. Assume that $a>\\sqrt{n}$. Then,\r\n\\[ n=ab\\ge a^2>(\\sqrt{n})^2=n \\]\r\na clear and easy contradiction. Thus, $a\\le\\sqrt{n}$.", "Solution_4": "10000th User i am really sure that when trying to find if a number n is prime or not,\r\nyou would try to see if there is a number d smaller than the square root of n so that\r\nd divides n.\r\ni'm really sure of that.\r\nabout your last post,for sufficiently big n the property works", "Solution_5": "I think I understand what you are trying to do. You want an algorithm to test whether $n$ is prime by testing all primes $p\\le\\sqrt{n}$.\r\n\r\nLet's get our ideas organized. From my previous post, the proof I gave was essentially that if a number is composite, then one of its positive factors is less than or equal to $\\sqrt{n}$. The statement I proved works for any positive integer $n$. \r\n\r\nNow, take the contrapositive of my statement. If none of the positive integers less than or equal to $\\sqrt{n}$ divide $n$, then $n$ is prime.\r\n\r\nSo it suffices to only test prime numbers less than or equal to $\\sqrt{n}$. If none of these divide $n$, then $n$ is prime.\r\n\r\nExample: let $n=101$. OK we know 101 is a prime number but for this example let's assume we don't know if it is prime or not. Now, $10<\\sqrt{101}<11$, so we only need to test the primes 2,3,5,7. After you test, you realize none of these prime numbers divide 101, so therefore 101 is prime.", "Solution_6": "so it follows that all prime divisors of n are in the interval $[1,[\\sqrt{n}]]$ ?", "Solution_7": "[quote=\"spx2\"]so it follows that all prime divisors of n are in the interval $[1,[\\sqrt{n}]]$ ?[/quote]No, this is something else different, check again the example I gave before, i.e. $n=10$. $2\\in [1,\\sqrt{n}]$ and $5\\in [\\sqrt{n},n]$. Nowhere in my previous posts says that all prime divisors of $n$ are in that interval, this is a false statement. Perhaps you want to know what happens to prime numbers larger than $\\sqrt{n}$ like 5|10 ?\r\n\r\nLook at the statement I wrote before. Now I challenge you this:\r\n\r\nLet $n$ be a product of two primes $p,q$ and given that $p\\sqrt{n}$.\r\n\r\nHere is a second related problem to this discussion:\r\n\r\nProve that if all the prime numbers in $[1,\\sqrt{n}]$ do not divide $n$ evenly, then so does all the prime numbers in \r\n$[\\sqrt{n},n]$. Then show that in this case $n$ is prime.\r\n\r\nIf you can do the above problems, you are good to go.", "Solution_8": "so the algorithm is based on this:\r\nif a number n is composite then it must have a factor in $[1,\\sqrt{n}]$.\r\n\r\nabout the first problem : \r\n$n=pq ,\\ pn \\Rightarrow q>\\sqrt{n}$\r\n\r\nabout the second problem what do you mean by \"divide n evenly \" ?", "Solution_9": "That's probably an American slang, sorry. It means \"divide $n$ exactly\". So if all primes $p\\le\\sqrt{n}$ do not divide $n$, then all primes $q>\\sqrt{n}$ do not divide $n$.", "Solution_10": "ok i understood..now about the 2nd problem,\r\nif there is a p prime number in $[\\sqrt{n},n]$ that divides n,\r\nthen $n/p$ must be in $[1,\\sqrt{n}]$ and it must also divide n.\r\nbut we started from the hipothesis that none of the numbers in\r\n$[1,\\sqrt{n}]$ divide n. so :) there it is.", "Solution_11": "Excellent job, and thus from 2nd problem, we conclude that $n$ is prime. :)\r\n\r\nIf there are any questions, post here, I'll check back." } { "Tag": [ "ratio" ], "Problem": "Of the following sets of data the only one that does not determine the shape of a triangle is:\r\n\r\n$ \\textbf{(A)}\\ \\text{the ratio of two sides and the included angle} \\\\\r\n\\qquad\\textbf{(B)}\\ \\text{the ratios of the three altitudes} \\\\\r\n\\qquad\\textbf{(C)}\\ \\text{the ratios of the three medians} \\\\\r\n\\qquad\\textbf{(D)}\\ \\text{the ratio of the altitude to the corresponding base} \\\\\r\n\\qquad\\textbf{(E)}\\ \\text{two angles}$", "Solution_1": "[hide=\"Click for solution\"]\n$ \\boxed{\\textbf{(D)}}$ is incorrect because we can \"shift\" the altitude over, thus changing the shape of the triangle.\n[/hide]", "Solution_2": "It can't be E, because then the triangles would be similar by AA." } { "Tag": [ "combinatorics unsolved", "combinatorics", "Combinatorial games", "IMO Shortlist", "game strategy", "game" ], "Problem": "Two players play alternatively on an infinite square grid. The first player puts an $X$ in an empty cell and the second player puts an $O$ in an empty cell. The first player wins if he gets $11$ adjacent $X$'s in a line - horizontally, vertically or diagonally. Show that the second player can always prevent the first player from winning.", "Solution_1": "It is known that a \"$k$-in-a-row\" game is a draw for $k\\geq 8$, with a simple pairing strategy for $k\\geq 9$. See [url]http://www.weijima.com/index.php?option=com_content&view=article&id=11&Itemid=15[/url]. The $5$-in-a-row is also called Gomoku, and its outcome is not known. $4$-in-a-row (and less) are a win for first player.", "Solution_2": "Divide the infinite grid into staggered columns of 3x3 grids, and treat them as separate tic-tac-toe games where you can force a draw. To get 11-in-a-row, it's clear that should it occur in a column or row, one must have 3-in-row in one of those 3x3 grids. After drawing it out, we can see that at most ten can be placed in a diagonal so that none of the 3x3 grids has 3 diagonally in a row. ", "Solution_3": "[hide = Solution] Our intuition leads us to first consider a finite grid. Most easy to work with being a square. Playing around with various squares, we begin to develop a strategy to tie in an odd square board. (Boards of the form $(2k + 1) \\times (2k + 1)$ for some integer $k$). \n\n[b][u]Lemma[/u]: [/b] We have a winning strategy for $(2k + 1) \\times (2k + 1)$ boards described as follows. If the first player $A$ puts an $X$ in a corner of a square board, then reflect over the line $\"y = x\"$. If $A$ marks an $X$ on an edge that is on the top or bottom of a square board, $B$ places an $O$ to the reflection of $X$ over the $y$-axis. If $A$ marks an $X$ on an edge that is on the left or right of a square board, $B$ places and $O$ to the reflection of $X$ over the $x$-axis. \n\n[hide = Proof of lemma] We induct on $k$. For $k = 1$, we actually have a slight variation in the figure below. If $k = 2$, then this strategy holds true. Now assume this is true for $k - 1$. We show it is true for $k$. If $A$ marks and $X$ anywhere that isn't an edge then this is simply the case for $k - 1$ or smaller. Now if $A$ marks an $X$ in a corner, $(\\pm 2k + 1) \\times (\\pm 2k + 1)$, then we just swap the signs and now it is guaranteed $A$ cannot make any diagonal of length $2k + 1$. \n\nIf $A$ marks an $X$ on an edge from the $x$-axis, $(x, \\pm 2k + 1), x \\neq 0$, then mirorr about the $y - axis$. This guarantees he cannot make a row of $2k + 1$ adjacent $x$'s along that row. If $x = 0$, then mirror about the $x$-axis.\n\nSimilarly if $A$ marks an $X$ on an edge from the $y$-axis, $(\\pm 2k + 1, y), y \\neq 0$, then mirror about the $x$-axis. If $y = 0$, mirror about the $y$-axis. \n\nThis always guarantees that $A$ can never fully make a row consisting of $2k + 1$ adjacent $X$'s. Thus we have that this strategy holds for the case of the $(2k + 1) \\times (2k + 1)$ board. $\\blacksquare$ [/hide]\n\nWith this lemma we can now implement the following strategy in the case of infinite boards: If $A$ places $X$ somewhere, $B$ should imagine the smallest board of the form $(2k + 1) \\times (2k + 1)$ centered at the origin such that $X$ lies on its edges or corners. We then have using the above lemma that we have a winning strategy. $\\blacksquare$ [/hide]\n\nSorry my solution is wrong. Don't read it.", "Solution_4": "Consider the following 10 x 10 square. There is one pair of numbered squares in every row, column and diagonal of length 5 or more. Repeat this square indefinitely. Then any line of 11 adjacent squares must include a numbered pair. Whenever the first player plays on a numbered square, the second player plays on the other number of the pair.\n\n n 1 2 3 4 5 6 6 7 n\n 8 8 1 2 3 4 5 9 7 10\n11 12 12 n 13 14 15 9 10 16\n11 17 18 18 13 14 15 n 16 19\n17 20 n 21 22 23 24 24 19 25\n20 26 n 21 23 22 27 27 25 28\n26 29 n n n n 30 30 28 31\n29 32 33 n n n n 34 34 31\n32 35 33 36 37 38 39 40 41 41\n n 35 42 42 36 37 38 39 40 n \n\n\n(put this numbers respectively in a 10*10 grid and follow the strategy that I have represented earlier.)\n(if you check the grid you'll see that it doesn't matter what you want to put instead of {n}.)\n[Note that to prove the result for 11 in this way the square can be at most 10 x 10 (not 11 x 11).]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[u][i] Problem: [/i][/u]\r\n [i] \n Solve the equation\n \n $ 4^{a} \\plus{} 9^{a} \\equal{} 1 \\plus{} 4 .6^{a \\minus{} 1 }$[/i] \r\n\r\n [i]Wait for a Nice - short and creative solution [/i] :wink:", "Solution_1": "This is the very interesting problem\r\n[hide]\nLet us take $ m\\equal{} 2^a$ and $ n\\equal{}3^a$\nthen our equation becomes $ m^2\\plus{}n^2\\equal{}1\\plus{}\\frac{2}{3}mn$\nwe can rewrite above as $ (m\\plus{}1)(m\\minus{}1)\\equal{}n(\\frac{2}{3}m\\minus{}n)$\nnow for $ a >\\equal{} 0$ R.H.S is always negative and L.H.S is positive,\nand for $ a < 0$ L.H.S is negative and R.H.S is positive.\nHence for no value of a the equation is satisfied.\n [/hide]", "Solution_2": "It is easy to prove, that equation had unique real solution $ \\minus{}0.5 0$\r\n\r\n\r\nThis is the correction. Thank you for showing an error.", "Solution_1": "$ n\\equal{}1 \\Rightarrow 1\\minus{}1^{2}\\minus{}\\frac{2(1)^{3}}{1\\plus{}1}\\minus{}\\frac{1^{4}}{(1\\plus{}1)^{2}}\\equal{}\\minus{}1.25< 0$\r\n\r\nIs there a typo with the inequality sign?", "Solution_2": "[quote=\"Arrange your tan\"]Let $ n$ belong to the set of positive integers.\n\nProve that:\n\n$ 1 \\minus{} n^2 \\plus{} \\frac {2n^3}{n \\plus{} 1} \\minus{} \\frac {n^4}{(n \\plus{} 1)^2} > 0$\n\n[/quote]\r\n\r\nThis is the correction. Thank you for showing that there's an error.", "Solution_3": "[hide=\"Proof\"]\nMultiply through by $ (n\\plus{}1)^{2}$ (a positive number) to get $ (n\\plus{}1)^{2}\\minus{}n^{2}(n\\plus{}1)^{2}\\plus{}2n^{3}(n\\plus{}1)\\minus{}n^{4}\\equal{}2n\\plus{}1>0$. \nSince $ 2n\\plus{}1>0$ when $ n>0$, we are done.[/hide]", "Solution_4": "[quote=\"Arrange your tan\"]Let $ n$ belong to the set of positive integers.\n\nProve that:\n\n$ 1 \\minus{} n^2 \\plus{} \\frac {2n^3}{n \\plus{} 1} \\minus{} \\frac {n^4}{(n \\plus{} 1)^2} > 0$\n\n\nThis is the correction. Thank you for showing an error.[/quote]\r\n--------------------------------------------------------------------------------------\r\n\r\nOr:\r\n\r\n$ 1 \\minus{} n^2[1 \\minus{} \\frac {2n}{n \\plus{} 1} \\plus{} \\frac {n^2}{(n \\plus{} 1)^2}] \\equal{}$\r\n\r\n$ 1 \\minus{} n^2(1 \\minus{} \\frac {n}{n \\plus{} 1})^2 \\equal{}$\r\n\r\n$ 1 \\minus{} n^2[\\frac {1}{(n \\plus{} 1)}]^2 \\equal{}$\r\n\r\n$ 1 \\minus{} (\\frac {n}{n \\plus{} 1})^2 > 1$" } { "Tag": [], "Problem": "Does anybody know how to use OpenGL in C# (under Windows) ? I've heard about / tried CsGL but couldn't make it work, do you guys have any experience with it or know about some other way to have OpenGL functionality in C#?", "Solution_1": "For [b]C++[/b], you have to download the header files \"gl.h\", \"glu.h\", \"glaux.h\" and link the libraries \"opengl32.lib\", \"glu32.lib\", \"glaux.lib\". These are found everywhere on the web.\r\n\r\nI don't know how to do it in C#, but I'm guessing it's pretty similar. You might want to post this on\r\n[url=http://www.gamedev.net/]www.gamedev.net/[/url]\r\nThis is a very good game programming forum and they'll probably be able to answer your question much better and faster.", "Solution_2": ".NET provides some stuff for openGL\r\n\r\nhttp://sourceforge.net/project/showfiles.php?group_id=33241\r\n\r\nAnd gamedev is also good as philip suggested.", "Solution_3": "Yes, I've played with OpenGL in C++ for some time now but the libraries in C# seem to be a little different. Doo posted a link to CsGL and some examples - that's pretty much the only resource on C# OpenGL that I know about too. I'll have a look on gamedev.net then... thanks to both of you." } { "Tag": [ "calculus", "integration", "induction", "trigonometry", "calculus computations" ], "Problem": "Evaluate\r\n\r\n\\[\\int_0^1 (1-x^2)^n dx\\ (n=0,1,2,\\cdots)\\]", "Solution_1": "[hide=\"hint\"]\nuse power series.[/hide]", "Solution_2": "You don't need to use power sreies, amirhtulsa :D", "Solution_3": "Well this is nice :lol: : $I_{n}=\\int_{0}^{1}\\left( 1-x^{2}\\right) ^{n}dx$\r\n\t\r\nFisrt let's make an integration by part :\r\n\t\r\n$I_{n+1}=\\int_{0}^{1}\\left( 1-x^{2}\\right) ^{n+1}dx=\\int_{0}^{1}\\left( x\\right) ^{\\prime }\\left( 1-x^{2}\\right) ^{n+1}dx$\r\n\t\r\n$=\\left[ x\\left( 1-x^{2}\\right) ^{n+1}\\right] _{0}^{1}+\\int_{0}^{1}2\\left( n+1\\right) x^{2}\\left( 1-x^{2}\\right) ^{n}dx$\r\n\t\r\n$=2\\left( n+1\\right) \\int_{0}^{1}x^{2}\\left( 1-x^{2}\\right) ^{n}dx$\r\n\t\r\n$\\Rightarrow $ $\\int_{0}^{1}x^{2}\\left( 1-x^{2}\\right) ^{n}=\\frac{I_{n+1}}{2n+2}$ (1)\r\n\t\r\nThen note that : $I_{n+1}=\\int_{0}^{1}\\left( 1-x^{2}\\right) \\left( 1-x^{2}\\right) ^{n}dx$\r\n\t\r\n$=\\int_{0}^{1}\\left( 1-x^{2}\\right) ^{n}dx-\\int_{0}^{1}x^{2}\\left( 1-x^{2}\\right) ^{n}dx =I_{n}-\\frac{I_{n+1}}{2n+2}$ (using (1) )\r\n\t\r\nHence $I_{n+1}=\\frac{2n+2}{2n+3}I_{n}$ for $n=0,1,2,...$\r\n\t\r\nThen we have : ( $I_{0}=1$ )\r\n\r\nand by easy induction we prove that : $I_{n}=\\frac{2}{3}\\times \\frac{4}{5}\\times ...\\times \\frac{2n}{2n+1}$\r\n\t\r\nand we are done.", "Solution_4": "You are right, erdos. :) \r\n\r\nLet $x=\\cos \\theta$, we have $\\int_0^1 (1-x^2)^n dx=\\int_0^{\\frac{\\pi}{2}} \\sin ^{2n+1} d\\theta\\ (n=0,1,2,\\cdots)$\r\n\r\nwhich is like Wallis' integral." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $a,b,c$ are the side lengths of a triangle, then\r\n\r\n$\\frac 1{5a+b+c}+\\frac 1{5b+c+a}+\\frac 1{5c+a+b}\\geq \\frac 2{7}(\\frac 1{b+c}+\\frac 1{c+a}+\\frac 1{a+b})$.", "Solution_1": "[quote=\"Vasc\"]If $a,b,c$ are the side lengths of a triangle, then\n\n$\\frac 1{5a+b+c}+\\frac 1{5b+c+a}+\\frac 1{5c+a+b}\\geq \\frac 2{7}(\\frac 1{b+c}+\\frac 1{c+a}+\\frac 1{a+b})$.[/quote]\r\nLet $a=y+z,$ $b=x+z$ and $c=x+y,$ where $x,$ $y$ and $z$ are positive numbers. Then\r\n$\\frac 1{5a+b+c}+\\frac 1{5b+c+a}+\\frac 1{5c+a+b}\\geq \\frac 2{7}(\\frac 1{b+c}+\\frac 1{c+a}+\\frac 1{a+b})\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{1}{6y+6z+2x}\\geq\\frac{2}{7}\\cdot\\sum_{cyc}\\frac{1}{2x+y+z}\\Leftrightarrow$\r\n$\\Leftrightarrow(7x+7y+7z)\\cdot\\sum_{cyc}\\frac{1}{3y+3z+x}\\geq(4x+4y+4z)\\cdot\\sum_{cyc}\\frac{1}{2x+y+z}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(3y+3z+x)\\cdot\\sum_{cyc}\\frac{1}{3y+3z+x}\\geq\\sum_{cyc}(2x+y+z)\\cdot\\sum_{cyc}\\frac{1}{2x+y+z}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\left(\\frac{3x+y+3z}{x+3y+3z}+\\frac{x+3y+3z}{3x+y+3z}-2\\right)\\geq\\sum_{cyc}\\left(\\frac{x+2y+z}{2x+y+z}+\\frac{2x+y+z}{x+2y+z}-2\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{4(x-y)^{2}}{(3x+y+3z)(x+3y+3z)}\\geq\\sum_{cyc}\\frac{(x-y)^{2}}{(2x+y+z)(x+2y+z)}\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(x-y)^{2}(x+y-z)(3x+3y+z)(x+y+2z)\\geq0.$\r\nLet $x\\geq y\\geq z.$ Then $S_{z}\\geq0,$ $S_{y}\\geq0$ and $S_{x}+S_{y}=$\r\n$=(x+y)(x-y)^{2}+6z^{3}+13(x+y)z^{2}+4(x^{2}+4xy+y^{2})z\\geq0.$\r\nId est, it's proved. :)", "Solution_2": "Nice solution Arqady. :lol: \r\nWe can also work directly with $a,b,c$ in SOS method. :)" } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "otro problemita : sea $n$ nuemeros donde hay $p$ de estos repetidos \u00bfde cuantas formas se pueden permutar $q$ elementos?", "Solution_1": "Creo que es asi: primero permuto los n numeros con p elementos repetidos, numero de formas es $n!$/$p!$, de los cuales es necesario solo escoger q de estos lluego el numero de formas es ${n!/p!\\choose q}$" } { "Tag": [ "advanced fields", "advanced fields theorems" ], "Problem": "The following is probably trivial... You may assume choice.\r\n\r\nLet S be a chain. Suppose that for every x, y in S the set T={u | x < u < y} is countable. Can S be uncountable?\r\n\r\nCheers.", "Solution_1": "The chain of all countable ordinals is uncountable." } { "Tag": [], "Problem": "20. A gum ball machine contains 9 red, 7 white and 8 blue gum balls. What is the least number of gum balls a person must buy to ensure getting four balls of each color?\r\n\r\nCan you guys post solutions with your answer?", "Solution_1": "For these kind of problems, u always need to think of the worst case scenario. So, the worst case scenario, is that u get all the 9 red balls first, then the 8 blue, and then finally 4 white. 9+8+4=21 gumballs", "Solution_2": "[quote=\"Quickster94\"]For these kind of problems, u always need to think of the worst case scenario. So, the worst case scenario, is that u get all the 9 red balls first, then the 8 blue, and then finally 4 white. 9+8+4=21 gumballs[/quote]\r\n\r\nOh... Okay, I got it.", "Solution_3": "[hide=\"My Solution\"]\nSo in order to make sure you get 4 of each color, you have to max out the 2 largest and get 4 in the smallest.\nSo $ 9\\plus{}8\\plus{}4\\equal{}21$.\nTherefore, the answer would be 21.\n[/hide]", "Solution_4": "[hide]Consider the worst possible case:\n\nYou get all the reds. After that you get all the blue's. after that all you need is 4 whites, so the maximum number is\n\n$ 9 \\plus{} 8 \\plus{} 4 \\equal{} \\boxed{21}$[/hide]\r\nedit: o.o..i was only typing for a second :|", "Solution_5": "[quote=\"BlackSwordsman\"][hide]Consider the worst possible case:\n\nYou get all the reds. After that you get all the blue's. after that all you need is 4 whites, so the maximum number is\n\n$ 9 \\plus{} 8 \\plus{} 4 \\equal{} \\boxed{21}$[/hide]\nedit: o.o..i was only typing for a second :|[/quote]\r\n\r\nI know.. They are really fast typers, aren't they?", "Solution_6": "Hi there:\r\nI agree with what Quickster94 said: to always think of the situation that makes you swerve the most, in which this case, you have to add all the gumballs, so its 9+7+8, which come up with your final answer:21", "Solution_7": "[quote=\"tpulak\"]Hi there:\nI agree with what Quickster94 said: to always think of the situation that makes you swerve the most, in which this case, you have to add all the gumballs, so its 9+7+8, which come up with your final answer:21[/quote]\r\n\r\nYou mean 9+8+4=21..right?", "Solution_8": "Yeah, sorry for my mistake :( , and thanks for the correction:so 9+8+4=21", "Solution_9": "[quote=\"tpulak\"]Yeah, sorry for my mistake :( , and thanks for the correction:so 9+8+4=21[/quote]\r\nNo problem.", "Solution_10": "[quote=\"DonkeyKong\"][quote=\"tpulak\"]Yeah, sorry for my mistake :( , and thanks for the correction:so 9+8+4=21[/quote]\nNo problem.[/quote]\r\nThese threads aren't a chat fourm, please don't write responses such as \"Ok\" \"Sure\" \"No Problem\"... etc.\r\n :lol:", "Solution_11": "Whoops, i'll make sure that doens't happen again :(", "Solution_12": "[quote=\"ZhangPeijin\"][quote=\"DonkeyKong\"][quote=\"tpulak\"]Yeah, sorry for my mistake :( , and thanks for the correction:so 9+8+4=21[/quote]\nNo problem.[/quote]\nThese threads aren't a chat fourm, please don't write responses such as \"Ok\" \"Sure\" \"No Problem\"... etc.\n :lol:[/quote]\r\n\r\nActually, they kinda are, but yeah, it isn't a good idea to use up one response to say one of those" } { "Tag": [], "Problem": "Three pencils and a jumbo eraser cost $ \\$1.24$. Five pencils and a jumbo eraser cost $ \\$1.82$. No prices include tax. In cents, what is the cost of a pencil?", "Solution_1": "Subtracting the first equation from the second gives 2 pencils cost 58 cents. Thus one pencil costs $ \\boxed{29}$ cents.", "Solution_2": "[hide]The answer is 29 Use elimination to solve this problem[/hide]", "Solution_3": "[hide=More graphic solution]\n[b]Step 1: Assign Variables[/b]\nHave $p$ be the cost of a pencil and $j$ be the cost of a jumbo eraser.\n[b]Step 2: Set up Equations[/b]\nWe have $2$ equations:\n$$3p+j=1.24$$\n$$5p+j=1.82$$\n[b]Step 3: Elimination[/b]\n$$3p+j=1.24$$\n$$5p+j=1.82$$\n$$\\implies (3p+j)-(5p+j)=1.24-1.82\\implies -2p= -.58\\implies \\frac{\\cancel{-2}p}{\\cancel{-2}}=\\frac{-.58}{-2}\\implies \\boxed{p=29}$$\n[/hide]" } { "Tag": [ "ratio", "geometry" ], "Problem": "What is the ratio of the area of a square that is inscribed in a regular octagon to that of the area of the regular octagon?? Express the final answer in simplified radicals.", "Solution_1": "If this is what you mean by inscribed, refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1398535#1398535]this[/url]." } { "Tag": [ "linear algebra", "matrix", "function", "calculus", "integration", "linear algebra unsolved" ], "Problem": "So I'm really stumped with this:\r\n\r\nSo I have some matrix $ A$ in $ S^m$ where $ A$ is negative semidefinite and has rank $ r < m$.\r\nNow consider matrices $ X$ in $ S^m$ in a neighborhood of $ A$ (not necessarily neg. sdf.)\r\nBy $ S(X)$ denote the eigenspace of $ X$ corresponding to its m-r largest eigenvalues.\r\n\r\nLet $ P(X)$ denote the orthogonal projection matrix of $ X$ onto $ S(X)$.\r\n($ P(X)$ can be given explicitly by $ P(X) = E(X)E(X)^T$, where $ E(X) = [e_p_ - _r_ + _1(X),...,e_p(X)]$, where $ e_i(X)$ is the eigenvector corresponding to the eigenvalue $ lambda_i(X)$)\r\n\r\nWhat I want to understand is why $ P(X)$ is supposedly a twice cont. differentiable function of $ X$ in a suff. small neighborhood of $ A$ (in fact even analytic, but I don't need this). Does this follow generally from $ P(X)$ being a symmetric operator? Can I see this directly from my explicit form for $ P$?", "Solution_1": "I'm not sure about how to prove $ C^2$ directly, but analyticity immediately follows from the fact that the projection in question is just the Cauchy integral of the resolvent over a small contour containing $ 0$. In general this gives you the spectral projection (i.e., the operator that is the identity on some eigenspace and $ 0$ on the complementary eigenspace, but, since all your matrices are self-adjoint, it is the same as the orthogonal projection you asked about).", "Solution_2": "Okay, thanks a lot. That was a big help.\r\n\r\nI looked a bit into your suggestion and it seems this is really the way this is done in the literature. Sadly, I couldn't find anything on a more elementary approach either.", "Solution_3": "I know this is old, but I still have a question on this actually:\r\n\r\nSo I understand that given some fixed matrix $ A$ I can express the orth. projection onto some eigenspace of $ A$ as a Riesz-integral (which is the Cauchy integral over a contour containing only the eigenvalues in question). I can also see easily that the resolvent is analytic in terms of the variable in the resolvent set of $ A$.\r\n\r\nNow, if for some operator $ T$ I formulate my perturbation as $ T(x) \\equal{} T \\plus{} xT' \\plus{} x^2T'' \\plus{} ...$, then for its resolvent I can show analycity in terms of $ x$, and use some kind of power series representation of this resolvent to prove analycity of $ P(T(x))$\r\n\r\nHowever, what I don't know is if this works for my actual problem, since my problem features arbitrary perturbations of $ A$. Any ideas?\r\n\r\nAlso, I would be interested in calculating the differential of such a projection $ P$. \r\n\r\nThank you for any help." } { "Tag": [], "Problem": "I got this forwarded to me from someone I know, with the line \"Why you don't tick off an Engineer who calls trying to straighten out his phone bill. \" In it was a scanned copy of a check, payable to a phone company for this amount:\r\n\r\n\\[\\$\\ .002+e^{i\\pi}+\\sum_{n=1}^{\\infty}\\frac1{2^{n}}\\]", "Solution_1": "LOL!\r\n\r\nmessage is too small . please make the message longer before submitting.", "Solution_2": "geez that's such a small phone bill :rotfl:", "Solution_3": "[quote=\"tjhance\"]geez that's such a small phone bill :rotfl:[/quote]\r\n\r\nI think that was the disparity between the actual phone bill and the bill according to the engineer?" } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Let $G$ be the centroid of a triangle $ABC$ and let $d$ be a line that intersects $AB$ and $AC$ at $B_{1}$ and $C_{1}$, respectively, such that the points $A$ and $G$ are not separated by $d$. \r\nProve that: $[BB_{1}GC_{1}]+[CC_{1}GB_{1}] \\geq \\frac{4}{9}[ABC]$.", "Solution_1": "We can transform the inequality $ [B B_{1} G C_{1}]$ + $ [C C_{1} G B_{1}]$ $ \\geq$ $ \\frac {4}{9} [A B C]$ \r\n$ LHS$ = $ 2 [G B_{1} C_{1}]$ + $ [B B_{1} C_{1}] \\plus{} [C C_{1} B_{1}]$ = $ 2[G B_{1} C_{1}]$ + $ 2[B_{1} M C_{1}]$ = $ 2 [G B_{1} M C_{1}]$ = $ \\frac {2}{3}[A B_{1} M C_{1}]$\r\nWe put $ \\delta$ = $ [B B_{1}M]$+ $ [C C_{1} M]$ M is the middle of $ BC$\r\nThe inequality becomes $ \\frac {2}{3}([A B C] \\minus{} \\delta)$ $ \\geq$ $ \\frac {4}{9}[A B C]$ $ \\Leftrightarrow$ $ [ABC] \\geq 3\\delta$\r\n taking the projections of $ B_{1} C_{1} on BC$ to be $ B_{2} C_{2}$ computing $ \\delta$ we will get $ \\delta \\equal{} \\frac {B_{1}B_{2} * \\frac {BC}{2}}{2}$ + $ \\frac {C_{1}C_{2} * \\frac {BC}{2}}{2}$ after this we replace $ B_{1} B_{2} \\plus{} C_{1} C_{2}$ with $ 2*D D'$ where $ D$ and $ D'$are the middle of $ B_{1} C_{1}$ and his projection on $ BC$ the only thing left to prove is that $ DD' \\leq \\frac {1}{3}h_{A} \\equal{} h_{G} , h_{A} height from A$ on $ BC , h_{G}$ similar not dificult,have in mind here that $ d$ passes under $ G$ , $ D$ will always be under the paralel to $ BC$ through $ G$", "Solution_2": "anyone has other solution?" } { "Tag": [ "probability", "number theory", "prime numbers" ], "Problem": "1. Several sets of prime numbers, such as ${7,83,421,659}$, use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?\r\n(A) 193\r\n(B) 207\r\n(C) 225\r\n(D) 252\r\n(E) 477\r\n\r\n2. A base-10 three-digit number $n$ is elected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of n are both three-digit numerals?\r\n(A) 0.3\r\n(B) 0.4\r\n(C) 0.5\r\n(D) 0.6\r\n(E) 0.7\r\n\r\n3. Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by 11.\r\n(A) 8180\r\n(B) 8181\r\n(C) 8182\r\n(D) 9000\r\n(E) 9090", "Solution_1": "[hide=\"Answer to 1\"]The hardest digits to use will be the non-prime digits (1, 4, 6, 8, 9). The even numbers will have to be 10's digits to create prime numbers. In the 40's, the first prime is 41, which also gets rid of the 1. In the 60's, the only prime is 67, which eliminates 6 and 7. In the 80's, there are 2 primes to consider: 83 and 89. Because the 3 can be used by itself, you want to make it 89. The remaining digits are all prime (2, 3, 5). That makes the sum of the numbers $2+3+5+41+67+89=207$. Therefore, the answer is $\\boxed{B}$.[/hide]\n\n[hide=\"Answer to 2\"]Find the lower limit, which will be $100_{11}$ in base 10. 100 is the lowest 3-digit number in base 11. $100_{11}=121_{10}$ Then find the upper limit, which is $888_9$. 888 is the largest 3-digit number in base 9. $888_9=728_{10}$. There are $728-(121-1)=608$ 3-digit numbers that work. There are a total of 900 3-digit numbers. $\\frac{608}{900}\\approx0.66$, which is closest to $0.7$. Therefore, the answer is $\\boxed{E}$.[/hide]\n\n[hide=\"Answer to 3\"]To find if a number is divisible by 11, add the alternating digits and see if the difference is divisible by 11. When you divide by 100 and take the remainder, you are asking for the last 2 digits. The first 3 digits are the quotient. The problem is really asking how many 3-digit numbers are divisible by 11. The lowest is 10,010. The highest is 99,990. Find the number of multiples of 11 in that range. $\\frac{99,990-(10,010-11)}{11}=\\frac{89991}{11}=8181$. That makes the answer $\\boxed{B}$.[/hide]", "Solution_2": "On Number 1, you messed up. I agree with you rlist of primes, but $2+3+5+41+67+89=207$ so the answer is B. You said that they summed to 225.", "Solution_3": "Thanks, I will edit my other post." } { "Tag": [ "summer program", "Mathcamp" ], "Problem": "How huge a disadvantage does a late application put you at? Due to problems with my teachers e-mail (and my laxadaisicalness in dealing with it), mine was just barely submitted. I solved 1-5,6b,7, and 9. I also had some insights on 8. Do I have any chance?", "Solution_1": "[quote=\"themonster\"]How huge a disadvantage does a late application put you at? Due to problems with my teachers e-mail (and my laxadaisicalness in dealing with it), mine was just barely submitted. I solved 1-5,6b,7, and 9. I also had some insights on 8. Do I have any chance?[/quote]\r\n\r\nHere's how we handle late applications. From the pool of on-time applicants, we will admit enough students to fill the camp. Inevitably, there are students who decide not to attend (Mathcamp\r\nis not the only program to which they're applying), and late applicants will be considered for these spots alongside the remainining on-time applicants.", "Solution_2": "I could not find out this answer in other topics I looked at, so here is question: Is it too late to apply late? Furthermore, is Monday too late to apply late? Thanks for the help.", "Solution_3": "[quote=\"JGeneson\"]I could not find out this answer in other topics I looked at, so here is question: Is it too late to apply late? Furthermore, is Monday too late to apply late? Thanks for the help.[/quote]\r\n\r\nAny time in May is generally not too late to apply late. However, I suspect we will take very very few late applicants this year.", "Solution_4": "Sorry for being clueless, but I have another late application question:\r\n\r\nIt seems the site does not want to upload my solutions file (at least it has not been happy the last couple days when I have tried it). My solutions are images, and I have compressed the file so that it is around 18 MB. Does anyone have any advice for how I could shrink this more or get the site to upload it? Is there a fax number that I have not been able to spot? Any help would be greatly appreciated. Thanks.", "Solution_5": "Have you tried to pdf your solutions?", "Solution_6": "There's a fax number somewhere on this forum. I'm not quite sure where it is, but there is one. And you can always snail mail it.", "Solution_7": "You could try to fit your file on an e-mail and send it that way.", "Solution_8": "When you say late applications, do you mean applying after April 26th, or just applying very late, but still before April 26th?", "Solution_9": "its only late after the dealine of April 26", "Solution_10": "[quote=\"Hellstar\"]When you say late applications, do you mean applying after April 26th, or just applying very late, but still before April 26th?[/quote]\r\n\r\nNote, you are replying to a thread from 2005. In fact, if I remember correctly, they couldnt even come close to accomodating all the on time applications last year, and no one got in who applied late.", "Solution_11": "[quote=\"lfm\"]In fact, if I remember correctly, they couldnt even come close to accomodating all the on time applications last year, and no one got in who applied late.[/quote]\r\n\r\nWe had to interpret the deadline somewhat generously, because of problems with the application site that prevented many people from submitting their applications by the precise deadline, but this is basically accurate. We've got our fingers crossed that the web application problems are fixed for this year, so one should probably take the deadline quite seriously this year." } { "Tag": [ "trigonometry" ], "Problem": "Solve the system of equations:\r\n$\\begin{cases}\\tan x-\\tan y-\\tan x.\\tan y=1\\\\ \\cos 2y+\\sqrt{3}\\cos 2x=-1\\end{cases}$", "Solution_1": "Hint: first equation rewrite: $sin{x}.cos{y}-sin{y}.cos{x}=cos{x}.cos{y}+sin{x}.sin{y}$ or $sin(x-y)=sin[\\frac{\\pi}{2}-(x-y)]$,..." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "geometry unsolved" ], "Problem": "Let $SABC$ be a tetrahedron with $SA\\perp SB\\perp SC$ satisfying the condition $SA+SB+SC=k$, $k$ is constant. Find the locus of the centroid $G$ of the triangle $ABC$.", "Solution_1": "The answer is the plane $x+y+z=\\frac{k}{3}.$", "Solution_2": "[quote=\"kunny\"]The answer is the plane $x+y+z=\\frac{k}{3}.$[/quote]\r\n\r\nDid you use analytic? :| Please post a full solution" } { "Tag": [ "trigonometry" ], "Problem": "What is the minimum value of $x \\sin x$ where $0 \\leq x < 2 \\pi$?\r\nMaximum?", "Solution_1": "Dude... this is beastly...\r\n\r\nI can't even get an exact answer using calculus...", "Solution_2": "Same, the answer doesn't come out nicely, more like a random question.\r\n\r\nWe have to solve the equation $x\\cos x+\\sin x=0$ and we get two roots $x_{1},x_{2}$. Playing around with trigonometry, the equality $x=-\\tan x$ yields\r\nthe max/min:\r\n$f(x_{i})=x_{i}\\sin x_{i}=\\frac{(-1)^{i+1}x_{i}^{2}}{\\sqrt{1+x_{i}^{2}}}$, with approximate values $x_{1}\\approx2.0287578381$ and $x_{2}\\approx4.9131804394$" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Is there a book or a site where i can find methods and/or lists to beyond Alg 1 factoring?\r\ne.g. \r\n(a^2)(c)+(b^2)(a)+(c^2)(b)-(a^2)(b)-(c^2)(a)-(b^2)(c)\r\n =(a-b)(b-c)(c-a)", "Solution_1": "$=a^{2}c+b^{2}a+c^{2}b-a^{2}b-c^{2}a-b^{2}c$\r\n$=a^{2}c-a^{2}b+ab^{2}-abc+bc^{2}-b^{2}c-c^{2}a+abc$\r\n$=a(ac-ab+b^{2}-bc)-c(b^{2}-cb+ac-ab)$\r\n$=a(b-a)(b-c)-c(a-b)(c-b)$\r\n$=a(b-a)(b-c)-c(b-a)(b-c)$\r\n$=(a-c)(b-a)(b-c)$\r\n$=(a-b)(b-c)(c-a)$\r\n\r\nhmm lol i expanded it first and saw that the term \"abc\" cancels out..but anyway the trick with this is to split up the terms and usually add and subtract another key term", "Solution_2": "[hide]One can observe, the expression is zero if $a=b$ so $a-b$ is a factor, similarly $(b-c)$ and $(c-a)$ are also factors. Since it's a third degree, we have\n$LHS=k(a-b)(b-c)(c-a)$ one can easily deduce that k=1 (For example, put a=0, b=1, c=-1, and you have 2=2k or k=1)[/hide]", "Solution_3": "[quote=\"Gyan\"][hide]One can observe, the expression is zero if $a=b$ so $a-b$ is a factor, similarly $(b-c)$ and $(c-a)$ are also factors. Since it's a third degree, we have\n$LHS=k(a-b)(b-c)(c-a)$ one can easily deduce that k=1 (For example, put a=0, b=1, c=-1, and you have 2=2k or k=1)[/hide][/quote]\r\n\r\nFactor Theorem? :clap: Quite elegantly done...\r\n\r\nIt's also a symmetric polynomial, isn't it?", "Solution_4": "[quote=\"mathnerd314\"]It's also a symmetric polynomial, isn't it?[/quote]\r\n\r\nNo. Intuitively, a polynomial that is symmetric shouldn't change when you permute the variables. So if you switch $a$ and $c$, you should not get another polynomial. That's not the case here.", "Solution_5": "In fact, it's anti-symmetric: transposing any two variables changes the sign of the polynomial.\r\n\r\nInteresting fact: if you take [i]any[/i] anti-symmetric polynomial in the variables $a, b, c$, it is equal to the product of this polynomial with some symmetric polynomial.", "Solution_6": "What if you are asked to factor a symmetric polynomial? is there any tricks or theorems that deal with that?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Given positive integer $n (n \\geq 2)$, find the largest positive integer $\\lambda$ satisfying :\r\nFor $n$ bags, if every bag contains some balls whose weights are all integer powers of $2$ (the weights of balls in a bag may not be distinct), and the total weights of balls in every bag are equal, then there exists a weight among these balls such that the total number of balls with this weight is at least $\\lambda$.", "Solution_1": "\\[\\lambda = \\lfloor \\frac{n}{2} \\rfloor + 1\\]", "Solution_2": "[hide = Solution]\nWe claim that the answer is $\\lambda = \\left \\lfloor \\frac{n}{2} \\right \\rfloor + 1$. First, we will establish that this is a lower bound. Firstly, we observe that any sum of $m$ as a sum of powers of two can be obtained from $m$'s binary representation by applying the operation $2^i \\rightarrow 2^{i-1} + 2^{i-1}$ repeatedly. For example, $7 = 2^1 + 2^1 + 2^1 + 2^0$ can be obtained from $7$'s binary representation $2^2 + 2^1 + 2^0$ by simply replace $2^2$ with $2^1 + 2^1$. The proof of this is easy by induction. Notice that this claim allows us to assume, WLOG, that the sum of the weights of the balls in every bag is a power of $2$. The reason we can do this is because if not, we can just take the balls which \"split off\" of one of the powers of two in the binary representation (as in the result above) of the common sum of the bags. So suppose that the sum of the balls in every bag has weight $2^k$. Then, we know that $\\lambda ( 2^k + 2^{k-1} + \\cdots + 2^1 + 2^0) \\ge n * 2^k \\Rightarrow \\lambda \\ge \\frac{2^kn}{2^k + 2^{k-1} + \\cdots + 2^1 + 2^0} > \\frac{n}{2}$, hence establishing that $\\lambda \\ge \\left \\lfloor \\frac{n}{2} \\right \\rfloor + 1$. \n\nNow, all that remains is to construct an example where $\\lambda = \\left \\lfloor \\frac{n}{2} \\right \\rfloor + 1$ is achieved for all $n \\in \\mathbb{N}$. Write $2^k \\le n < 2^{k+1}$, and let $\\lambda' = \\left \\lfloor \\frac{n}{2} \\right \\rfloor + 1$. Then, consider writing $2^k$ $\\lambda'$ times in a row, and then $2^{k-1}$ $\\lambda'$ times in a row, $\\cdots$, and finally writing $2^0$ $\\lambda'$ times in a row. Then, let's draw a divider so that between any two dividers the sum is zero. It's easy to show that this \"works\" since $2^i \\lambda' > n$ for $i \\in \\mathbb{N}$ and $\\lambda' (2^0 + 2^1 + \\cdots + 2^k) = \\lambda' (2^{k+1} - 1) \\ge \\frac{n+1}{2} (2^{k+1} - 1) \\ge n * 2^k$. The last part follows since $\\frac{n+1}{2n} = \\frac12 + \\frac{1}{2n} \\ge \\frac12 + \\frac{1}{2(2^{k+1}-1)} = \\frac{2^k}{2^{k+1}-1}$. \n\nExample:\n\n$$8|8|8|8|8|44|44|422|22211|111$$\n$\\square$ \n[/hide]" } { "Tag": [ "linear algebra", "matrix", "probability", "LaTeX", "vector", "probability and stats" ], "Problem": "Given the transition matrix below, find all communicating classes, classify all states, and discuss the\r\nlong run behaviour.\r\n\r\n(1/4) (1/8) (0) (5/8)\r\n(1/3) (0) (0) (2/3)\r\n(1) (0) (0) (0)\r\n(1/2) (0) (1/2) (0)", "Solution_1": "States 1,2 and 4 communicate as well as they can. (In fact, each one can go to each of the others in one step.) But state 3 is an absorbing state. And since there is a pathway (specifically from state 4) from the set {1,2,4} to state 3, the result will be that in the long run, the system will wind up in state 3 with probability 1.\r\n\r\nUseful exercise: treat each positive value in that matrix as if it were 1 and draw a picture of the resulting directed graph.", "Solution_2": "So, the classes are {1,2,4} and {3}. Are all of these states ergodic?", "Solution_3": "Kent, it looks to me like you've misread the transition matrix. (Either that, or I'm misreading it.) In particular, the way I read it, state 3 immediately empties into state 1, and it's not possible to get from state 4 to state 2 in only one step.\r\n\r\nBrainMan, it would be very helpful if you learned a little bit of [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ).", "Solution_4": "Oops, yes I misread the matrix. In the long run, all four states will be populated. The long-term behavior is to approach the vector\r\n\r\n$ \\begin{pmatrix}\\frac{48}{105}&\\frac{6}{105}&\\frac{17}{105}&\\frac{34}{105}\\end{pmatrix}.$" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Is there a certificate of participation for those who participate in the USAMO?", "Solution_1": "Yes, the AMC office creates and sends a Certificate of Participation with the USAMO results and the solutions. This material will be sent sometime in late May, after the grading is complete and all results have been compiled.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "function", "LaTeX", "calculus", "derivative", "analytic geometry", "quadratics", "algebra" ], "Problem": "Assume that a function $f(x)=\\frac{3x+a}{x^2+3}$ has maximal value $\\frac{3}{2}.$\r\n\r\n(1) Find the value of constant number $a$.\r\n(2) Find the range of the value of $b$ for which the equation for $x,\\ f(x)=b$ has two distinct real roots.", "Solution_1": "[hide] I don't know LATEX so I am just going to describe the process I used. First, differentiate using the quotient rule. Set the derivative equal to zero and then solve for \"a\". Substitute this back into the original and set the original equal to 3/2. This will give you the x-coordinate of the maximum. Substitute this value into the original equation that still contains the \"a\" value and solve for \"a\". I have not yet tried the second half of your problem...I will try to post soon.[/hide]", "Solution_2": "That was way too much to Latex...\r\n[hide][img]http://img.photobucket.com/albums/v194/mukmasterxps/solnsc2.png[/img][/hide]", "Solution_3": "PART A: Your answer is right.\r\nComment: Let $p$ be the value such that $f'(p)=0,$ maximal value is $\\frac{3}{2p}.$\r\n\r\nPART B: Is $bx^2-3x+3(b-1)=0$ always quadratic equation? :( \r\n\r\nkunny", "Solution_4": "Since b is a constant, wouldn't it be? :? :?", "Solution_5": "[quote=\"The Zuton Force\"]Since b is a constant, wouldn't it be? :? :?[/quote]\r\n\r\nYes, $b$ is costant, but in case of $b=0,$ you can't use discriminant." } { "Tag": [], "Problem": "I practice solving Olympiad problems, and when I tried this German problem I downloaded it from internet, I failed. Can anyone tell me how to solve this problem?\r\n\r\nQuestion:\r\nLet x be a non-zero real number satisfying the equation $ ax^2+bx+c=0$. Furthermore, let a, b, and c be integers satisfying $ |a|+|b|+|c|>1$. Prove that\r\n\r\n$ |x| \\ge \\displaystyle \\frac{1}{|a|+|b|+|c|-1} $", "Solution_1": "Just try dividing equation with x. And something else :)" } { "Tag": [ "quadratics", "function", "algebra unsolved" ], "Problem": "Let a,b,c be integers, a>0.If the quadratic $ax^{2}+bx+c=0$ has 2 distinct roots in $(n;n+1)$,where \" n\" is an integer,prove that \\[a\\geq5\\] \\[c\\geq5n^{2}+1\\]", "Solution_1": "no answeres....?", "Solution_2": "Not yet :wallbash: \r\nSome little hint ? :oops:", "Solution_3": "i don't know how to do it either...but if i find out,i'll post it. Lets see who solves it first. :lol:", "Solution_4": "$\\frac{-b\\pm \\sqrt{b^{2}-4ac}}{2a}=n,n+1$\r\n$\\frac{-b-\\sqrt{b^{2}-4ac}}{2a}=n$\r\n$\\frac{-b+\\sqrt{b^{2}-4ac}}{2a}=n+1$\r\n-b/a=2n+1", "Solution_5": "does this help?\r\n\r\nn^2< c/a<(n+1)^2\r\n\r\n2n<-b/a<2n+2", "Solution_6": "[quote=\"Altheman\"]does this help?\n\nn^2< c/a<(n+1)^2\n\n2n<-b/a<2n+2[/quote]\r\n\r\nThat' s all I've got too ! :mad:", "Solution_7": "[quote=\"Tinu\"]Let a,b,c be integers, a>0.If the quadratic $ax^{2}+bx+c=0$ has 2 distinct roots in $(n;n+1)$,where \" n\" is an integer,prove that \\[a\\geq5\\] \\[c\\geq5n^{2}+1\\] [/quote]\r\n\r\nuh, are you sure $a \\ge 5$? a quadratic with distinct roots $5,6$ is just $x^{2}-11x+30$.", "Solution_8": "The problem says \"in $(n; n+1)$\" and not \"in $[n;n+1]$\" - meaning, you can have $n1$ and $g(1)>1$\n[color=red]Case 1:[/color] $-\\frac{b}{2a}\\le \\frac12$\n then $g(0)=c+\\frac{b^{2}-4ac}{4a}=\\frac{b}{4a}=a*(\\frac{-b}{2a})^{2}\\le \\frac{a}4>1$ so $a>4$ or $a \\ge 5$.\n[color=red]Case 2:[/color]$-\\frac{b}{2a}>\\frac12$ so $1-(\\frac{-b}{2a})=1+\\frac{b}{2a}< \\frac12$ \nthen $g(1)=a+b+c+\\frac{b^{2}-4ac}{4a}=\\frac{4a^{2}+4ab+b^{2}}{4a}=a*(1+\\ac{b}{2a})^{2}<\\frac{a}{4}>1$ and again $a \\ge 5$ :D[color=green][/hide]Part 2[/color]\n[hide]Let $f(x)=ax^{2}+bx+c$ has real roots in $(n;n+1)$ then $g(x)=f(x+n)$ has real roots in $(0;1)$ and obviosly $g(x)=ax^{2}+...$ so for any $(n,n+1)$ $a \\ge 5$[color=green][/hide]\n\nPart 3[/color][hide]$f(x)=ax^{2}+bx+c$ has real roots in $(n;n+1)$ then $n<-\\frac{b}{2a}0$ this means that $c = an^{2}+1$ only when $n=0$ else $c \\ge an^{2}+n+1$\n[color=darkred]Remark 2:[/color] when $n<0$ the limits of c are not the same. For examlple $f(x)=6x^{2}+5x+1=0$ has real roots $x=-\\frac12$ and $x=-\\frac13$ but $1=c<6(-1)^{2}+1$.\nFor negative n: $n \\ge a(n+1)^{2}+1$\nAll the ideas of this proof was given to me from a friend :) [size=150][b]Rosnik[/b][/size][/hide]" } { "Tag": [], "Problem": "What is the product of all possible digits $ x$ such that the six-digit number $ 341,4x7$ is divisible by 3?", "Solution_1": "Add the digits up (not including x).\r\n\r\nYou should get 19.\r\n\r\nThen there are three possible digits that result in a number divisible by 3.\r\n\r\nThose numbers are 2, 5, and 8.\r\n\r\nMultiply, and you should clearly see that the answer is [b]80[/b]." } { "Tag": [], "Problem": "The remainder, when [i]f[/i]([i]x[/i]) is divided by ([i]x[/i] - [i]a[/i])([i]x[/i] - [i]b[/i]), is written in the form [i]A[/i]([i]x[/i] - [i]a[/i]) + [i]B[/i]([i]x[/i] - [i]b[/i]); prove that\r\n\r\n$ A\\equal{}\\frac{f(b)}{b\\minus{}a}, B\\equal{}\\frac{f(a)}{a\\minus{}b}$", "Solution_1": "Write $ f(x) \\equal{} (x \\minus{} a)(x \\minus{} b) g(x) \\plus{} A(x \\minus{} a) \\plus{} B(x \\minus{} b)$ and substitute the two obvious values.", "Solution_2": "thanks t0r, I should've seen that." } { "Tag": [], "Problem": "When we start writing in cotniuing Arabic numbers, up to n.If totally we use 3289 Arabic numericals to complete , then what is the number of n?\r\n\r\n\r\n$ \\mathsf{12345678910111213.....}$", "Solution_1": "[hide]\nNumbers with less than or equal to one digit $ \\equal{} (d \\le 1) \\equal{} 9$\n\n$ d \\le 2 \\equal{} 99 \\minus{} 9 \\equal{} 90$\n$ d \\le 3 \\equal{} 999 \\minus{} 99 \\equal{} 900$...\n\n$ \\Rightarrow S_n \\equal{} 1(9) \\plus{} 2(90) \\plus{} 3(900) \\plus{} 4(9000)...$\n\n$ S_n \\le 3289 \\Rightarrow S_n \\equal{} 1(9) \\plus{} 2(90) \\plus{} 3(900) \\equal{} 9(321) \\equal{} 2889$\n\n$ \\Rightarrow 3289 \\minus{} 2889 \\equal{} 400$\n\nThere are $ 400$ digiths worth of $ 4$ digit numbers $ \\Rightarrow \\dfrac{400}{4} \\plus{} 999 \\equal{} 1099$\n\nTherefore $ n \\equal{} 1099$[/hide]" } { "Tag": [ "function", "inequalities", "parameterization", "trigonometry", "limit", "logarithms", "calculus" ], "Problem": "This theorem addresses the behavior of a meromorphic function $ f(z)$ in a deleted neighborhood surrounding an essential singularity, $ a$. In essence it states that for any positive $ \\epsilon$, the inequality:\r\n\r\n$ |f(z) \\minus{} c| < \\epsilon$ holds for any $ c$ an infinite number of times in $ 0 < |z \\minus{} a| < \\delta$\r\n\r\nAnswer the following:\r\n\r\n(1) For a given $ c$, can I find an infinite set of concentric rings around the singularity $ 0 < . . . r_3 < r_2 < r_1 < k$ for some $ k$ say $ k \\equal{} 1$ such that between succeeding rings, I can always find only one point $ z_n$ satisfying $ |f(z_n) \\minus{} c| < \\epsilon$ for $ r_{n \\plus{} 1}\\leq|z \\minus{} a|\\leq r_n$\r\n\r\n(2) If (1) is not true no matter how small I make the difference between successive rings, then how do the multiple points accumulate? What are their distribution and is this distribution dependent on the type of essential singularity?\r\n\r\n(3) Would the distribution of points depend on the function or would it be a general property of essential singularities?\r\n\r\n(4) Is the qualitative distribution independent of the value $ c$ and $ \\epsilon$?\r\n\r\n(5) If (4) is not true, how would the distribution depend on these parameters?", "Solution_1": "Since the set $ \\{z\\colon |f(z)\\minus{}c|<\\epsilon\\}$ is open, you should rethink your approach to the problem.", "Solution_2": "Great . . . I don't entirely understand that which means of course I don't understand the theorem. I think you mean in any small neighborhood of a point $ z_n$ satisfying the inequality, there is an open set of points surrounding it that also satisfies the inequality. I'll continue to look into it.\r\n\r\nPerhaps I should change the inequality to \"equality\" since that is what the theorem is saying: \". . . the value c will be taken at some point $ z_1\\in N_{\\delta_1}(a)$ where $ N_{\\delta_1}(a)\\equal{}\\{z : 0<|z\\minus{}a|<\\delta_1$\". So now, let $ \\delta_2<|z_1|$ then c will be taken at some point $ z_2\\in N_{\\delta_2}(a)$. Continue in this manner with $ \\delta_n<|z_{n\\minus{}1}|$\r\n\r\nI'll end up with the sequence $ \\{\\delta_n\\}$. Now, can this set be used to characterize essential singularities?", "Solution_3": "More precisely, any point $ z_0$ for which $ |f(z_0)\\minus{}c|<\\epsilon$ has a neighborhood in which [b]all[/b] points satisfy $ |f(z)\\minus{}c|<\\epsilon$. So there is no way how you can have just one such point in a circular ring. \r\n\r\nIt's convenient to assume that the point of essential singularity is $ \\infty$. Thus one is led to study the behavior of [url=http://en.wikipedia.org/wiki/Entire_function]entire functions[/url] at infinity. Things like order, number of asymptotic values, Nevanlinna theory, etc. There are many thousands of pages on this topic. \r\n\r\nFinally, there is not that much difference between $ \\exp$ and $ \\sin$ in the complex plane: $ \\sin z\\equal{}(\\exp(iz)\\minus{}\\exp(\\minus{}iz)/(2i)$.", "Solution_4": "[quote=\"mlok\"]More precisely, any point $ z_0$ for which $ |f(z_0) \\minus{} c| < \\epsilon$ has a neighborhood in which [b]all[/b] points satisfy $ |f(z) \\minus{} c| < \\epsilon$. \n[/quote]\r\n\r\nSo then I have a \"region\" which satisfies the inequality within an annulus $ \\{z: r_2\\leq |z \\minus{} a|\\leq r_1\\}$. Then between the annulus $ \\{z: r_3\\leq |z \\minus{} a|\\leq r_2\\}$ I'd expect to find another set of regions since by the theorem \". . . c will be taken at some point $ z_2\\in N_{\\delta_2}(a)$ with $ N_{\\delta_2}(a) \\equal{} \\{z: 0 < |z \\minus{} a| < r_2\\}$. Continuing in this manner, I'd expect then to find a \"smattering\" of regions in each annulus as I head towards the singularity. I wonder then how these regions are distributed throughout the neighborhood of $ a$ . . . thousands of pages, huh?", "Solution_5": "That makes more sense. Still, you should be aware of situations like $ f(z)\\equal{}e^{1/z}$ and $ c\\equal{}0$ (i.e., $ c$ is an omitted value of $ f$), in which case $ \\{z\\colon |f(z)\\minus{}c|<\\epsilon\\}$ is just one region (namely, an open disk whose boundary contains $ 0$.)", "Solution_6": "Thought this was interesting: In the plots below I've plotted the Casorati-Weierstrass regions for the sets $ \\{z: |f(z) \\minus{} (1 \\plus{} i)| < 1\\}$ for the two functions $ f(z) \\equal{} e^{1/z}$ and $ f(z) \\equal{} Sin(1/z)$. The plots give you the impression of \"scaling\" suggesting the trends continue ad infinitum towards the singularity like a fractal. However this may be due to the limitations of the graphics or code below since the functions are oscillating with increasing amplitude and frequency as they approaches the singularity.\r\n\r\n[img]http://img253.imageshack.us/img253/8021/cwregionsiv1.jpg[/img]\r\n\r\nFor reference, here's my Mathematica code for the first plot:\r\n\r\n[code]f1[z_] := Exp[1/z]; \nz0 = 1 + I; \nzmax = 0.2; \ndz = 0.0011; \nplist = Flatten[Table[{x, y, \n Abs[f1[z] - z0] /. z -> x + I*y}, \n {x, -zmax, zmax, dz}, {y, -zmax, zmax, \n dz}], 1]; \nsel = Select[plist, #1[[3]] < 1 & ]; \npoints = Point /@ ({#1[[1]], #1[[2]]} & ) /@ \n sel; \np1 = Show[Graphics[points], Axes -> True, \n PlotRange -> {{-zmax, zmax}, \n {-zmax, zmax}}][/code]", "Solution_7": "It's not really a fractal (self-similar) structure: it's a periodic pattern (both $ e^z$ and $ \\sin z$ are periodic functions) after Mobius transformation $ z\\mapsto 1/z$.", "Solution_8": "Note on names- it's [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Casorati.html]Casorati[/url] with an a.", "Solution_9": "Great . . . seems to me I could have accomplished the same result above by simply using the \"multi-inverse-functions\" defined by:\r\n\r\n$ f^{ \\minus{} 1}(w) \\equal{} \\frac {1}{ln(r_w) \\plus{} i(\\Theta_w \\plus{} 2n\\pi)}$\r\n\r\n$ f^{ \\minus{} 1}(w) \\equal{} \\frac {1}{ \\minus{} iLog[iw \\plus{} (1 \\minus{} w^2)^{1/2}]}$\r\n\r\nGranted questions of analyticity come into play with the inverse functions. Suppose we ignore that for now and consider: Can a function have an essential singularity without it's multi-inverse involving in some way, the multi-valued Log function? Why do I think the answer to that question is no.\r\n\r\nIf so, then it just seems more natural then to re-cast Casorati-Weierstrass in terms of such multi-inverses. Here's a start:\r\n\r\nIf $ f^{ \\minus{} 1}(w)$ contains an infinitely-valued multi-function in some particular way that I currently can't define, then $ f(z)$ will contain an essential singularity at each of the branch points of $ f^{ \\minus{} 1}(w)$ with $ f(z)$ \"taking\" all values of the inverse multi-function in neighborhoods surrounding those branch points. No doubt there are some problems with this though.\r\n\r\n(I corrected the spelling above, thanks. I'm particular about that too).", "Solution_10": "[quote]Can a function have an essential singularity without it's multi-inverse involving in some way, the multi-valued Log function?[/quote]\r\nOf course it can. Choose any sequence of nonzero complex numbers $ c_n$ ($ n\\ge 0$) such that $ \\lim_{n\\to \\infty}|c_n|^{1/n} \\equal{} 0$. This sequence defines an entire function $ f(z) \\equal{} \\sum_{n \\equal{} 0}^{\\infty}{c_nz^n}$ which has an essential singularity at infinity. There is no reason for a random power series like that to be related to logarithms. You can consider $ f(1/z)$ if you prefer to have a singularity at $ 0$.\r\n\r\nThe reason logarithms come up in the most obvious examples is that $ \\exp$ and its relatives $ \\sin$ and $ \\cos$ are the first transcendental entire functions that come to mind.", "Solution_11": "Ok. Figured I was stretching my understanding of all this. Actually this stems from my wish to understand the proof of Picard's Great Theorem. Others wishing to review the proof can find it in \"Complex Analysis\" by M. Gonzalez (vol 2), and \"The Theory of Functions\" by Titchmarsh. It looks accessible although can't say I follow it yet but I'll spend time with it as well as this theory. I still like the idea of the multi-inverse function and will also play around with those in Mathematica. For me, Mathematica is an integral part of understanding mathematics. Just my own personal style, that's all. :)" } { "Tag": [ "LaTeX" ], "Problem": "hi! I want to customizing the style of a document. In my cls file I wrote\r\n\r\n\\newcommand{\\name}[1]{\\center{\\bf{#1}}}\r\n\r\nmy problem is that in the .tex file when I do \\name{blabla} it centers not only blabla but the rest of the text that I wrote after blabla. Any ideas??\r\n\r\nAnd I also would like to know if there's any difference between writing \r\n\r\n\\newcommand{\\name}[1]{\\center{\\bf{#1}}} and \\newcommand{\\name}[1]{\\center\\bf{#1}}\r\n\r\nbecause I've seen that not always it's used the { between \\\r\n\r\nthanks in advance!!!", "Solution_1": "Centred text uses an environment rather than a command. Try\r\n[code]\\newcommand{\\name}[1]{\\begin{center} \\bf{#1} \\end{center}}[/code]", "Solution_2": "it worked!!\r\nthanks again stevem!!" } { "Tag": [ "geometry" ], "Problem": "Today was my geometry regent, and there was one question where we had to construct a perpindicular line through a point OFF the line.\r\n\r\nI made 2 arcs {like constructing a perpindicular bisector} and made the arcs intercept the point. Then I joined the points. Will I be given full credit for this...? :(", "Solution_1": "It depends on the person grading it. :P \r\n\r\nI didn't understand the question. Is it like this?:\r\n\r\n [geogebra]b45c6ebddfc5646e395be20069394cad4b9395f7[/geogebra] \r\n\r\nIf so, what information do you have?", "Solution_2": "I would think...\r\n [geogebra]fd5ea0d23316e3dc2e2985c0837b07e06a165490[/geogebra] \r\n(The final line should be perpendicular, if I wasn't using an online drawing tool)", "Solution_3": "The diagram is exactly what you gave pi, but the point is moved over past the midpoint. I drew 2 arcs, {like when you construct a perpindicular bisector} and made them intersect that point, then I made a line. Is that valid?", "Solution_4": "Use geogebra to show what you did, then I'll tell you whether or not it is valid.", "Solution_5": "http://www.handprint.com/HP/WCL/IMG/LPR/perp.gif It looks like this, however the point is NOT on the line, AND its shifted a little to the right of the midpoint. Same concept tho.", "Solution_6": "I am pretty sure it is correct, but you have to make sure the two blue circles have the same radius (very important). A faster way is to just find half the distance between the two center points of the blue circles. The circle way works too because half the diameter is the radius. Nice technique though!", "Solution_7": "[hide=\"what I did\"]\nI drew two arcs intersecting the line from the point above the line. Setting any radius with the compass such that it reaches farther than half the distance from the two points on the line, draw an arc below the line. Keeping the radius, place the pointy end on the other point on the line and draw an arc that intersects the previous arc. Now just connect the intersection and the point above the line.\n(sorry if this is a little hard to follow)[/hide]", "Solution_8": "A fast method:\r\n\r\n [geogebra]9b6dbb650c61c06438eaae55f03d3bf524056b32[/geogebra] \r\n\r\nWhere the intersection point by the right angle mark is the midpoint of the line formed by the points of intersection of the circle and the original line.", "Solution_9": "Actually, that wouldn't be any faster. Assuming you're only given a compass and straightedge, you would still have to construct the midpoint, which does not take less time than constructing an intersection of two arcs." } { "Tag": [], "Problem": "Ok, I've been trying to solve this question \r\n\r\nA line is a tangent to a circle at B. Points A and C are on the line on opposite sides of B. A chord MN is parallel to the tangent. Prove that triangle MBN is isosceles.\r\n\r\n I'm suppose to use the Tangent-Chord Theorem, but i must prove it first in order to use it. I spent the past hour and i still can't figure out the Tangent Chord Theorem so I can use it in my proof :( i've tried using the alternate angle theorem to prove that $\\angle NOB=x$\r\n\r\nsince it is a central angle, arc $NB=x$, since arc $NB$ and arc $MB$ are equal, the chords $MB$ and $NB$ are equal, making $\\triangle MNB$ isosceles", "Solution_2": "I understand the steps and process you went through, but I'm unsure how you got the angle MNB = x/2\r\n\r\nAlso, is there a way to use the tangent chord theorem to prove it?", "Solution_3": "an inscribed angle is half the arc it intercepts, the proof of this is not that hard, but you have to consider multiple cases" } { "Tag": [ "trigonometry", "geometry unsolved", "geometry" ], "Problem": "$A,\\ B,\\ C,\\ D,\\ E$ and $F$ lie (in that order) on the circumference of a circle. The chords $AD,\\ BE$ and $CF$ are concurrent. $P,\\ Q$ and $R$ are the midpoints of $AD,\\ BE$ and $CF$ respectively. Two further chords $AG \\parallel BE$ and $AH \\parallel CF$ are drawn. Show that $PQR$ is similar to $DGH$.", "Solution_1": "JUST ANGLE CHASING\r\nlet the point of concurrent be $X$,its easy to see that$PQRX$ is cyclic quadrilateralm so:\r\n\r\n$\\angle HDG =\\angle HAG=\\angle QXC=\\angle QPR \\ ,\\ \\angle PQR=\\angle PXR=\\angle PAH=\\angle DGH$", "Solution_2": "It is just angle chasing after the quadrilateral PQRX is known to be cyclic, but I could not see it (not that it is cyclic, but why).\r\n\r\nLet X be a fixed point inside a circle (O) of radius R at the distance d = OX from the circle center. Let AB be an arbitrary chord passing through the point X and M the midpoint of this chord. The locus of midpoints M is a circle with the diameter OX.\r\n\r\nDenote a = XA, b = XB, d = XO. WLOG, assume $a \\ge b$. Denote $\\theta = \\angle AXO$. Then $\\angle BXO = 180^\\circ - \\theta$. Using the cosine theorem for the triangles $\\triangle AXO, \\triangle BXO$,\r\n\r\n$XA^2 + XO^2 - 2\\ XA \\cdot XO \\cos \\angle AXO = OA^2$\r\n\r\n$XB^2 + XO^2 - 2\\ XB \\cdot XO \\cos \\angle BXO = OB^2$\r\n\r\nRewriting:\r\n\r\n$a^2 + d^2 - 2ad \\cos \\theta = R^2$\r\n\r\n$b^2 + d^2 + 2bd \\cos \\theta = R^2$\r\n\r\nSubtracting these 2 equations\r\n\r\n$a^2 - b^2 - 2d(a + b) \\cos \\theta = 0$\r\n\r\nReducing by $a + b \\neq 0$:\r\n\r\n$d \\cos \\theta = \\dfrac{a - b}{2} = \\dfrac{a + b}{2} - b$\r\n\r\n$XO \\cos \\theta = \\dfrac{AB}{2} - XB = MB - XB = XM$\r\n\r\nThe proposition follows." } { "Tag": [ "puzzles" ], "Problem": "Suppose you have two cars, A and B, approaching the finish line of a race.\r\n\r\nThe front of A crosses over the finish line first, but then it either just slows, comes to a complete stop, or goes backward by whatever combination of\r\nforces applied inside to its controls and/or to the outside of it. At any rate \r\nit turns out that the entirety of car B makes it past the finish line before car A\r\ndoes, or it is the case (although car A initially crossed the finish line, car A never does continue moving across).\r\n\r\nI don't know the answer to this question:\r\n\r\nWhich car has won the race, the first to get the front part of its body across the finish line, or the first car to get its entire body past the finish line (and presumably stay there, and not move backwards past the finish line)?", "Solution_1": "It depends what the rules of the particular race are. But I suppose you mean what would they do if no rules were defined beforehand...\r\n\r\nIn that case, I believe this would belong in the Round Table...\r\n\r\nPersonally, I think that\r\n\r\nif you have to get \"across\" the finish line, you have to completely pass it.\r\nIf you have to get \"to\" the finish line, you have to reach it.\r\nif no rules are defined, you have to get your center of mass to cross the finish line first." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "What is the smallest positive perfect cube that can be written as\nthe sum of three consecutive integers?", "Solution_1": "If a number is the sum of 3 consecutive cubes, it must be div 3. The smallest cube div 3 is 3^3=27" } { "Tag": [ "geometry", "trapezoid", "ratio" ], "Problem": "A isosceles trapezoid has a longer base, $ a$ and a shorter base, $ b$. If $ a$ is also equal to the diagonal of the trapezoid and $ b$ the height of the trapezoid, then what is the ratio $ \\frac{b}{a}$", "Solution_1": "As it is isoceles, we can see that $ b^{2}+(\\frac{a+b}{2})^{2}= a^{2}$.\r\n\r\n$ 4b^{2}+b^{2}+2ab+a^{2}=4a^{2}$\r\n\r\n$ 5b^{2}+2ab-3a^{2}=0$\r\n\r\n$ (5b-3a)(b+a)=0$\r\n\r\nIt follows that $ \\frac{b}{a}=\\frac35$." } { "Tag": [], "Problem": "Prove that $ x^{2}+y^{2}=133703$ has no integer solutions for $ (x,y)$.", "Solution_1": "$ k^{2}\\equiv0,1\\mod4$\r\n\r\nTherefore $ a^{2}+b^{2}$ cannot be equivalent to $ 3\\mod4$. However, $ 133703\\equiv3\\mod4$" } { "Tag": [ "AoPSwiki" ], "Problem": "This thread is for the Team Chess Tournament Round 1: Team Beta (White) vs. Team Eta (Black). Play begins on December 7th. Any questions may be asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]here[/url]. All information/FAQ is being updated [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]here[/url].\r\n\r\n[hide=\"Team Beta\"]\n 1. Temperal \n 2. CircleSquared \n 3. abacadaea\n 4. miyomiyo\n 5. davidyko \n[/hide]\n\n[hide=\"Team Eta\"]\n 1. dbkarp\n 2. rem\n 3. ZeroFive1\n 4. moogra\n 5. Patterns_34 \n[/hide]\r\n\r\nEach team will have 3 days for each move. \r\n\r\nGood luck!", "Solution_1": "[b]Move 1 (White): d4[/b]\r\n\r\n[img]http://i2.tinypic.com/85bepug.jpg[/img]", "Solution_2": "[b]Move 1 (Black): f5[/b]", "Solution_3": "I know that I am not in the game, but...\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2\r\n\r\nSame game, different info. Is this a mistake?", "Solution_4": "Yes that was a mistake: it should've been [b]f[/b]5, not [b]d[/b]5.\r\nThank you for the catch 1=2. :)\r\n\r\n[b][Edit][/b] I think my method of entering the list must've been misinterpreted. I was planning on creating the list of moves given so far and building on it as the game progresses. Example (this isn't the real game, just illustrating an example):\r\n[b]Black's 3rd move:[/b]\r\n[list]1. e4 c5\n2. Nf3 Nc6\n3. d4 $ \\boxed{\\text{cxd4}}$[/list]\r\nAnd so on. \r\nI'll re-edit it and from now on enter the moves in as just the move one side made so it isn't misinterpreted.\r\nI apologize for the inconvenience that temporarily occurred today.", "Solution_5": "Any error in the Wiki page will be amended by Temperal :) so do not stress about that.", "Solution_6": "[quote=\"1=2\"]I know that I am not in the game, but...\n\nhttp://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2\n\nSame game, different info. Is this a mistake?[/quote]\n\nThanks for the notification. :) I'll correct that.\n\n[quote]Any error in the Wiki page will be amended by Temperal :) so do not stress about that.[/quote]\r\n\r\nThat's right. :D", "Solution_7": "[b]White: 2. g3[/b]", "Solution_8": "Black's Second Move: [b]Nf6[/b]", "Solution_9": "[b]3. Bg2[/b]", "Solution_10": "[quote=\"Temperal\"][quote=\"mz94\"]ok, like, do we win by forfeit? idk how much time they have but its been at least a week or 2[/quote]\n\nTeam [color=red]Eta[/color] now has a warning for exceeding the time limit.\n\nYou can win by any normal chess methods, or if the game is taking too long, the TDs will decide.[/quote]\r\nReferring to us or Team Zeta?\r\nWe apologize for not posting our move sooner; we thought the time limit was $ 3$ days instead of $ 2$ for a moment (more technically, we were $ 2$ hours away from reaching the 3 day limit).\r\n\r\nRegardless, this is our move: [b]3. ... e6[/b]. \r\nAgain, apologies for the delay.", "Solution_11": "[quote=\"Patterns_34\"][quote=\"Temperal\"][quote=\"mz94\"]ok, like, do we win by forfeit? idk how much time they have but its been at least a week or 2[/quote]\n\nTeam [color=red]Eta[/color] now has a warning for exceeding the time limit.\n\nYou can win by any normal chess methods, or if the game is taking too long, the TDs will decide.[/quote]\nReferring to us or Team Zeta?\nWe apologize for not posting our move sooner; we thought the time limit was $ 3$ days instead of $ 2$ for a moment (more technically, we were $ 2$ hours away from reaching the 3 day limit).\n\nRegardless, this is our move: [b]3. ... e6[/b]. \nAgain, apologies for the delay.[/quote]\r\n\r\ni think its Zeta. its probably a typo on Temperal's behalf. if he wanted to give you a warning, he probably wouldve posted it here", "Solution_12": "Yeah, sorry, typo.", "Solution_13": "[b]4. Nf3[/b]", "Solution_14": "[b]4. ... Be7[/b]", "Solution_15": "[b]5. 0-0[/b]", "Solution_16": "[b]5. ... 0-0[/b]", "Solution_17": "[b]6. Nc3[/b]", "Solution_18": "[b]6. ... d5[/b]", "Solution_19": "[b]7. Bf4[/b]", "Solution_20": "This topic is dead... :( I wanted to play...", "Solution_21": "Patterns sent me a message saying how his entire team is inactive due to school work (finals, I'm sure). As the current active team in this game, you are allowed to accept their temporary vacation or penalize them. \r\n\r\n[hide=\"Psst\"]It's best in everyone's interest to accept the vacation; be a good sport =P. If no one responds by February it is an automatic forfeit and Team Beta gets a win while Team Eta gets the draw.[/hide]", "Solution_22": "I apologize that we're not responding.\r\n\r\nAt this moment, most of us are still having finals, for this [i][u]is[/u][/i] the week some of us are taking it (I will be available to respond a move for the team by Monday or so after finishing up my finals AND taking the SAT this Saturday).\r\n\r\nAs a person who is still active for the team, I ask those of Team Beta to be patient for just a little longer, for they will get a respond and a move from us by at most Monday. :wink:", "Solution_23": "I apologize I didn't post a move on the date I promised.\r\nOn behalf of my inactive team:\r\n\r\n[b]7. ... Nd7[/b]", "Solution_24": "I guess this is what my team wanted: :maybe:\r\n\r\n[b]8. Qd3[/b]", "Solution_25": "Team $ \\beta$ has a replacement?\r\nIf so, it would be helpful if the opposing team (like us) was also aware. :) \r\n\r\nIn addition, just to let the moderators know, the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2]AoPSWiki page[/url] of our game hasn't been amended yet.", "Solution_26": "[quote=\"Patterns_34\"]Team $ \\beta$ has a replacement?\nIf so, it would be helpful if the opposing team (like us) was also aware. :) \n\nIn addition, just to let the moderators know, the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2]AoPSWiki page[/url] of our game hasn't been amended yet.[/quote]\r\n\r\nTemperal is taking a few weeks off. I don't think he needs a replacement, we have four people left. I guess I will amend the Wiki page.", "Solution_27": "I certainly don't need a replacement. I'm just preparing for state MC and AMC 12...", "Solution_28": "I apologize we responded late. As for our move:\r\n[b]8. ... a6[/b]", "Solution_29": "[b]9. Qd2[/b]", "Solution_30": "*nudge*\r\n\r\n\r\n[size=9]random text here to make the message longer[/size]", "Solution_31": "[b]9. ... b5[/b]\r\nI greatly apologize for the delay. :(", "Solution_32": "[b]10. h4[/b]" } { "Tag": [], "Problem": "If $ x \\plus{} y \\equal{} 5$ , $ x \\plus{} z \\equal{} 8$ and $ y \\plus{} z \\equal{} 1$ , what is the value of\n$ x \\plus{} y \\plus{} z$ ?", "Solution_1": "\\begin{eqnarray*} x + y + z & = & \\frac {1}{2}(2(x + y + z)) \\\\\r\n& = & \\frac {1}{2}\\left((x + y) + (y + z) + (z + x)\\right) \\\\\r\n& = & \\frac {1}{2}(5 + 1 + 8) \\\\\r\n& = & \\frac {1}{2} \\times 14 \\\\\r\n& = & \\boxed{7} \\end{eqnarray*}" } { "Tag": [], "Problem": "\u0395\u03b9\u03c0\u03b1 \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b7 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b1\u03bd\u03bf\u03b9\u03be\u03c9 \u03b1\u03c5\u03c4\u03bf \u03c4\u03bf \u03c4\u03bf\u03c0\u03b9\u03ba \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b2\u03b1\u03bb\u03c9 \u03c4\u03bf \u03b5\u03c1\u03c9\u03c4\u03b7\u03bc\u03b1 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c4\u03bf\u03c0\u03b9\u03ba \u03c0\u03bf\u03c5 \u03b5\u03c7\u03b5\u03b9 \u03b1\u03bd\u03bf\u03b9\u03be\u03b5\u03b9 \u03bf \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c4\u03bb\u03bf \u0398\u03b5\u03c9\u03c1\u03b9\u03b1 \u0393\u03c1\u03b1\u03c6\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd. \u0395\u03b9\u03b4\u03b1 \u03bf\u03bc\u03c9\u03c2 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03b4\u03c9\u03b8\u03b5\u03b9 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03c5\u03bc\u03b5\u03bd\u03bf \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03c7\u03b5\u03b9 \u03b2\u03b1\u03bb\u03b5\u03b9, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b2\u03b1\u03b6\u03c9 \u03c4\u03b7\u03bd \u03b1\u03c0\u03bf\u03c1\u03b9\u03b1 \u03bc\u03bf\u03c5 \u03b5\u03b4\u03c9 \u03ba\u03b1\u03b9 \u03b1\u03bd \u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bc\u03bf\u03bd\u03c4\u03c2 \u03c3\u03c9\u03c3\u03c4\u03bf \u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03ba\u03b9\u03bd\u03bf\u03c5\u03bd\u03b5 \u03b1\u03c1\u03b3\u03bf\u03c4\u03b5\u03c1\u03b1. \u039f\u03c1\u03b9\u03c3\u03c4\u03b5 \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03c1\u03c9\u03c4\u03b9\u03bf\u03bc\u03bf\u03c5\u03bd...\r\n\r\n\u03a5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b5\u03c5\u03b8\u03b5\u03b9\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3\u03c4\u03bf \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 \u039d\u03c4\u03b9\u03c1\u03b1\u03ba (\u03bf\u03c7\u03b9 \u03b1\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03c3\u03b5 \u03b1\u03c4\u03bf\u03c0\u03bf, \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b7 \u03ba\u03c4\u03bb.) \u03c0\u03bf\u03c5 \u03bb\u03b5\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03b1\u03bd \u03c3\u03b5 \u03b5\u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03b8\u03b5 \u03ba\u03bf\u03c1\u03c5\u03c6\u03b7 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03bd\u03c9\u03bc\u03b5\u03bd\u03b7 \u03bc\u03b5 \u03bd/2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03b5\u03c2 \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd, \u03c4\u03bf\u03c4\u03b5 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 hamiltonian circuit.\r\n\r\n\u0391\u03b1\u03b1, \u03ba\u03b1\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03c7\u03c9\u03c1\u03b9\u03c3\u03bc\u03bf \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b5\u03c9\u03bd, \u03c7\u03c1\u03b7\u03c3\u03b7 \u03b1\u03bb\u03bb\u03c9\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd \u03ae \u03bb\u03b7\u03bc\u03bc\u03b1\u03c4\u03c9\u03bd...", "Solution_1": "\u039a\u03ac\u03c0\u03c9\u03c2 \u03b2\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac: \r\n\r\n\u0397 \u03c3\u03c5\u03bd\u03b7\u03b8\u03b9\u03c3\u03bc\u03ad\u03bd\u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 Dirac \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c4\u03b5\u03af \u03c7\u03c9\u03c1\u03af\u03c2 \u03ac\u03c4\u03bf\u03c0\u03bf: \u0391\u03c0\u03cc \u03ba\u03ac\u03b8\u03b5 \u03bc\u03bf\u03bd\u03bf\u03c0\u03ac\u03c4\u03b9 \u03c3\u03b5 \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03bc\u03bf\u03bd\u03bf\u03c0\u03ac\u03c4\u03b9 (\u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03bc\u03b1\u03c2 \u03b5\u03be\u03b7\u03b3\u03b5\u03af \u03c0\u03c9\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03bd \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5) \u03ba\u03b1\u03b9 \u03c4\u03ad\u03bb\u03bf\u03c2, \u03ba\u03ac\u03b8\u03b5 \u03bc\u03bf\u03bd\u03bf\u03c0\u03ac\u03c4\u03b9 \u03c3\u03b5 \u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03c4\u03c1\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03ba\u03cd\u03ba\u03bb\u03bf Hamilton (\u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03bc\u03b1\u03c2 \u03bb\u03ad\u03b5\u03b9 \u03c0\u03c9\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5).", "Solution_2": "A\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b3\u03c1\u03b1\u03c8\u03b5 \u03bc\u03bf\u03c5 \u03c4\u03b7\u03bd \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b7 \u03ae \u03b4\u03c9\u03c2' \u03bc\u03bf\u03c5 \u03b5\u03bd\u03b1 link \u03b3\u03b9\u03b1\u03c4\u03b9 \u03b8\u03b5\u03bb\u03c9 \u03bd\u03b1 \u03c4\u03b7 \u03bc\u03b1\u03b8\u03c9 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b2\u03c1\u03b9\u03c3\u03ba\u03c9 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b1...", "Solution_3": "\u039a\u03b5\u03c6\u03ac\u03bb\u03b1\u03b9\u03bf 4.2 [url=http://www.ecp6.jussieu.fr/pageperso/bondy/books/gtwa/gtwa.html]\u03b5\u03b4\u03ce[/url].\r\n\r\n\u0397 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03ac\u03c4\u03bf\u03c0\u03bf \u03b1\u03bb\u03bb\u03ac \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03c6\u03b5\u03c5\u03c7\u03b8\u03b5\u03af \u03cc\u03c0\u03c9\u03c2 \u03b5\u03be\u03b7\u03b3\u03ce \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf post.", "Solution_4": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03c9, \u03b8\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03b2\u03b1\u03c3\u03c9." } { "Tag": [ "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "when performing partial differentiation with respect to one variable, all the others are treated as constant. But what about numbers unrelated to variables. \r\n\r\nfor example 1 + x y^-1\r\n\r\ndoes the 1 disappear as usual when diff w.r.t y?\r\n\r\nits been so long since i've done this!\r\n\r\nthanks", "Solution_1": "Yes. It's constant in $y$, so it vanishes under differentiation.", "Solution_2": "brilliant, thanks for the clarification" } { "Tag": [ "floor function", "number theory", "relatively prime", "rational numbers" ], "Problem": "Prove that there is no positive rational number $x$ such that \\[x^{\\lfloor x\\rfloor }=\\frac{9}{2}.\\]", "Solution_1": "Put $ x\\equal{}\\frac{p}{q}$ with $ p,q$ relatively prime, and $ \\lfloor x\\rfloor\\equal{}n$. Then we want that $ 2p^{n}\\equal{}9q^{n}$, so $ 2|q$, for $ n>1$ this means the RHS has more than one factor $ 2$ so $ 2|p$, contradiction with relative prime.\r\n\r\nSo $ n\\equal{}1$ (since $ n\\equal{}0$ is impossible), so $ 2p\\equal{}9q$, so $ x\\equal{}\\frac{9}2$, but then $ n>1$, contradiction.", "Solution_2": "Another way (though not as good):\n\nSuppose $x$ is a positive rational. If $x<1 $ then $x^{\\lfloor x\\rfloor } = x^0 = 1 \\neq 9/2 $. If $1\\leq x < 2 $, $x^{\\lfloor x\\rfloor } = x^1= x \\neq 9/2 $ since $9/2 > 2$. If $ 2\\leq x < 3 $ then $x^{\\lfloor x\\rfloor } = x^2 \\neq 9/2 $ since $3/\\sqrt{2}$ is irrational. If $ 3\\leq x $ then $ x^{\\lfloor x\\rfloor } \\geq 3^3 = 27 > 9/2 $. So there are no positive rational solutions to $x^{\\lfloor x\\rfloor }= \\frac{9}{2}$.", "Solution_3": "$ My $ $ solution $\n\n$ (a,b)=1 $\n$ x=\\frac{a}{b} $\n$ \\lfloor x\\rfloor=c $\n$$ x^{\\lfloor x\\rfloor}=\\frac{a^c}{b^c}=\\frac{9}{2}\\implies a^c=9,b^c=2\\implies c=1,a=9,b=2$$\n$ Then $\n$ \\lfloor x\\rfloor=1,x=\\frac{9}{2} $ No solution" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "circumcircle", "geometry unsolved" ], "Problem": "Let ABCD be a tetrahedron whose 4 altitudes are concurrent and let R be the radius of its circumcircle (sphere). Let $ h_1;h_2;h_3;h_4$ be the lengths of the altitudes corresponding A,B,C,D and $ R_1;R_2;R_3;R_4$ be the circumcircle's radius of the opposite faces of A,B,C,D. Prove that\r\n$ \\frac{1}{h_1\\plus{}2\\sqrt{2}R_1}\\plus{}\\frac{1}{h_2\\plus{}2\\sqrt{2}R_2}\\plus{}\\frac{1}{h_3\\plus{}2\\sqrt{2}R_3}\\plus{}\\frac{1}{h_4\\plus{}2\\sqrt{2}R_4} \\geq \\frac{1}{R}$.\r\nWhen does equality occurs?", "Solution_1": "What's a [b]Tetrahedron[/b]? I don't know this definition!", "Solution_2": "4 points in a space define a Tetrahedron!:D" } { "Tag": [ "geometry", "calculus", "integration" ], "Problem": "[i]A circle is placed within another circle, not necessarily the centre. Find the area of all the circles that can be placed between the smaller and the larger circle, while touching both circles.[/i]\r\n\r\nMasoud Zargar", "Solution_1": "The problem isn't clear. If what's wanted is the SUM of the areas of all such circles, then it's obviously infinite, since in the special case of concentric circles we have infinitely many circles with constant area (general case can be solved via integral calculus, but the resulting integral also diverges).\r\n\r\nWhat's exactly wanted in the problem?", "Solution_2": "Well, let's imagine we have a circle $C_1$ placed somewhere in $C_2$. Between the two circles, there is a gap with a certain area. Now, between the area, we construct as many circles at possible that are tangent to both circles. All the surrounding circles are also linked. What is the area of these small tangent circles?\r\n\r\nMasoud Zargar", "Solution_3": "Well then, it should've been said that each circle ALSO touches the two adjacent ones, besides the two given circles :)", "Solution_4": "Well, I usually lose lots of marks on math tests for communication. :blush: :). Now, let's begin solving for I do not have a solution.", "Solution_5": "But there's also another problem: such construction may not be possible if we want the added circles to be connected throughout. It can be easily shown in the case of concentric circles: if the center of the given circles is $O$, their radii $R$ and $r$, and the two adjacent new circles have centers $C_1$ and $C_2$, then (proof will be supplied if needed) $\\angle C_1OC_2=2\\arcsin \\frac{R-r}{R+r}$, and $\\frac{\\pi}{\\arcsin \\frac{R-r}{R+r}}$ may not be integer." } { "Tag": [ "calculus", "integration", "trigonometry", "complex analysis", "complex analysis unsolved" ], "Problem": "How to solve the integral $ \\int_{0}^{\\plus{}\\infty}\\frac{sen(x)}{x^p}dx$ for $ 0 < p < 2$?\r\n\r\n Please, I need some help!!!!!", "Solution_1": "Maple says\r\n\\[ \\int _{0}^{\\infty }\\!{\\frac {\\sin \\left( x \\right) }{{x}^{p}}}{dx}\\equal{}{\r\n\\frac {\\sqrt {\\pi }{2}^{\\minus{}p}\\Gamma \\left( 1\\minus{}1/2\\,p \\right) }{\\Gamma \r\n \\left( 1/2\\plus{}1/2\\,p \\right) }}\\]" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "A. 342. Prove that for any p prime of the form 4k+1\r\n[sqrt(p)]+[sqrt(2p)]+..[sqrt((p-1)p)]=(p-1)(2p-1)/3.\r\n\r\nI know a similar problems was published here, and pbornstein gave a solution, but I can't remember it and can't find it on the forum.", "Solution_1": "Let $LPR(x)$ denote the least positive residue of $x$ modulo $p$. \r\n\r\nFirst, observe that for all primes $p == 1 (mod 4)$, $(\\frac{-x}{p}) = (-1)^{\\frac{p-1}{2}} (\\frac{x}{p}) = (\\frac{x}{p})$. Therefore, $\\{LPR(x^2)\\}_{x=1}^{\\frac{p-1}{2}}$ can be paritioned into $\\frac{p-1}{4}$ pairs of residues of the form $\\{k, p-k\\}$. It follows that ${\\sum_{1}^{p-1}\\{\\frac{x^2}{p}\\} = \\frac{2}{p} \\sum_{1}^{\\frac12 (p-1)} LPR(x^2)} = \\frac{2}{p} * p(\\frac{p-1}{4}) = \\frac{p-1}{2}$. Therefore,\r\n$S = \\sum_{1}^{p-1} [pk]$ \r\n$=(p-1)^2 - \\sum_{1}^{p-1} [\\frac{y^2}{p}]$ (because both equal the number of lattice points under the graph $y=\\sqrt{xp}, 1 \\leq x \\leq p-1$)\r\n$= (p-1)^2 - \\sum_{1}^{p-1}\\frac{y^2}{p} + \\sum_{1}^{p-1}\\{\\frac{y^2}{p}\\}$\r\n$=(p-1)^2 - \\frac{(p-1)(2p-1)}{6} + \\frac{p-1}{2}$, by our above remarks,\r\n$= \\frac{(p-1)(2p-1)}{3}$, after simplification, which proves the required.\r\n\r\nNeat problem." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a_{1},\\ a_{2},\\ldots,\\ a_{n}$ are positive real numbers such that \\[ \\frac{1}{1+a_{1}}+\\frac{1}{1+a_{2}}+\\cdots+\\frac{1}{1+a_{n}}=1\\] Show that \\[ \\sqrt{a_{1}}+\\sqrt{a_{2}}+\\cdots+\\sqrt{a_{n}}\\geq (n-1) \\left(\\frac{1}{\\sqrt{a_{1}}}+\\frac{1}{\\sqrt{a_{2}}}+\\cdots+\\frac{1}{\\sqrt{a_{n}}}\\right).\\]", "Solution_1": "there exist positive $ x_{i}$ for $ i=1,2,...,n$ with sum $ S$ such that $ a_{i}=\\frac{S-x_{i}}{x_{i}}$\r\nnow $ \\left(\\sum \\sqrt{\\frac{S-x_{i}}{x_{i}}}\\right)^{2}\\geq n(n-1)^{2}+\\sum \\frac{S-x_{i}}{x_{i}}\\geq (n-1)^{2}\\left(n+\\sum \\frac{x_{i}}{S-x_{i}}\\right)$\r\nthus we are left with showing \r\n$ n\\geq \\sum \\sqrt{\\frac{x_{i}x_{j}}{(S-x_{i})(S-x_{j})}}$ :wink:" } { "Tag": [ "inequalities", "quadratics", "geometry", "inequalities unsolved" ], "Problem": "Let $ a_{1}$, $ a_{2}$, $ a_{3}$, $ b_{1}$, $ b_{2}$, $ b_{3}$ be six positive real numbers. Prove the inequality\r\n\r\n$ \\left(a_{1}b_{2} \\plus{} a_{1}b_{3} \\plus{} a_{2}b_{3} \\plus{} a_{2}b_{1} \\plus{} a_{3}b_{1} \\plus{} a_{3}b_{2}\\right)^{2}\\ge 4\\left(a_{1}a_{2} \\plus{} a_{1}a_{3} \\plus{} a_{2}a_{3}\\right)\\left(b_{1}b_{2} \\plus{} b_{1}b_{3} \\plus{} b_{2}b_{3}\\right)$,\r\n\r\nand prove that equality happens if and only if $ \\frac {a_{1}}{b_{1}} \\equal{} \\frac {a_{2}}{b_{2}} \\equal{} \\frac {a_{3}}{b_{3}}$.", "Solution_1": "I see that the inequality resembles discriminat of a quadratic, but experimenting, I couldn't solve it. \r\n\r\nI also wonder if Schur (generalized perhaps) could solve it, \r\n\r\nor perhaps just multiplying it all out? (I tried to use brute force, but I couldn't get it) :blush:", "Solution_2": "I find the inequlaity very hard and unfortunately cannot really solve it.\r\nNow the fact that the inequlaity resembles a quadratic means that if you can find at least one real solution for $ x$ in the equation:\r\n$ (a_{1}a_{2}\\plus{}a_{1}a_{3}\\plus{}a_{2}a_{3})x^{2}\\plus{}(a_{1}b_{2}\\plus{}a_{1}b_{3}\\plus{}a_{2}b_{1}\\plus{}a_{2}b_{3}\\plus{}a_{3}b_{1}\\plus{}a_{3}b_{2})x\\plus{}(b_{1}b_{2}\\plus{}b_{1}b_{3}\\plus{}b_{2}b_{3}) \\equal{} 0$\r\nyou will be finished since that would mean the discriminant(which is the LS of ineq) is non-negative...\r\nHowever I don't know how to prove that yet...\r\nwhere is this from?", "Solution_3": "[quote=\"rem\"]Now the fact that the inequlaity resembles a quadratic means that if you can find at least one real solution for $ x$ in the equation:\n$ (a_{1}a_{2}\\plus{}a_{1}a_{3}\\plus{}a_{2}a_{3})x^{2}\\plus{}(a_{1}b_{2}\\plus{}a_{1}b_{3}\\plus{}a_{2}b_{1}\\plus{}a_{2}b_{3}\\plus{}a_{3}b_{1}\\plus{}a_{3}b_{2})x\\plus{}(b_{1}b_{2}\\plus{}b_{1}b_{3}\\plus{}b_{2}b_{3}) \\equal{} 0$\nyou will be finished since that would mean the discriminant(which is the LS of ineq) is non-negative...\nHowever I don't know how to prove that yet...\nwhere is this from?[/quote]\r\n\r\nIt is from IMO Longlist 1987 #4, proposed by Australia. All the other Australia problems were easy one-liners that year, so I wonder why this one is so hard. \r\n\r\nAnyway, I tried letting $ a_{1}\\equal{} b_{1}k_{1}$, $ a_{2}\\equal{} b_{2}k_{2}$, $ a_{3}\\equal{} b_{3}k_{3}$, and substituting $ k_{1}$, $ k_{2}$, $ k_{3}$ into the quadratic, and I couldn't get anything very good, but problably I was to tired. However, I did get something that ended up looking alot like generalized Schur....\r\n\r\nThen I tried to multiply out the entire L.H.S. and R. H. S. and see if I could do anything to it. I got something else that looked like Schur....\r\n\r\nI couldn't combine the two methods together, or make them work against each other to solve the problem so.... anyway, I was probably too tired", "Solution_4": "[quote=\"sunchips\"]Given $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}>0$, prove that \n\n$(a_{1}b_{2}+a_{1}b_{3}+a_{2}b_{1}+a_{2}b_{3}+a_{3}b_{1}+a_{3}b_{2})^{2}\\ge 4(a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3})(b_{1}b_{2}+b_{1}b_{3}+b_{2}b_{3})$\n\nand prove that equality happens iff $\\frac{a_{1}}{b_{1}}=\\frac{a_{2}}{b_{2}}=\\frac{a_{3}}{b_{3}}$[/quote]\r\n\r\nLet's call $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ as $a,b,c,d,e,f$ for convenience. Then the original inequality is \\[((a+b+c)(d+e+f)-(ad+be+cf))^{2}\\geq 4(ab+bc+ca)(de+ef+fd).\\] First we know that $(ad+be+cf)^{2}\\leq (a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})$ by Cauchy-Schwarz inequality. So \\[(LHS)=((a+b+c)(d+e+f)-(ad+be+cf))^{2}\\] \\[\\geq ((a+b+c)(d+e+f)-\\sqrt{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})})^{2}.\\] Now let $A=a+b+c, B=d+e+f, C=\\sqrt{a^{2}+b^{2}+c^{2}}, D=\\sqrt{d^{2}+e^{2}+f^{2}}$. Then $(LHS) \\geq (AB-CD)^{2}\\geq (A^{2}-C^{2})(B^{2}-D^{2})$, which is true as this is equivalent to $(AD-BC)^{2}\\geq 0$. So \\[(LHS) \\geq (A^{2}-C^{2})(B^{2}-D^{2})=(2(ab+bc+ca))(2(de+ef+fd))\\] \\[=4(ab+bc+ca)(de+ef+fd)=(RHS),\\] and the proof completes.\r\n\r\nEquality holds iff $\\frac{a}{d}=\\frac{b}{e}=\\frac{c}{f}$ and $BC=AD$, but $BC=AD$ is needless as $\\frac{a}{d}=\\frac{b}{e}=\\frac{c}{f}$ infers it.", "Solution_5": "This is, indeed, a \"one-liner\" (in the sense of: one idea and it's trivial) assuming as known some facts about geometrical inequalities. The proof goes as follows:\r\n\r\nI rewrite the problem using simpler notations:\r\n\r\n[color=blue][b]Problem.[/b] Let a, b, c, x, y, z be nonnegative reals. Prove the inequality\n\n$ \\left(ay \\plus{} az \\plus{} bz \\plus{} bx \\plus{} cx \\plus{} cy\\right)^{2}\\geq 4\\left(bc \\plus{} ca \\plus{} ab\\right)\\left(yz \\plus{} zx \\plus{} xy\\right)$,\n\nwith equality if and only if $ a: x \\equal{} b: y \\equal{} c: z$.[/color]\r\n\r\n[i]Solution.[/i] According to the Conway substitution theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url]), since a, b, c are nonnegative reals, there exists a triangle ABC with area $ S \\equal{} \\frac12\\sqrt {bc \\plus{} ca \\plus{} ab}$ and with its angles A, B, C satisfying $ \\cot A \\equal{} \\frac {a}{2S}$, $ \\cot B \\equal{} \\frac {b}{2S}$, $ \\cot C \\equal{} \\frac {c}{2S}$ (note that we [b]cannot[/b] denote the sidelengths of triangle ABC by a, b, c here, since a, b, c already stand for something different). Similarly, since x, y, z are nonnegative reals, there exists a triangle XYZ with area $ T \\equal{} \\frac12\\sqrt {yz \\plus{} zx \\plus{} xy}$ and with its angles X, Y, Z satisfying $ \\cot X \\equal{} \\frac {x}{2T}$, $ \\cot Y \\equal{} \\frac {y}{2T}$, $ \\cot Z \\equal{} \\frac {z}{2T}$. Now, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url] Theorem [b](c)[/b], we have\r\n\r\n$ \\cot A\\cot Y \\plus{} \\cot A\\cot Z \\plus{} \\cot B\\cot Z \\plus{} \\cot B\\cot X \\plus{} \\cot C\\cot X \\plus{} \\cot C\\cot Y$\r\n$ \\geq 2$,\r\n\r\nwhat rewrites as\r\n\r\n$ \\frac {a}{2S}\\cdot\\frac {y}{2T} \\plus{} \\frac {a}{2S}\\cdot\\frac {z}{2T} \\plus{} \\frac {b}{2S}\\cdot\\frac {z}{2T} \\plus{} \\frac {b}{2S}\\cdot\\frac {x}{2T} \\plus{} \\frac {c}{2S}\\cdot\\frac {x}{2T} \\plus{} \\frac {c}{2T}\\cdot\\frac {y}{2T}\\geq 2$,\r\n\r\nwhat simplifies to\r\n\r\n$ ay \\plus{} az \\plus{} bz \\plus{} bx \\plus{} cx \\plus{} cy\\geq 2\\cdot 2S\\cdot 2T$\r\n$ \\equal{} 2\\cdot 2\\cdot\\frac12\\sqrt {bc \\plus{} ca \\plus{} ab}\\cdot 2\\cdot\\frac12\\sqrt {yz \\plus{} zx \\plus{} xy} \\equal{} 2\\sqrt {\\left(bc \\plus{} ca \\plus{} ab\\right)\\left(yz \\plus{} zx \\plus{} xy\\right)}$.\r\n\r\nUpon squaring, this becomes\r\n\r\n$ \\left(ay \\plus{} az \\plus{} bz \\plus{} bx \\plus{} cx \\plus{} cy\\right)^{2}\\geq 4\\left(bc \\plus{} ca \\plus{} ab\\right)\\left(yz \\plus{} zx \\plus{} xy\\right)$,\r\n\r\nqed. (the equality case is left to the reader).\r\n\r\n[b]EDIT:[/b] The same problem has been discussed at http://www.mathlinks.ro/viewtopic.php?t=151682 .\r\n\r\n Darij", "Solution_6": "Thanks for everyone sharing their crazy smartness! :omighty: \r\n\r\nBefore I came back to check on this post, I was playing around with the discriminant stuff, and I think I have a solution based on that theme. \r\n\r\nSo we must find a value for $t$ such that the quadratic (using Darij's notation), \r\n\r\n$f(t)=(ab+ac+bc)t^{2}+(ay+az+bx+bz+cx+cy)t+(xy+xz+yz)$ is negative (or non-positive), since very big $t$ will result in a positive value for the quadratic.\r\n\r\nThis is accomplished by: \r\n\r\nif $\\frac{x}{a}=m$, $\\frac{y}{b}=n$, $\\frac{z}{c}=p$, and (without loss of generality) $m \\geq n \\geq p$, then $t=n$ suffices. The proof is just an easy matter of cancelling terms, and reduces to $0 \\geq ac(n-m)(n-p)$\r\n\r\nSunchips :) \r\n\r\nP.S. Darij-how did you learn all that crazy math, what is your secret? :wow:" } { "Tag": [ "geometry", "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "Find the area bounded by $ e^{ \\minus{} |x| } \\minus{} \\frac {1}{2}$ and $ |x| \\plus{} |y| \\equal{} \\ln2$.\r\nplease help me to find the solution.", "Solution_1": "You might want to specify which area you want to find. \r\n\r\n$ |x| \\plus{} |y| \\equal{} \\ln 2$ itself encloses a region and the equation $ y_1 \\equal{} e^{ \\minus{} |x|} \\minus{} \\frac {1}{2}$ slices through that closed region. The question is which piece are we interested in: the portion above or below $ y_1$?", "Solution_2": "Looking at it graphically, there are two possible interpretations of this, the first of which is more likely:\r\n\r\nThe first is equal to:\r\n\r\n$ 2\\left (\\int^0_{ \\minus{} \\ln 2} e^x \\minus{} \\frac {1}{2}dx \\plus{} \\frac {(\\ln 2)^2}{2}\\right )$\r\n\r\nThe second is equal to $ 2 (\\ln 2)^2$ subtract the first.", "Solution_3": "can somebody tell me what are the points of intersections ??", "Solution_4": "They intersect at two points: $ (\\ln 2,0),( \\minus{} \\ln 2, 0)$", "Solution_5": "There are two domain. The graph of $ y \\equal{} e^{ \\minus{} |x|} \\minus{} \\frac {1}{2}$ is like tractrix,\r\n\r\nThe small area $ S_1\\equal{}2*\\left(\\frac {1}{2}(\\ln 2))^2 \\minus{} \\int_0^{\\ln 2} \\left(e^{ \\minus{} x} \\minus{} \\frac {1}{2}\\right)\\ dx\\right) \\equal{} \\boxed{(\\ln 2)^2 \\plus{} \\ln 2 \\minus{} 1}$\r\n\r\nThe large area $ S_2\\equal{}(\\sqrt {2}\\ln 2)^2 \\minus{} S_1 \\equal{} \\boxed{(\\ln 2)^2 \\minus{} \\ln 2 \\plus{} 1}$.", "Solution_6": "@ eastymorton.\r\n\r\nI agree that they intersect at $ (\\ln 2,0),( \\minus{} \\ln 2, 0)$ but i feel that that the point $ \\ \\minus{} ln2 , 0$ is satisfied only when $ x,y < 0$ ie it is in the third quadrant with $ \\minus{} y \\equal{} e^x \\plus{} 1/2$ and $ x \\plus{} y \\equal{}\\minus{} \\ ln2$ . the curve $ \\minus{} y \\equal{} e^x \\plus{} 1/2$ intersects the y baxis only at $ \\minus{} 1/2$ .\r\n\r\nso i fell the the other curve which is now in the 2nd quadrant should be there in the 3rd quadrant,but now there is no area enclosed .\r\n\r\nso how do we find it ..please help ansd pardon me if i am wrong" } { "Tag": [ "factorial" ], "Problem": "What is the units digit of $ 1! \\plus{} 3! \\plus{} 5! \\plus{} 7! \\plus{} 9! \\plus{} 11!$ ?", "Solution_1": "Every factorial after $ 5!$ ends in 0, so we just have $ 1 \\plus{} 6 \\equal{} 7$.", "Solution_2": "$ \\ 1!$ ends in 1\r\n3 ends in 6\r\n5 ends in 0\r\n7 ends in 0\r\nand so on for 9, and 11.\r\nso 6+1=7\r\nAnswer is 7\r\nEDIT: Beaten(why am i always beaten :( )" } { "Tag": [ "function" ], "Problem": "If f(0)=2, f(1)=5, f(n+1)=f(n)-f(n-1) for all integers n such that n$\\geq$ 1 find the value of f(7)", "Solution_1": "[hide] 2 5 3 -2 -5 -3 2, f(7)=2[/hide]", "Solution_2": "ok answer is really 5, you missed one more but that's ok. is it possible to write explicit formula for f(n)?", "Solution_3": "[quote=\"yeppyyep\"]ok answer is really 5, you missed one more but that's ok. is it possible to write explicit formula for f(n)?[/quote]\r\n\r\n[hide=\"Y Not?\"]$f(n) = \\left( 1-\\frac{4i\\sqrt{3}}{3}\\right)\\left( \\frac{1+i\\sqrt{3}}{2}\\right)^{n}+\\left( 1+\\frac{4i\\sqrt{3}}{3}\\right)\\left( \\frac{1-i\\sqrt{3}}{2}\\right)^{n}$\n\nwhere $i=\\sqrt{-1}$[/hide]", "Solution_4": "sorry, but how to do that?", "Solution_5": "[url]http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf[/url]\r\n\r\nBut that's definitely way more trouble then it's worth in this case." } { "Tag": [ "calculus", "derivative", "trigonometry", "function", "calculus computations" ], "Problem": "Um...For what values of $ x$ does the graph of $ f(x) \\equal{} x\\plus{}2\\sin{x}$ have a horizontal tangent?\r\n\r\nSo, the derivative of that function is $ 1\\plus{}2\\cos{x}$, so $ \\cos{x}\\equal{}\\frac{\\minus{}1}{2}$. Solving for $ x$, I get $ \\frac{2\\pi}{3}$.\r\n\r\nAt this point, I understand that there are infinite solutions, seperated by some interval related to $ \\pi$ in some way, but I don't remember how to express them the correct way. Help please?", "Solution_1": "$ \\cos (x)\\equal{}\\minus{}\\frac{1}{2}\\equal{}\\cos\\left(x\\plus{}2\\pi c_{1}\\right)$\r\n\r\n$ x\\equal{}2\\pi c_{1}\\minus{}\\frac{2\\pi }{3}\\lor x\\equal{}2\\pi c_{1}\\plus{}\\frac{2\\pi }{3}$\r\n\r\nFor periodic functions, just find trivials solutions and add a integer multiple of the period.", "Solution_2": "My book says the solution is $ (2n\\plus{}1)\\pi\\pm\\frac{\\pi}{3}$ That's the same as your solution, right?" } { "Tag": [ "floor function", "number theory proposed", "number theory" ], "Problem": "Let $ a_i$ be a non-negative integer sequence such that\r\n\\[ a_{k \\plus{} 1} \\equal{} \\lfloor a_k(1 \\minus{} \\frac {1}{p_k})\\rfloor\\ \\forall k\\in \\mathbb N\r\n\\]\r\nSuppose $ n\\in \\mathbb N,a_1 \\equal{} p_n \\minus{} 2$, then prove that $ a_n \\equal{} 0$\r\n\r\n[hide=\"Terminology\"]1) $ p_m$ denotes the $ m$-th prime number.\n2) $ \\lfloor x\\rfloor$ denotes the maximum integer $ \\le x$[/hide]\n[hide=\"Examples\"]\n$ a_1\\equal{}p_3\\minus{}2\\equal{}3,a_2\\equal{}1,a_3\\equal{}0$\n\n$ a_1 \\equal{} p_9 \\minus{} 2 \\equal{} 21,10,6,4,3,2,1,0,0 \\equal{} a_9$\n\n$ a_1 \\equal{} p_{13} \\minus{} 2 \\equal{} 39,19,12,9,7,6,5,4,3,2,1,0,0 \\equal{} a_{13}$\n\n$ a_1 \\equal{} p_{25} \\minus{} 2 \\equal{} 95,48,32,25,21,19,17,16,15,14,13,\\cdots ,3,2,1,0,0 \\equal{} a_{25}$[/hide]", "Solution_1": "No one interested? :maybe: \r\nI can tell you that it's not a random conjecture..., I have some ideas, but I want to know your solutions, opinion.", "Solution_2": "O.K. I am joining! :lol: \r\n\r\nI found a recursion like expression, probably you have already found it....\r\nBut it is hard to continue with it because of the greatest integer function. I think it will be better for me to think over it if you post the idea from which you generated this extraordinary problem.\r\n\r\n[hide=\"click here\"]P/s: Tahole dekhbe besh kichu lok agrohi hoye ei topic e join koreche.....tumi ki etar opor kono seminar e bolbe? Tahole beshi na janano ekdike valo...kintu ekdom kono idea nei to....fole kotha theke suru korbo kichui bujhte parchina, atleast tomar kivabe mone holo seta bolo...[/hide]" } { "Tag": [], "Problem": "Find $n$ such that....\r\n$-5^4+5^5+5^n$ is a perfect square.", "Solution_1": "[quote=\"shadysaysurspammed\"]Find $n$ such that....\n$-5^4+5^5+5^n$ is a perfect square.[/quote]\r\n\r\nN=5.\r\n\r\n-5^4 + 5^5 + 5^5 = 75^2", "Solution_2": "[hide]$-5^{4}+5^{5}+5^{n}=x^{2}$, for some integer $x$. Clearly $n$ must be a nonnegative integer, or the expression would not be an integer.\n\n$5^{4}(-1+5)+5^{n}=x^{2}$\n$50^{2}+5^{n}=x^{2}$\n$x^{2}-50^{2}=5^{n}$\n$(x+50)(x-50)=5^{n}$\n\nSearching for two powers of 5 with a difference of 100, we find that 25 and 125 are the only ones. Therefore, $5^{n}=(25)(125)\\Rightarrow n=5$, and there are no other solutions.[/hide]", "Solution_3": "i did this the same way as matt276eagles.\r\nanother problem like that is,\r\nfind $n$ such that $2^4+2^7+2^n$ is a perfect square. (i found this and the previously posted problem in the art and craft of problem solving.)", "Solution_4": "[quote=\"maokid7\"]\nfind $n$ such that $2^4+2^7+2^n$ is a perfect square. [/quote]\r\n\r\nAgain...\r\n$2^n=(m+12)(m-12)$\r\n8 and 32 are 2 powers of 2 which differ by 24....so\r\n$2^n=2^3*2^5$\r\n$n=8$" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "geometry", "3D geometry", "trigonometry", "calculus" ], "Problem": "Can someone post the solutions to the following problems?\r\n1. A graphing calculator has on its display pixels that are 0.01 apart horizontally. Find the value of n such that the graph of sin(nx) will be graphed as a horizontal line: $n=\\frac{100}{\\pi}$, $100$, $100\\pi$, $\\frac{200}{\\pi}$, $200\\pi$.\r\n\r\n2. $\\displaystyle N=\\{(a,b) | b < \\frac{1}{1+a^2}\\}$ is a subset of $R^2$. Given any line contained in N, what is its maximum slope?\r\n\r\n3. John is happily making a right circular cone of volume 10 cubic feet, with a base of radius $r$ and a perpendicular height $h$. What is the approximate dimension (r,h) of the cone which would minimize the construction costs if the material cost for the side of the cone is 34 cents per square foor, and for the bottom the cost is 44 cents per square foot.", "Solution_1": "[hide=\"1.\"]\nIf it appears as a horizontal line, then $\\sin{(nx)}$ has a period of $.01$, or $\\sin{(nx)}=\\sin{(n(x+0.01))}$ so $nx+2\\pi=nx+0.01n \\Rightarrow n=200\\pi$.[/hide]", "Solution_2": "Oh, good solution. Any solutions for the other two?", "Solution_3": "For #2, does it have to be a whole line (extending to infinity) or can it be a line segment?", "Solution_4": "I think they mean a whole line.", "Solution_5": "3)\r\n[hide]\nI'm not sure if having different costs for material will affect anything. This may be completely wrong but:\n$V = \\pi r^2h = 10$\n$h = 10/(\\pi r^2)$\n\n$A = 2\\pi r^2 + 2r\\pi h$\n$A = 2\\pi r^2 + 2r\\pi(10/(\\pi r^2))$\nFrom here I think you'd need calculus or a graphing calculator to solve the problem.\nHere's a calculus solution just for fun. I hid it since we're not suppose to post calculus in this forum:\n[hide]\n$\\frac{dA}{dx} = 4\\pi r - 20r^{-2}$\nYou want to find critical numbers:\n$0 = 4\\pi r - 20r^{-2}$\n$0 = 4r(\\pi - 5r^{-3})$\n$4r = 0$\n$r = 0$\n$\\pi - 5r^{-3} = 0$\n$\\pi = 5r^{-3}$\n$\\frac{\\pi}{5} = \\frac{1}{r^3}$\n$\\sqrt[3]{(\\frac{\\pi}{5})} = \\frac{1}{r}$\n$\\frac{1}{\\sqrt[3]{(\\frac{\\pi}{5})}} = r$\nThe function decreases on the interval from 0 to r and then increases from r to infinity so this must be a minimum.\n$h = 10/(\\pi * r^2)$\n$h = 2.335$\n$(r,h) = (1.168, 2.335)$\n[/hide]\n[/hide]", "Solution_6": "Thanks, SnowStorm; however, since it's a cone, it should be V=(1/3)(pi*r^2*h)=10.\r\n\r\nDoes anyone have a solution to #2? The answer, by the way, is:\r\n[hide]0[/hide]\r\n\r\nAlso, here's another problem I need a solution to :) :\r\nHow many numbers are there that satisfy the following equation?\r\n$\\displaystyle n+n^2+\\frac{1}{n}+\\sqrt{n}=10$\r\n(these are just numbers, not integers or postive, etc.)", "Solution_7": "I got 2, but i'm not sure how to prove it without using a quasi-calculus solution.", "Solution_8": "[hide=\"2.\"]\nConsider the graph $b = \\frac{1}{1+a^2}$. It has a horizontal asymptote at $b=0$. If $b < \\frac{1}{1+a^2}$, the line must be under the horizontal asymptote for any $a$. However, if a line has a non-zero slope, its range is all reals and it won't satisfy the inequality. Thus the only slope of a line contained in $N$ is $0$.[/hide]", "Solution_9": "so, yea, i turned it into a 6th order polinomial (your last question). From there, i decided to let my calculator solve it for me....both of these are estimates:\r\n\r\nn= 1.62476\r\nn= 1.98750\r\n\r\ncomplex:\r\nn= -1.22952+1.56676i\r\nn= -1.22952-1.56676i\r\nn= -1.57665+1.95906i\r\nn= -1.57665-1.95906i\r\n\r\nim sure i didnt help at all...but...im gonna post anyway" } { "Tag": [ "geometry", "symmetry", "incenter", "angle bisector" ], "Problem": "The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \\cap BD \\equal{} P$ and $ AC\\cap B'D \\equal{} Q$ then prove that $ PQ \\perp AC$", "Solution_1": "call $ E: AB \\cap CD$ so $ \\angle BPC \\equal{} \\angle BCA' \\minus{} \\angle CBD \\equal{} \\angle ADB \\minus{} \\angle CBD \\equal{}$\r\n$ \\equal{} 90 \\minus{} \\angle ABD \\minus{} \\angle CBD \\equal{} 90 \\minus{} \\angle ABC \\equal{} \\angle BEC$.\r\nThen BCPE is cyclic and $ EB \\perp BD$.\r\n\r\nThen the problem is equal to this:\r\n\r\nH is the orthocenter of ABC and AH,BH,CH meet respectively BC,CA,AB on D,E,F. Call H' the symmetric of H w.r.t. ED and ED meet AF on G and AH' on K. Then $ KF \\perp ED$.\r\n\r\nWe can prove it in this way: F is the harmonic cojugate of G w.r.t. AH, but G on $ \\triangle AKH$ is the feet of the internal angle bisector of $ \\angle AKH$, so F is the feet of the external angle bisector of $ \\angle AKH$ and then $ FK \\perp KG$.", "Solution_2": "Denote AC meet BD at G, B.G.D.P is harmonic division GC is bisector of angle BQC", "Solution_3": "not a hard one\nHint : $AC \\cap BD=T$ and $BB' \\cap AC=L$ . now use Menelaus theorem for$\\bigtriangleup BB'D$ and point $Q$ and use sin law for $\\bigtriangleup BCD$ and point $P$ then you will see that $\\frac {PB}{PD}=\\frac {QB'}{QD}=\\frac {BT}{TD}$ so done.", "Solution_4": "By symmetry, $\\Delta AB'Q=\\Delta ABQ\\implies \\angle BQT=\\angle DQT$, so we need to show that $\\frac{BP}{PD}=\\frac{BT}{TD}$, where $T=AC\\cap BD$.\nWe see easily that $D$ is the incenter of $\\Delta ACP$, so $D, B$ are harmonic conjugate w.r.t. $\\overline{PT}$, hence $\\frac{BT}{BP}=frac{TD}{PD}\\iff\\frac{PD}{PB}=\\frac{TD}{BT}$; since $TQ$ is internal angle bisector of $\\angle BQD$, it follows that $PQ$ is the external bisector of $\\angle BQD$ and $PQ\\bot AC$.\n\nBest regards,\nsunken rock", "Solution_5": "Let $AC\\cap BD= E $ and $BB'\\cap AC = M $. Also, let $BB'\\cap PQ = X $. \nIt is easy to see that $A'$ lies on $\\odot (ABCD) $. As, $BA = BA'$, so, $BC $ bisects $\\angle PCQ $. Thus, $(-1) = (P,E;B,D)\\stackrel{Q}{\\doublebarwedge}(P,E;B,D)\\stackrel{Q}{\\doublebarwedge}(X,M;B,B')$. But, $M$ is the midpoint of $BB'$. Hence, $X $ is a point at infinity and so, $BB' || PQ $. This gives $\\angle PQC = 90^{\\circ} $. ", "Solution_6": "Let $\\{L\\}$=$BA'\\cap CD$, $\\{T\\}$=$AC\\cap BD$, $\\{Q\\}$=$BC \\cap A'D$.\nSimple angle chasing gives $T,Q,L$ colliniar and $LQ \\perp BD$.\n\nBy a well, known lemma, we get that $QT$ is the polar of $P$, so $P$ is on the polar of $T$ which gives $(P,T,B,D)=-1$.\n\nNow, all we have to prove is that $\\frac{BT}{TD}=\\frac{QB'}{B'D}$, which is Menelaus' theorem in triangle $BB'D$ with line $QT$ $\\blacksquare$\n\n\n", "Solution_7": "Well though the computations took a while \nThey were pretty straightforward and simple\nAnd ended nicely\nI blindly used complex numbers and solved it. " } { "Tag": [], "Problem": "A random pole:\r\n\r\nDo you usually type in the BBCode or do you click the icons?", "Solution_1": "Its becoming more popular these days to let people type their own HTML in their messages, (of course, drastically limited/cleaned etc). I quite like that", "Solution_2": "Haha, icons are for n00bs.", "Solution_3": "I use both. I type the bold tags and the easy ones. When I use hide, I click it. To date I've only used bold, underline, and hide on this forum (as far as I remember). I've used quote but only quoting from the thread", "Solution_4": "I got so confused once while typing my biology lab in Microsoft Word I started to type \"[ b ]\" instead of Ctrl B. :D" } { "Tag": [ "limit", "real analysis", "real analysis solved" ], "Problem": "Denote by p_n te sequence of prime numbers.\r\nCalculate lim n->oo from [product from k=1 to n from (n+k)^1/p_(n+k)]", "Solution_1": "Let Q(n) = prod_(k=1 to n) (n+k)^(1/p_(n+k)) \r\nx(n) = lnQ(n) = \\sum _(k= 1 to n) ln(n+k)/p_(n+k)\r\nx(n) = n.(x(n)/n)\r\n\r\nI am going to get an equivalent of x(n)/n with a mixture \r\nequivalent technic and Cesaro-Stolz technic that I learned on \r\nmathlinks :D \r\n\r\n \r\n\r\n(x(n) - x(n-1))/(n - (n-1)) = ln(2n)/p_(2n) but the n-th prime p_n is equivalent when n tends to infinity to n.lnn\r\n \r\n(x(n) - x(n-1))/(n - (n-1)) ~ ln(2n)/(2n.ln(2n)) ~ 1/(2n) \r\n\r\nx(n) = n.(x(n)/n) ~ n(x(n) - x(n-1))/(n - (n-1)) ~ n/(2n) \r\n\r\nand in 6th grade I learned n/(2n) = 1/2 (if n=/=0)\r\n\r\n \\lim x(n) = 1/2 \r\n \\lim Q(n) = \\sqrt e" } { "Tag": [ "integration", "trigonometry", "function", "calculus", "derivative", "LaTeX", "absolute value" ], "Problem": "$ \\int_0^\\infty\\mid e^{\\minus{}x}\\sin x\\mid\\,dx$", "Solution_1": "Split it into the intervals $ [n \\pi, (n \\plus{} 1)\\pi]$. The function has a nice enough anti-derivative between points where the sign changes, so you can compute on each interval with no problem. Then sum the resulting series (for which it will help to know that $ \\sin x \\equal{} \\frac {e^{ix} \\minus{} e^{ \\minus{} ix}}{2i}$).\r\n\r\nFor purposes of calculus, absolute value is just another piecewise linear function, and piecewise linear functions are best dealt with by splitting at the joints and working separately on each piece.", "Solution_2": "I have\r\n\r\n$ \\int_0^\\infty\\mid e^{ - x} \\sin x \\mid \\, dx$\r\n\r\n$ = \\int_0^\\pi e^{ - x} \\sin x\\, dx - \\int_\\pi^{2\\pi} e^{ - x} \\sin x\\, dx + \\int_{2\\pi}^{3\\pi} e^{ - x} \\sin x\\, dx - ...$\r\n\r\n$ = \\sum_{n = 0}^\\infty (\\int_{2n\\pi}^{(2n + 1)\\pi} e^{ - x} \\sin x\\, dx - \\int_{(2n + 1)\\pi}^{(2n + 2)\\pi} e^{ - x} \\sin x\\, dx)$\r\n\r\n$ = \\sum_{n = 0}^\\infty (\\frac { - e^{ - x}}{2} (\\cos x + \\sin x)]_{2n\\pi}^{(2n + 1)\\pi} - (\\frac { - e^{ - x}}{2} (\\cos x + \\sin x)]_{(2n + 1)\\pi}^{(2n + 2)\\pi})$\r\n\r\n$ = \\sum_{n = 0}^\\infty (\\frac {e ^{ - 2n\\pi}}{2} - \\frac {e ^{ - (2n + 2)\\pi}}{2})$\r\n\r\n\r\n$ = \\frac {1}{2}( 1 - e ^{ - 2\\pi} + e ^{ - 2\\pi} - ...)$\r\n\r\n$ = \\frac {1}{2}.$\r\n\r\nSome intermediate steps not written as it took about three minutes to work out (and is probably wrong) and over twelve minutes to $ \\text{\\LaTeX}$ not very well, during which time (I've just spotted) JBL posted. I must learn to $ \\text{\\LaTeX}$.", "Solution_3": "Well, actually the summing part is simpler than I suggested (I somehow didn't notice that the values of sine and cosine are all at integer multiples of pi :| ) Somewhere you lost the sign you introduced, I think, but I'm not sure where -- $ \\frac{1}{2}$ is the value of the integral [i]without[/i] the absolute value.", "Solution_4": "\\[ (1\\plus{}e^{\\minus{}\\pi})^2 \\sum_{n \\equal{} 0}^\\infty e^{\\minus{}2n\\pi}\\]", "Solution_5": "Thanks, Immanuel.\r\n\r\nI got a sign wrong and now make it \r\n\r\n$ \\frac {(1 \\plus{} e^{ \\minus{} \\pi})^{2}}{2} \\sum_{n \\equal{} 0}^\\infty e^{ \\minus{} 2n\\pi}$, \r\n\r\n(that is, half your answer) \r\n\r\n$ \\equal{} \\frac {(1 \\plus{} e^{ \\minus{} \\pi})^{2}}{2}(\\frac {1}{1 \\minus{} e ^{ \\minus{} 2\\pi}})$ \r\n\r\n$ \\equal{} \\frac {(1 \\plus{} e^{\\minus{}\\pi})}{2(1 \\minus{} e^{ \\minus{} \\pi})}$\r\n\r\nwhich is approximately $ 0.545$.\r\n\r\nHave I still got it wrong?", "Solution_6": "You're welcome. Sorry forgetting the half..." } { "Tag": [ "MATHCOUNTS", "videos" ], "Problem": "Is there any place to get the National Competition videos online?\r\n\r\nThanks.", "Solution_1": "The 2006 competition wasn't broadcasted but you can find a short 2005 clip [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=765&z=71]here[/url]", "Solution_2": "I have videos of myself in countdown somewhere...", "Solution_3": "can u put them online?", "Solution_4": "http://www.mathcounts.org/webarticles/anmviewer.asp?a=765&z=71", "Solution_5": "But doesn t that only have the final round and interviews?", "Solution_6": "its the only one published\r\n\r\nhttp://www.mathcounts.org/webarticles/anmviewer.asp?a=765&z=71", "Solution_7": "http://www.msnbc.msn.com/id/12772703/\r\n\r\nClick on the picture for the video, that's all we've got, really...", "Solution_8": "[quote=\"Klebian\"]http://www.msnbc.msn.com/id/12772703/\n\nClick on the picture for the video, that's all we've got, really...[/quote]\r\n\r\nisnt it the same video", "Solution_9": "no, it s not.\r\n\r\nThere s another video.\r\n\r\n[url]http://www.wusatv9.com/video/player.aspx?aid=31628[/url]", "Solution_10": "I found another one:\r\nhttp://www.raytheon.com/about/contributions/mathmovesu/\r\n\r\nAlso, I tried Google Video and YouTube, but neither have MATHCOUNTS Videos.", "Solution_11": "Thank You very much. We appreciate the videos you guys put.\r\n\r\nI only saw the video online at MathCounts, and that was pretty short.", "Solution_12": "The one on the MathCounts site is for 2005 and the one on the MathMovesU site is for 2004.", "Solution_13": "[quote=\"Ignite168\"]The one on the MathCounts site is for 2005 and the one on the MathMovesU site is for 2004.[/quote]\r\n\r\nyes. i should have mentioned earlier. :oops:", "Solution_14": "Is there a Mathcounts Movie? Mr. Rusczyk gave a link to something on idmb that said he starred in the Mathcounts Movie 2004 adn 2005...", "Solution_15": "It's the 60 minute show that was shown on ESPN with highlights of the National Competition in 2004 and 2005 not a movie." } { "Tag": [ "MIT", "college" ], "Problem": "It's almost here.\r\n\r\nWho do you think/believe will win? who are you cheering for?\r\n\r\n\r\nMe: cheering for Agassi(if he plays), Safin, Nadal, Sharapova, Dementieva, Vaidisova and Clijsters", "Solution_1": "safin, roddik, and hewitt", "Solution_2": "I heard Nadal is not playing. :(", "Solution_3": "this is going to be the most boring aus open ever...\r\nnadal, agassi and safin are out...\r\n\r\nthats what i heard\r\n\r\nfederer can win it playing with his left hand", "Solution_4": "Maybe left foot..", "Solution_5": "Kim Clijsters injured her foot, don't know if she will play or not.\r\n\r\nthis is going to be the boringest Aus open ever.", "Solution_6": "[quote=\"manuel\"]this is going to be the most boring aus open ever...\nnadal, agassi and safin are out...\n\nthats what i heard\n\nfederer can win it playing with his left hand[/quote]\r\n\r\nyou never know...federer has lost already once this year. and a sleeper might come out and beat federer. it's possible (hopefully andy roddick :D )", "Solution_7": "i love these threads...\r\ni want roddick to win, or hewitt, but i think federer will win", "Solution_8": "by the way, this is a little off topic, but any seniors here gonna play college tennis?\r\n\r\nim going to play in mit yayyyy... practiced with the team last month..", "Solution_9": "wow, what a match, venus gone", "Solution_10": "I heard dementieva went as well......less star studded aus open this year,sad burt it'll be awesome if ivan ljubicic wins,my fav out of the playing lot.Safin isn't playin :( He's so cool.\r\n\r\nHow many say federe will win?", "Solution_11": "Dissappointing, but not terribly suprising. I mean, she's played like what, 3 matches since Wimbledon?", "Solution_12": "ok ,hewitt kicked out, chela in . :)", "Solution_13": "who wud have saw that coming? \r\nL. Hewitt beat by Chena? Ouch! :stretcher:", "Solution_14": "Hantuchova over Serena :D", "Solution_15": "Roddick gone :( \r\n\r\nI think american men's tennis era is probably over, women too.", "Solution_16": "Oh yeah!!!I like the guy who beat roddick.Seen him for the first time.", "Solution_17": "ok , fed's won already. :( all gone , hewitt ,roddick..... maybe ljubicic can do something :huh:", "Solution_18": "theres always tommy has.", "Solution_19": "tribute to haas , he had fed really screwed up, too baaadd fed won :( \r\n\r\nnice goin haas :coolspeak:", "Solution_20": "[b]Alright guys..BIG TIME UPDATE!!![/b]\r\nBAGDATIS is in the FInal!!!Wooohoooooo...he beat nalbandian.Brought him down to earth bagdatis.Federer's in the final.\r\n\r\nAnd in the [i]ladies[/i]..Our new champ....Amelie.. .(Finally she got a title).mauresmo!!!!!\r\n\r\nat the moment i'm seeing paes/damm vs buggin brothers..oops bryan bros...first set paes/damm second bryans and third its 2 games all.", "Solution_21": "I feel bad for Mauresmo, not being able to win it, to have that moment. But great sportsmanship when Amelie went up to sit by Justine to find out how she was feeling. That was nice to see.", "Solution_22": ":ewpu: sounds confusinggg, smelly fart sayin mauresmo wins the title , while Master says \"not being able to win it.\" .....hey guys , so who won? i missed the match and now i cant find score on TV anywhere :(", "Solution_23": "Sorry, maybe I didn't include as many details as I should have.\r\n\r\nBasically, Henin-Hardenne retired because she was not feeling well. Henin said it was a stomach-ache. That's why I said I feel sorry for Mauresmo, not having that moment, not being able to 'win' it, but it kind of being handed to her.\r\n\r\nMany people believe that Henin should have just 'sucked it up', and played through the pain, at least finish the match.", "Solution_24": "Federer won at last :first: \r\n\r\nHowever it was some real good tennis from Bagdatis :winner_second: \r\n\r\n5-7:7-5:6-0:6-2\r\nWow!!!", "Solution_25": "[quote=\"The Master\"]Sorry, maybe I didn't include as many details as I should have.\n\nBasically, Henin-Hardenne retired because she was not feeling well. Henin said it was a stomach-ache. That's why I said I feel sorry for Mauresmo, not having that moment, not being able to 'win' it, but it kind of being handed to her.\n\nMany people believe that Henin should have just 'sucked it up', and played through the pain, at least finish the match.[/quote]\r\n\r\nI'm disgusted by the opinions that the pundits are throwing out. Before this match, anyone who had watched even a tiny amount of women's tennis in the last three years would call her the grittiest player in the womens' game, but she retires from a big match after playing a pair of tight three setters the previous couple of matches and everyone throws her under the bus.", "Solution_26": "[quote=\"blahblahblah\"]I'm disgusted by the opinions that the pundits are throwing out. Before this match, anyone who had watched even a tiny amount of women's tennis in the last three years would call her the grittiest player in the womens' game, but she retires from a big match after playing a pair of tight three setters the previous couple of matches and everyone throws her under the bus.[/quote]Ya, sometimes people are too quick to judge without thinking. I don't think any professional athlete would retire from a match like that without very good reason. It is not in their nature." } { "Tag": [ "number theory", "greatest common divisor", "algebra unsolved", "algebra" ], "Problem": "let $ a_n \\equal{} n^2 \\plus{} 20$ and $ d_n$ denotes the greatest common divisor of $ a_n$ and $ a_{n \\plus{} 1}$\r\n then prove that $ d_n$ divides $ 81$.", "Solution_1": "gcd(n^2+2n+21, n^2+20) = gcd(2n+1, n^2+20). Since 2n+1 is odd, this equal to gcd(2n+1, 2n^2+40)=gcd(2n+1, 2n^2+40 - n(2n+1)) = gcd(2n+1, 40-n)= gcd(2n+1-2(40-n),40-n)=gcd(81,40-n), so the gcd divides 81." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "trigonometry", "geometry proposed" ], "Problem": "In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.", "Solution_1": "Since the reflection $ D$ of $ H$ in $ B_{0}$ is the second intersection of $ BO$ with the circumcircle, the lines $ HB_{0}$ and $ PQ$ are parallel if and only if the pencil $ (PQ, PH, PB_{0}, PD)$ is harmonic. The conclusion follows from the fact that $ BH \\cap (PQ, PH, PB_{0}, PD) \\equal{} (B, Q, H, B_{1}) \\equal{} \\minus{} 1$, where $ B_{1}$ is the foot of the $ B$-altitude in triangle $ ABC$.", "Solution_2": "Just a question to clarify the brilliant solution of Pohoatza: is point D the second intersection of BO with the circumcircle? Thanks", "Solution_3": "Indeed. I've edited now. Thanks.", "Solution_4": "My solution is metric. If B_1 is the foot of the altitude from B, we can consider triangles B_1B_0H and B_1PQ. Then it suffices to prove the equality B_1Q/B_1H = B_1P/B_1B_0 . Both quotients are equal to 2sinAsinC/cos(A-C).", "Solution_5": "If $ \\{T\\}\\equal{}HB_0\\cap BO$ then $ \\frac{BH}{BT}\\equal{}\\frac{2R\\cos{B}}{2R}\\equal{}\\cos{B}$.\r\nIn addition, $ \\triangle{BC_1Q}\\sim \\triangle{BCP}$, thus $ \\frac{BQ}{BP}\\equal{}\\frac{BC_1}{BC}\\equal{}\\cos{B}$.\r\nThus, $ \\frac{BH}{BT}\\equal{}\\frac{BQ}{BP} \\Longleftrightarrow PQ\\parallel{}HB_0$.", "Solution_6": "Construct the diameter $ BOT$. In fact that the quadrilateral $ AHCT$ is a parrallelogram. We have :\r\n$ \\frac{BP}{PH}\\equal{}\\frac{[BMN]}{[HMN]}\\equal{}\\frac{BM.BN}{HM.HN}$\r\n$ \\frac{BQ}{BT}\\equal{}\\frac{[BAC]}{[TAC]}\\equal{}\\frac{BA.BC}{HA.HC}$\r\nEasy to proof that $ \\frac{BM.BN}{HM.HN}\\equal{}\\frac{BA.BC}{HA.HC}$\r\nHence, by Thales theorem we get that $ PQ // HT \\Rightarrow PQ // HB_{0}$", "Solution_7": "Let BT be a diameter of (ABC); it\u2019s easy to see that $ \\triangle BAT \\sim \\triangle BA_{1}H$ and P and Q are homologues points, hence the conclusion follows.\r\n\r\nBest regards,\r\nsunken rock", "Solution_8": "[quote]In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.[/quote]\n[b]Solution:[/b] Let $BP \\cap HB_0=X$. Let $\\ell || HB_0$ through $P$ intersects $BB_1$ at $Y$\n$$-1=(H, ~ X ; ~B_0, ~ \\infty_{HX}) \\overset{P}{=} (H, ~ B ; ~ B_1, ~Y)$$\nBut, $$-1=(A , ~ C ; ~ A_1C_1 ~ \\cap ~ AC, ~B_1) \\overset{B}{=} (C_1, ~ A_1 ; ~A_1C_1 ~ \\cap ~ AC , ~ Q) \\overset{C}{=} (H , ~ B ; ~B_1, ~Q)$$\nHence, $Y \\equiv Q$", "Solution_9": "Just compute everything!\n\nLet $D$ be the foot of the $A$-altitude, and let $A'$ be the $A$-antipode; it suffices to show $\\frac{AQ}{AP} = \\frac{AH}{AA'}$. Notice that $\\angle AQB_1 = 90^\\circ - B + C$, so $$\\frac{AQ}{AB_1} = \\frac{\\sin B}{\\cos(B-C)} \\implies AQ = \\frac{2R \\cos A \\sin B\\sin C}{\\cos(B-C)}.$$ Also, $$AP = \\frac{AD}{\\cos(B-C)} = \\frac{2R\\sin B\\sin C}{\\cos(B-C)}.$$ It is evident now that $\\frac{AQ}{AP} = \\cos A = \\frac{AH}{AA'}$, as needed.\n\n[color=#9a00ff][b]Remark.[/b][/color] It turns out that $\\overline{AQ}$ and $\\overline{AP}$ are similar parts in the inverse similarity $\\triangle AB_1C_1 \\sim \\triangle ABC$, which explains a lot.", "Solution_10": "huh\n[hide = solution]\nConsider the map $\\tau$ which reflects a point in the bisector of $\\angle ABC$ and dilates with a factor of $\\cos B$ wrt to $B$. Let $B_1$ denote the $B$-antipode of $\\triangle BAC$. Remark that under $\\tau$ the segment $BB_1$ maps to $BH$ since triangle $\\triangle BAC$ goes to $\\triangle BA_1C_1$. In particular $P = BB_1 \\cap AC$ goes to $Q = BH \\cap A_1C_1$. Thus $BP/BB_1 = BQ/BH$ from where it follows that $PQ \\parallel HB_1$. Since $H, B_0, B_1$ are collinear this finishes.\n[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "I have a problem getting set up with latex i dont know what file to chose for the Distrabution Directory could someone please help?", "Solution_1": "On top of your screen is a button leading you to an explanation site : \r\n\r\n[url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]\r\n\r\npart of the forum is the latex forum where questions like this belong :\r\n[url]http://www.mathlinks.ro/Forum/index.php?f=123[/url]", "Solution_2": "Sorry didnt see that part of the forum. And ive looked at the site and it dosent help \r\nThank though" } { "Tag": [], "Problem": "2. Al gets the disease algebritis and must take one green pill and one pink pill each\r\nday for two weeks. A green pill costs $ \\$1$ more than a pink pill, and Al's pills\r\ncost a total of $ \\$546$ for the two weeks. How much does one green pill cost?\r\n\r\n$ (A) \\$7 \\qquad (B) \\$14 \\qquad (C) \\$19 \\qquad (D) \\$20 \\qquad (E) \\$39$", "Solution_1": "[hide]If it costs 546 dollars for 14 days, then it should cost 39 dollars per day. This means that a green pill is 20 dollars, while a pink pill is 19. [b]A green pill costs 20 dollars. [size=150]D[/size][/b][/hide]", "Solution_2": "Let $ x$ be the cost of a green pill; then $ x \\plus{} (x\\minus{}1) \\equal{} \\frac{546}{2 \\times 14} \\equal{} 39 \\Longrightarrow x \\equal{} 20$.\r\n\r\nI actually got this question wrong when taking a practice test by misreading. :oops:" } { "Tag": [ "function", "algebra", "domain", "complex analysis" ], "Problem": "Let $ f$ be analytic in a neighborhood of the set $ Im(z)\\geq 0$ and assume $ f(i)=0$ \r\nand that $ f(z)\\leq 1$ when $ Im(z)=0$. Show that\r\n\\[ |f(z)| \\leq \\left|\\frac{ (z-i) }{ (z+i) }\\right| \\]\r\nfor all $ z$ with $ Im(z)\\geq 0$", "Solution_1": "Basically you just want to use Schwarz's Lemma, but to put it into the correct form you need Cayley's transform.\r\n\r\nSo note, that $ M(z)=\\frac{z-i}{z+i}$ (Cayley's transform) maps the upper half-plane conformally onto the unit disk. Then let $ g(z)=f(M^{-1}(z))$ and note that $ g: D \\rightarrow D$. That is actually a little tricky since you were only given $ f(z)\\leq1$ when $ \\Im z=0$, but you can deduce it by realizing that conformal maps map boundaries into boundaries and knowing that $ f(i)=0$ maps an interior point into an interior point.\r\n\r\nNow we have $ |g(z)| \\leq 1$ and $ g(0)=0$. Therefore, $ |g(z)| \\leq |z|$ by Schwarz's Lemma. That is equivalent to: $ |f(M^{-1}(z))| \\leq |z|$. Replacing $ z$ by $ M(z)$ gives $ |f(z)| \\leq \\left| \\frac{z-i}{z+i}\\right|$ for $ \\Im z \\geq 0$.", "Solution_2": "[quote]That is actually a little tricky[/quote]\r\nThat part is so tricky that there is a counterexample: $ f(z)=\\frac{e^{-i z}-e}{1+e}$.", "Solution_3": "Thanks Mlok for the counterexample.\r\n\r\nSo it looks like if we change Color's condition that $ |f(z)| \\leq 1$ when $ \\Im z =0$ to $ |f(z)| \\leq 1$ when $ \\Im z \\geq 0$, then the proof I provided will work.\r\n\r\nThis does raise a question of my own though. For a function $ f$ satisfying the original conditions of the problem, I mentally imagined the boundary curve (the real axis) being conformally mapped by $ f$ to a simple closed curve contained in the unit disk or possibly being the boundary of the unit disk itself. Since I knew one interior point of the upper half-plane mapped into the interior of this closed curve, I thought all of the points would as well via a Jordan curve argument. \r\n\r\nSo my question is, under what conditions can we be assured that will be true? Or do I need something like my boundary curve not containing the point at $ \\infty$?", "Solution_4": "[quote]Since I knew one interior point of the upper half-plane mapped into the interior of this closed curve[/quote]\r\nBut we don't know that. In my counterexample the image of the real axis is indeed a simple closed curve [namely a circle] which lies in the unit disk -- but $ 0$ is outside of the curve. \r\n\r\nFor unbounded domains the maximum principle is replaced by the [url=http://en.wikipedia.org/wiki/Phragm%C3%A9n-Lindel%C3%B6f_principle]Phragm\u00e9n-Lindel\u00f6f principle[/url]. The wikipedia page gives a rather sketchy special case. When working with the sector $ \\{z\\colon |\\arg z|<\\beta\\}$, you need $ |f(z)|$ to be majorized by $ \\exp(c|z|^{\\alpha})$ with $ \\alpha<\\frac{\\pi}{2\\beta}$. \r\n\r\nAn accessible reference: [i]Potential Theory in the Complex Plane[/i] by T. Ransford." } { "Tag": [ "MATHCOUNTS" ], "Problem": "The room rate at the Detroit Marriott hotel for Guests at MATHCOUNTS Nationals is 129 dollars/night + 15% tax. How much money could your Dad/Uncle/Grandpa save if they share a room with me? How much if two people share a room with me?\r\n\r\nOK, so this isn't a very interesting math question, but I'm talking real money. I'm the father of a nationals competitor and I've reserved a room for four nights (arriving May 4th). I am looking for someone (adult male) to share a room and hotel expenses for some or all of those nights. So, if you have any relatives who are coming and want to ease the cost, PM or e-mail me please.\r\n\r\nThanks, and good luck to you all at nationals!\r\n\r\nDavid", "Solution_1": "So, first of all, this 129 figure, is this for sharing, or a separate room?", "Solution_2": "[quote=\"dwx314\"]So, first of all, this 129 figure, is this for sharing, or a separate room?[/quote]\r\n\r\nThat's the room rate. One extra person is no extra charge. I didn't ask about beyond double occupancy. The room has two beds, and is nonsmoking. So I'm looking at roughly 600 dollars for the single occupancy room for four nights. That's a lot more than the airfare, so I'd love to share expenses.", "Solution_3": "but don't you have to room with a fellow teammate though?", "Solution_4": "[quote=\"dwx314\"]but don't you have to room with a fellow teammate though?[/quote]\r\n\r\nIt should be apparent I'm a parent. (See first post, and sorry for the pun) Competitors all room with other competitors, and there is no charge for them of course. But guests (parents, etc.) pay for rooms.", "Solution_5": "So, a night =159.85\r\n4 nights = 639.4\r\nWow. Thats a lot to save.\r\n\r\nSorry for confusing you with a competitor. (entropy2?! how obvious could it get?)\r\n\r\nGood luck to your son/daughter. Too bad I didn't go. (can you tell I'm a student)", "Solution_6": "[hide=\"the answer\"]593.40 is the total cost for 4 nights, so divide by 2. . . 296.70![/hide]", "Solution_7": "Thanks for wishing me luck, dwx. I stole my dad's username (meaning he uses entropywins other places) so he had to use a new one. We were in the mathcounts class together btw :D" } { "Tag": [ "geometry", "3D geometry", "AMC 10" ], "Problem": "I need really good help on painting cubes here.\r\n\r\nI have like really few senses on cubes and whenever this appears on AMC-10, it's like dead killer to me.\r\n\r\nCan anyone post some introduction and then some intermediate and so forth so I can increase my cubic ability?\r\n\r\nThank you very much!!!", "Solution_1": "Well.. here are the VERY basics.. (this is all I know)\r\n\r\nFor a painted cube divided into n^3 smaller cubes...\r\nNumber of smaller cubes with [b]NO [/b]paint on them = (n-2)^3.\r\n(Imagine a smaller cube shielded from paint by the cubes on the surface)\r\n\r\nNumber of smaller cubes with paint on [b]one [/b]side = 6(n-2)^2\r\nThese cubes are on the faces of the large one,\r\nbut we can't count the cubes on the corners or edges,\r\nso imagine a smaller square with side length n-2 for each of the 6 faces of the original cube.\r\n\r\nNumber of smaller cubes with paint on [b]two[/b] sides = 12(n-2)\r\nA cube has 12 edges, and each edge has n smaller cubes, but we're not counting the corners.\r\n\r\nNumber of smaller cubes with paint on [b]three[/b] sides = 8 (on the corners)\r\n\r\nOh, and total surface area of n^3 unit cubes = 6n^3\r\nThere are n^3 cubes, each with surface area 6.", "Solution_2": "Tip: whenever you see such a problem, draw a diagram. Make a big cube and put lines on it to show where the unit cubes are. If the problem doesn't give a specific number of unit cubes, arbitrarily set the side length of the big cube to whatever you think would be most helpful in visualizing it.", "Solution_3": "That is SOOOOOOOO useful!\r\n\r\nTo be honest, I didn't know any of them. :( \r\n\r\nHmm.. Thanks! :D", "Solution_4": "Most number of cubes visible at once:\r\n\r\nn:^3: - (n - 1):^3:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all finite set $S\\subset {\\mathbb Z}_+$ which satisfying both the following conditions:\r\n1) There is any power of two in $S$, ie... there is $k\\in{\\mathbb Z}_+$ which $2^k\\in S$.\r\n2) Any $x,y\\in S$ which $x\\neq y$ then $\\frac{x+y}{\\gcd(x,y)}\\in S$.\r\n\r\nEnjoy~ :)", "Solution_1": "Assume that there is a set $S$ satisfying (1) and (2) with more than one element.\r\n\r\nIt's not clear from the notation whether the power of two that's in the set $S$ could be $1 = 2^0$. If so, then putting $y=1$ we have $x \\in S \\Rightarrow x+1 \\in S$ and so $S$ is infinite.\r\n\r\nIf now there's a power of 2 other than 1 in $S$ then there is an even number in $S$. If $x$ is odd and $y$ is a power of 2 then $\\mathrm{hcf}\\{x,y\\} = 1$, so $x \\in S \\Rightarrow x+y \\in S$ and again $S$ is infinite.\r\n\r\nNow suppose that all the elements of $S$ are even. Let $x$ and $y$ be the first two elements of $S$ in order. Then $\\mathrm{hcf}\\{x,y\\} \\ge 2$ and so $z = \\frac{x+y}{\\mathrm{hcf}\\{x,y\\}} \\le \\frac{x+y}{2} < y$, a contradiction.\r\n\r\nSo the only possibility is that $S = \\{ 2^k \\}$ for some $k$.", "Solution_2": "the answer's wrong!", "Solution_3": "namely $S=\\{2^k;2^k(2^k-1)\\}$ for $k>0$ ..." } { "Tag": [ "inequalities", "trigonometry", "function", "triangle inequality", "inequalities unsolved" ], "Problem": "let ABC be a triangle,and x,y,z>0 the following inequality is true or not?\r\n$ xtan \\frac A2 + ytan \\frac B2 +ztan \\frac C2 \\geq \\sqrt {xy+yz+zx} $", "Solution_1": "[quote=\"ityaa\"]let ABC be a triangle,and x,y,z>0 the following inequality is true or not?\n$ xtan \\frac A2 + ytan \\frac B2 +ztan \\frac C2 \\geq \\sqrt {xy+yz+zx} $[/quote]\r\n\r\nat least, this is not true if z is very big (compared with x and y), and the triangle ABC is a degenerate pseudo-triangle with angles A = 90\u00b0, B = 90\u00b0, C = 0\u00b0. who knows, maybe it will hold if the angles A, B, C of the triangle ABC and the numbers x, y, z are equally sorted, so, for instance, $A\\geq B\\geq C$ and $x\\geq y\\geq z$. the problem is, i have no time to check right now.\r\n\r\ngreetings from saarbr\u00fccken this time (tomorrow will be the first demo exam)\r\n\r\ndarij", "Solution_2": "thanks,But How about the follow one?\r\n$ (y+z)tan \\frac A2 + (z+x)tan \\frac B2 +(x+y)tan \\frac C2 \\geq \\ 2 \\sqrt {xy+yz+zx} $", "Solution_3": "Your new conjecture is right. Also, the old conjecture is right for the case when the number arrays (x; y; z) and (A; B; C) are equally sorted. I will prove this below.\r\n\r\nAt first, I change the notations:\r\n\r\n[color=blue][b]Problem.[/b] Let $\\alpha$, $\\beta$, $\\gamma$ be the three angles of a triangle, and let x, y, z be three nonnegative real numbers.\n\n[b](a)[/b] Prove the inequality\n\n$\\left(y+z\\right)\\tan\\frac{\\alpha}{2}+\\left(z+x\\right)\\tan\\frac{\\beta}{2}+\\left(x+y\\right)\\tan\\frac{\\gamma}{2}\\geq 2\\sqrt{yz+zx+xy}$.\n\n[b](b)[/b] If, additionally, the number arrays (x; y; z) and $\\left(\\alpha;\\;\\beta;\\;\\gamma\\right)$ are equally sorted, then prove the inequality\n\n$x\\tan\\frac{\\alpha}{2}+y\\tan\\frac{\\beta}{2}+z\\tan\\frac{\\gamma}{2}\\geq\\sqrt{yz+zx+xy}$.[/color]\r\n\r\n[i]Solution.[/i] Let $A=90^{\\circ}-\\frac{\\alpha}{2}$, $B=90^{\\circ}-\\frac{\\beta}{2}$, $C=90^{\\circ}-\\frac{\\gamma}{2}$. Since $\\alpha$, $\\beta$, $\\gamma$ are the three angles of a triangle, we have, for instance $\\alpha<180^{\\circ}$, what yields $A=90^{\\circ}-\\frac{\\alpha}{2}>90^{\\circ}-\\frac{180^{\\circ}}{2}=0^{\\circ}$, and similarly we get B > 0\u00b0 and C > 0\u00b0, and we also get $\\alpha+\\beta+\\gamma=180^{\\circ}$ by the sum of the angles in the triangle, what yields\r\n\r\n$A+B+C=\\left(90^{\\circ}-\\frac{\\alpha}{2}\\right)+\\left(90^{\\circ}-\\frac{\\beta}{2}\\right)+\\left(90^{\\circ}-\\frac{\\gamma}{2}\\right)=270^{\\circ}-\\frac{\\alpha+\\beta+\\gamma}{2}$\r\n$=270^{\\circ}-\\frac{180^{\\circ}}{2}=180^{\\circ}$.\r\n\r\nThus, the angles A, B, C are all positive, and add up to 180\u00b0. Hence, A, B, C are the angles of a triangle. Since $A=90^{\\circ}-\\frac{\\alpha}{2}$, we have $\\cot A=\\tan\\frac{\\alpha}{2}$, and similarly $\\cot B=\\tan\\frac{\\beta}{2}$ and $\\cot C=\\tan\\frac{\\gamma}{2}$. Also, since $A=90^{\\circ}-\\frac{\\alpha}{2}$, $B=90^{\\circ}-\\frac{\\beta}{2}$, $C=90^{\\circ}-\\frac{\\gamma}{2}$, the number arrays $\\left(\\alpha;\\;\\beta;\\;\\gamma\\right)$ and (A; B; C) are oppositely sorted; but since the function f(t) = cot t is monotonically decreasing on the interval ]0\u00b0;\u00a0180\u00b0[, the number arrays (A; B; C) and (cot A; cot B; cot C) are oppositely sorted. Hence, the number arrays $\\left(\\alpha;\\;\\beta;\\;\\gamma\\right)$ and (cot A; cot B; cot C) are equally sorted. Thus, requiring that the number arrays (x; y; z) and $\\left(\\alpha;\\;\\beta;\\;\\gamma\\right)$ are equally sorted is equivalent to requiring that the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted. Hence, the above problem follows from the following theorem:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let A, B, C be the three angles of a triangle ABC, and let x, y, z be three nonnegative real numbers.\n\n[b](a)[/b] We have the inequality\n\n$\\left(y+z\\right)\\cot A+\\left(z+x\\right)\\cot B+\\left(x+y\\right)\\cot C\\geq 2\\sqrt{yz+zx+xy}$.\n\n[b](b)[/b] If, additionally, the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted, then we have the inequality\n\n$x\\cot A+y\\cot B+z\\cot C\\geq\\sqrt{yz+zx+xy}$.[/color]\r\n\r\nAnd the [i]proof of Theorem 1[/i] can be done as follows:\r\n\r\nFirst we will prove Theorem 1 [b](a)[/b]. The inequality in question,\r\n\r\n$\\left(y+z\\right)\\cot A+\\left(z+x\\right)\\cot B+\\left(x+y\\right)\\cot C\\geq 2\\sqrt{yz+zx+xy}$,\r\n\r\nis homogeneous in the three variables x, y, z (in fact, these three variables occur in the power 1 on the left and on the right hand side of this inequality). Thus, we can apply dehomogenization by WLOG assuming that yz + zx + xy = 1.\r\n\r\nSince the numbers x, y, z are nonnegative, there exist angles X, Y, Z in the interval ]0\u00b0; 90\u00b0] satisfying x = cot X, y = cot Y, z = cot Z.\r\n\r\nThe equation yz + zx + xy = 1 is equivalent to zx + xy = 1 - yz, what rewrites as x (y + z) = 1 - yz. Thus, $x=\\frac{1-yz}{y+z}$. Since x = cot X, y = cot Y, z = cot Z, we thus have $\\cot X=\\frac{1-\\cot Y\\cot Z}{\\cot Y+\\cot Z}$. But by the addition formula for cotangents, $\\frac{\\cot Y\\cot Z-1}{\\cot Y+\\cot Z}=\\cot\\left(Y+Z\\right)$. Thus, $\\cot X=\\frac{1-\\cot Y\\cot Z}{\\cot Y+\\cot Z}=-\\frac{\\cot Y\\cot Z-1}{\\cot Y+\\cot Z}=-\\cot\\left(Y+Z\\right)=\\cot\\left(180^{\\circ}-\\left(Y+Z\\right)\\right)$. Since all three angles X, Y, Z lie in the interval ]0\u00b0; 90\u00b0], this yields X = 180\u00b0 - (Y + Z). Thus, X + Y + Z = 180\u00b0. Hence, the angles X, Y, Z lie in the interval ]0\u00b0; 90\u00b0] and add up to 180\u00b0. Thus, these angles X, Y, Z are the angles of some triangle XYZ.\r\n\r\nNow, applying http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3 Theorem [b](c)[/b] to the triangles ABC and XYZ, we get the inequality\r\n\r\n$\\cot A\\cot Y+\\cot A\\cot Z+\\cot B\\cot Z+\\cot B\\cot X+\\cot C\\cot X+\\cot C\\cot Y\\geq 2$.\r\n\r\nSince x = cot X, y = cot Y, z = cot Z, this becomes\r\n\r\n$\\cot A\\cdot y+\\cot A\\cdot z+\\cot B\\cdot z+\\cot B\\cdot x+\\cot C\\cdot x+\\cot C\\cdot y\\geq 2$.\r\n\r\nIn other words,\r\n\r\n$\\left(y+z\\right)\\cot A+\\left(z+x\\right)\\cot B+\\left(x+y\\right)\\cot C\\geq 2$.\r\n\r\nSince yz + zx + xy = 1, this becomes\r\n\r\n$\\left(y+z\\right)\\cot A+\\left(z+x\\right)\\cot B+\\left(x+y\\right)\\cot C\\geq 2\\sqrt{yz+zx+xy}$,\r\n\r\nand Theorem 1 [b](a)[/b] is proven.\r\n\r\nThe proof of Theorem 1 [b](b)[/b] is substantially simpler: Since the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted, the Chebyshev inequality yields\r\n\r\n$\\frac{x\\cot A+y\\cot B+z\\cot C}{3}\\geq\\frac{x+y+z}{3}\\cdot\\frac{\\cot A+\\cot B+\\cot C}{3}$.\r\n\r\nUpon multiplication with 3, this becomes\r\n\r\n$x\\cot A+y\\cot B+z\\cot C\\geq\\left(x+y+z\\right)\\cdot\\frac{\\cot A+\\cot B+\\cot C}{3}$.\r\n\r\nNow, since $\\left(x+y+z\\right)^2-3\\left(yz+zx+xy\\right)=\\frac12\\left(\\left(y-z\\right)^2+\\left(z-x\\right)^2+\\left(x-y\\right)^2\\right)\\geq 0$, we have $\\left(x+y+z\\right)^2\\geq 3\\left(yz+zx+xy\\right)$, so that $x+y+z\\geq\\sqrt{3\\left(yz+zx+xy\\right)}$. Also, by a well-known triangle inequality, $\\cot A+\\cot B+\\cot C\\geq\\sqrt3$. Thus,\r\n\r\n$x\\cot A+y\\cot B+z\\cot C\\geq\\left(x+y+z\\right)\\cdot\\frac{\\cot A+\\cot B+\\cot C}{3}$\r\n$\\geq\\sqrt{3\\left(yz+zx+xy\\right)}\\cdot\\frac{\\sqrt3}{3}=\\sqrt{yz+zx+xy}$,\r\n\r\nand Theorem 1 [b](b)[/b] is proven.\r\n\r\n darij", "Solution_4": "darij grinberg,thank you for your nice solution.\r\nyou are a kind and careful man", "Solution_5": "thanks very much for your solution.nice." } { "Tag": [], "Problem": "Each of the ten volumes of the collected works of Theodore Sturgeon is available in paperback for $ \\$$15 or in hardcover for $ \\$$25. Theresa buys a copy of each of the ten volumes for a total of $ \\$$220. How many hardcover volumes did she buy?", "Solution_1": "Assume she bought all paperback. This would be 15*10=150 dollars. Each of the hardbacks is 10 dollars more expensive. She paid 70 dollars more than 150 dollars. 70/10= $ \\boxed{7}$ hardbacks" } { "Tag": [ "geometry", "rectangle", "parallelogram", "rhombus" ], "Problem": "Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:\r\n$ \\textbf{(A)}\\ \\text{a square} \\qquad\\textbf{(B)}\\ \\text{a rectangle} \\qquad\\textbf{(C)}\\ \\text{a parallelogram with unequal sides} \\\\\r\n\\textbf{(D)}\\ \\text{a rhombus} \\qquad\\textbf{(E)}\\ \\text{a quadrilateral with no special properties}$", "Solution_1": "[hide=\"Click for solution\"]\nThe diagonals of the quadrilateral formed lie along the two perpendicular lines joining the midpoints of the opposite sides of the rectangle. These diagonals are of different lengths, and are perpendicular bisectors of each other. Therefore, the figure is a rhombus, or $ \\boxed{\\textbf{(D)}}$.[/hide]", "Solution_2": "But a kite also has diagonals of different length that are perpendicular" } { "Tag": [], "Problem": "Does anyone else participate in UIL math/number sense?", "Solution_1": "Hmmm...I do", "Solution_2": "Hmmm...I do [size=59]not[/size]", "Solution_3": "Hmmm...I do wish I still did. (Un)fortunately, I don't live in Texas anymore?", "Solution_4": "(Un)fortunately, I have never lived in TX except at a Holiday Inn Hotel...", "Solution_5": "i did district uil, but i didn't do state.", "Solution_6": "dude jason zhang", "Solution_7": "I do.\nAnyone going to state?\nIt'd be cool to see some people who do real math.\nI'm going in Math/Number Sense/Calculator/Science/Computer Science.\nSo PM me if you are going in any of those and perhaps we'll meet up.", "Solution_8": "dude shri g" } { "Tag": [ "\\/closed" ], "Problem": "When I press back I always see this\r\nFatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 40 bytes) in /data/home/mathlinks/www/includes/functions_cache.php on line 339\r\n\r\nwt's wrong?\r\n\r\nbtw, can we discuss apmo questions now?", "Solution_1": "did the apmo 2004 take place yet? \r\n\r\nthat is because sometimes the memory is full. it's a strange phenomena, because I have extended the memory from 8Mb to 16Mb :?", "Solution_2": "We're not supposed to discuss the APMO questions until they appear on the CMS site (which should be in a week's time or so, I suppose). I think it will be safe to discuss APMO questions after Friday or so, because we certainly don't want this year's marks to all be voided like it happened in 2001 and 2002.", "Solution_3": "could you please be more explicit about this rule? what's its purpose? didn't you guys already took the APMO ?! :?", "Solution_4": "Well, the APMO takes place at different times for participants from different countries. Before 2003 it could be done over a week, but due to much question-leaking in 2001 and 2002 the results of the APMO were voided in those 2 years.\r\n\r\nAs a result above all the APMO question sheets there are instructions which specifically say \"Do not reveal the contents of this exam to anyone, especially over the Internet, until the exam is posted on the official APMO website.\" It does take a bit long for the questions to get uploaded though, so the general rule is that you wait 3 or 4 days (and in that time, the questions have already been marked :D ) then it should be safe.", "Solution_5": "well, I do not understand why not all people take apmo in the same time ... is that really that hard? :S I mean I am confused about the \"week\" difference ...", "Solution_6": "Well, there are people in USA and Latin America, as well as Hong Kong, Taiwan, Singapore etc. taking APMO. Clearly it's not possible to have it all at the same time, so the Americas takes it on the afternoon of 15 March and the Western Pacific Rim takes it on the morning of 16th March.\r\n\r\nI guess in the past countries just had more flexibility about which day it would be on.", "Solution_7": "Well, a week is still a bit too much, isn't it?" } { "Tag": [], "Problem": "1) Going across a period on the periodic table, what is the relationship between electron shielding (from the nucleus's pull) and first ionization energy?\r\n\r\n2) What is exactly meant by \"base strength\" and \"acid strength\"? \r\n\r\n3) What structural features must a compound have to be able to undergo autoionization? \r\n\r\n4) If $HClO_{4}$ and $H_{2}NNH_{3}^{+}$ react, what would the products be?", "Solution_1": "2) Isn't that how completely the acid/base dissociates in water? Like it would be a weak acid/base if only a small fraction of it dissociates and strong acid/base if all or most of it dissociates:\r\n[url]http://en.wikipedia.org/wiki/Strong_acid[/url]\r\n[url]http://en.wikipedia.org/wiki/Strong_base[/url]" } { "Tag": [ "LaTeX", "limit" ], "Problem": "let Pk denonate the nth prime \r\nprove that $lim_{\\infty}\\prod\\left(1 - \\frac{1}{p_n}\\right) = 0$\r\n-edit- latex added in", "Solution_1": "Is that equality correct?\r\n\r\nIf it is so, then one of the terms of the product must be zero, and I can't see which one this might be.", "Solution_2": "obviously the series converge to $0$ if the product is to infinity ....", "Solution_3": "True, but I don't think that's what the problem is attempting to ask.", "Solution_4": "I think the questions asks us to prove that \\[\\lim_{\\infty}\\prod\\left(1 - \\frac{1}{p_n}\\right) = 0.\\]", "Solution_5": "[quote=\"shyong\"]obviously the series converge to $0$ if the product is to infinity ....[/quote]\r\nnot necessarily. (1-1/4)(1-1/9)(1-1/16)... converges to 1/2, though the product is to infinity", "Solution_6": "It suffices to prove that $\\prod\\frac{1}{1-p_i}$ goes to $\\infty$. But because $\\frac{1}{1-p}=1+p+p^2+...$ we have that this product like the harmonic series.", "Solution_7": "[quote=\"Magnara\"]It suffices to prove that $\\prod\\frac{1}{1-p_i}$ goes to $\\infty$.[/quote]How do you know this? I would understand if you said $\\prod\\frac{p_k}{p_k-1}$, but then again I have almost no familiarity with infinite products. [quote=\"Magnara\"]But because $\\frac{1}{1-p}=1+p+p^2+...$ we have that this product like the harmonic series.[/quote]I thought that $\\frac{1}{1-x}=1+x+x^2+\\ldots$ was only valid for something like $|x|<1$.", "Solution_8": "[quote=\"AntonioMainenti\"][quote=\"Magnara\"]It suffices to prove that $\\prod\\frac{1}{1-p_i}$ goes to $\\infty$.[/quote]How do you know this? I would understand if you said $\\prod\\frac{p_k}{p_k-1}$, but then again I have almost no familiarity with infinite products. [quote=\"Magnara\"]But because $\\frac{1}{1-p}=1+p+p^2+...$ we have that this product like the harmonic series.[/quote]I thought that $\\frac{1}{1-x}=1+x+x^2+\\ldots$ was only valid for something like $|x|<1$.[/quote]\r\nevery individual p is less than 1, so his proof still holds", "Solution_9": "er...primes are rarely smaller than one; perhaps y'all are talking about this infinite product: $\\prod_{i=1}^{\\infty} \\frac{1}{1-\\frac{1}{p_i}} = \\prod_{i=1}^{\\infty} \\sum_{k=0}^{\\infty} \\frac{1}{p^k} = \\sum_{i=1}^{\\infty} \\frac{1}{i}$, which is indeed the harmonic series, eh?", "Solution_10": "[quote=\"MysticTerminator\"]$\\prod_{i=1}^{\\infty} \\frac{1}{1-\\frac{1}{p_i}} = \\prod_{i=1}^{\\infty} \\sum_{k=0}^{\\infty} \\frac{1}{p_i^k}$.[/quote]How'd you get that?", "Solution_11": "sum of geometric series, right?", "Solution_12": "Yep, that's a well - known equality, it has a name but I can't remember it right now.", "Solution_13": "[quote=\"MysticTerminator\"]sum of geometric series, right?[/quote]Obviously.. :blush: I was thinking it was some kind of sophisticated computation.\r\n\r\nSo basically to show that $\\lim\\prod_i\\left(1-\\frac{1}{p_i}\\right)=0$, you show that $\\lim\\prod_i\\left(1-\\frac{1}{p_i}\\right)^{-1}=\\lim\\prod_i\\frac{1}{1-\\frac{1}{p_i}}=\\lim\\prod_i\\sum_k\\frac{1}{p_i^k}\\ge\\lim\\sum_i\\frac{1}{i}$, which is infinite. I think I'll have remember that one; very nice.", "Solution_14": "This fact is a corrolary of Euler's proof of the divergence of [tex]\\sum_{p \\in \\mathbb{P}}^\\infty \\frac{1}{p} = \\infty[/tex] where [tex]\\mathbb{P}[/tex] is the set of prime numbers.", "Solution_15": "How does one get $\\lim\\prod_i\\sum_k\\frac{1}{p_i^k}\\ge\\lim\\sum_i\\frac{1}{i}$? I had to use holder.", "Solution_16": "[quote=\"Singular\"]How does one get $\\lim\\prod_i\\sum_k\\frac{1}{p_i^k}\\ge\\lim\\sum_i\\frac{1}{i}$? I had to use holder.[/quote]The left hand side is \\[\\left(1+\\frac{1}{2}+\\frac{1}{2^2}+\\ldots\\right)\\left(1+\\frac{1}{3}+\\frac{1}{3^2}+\\ldots\\right)\\left(1+\\frac{1}{5}+\\frac{1}{5^2}+\\ldots\\right)\\cdots\\]When this is expanded (theoretically..) you'll get every $\\frac{1}{n}$ term for $n\\ge1$. At first I thought you would get repeats but I think I was wrong, so just $\\lim\\prod_i\\sum_k\\frac{1}{p_i^k}=\\lim\\sum_i\\frac{1}{i}$.", "Solution_17": "Not getting repeats is exactly the fundamental theorem of arithmetic!", "Solution_18": "[quote=\"blahblahblah\"]Not getting repeats is exactly the fundamental theorem of arithmetic![/quote]That's precisely what made me change my mind. ;) :D" } { "Tag": [ "limit", "trigonometry", "function", "logarithms", "inequalities", "continued fraction", "number theory" ], "Problem": "Hi,\r\nI have a question about a trigonometric limit which is the following one:\r\nDoes the following limit exists \r\n$\\lim_{n\\rightarrow \\infty}n(1-\\sin n)$,\r\nand in case it does is it $\\infty$?\r\nDidilica", "Solution_1": "The function is unbounded but is also zero infinitely often. The limit doesn't exist.", "Solution_2": "(Useful) [b]Theorem:[/b] Let a function $f$ be defined on a neighbourhood $J$ of a point $a,$ except possibly at the point $a$ itself. Then, \\[\\lim_{x \\to a}f(x) = L\\] if and only if \\[\\lim_{n \\to \\infty}f(x_{n}) = L\\] for every sequence $x_{1},\\ x_{2},\\ x_{3}, \\ldots$ of points in $J$ such that $x_{n}\\not= a$ for any positive integer $n$ and $x_{n}\\to a$ as $n \\to \\infty.$\r\n---------------------------------------------\r\nComing to our problem, make the substitution $x = \\frac1{n}.$\r\nSo, $\\lim_{n \\to \\infty}n(1-\\sin n) = \\lim_{x \\to 0}\\frac{(1-\\sin \\frac1{x})}{x}$\r\n\r\nNow, if we pick two sequences $\\{a_{n}\\}$ and $\\{b_{n}\\}$ such that $a_{n}= \\frac1{2\\pi n}$ and $b_{n}= \\frac1{(2n+\\frac12)\\pi},$ then we note $a_{n}\\to 0$ and $b_{n}\\to 0.$ Also, $a_{n},\\ b_{n}\\not= 0$ for any $n.$\r\n\r\nBut, $\\lim_{n \\to \\infty}\\frac{1-\\sin 2\\pi n}{\\frac1{2\\pi n}}= \\infty,$ whereas\r\n\r\n$\\lim_{n \\to \\infty}\\frac{1-\\sin (2n+\\frac12)\\pi }{\\frac1{(2n+\\frac12)\\pi }}= 0.$\r\n\r\nHence, $\\lim_{n\\rightarrow \\infty}n(1-\\sin n)$ doesn't exist.", "Solution_3": "FieryHydra,\r\nWhat you have shown is that $\\lim_{x\\rightarrow 0}\\frac{1-\\sin\\frac{1}{x}}{x}$ does not exist. Next you plugged instead of $x$ the sequence $\\frac{1}{n}$ and have concluded that $\\lim_{n \\rightarrow \\infty}n(1-\\sin(n))$ does not exist.\r\nLet`s apply your argument to the following problem.\r\nIt is known that $\\lim_{x\\rightarrow \\infty}\\sin x$ does not exist. Next by using your ideea we take $x_{n}=2\\pi n$ which is a sequence which converges to ${\\infty}$ and we should conclude that $\\lim_{n\\rightarrow \\infty}\\sin(2\\pi n)=0$ does not exist, which is false.\r\nI do not think that your argument is correct. To prove that the limit does not exits by using the theorem you quoted you have to pick two sequences of natural numbers and show that the function evaluated at these two sequences converge to two different limits....\r\nDidi", "Solution_4": "[quote=\"didilica\"]FieryHydra,\nWhat you have shown is that $\\lim_{x\\rightarrow 0}\\frac{1-\\sin\\frac{1}{x}}{x}$ does not exist. Next you plugged instead of $x$ the sequence $\\frac{1}{n}$ and have concluded that $\\lim_{n \\rightarrow \\infty}n(1-\\sin(n))$ does not exist.\nLet`s apply your argument to the following problem.\nIt is known that $\\lim_{x\\rightarrow \\infty}\\sin x$ does not exist. Next by using your ideea we take $x_{n}=2\\pi n$ which is a sequence which converges to ${\\infty}$ and we should conclude that $\\lim_{n\\rightarrow \\infty}\\sin(2\\pi n)=0$ does not exist, which is false.\nI do not think that your argument is correct. To prove that the limit does not exits by using the theorem you quoted you have to pick two sequences of natural numbers and show that the function evaluated at these two sequences converge to two different limits....\nDidi[/quote]\r\ndidilica, if you read what I wrote earlier carefully, you will see that the theorem assumes that $f$ is defined on some neighbourhood of a point $a.$ In your example, $a = \\infty$ is not a valid point because you cannot construct a neighbourhood around $\\infty.$ We must take care of that first. \r\n\r\nSo, we write $\\lim_{x\\rightarrow \\infty}\\sin x = \\lim_{x \\to 0}\\sin \\frac1{x}.$\r\n\r\nNow, we consider two sequences $a_{n}$ and $b_{n}$ such that $a_{n},\\ b_{n}\\to 0,$ and show that $\\lim_{n \\to \\infty}\\sin \\frac1{a_{n}}\\not = \\lim_{n \\to \\infty}\\sin \\frac1{b_{n}}.$ In this case, we can choose $a_{n}= \\frac1{2n\\pi}$ and $b_{n}= \\frac1{(2n+\\frac12)\\pi}.$\r\n\r\nLastly, I can't see how my argument implied in anyway that for $x_{n}= 2n\\pi,\\ \\lim_{n\\rightarrow \\infty}\\sin(x_{n})$ does not exist.", "Solution_5": "FieryHydra, you make the substitution $x = \\frac1{n}$ and then use the sequence $a_{n}= \\frac1{2 \\pi n}$, whose (multiplicative) inverse is $2 \\pi n$, which is not an integer!\r\nLook one more time at the theorem you stated: It says that $\\lim f(x) = L$ iff $\\lim f(x_{n}) = L$ happens for all sequences, not that $\\lim f(x)$ does not exist iff $\\lim f(x_{n})$ does not exist for all sequences.\r\n\r\n[i]Remark.[/i] The neighbourhoods of $\\infty$ contain intervals of the form $\\left( a, \\infty \\right)$, in which case your theorem does hold :!:\r\n\r\nAre we clear now?\r\n\r\nAs for a solution to the problem, I was thinking of using the fact the $\\left\\{ 1,2,\\ldots \\right\\}$ is dense on the trigonometric circle and some estimates on how big is the $n$ which is very close to $\\frac{\\pi}2$. But I got stuck... I'll let somebody else :)", "Solution_6": "You want to show that there is a constant $c>0$ such that there are infinitely many pairs $m,n$ with $|n-m\\pi|<\\frac{c}{n}$. (then $n|\\sin n|=n|\\sin(n-m\\pi)| \\frac{n}{2}$ for arbitrarily large $n$, and so no limit exists.", "Solution_8": "That's not good enough; you've shown that our sequence doesn't have a finite limit, but we'd rather show that the sequence takes small values infinitely often and doesn't tend to $\\infty$.\r\n\r\nI was misreading the question slightly; my statement works well for showing that $n\\sin n$ takes small values, but we need the altered version below for this sequence:\r\nShow that there are infinitely many pairs $m,n$ and some constant $c$ such that $|n-(2m+\\frac12)\\pi|<\\frac{c}{n}$.\r\n\r\nThen $\\sin n>1-\\frac{c^{2}}{2n^{2}}$ and $n(1-\\sin n)<\\frac{c^{2}}{2n}$; we have a subsequence tending to zero.", "Solution_9": "Oh I see. So we want to show that there must be some M such that there is no N s.t. all n greater than N must have n(1-sin n) > M? So we have to actually show a bounded sequence. hmmm,\r\n\r\nWell, continuing from your arguement, all we really need to show is that we can find $|n-(2m+\\frac12)\\pi|<\\frac{c}{n}$ for infinitely many $n,m$. Why don't we instead find the equivalent: \r\n$|\\frac{2n}{4m+1}-\\pi| < \\frac{2c}{n(4m+1)}$. \r\nIt is well known that if you take rational approximations to $\\pi$ given to us by continued fractions then the sequence of approximations $\\frac{p_{n}}{q_{n}}$ differ by less than $\\frac{1}{q_{n}^{2}}$ from $\\pi$. So if we knew that we could set our fraction $\\frac{2n}{4m+1}$ to be one of these approximations we would have that: \r\n$|\\frac{2n}{4m+1}-\\pi| < \\frac{1}{(4m+1)^{2}}$.\r\n\r\nBut the bound is not quite what we want, we want an $n$ in there, but notice that since this is supposed to be a good approximation to $2n < D(4m+1)$, where I'm too lazy to check what D might be, probably 4 or something. So we have that certainly: \r\n\r\n $|\\frac{2n}{4m+1}-\\pi| < \\frac{1}{(4m+1)^{2}}< \\frac{D}{n(4m+1)}$\r\n\r\nNow the issue is what if $P_{n}$ is not even, and $Q_{n}$ is not 1 mod 4. Well, first of all, we know the two are relatively prime, so we need only worry about what $Q_{n}$. First we make sure that is odd infinitely often. This is true because in fact: $(Q_{n-1},Q_{n}) =1$. So we know that no two consecutive denominators are even. Given any of these odd denominators, what if we get unlucky and we get only things that are 3 mod 4. That's okay, why then just set: $2n = 3P_{n}$ and $4m+1 = 3Q_{n}$. In this case both $n$ and $4m+1$ go up by a factor of three, so just in case we need to do this why don't we just add a factor of 10 into D. So we have our infinite pairs.", "Solution_10": "That's it. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90103]here[/url] for another proof of this result.", "Solution_11": "Hi,\r\nHere is (in my opinion ) a very nice connection:\r\n\r\nDo you remember that the following problem was open:\r\n\r\nStudy the convergency of the following series: $\\sum_{n=1}^{\\infty}\\frac{1}{n}\\left(\\frac{2+\\sin n}{3}\\right)^{n}$.\r\n\r\nClaim that $\\sum_{n=1}^{\\infty}\\frac{1}{n}\\left(\\frac{2+\\sin n}{3}\\right)^{n}=\\infty$.\r\nProof: By way of contradiction we assume that the sum converges. Then, by a well known theorem of Olivier (since the series has positive terms) we get that $\\lim_{n\\rightarrow \\infty}n\\frac{1}{n}\\left(\\frac{2+\\sin n}{3}\\right)^{n}=\\lim \\left(\\frac{2+\\sin n}{3}\\right)^{n}=0$. This implies that $\\lim \\left(\\frac{3}{2+\\sin n}\\right)^{n}=\\infty$, or equivalently\r\n$\\lim n\\ln \\frac{3}{2+\\sin n}=\\infty$.\r\nNow we have, via the inequality $\\ln (1+x)\\leq x$, $\\forall x \\geq 0$ that \r\n$\\lim n \\frac{1-\\sin n}{2+\\sin n}=\\infty$. \r\nOn the other hand since $n \\frac{1-\\sin n}{2+\\sin n}\\leq n \\left(1-\\sin n\\right)$ , we get that $\\lim_{n \\rightarrow \\infty}n\\left(1-\\sin n\\right)=\\infty$, which is a contradiction. \r\nThus the series diverges. \r\nWhat do you think?\r\nI think that the same statement holds for the following series\r\n$\\sum_{n=1}^{\\infty}\\frac{1}{n}\\left(\\frac{2+\\sin(\\beta n)}{3}\\right)^{n}=\\infty$, where $beta$ is any irrational number.", "Solution_12": "That's not a proof of anything; the \"theorem of Olivier\" you quote is not true without additional hypotheses. Some terms of the series are very close to $\\frac1n$, but most are far smaller.\r\n\r\nSee this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]old thread[/url] for a correct proof that the series converges.", "Solution_13": "you are right, the terms should be nonincreasing to apply that theorem, I missed that.", "Solution_14": "this question and other similar questions can easily be solved if use the fact that sinn is dense in [-1,1]", "Solution_15": "[quote=\"zdyzhj\"]this question and other similar questions can easily be solved if use the fact that sinn is dense in [-1,1][/quote]\r\nIt's not that easy. \r\n\r\nConsider this example: define the sequence $ x_n$ defined by\r\n\r\n$ \\frac12,\\frac34,\\frac14,\\frac78,\\frac58,\\frac38,\\frac18 ,\\frac{15}{16},\\dots$\r\n\r\nWe do have that $ x_n$ is dense in $ [0,1].$ Does it follow that $ \\liminf nx_n\\equal{}0?$\r\n\r\nWell, no. The best we can do is the following subsequence:\r\n\r\n$ 1\\cdot x_1\\equal{}\\frac12.$\r\n\r\n$ 3\\cdot x_3\\equal{}3\\cdot 14\\equal{}\\frac34.$\r\n\r\n$ 7\\cdot x_7\\equal{}7\\cdot18\\equal{}\\frac78.$\r\n\r\nIn general, $ (2^k\\minus{}1)x_{(2^k\\minus{}1)}\\equal{}\\frac{2^k\\minus{}1}{2^k}$ and this doesn't tend to zero. In fact, $ \\liminf nx_n\\equal{}1.$" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Show that the automorphisms of the region complex plane minus 2 points are all mobius transformation.", "Solution_1": "I'm looking forward to see the answer to this.", "Solution_2": "This is not true. Say, if $R$ is a simply connected proper subset of $\\mathbb C$, then there exist exactly two conformal automorphisms of $R \\setminus \\{x, y\\}$ (where $x$ and $y$ are two distinct points in $R$). One is the identity map. The other is the automorphism of $R$ that exchanges $x$ and $y$. Clearly, the latter map doesn't have to be a M\u00f6bius transformation.\r\n\r\nUpdate: as I understand the question, it asks about automorphisms of the set that equals a region (of the complex plane) with two points removed", "Solution_3": "Yeah, but now we are talking about the whole plane. \r\n\r\nFirst of all note that our automorphism cannot have essential singularities at infinity or any of the two removed points, since that would contradict bijectivity. So, removing 3 singular parts we get a costant function by Liouville's theorem, and hence the automorphism is a rational function. Now just check which ones work." } { "Tag": [], "Problem": "What is the 13th number in the arithmetic sequence:\n$ 7,10,13,\\ldots$ ?", "Solution_1": "$ a_n \\equal{} 7\\plus{}(n\\minus{}1)3 \\equal{} 3n\\plus{}4$.\r\nThus, $ a_{13} \\equal{} 3 \\cdot 13 \\plus{} 4 \\equal{} \\boxed{43}$." } { "Tag": [ "Stanford", "college", "calculus", "integration", "function", "real analysis", "linear algebra" ], "Problem": "So, I look at the sample demos of the Stanford EPGY online math courses. And the real analysis course. And behold - the explanations that you have to udnderstand look the same as the textbook of a standard calculus course. Besides, most college math courses take several quarters to complete, whereas Stanford EPGY only require one course for each (abstract alg, real analysis, etc..). So are the stanford EPGY math courses just brief versions of what you would get in a college math course? And would one of them ever count for credit for a state university one year sequence of say, abstract algebra or real analysis?\r\n\r\nhttp://epgy.stanford.edu/courses/ma...115lecture.html\r\n\r\n...Then again, explanations are supposed to simplify what the textbook is saying. So they tend to be more informal. Still, why not let the textbook be more informal itself? \r\n\r\n...\r\n\r\nAnd why in the heck do they split up multivariable calculus into integral and differential???[/code]", "Solution_1": "That is pretty standard for a basic course in real analysis. A lot of universities have a first real analysis course similar to that one, whereas a second real analysis course would introduce metric spaces, equicontinuity, maybe fourier series, etc. It really varies by university, but I imagine that most institutions would give you a quarter's worth of credit for the material that that course claims to cover.", "Solution_2": "So here is my university's 3 quarter real ana sequence:\r\n\r\nMATH 424 Fundamental Concepts of Analysis (3) NW \r\nThe real number system; field, order, and LUB axioms. Metric spaces: Euclidean space. Bolzano-Weierstrass property. Sequences and limits of Sequences. Cauchy sequences and completeness. The Heine-Borel Theorem. Uniform continuity. Connected sets and the intermediate value theorem. Prerequisite: either 2.0 in MATH 328 or 2.0 in MATH 335. Offered: A. \r\n\r\nMATH 425 Fundamental Concepts of Analysis (3) NW \r\nOne-variable differential calculus: chain rule, inverse function theorem, Rolle's theorem, intermediate value theorem, Taylor's theorem, and intermediate value theorem for derivatives. Multivariable differential calculus: mean value theorem, inverse and implicit function theorems, and Lagrange multipliers. Prerequisite: either 2.0 in MATH 326 or 2.0 in MATH 335; 2.0 in MATH 424. Offered: W. \r\n\r\nMATH 426 Fundamental Concepts of Analysis (3) NW \r\nLebedgue measure on the reals. Construction of the Lebesgue integral and its basic properties. Monotone Convergence Theorem, Fatou's Lemma, and Dominated Convergence Theorem. Integration of series. Continuity and differentiability theorems for functions defined by integrals. Introduction to general measures and integration. Prerequisite: 2.0 in MATH 425. Offered: Sp.\r\n\r\n\r\nSo... It would actually seem to cover some of all three. But as for rigor - that depends on professor if we're comparing it to uni. And checking the syllabus on http://epgy.stanford.edu/courses/math/M115/M115summary.pdf, they definitely don't cover what a lot of the UWashington course covers - they mostly cover what we already heard about in freshman calculus, only more rigorously. But nothing that we don't hear about in freshman calculus, such as \"Heine-Borel Theorem\", which is in the Uwashington course.\r\n\r\nIt's REALLY strange - they give MORE credit for linear algebra and JUST multivariable differential calculus than REAL ANALYSIS. Even though real analysis is supposed to have more.", "Solution_3": "i took a multivariable calculus course on epgy. it was very thorough. they don't cut any corners.", "Solution_4": "Many places have less real analysis than that three-quarter sequence standard. I went to UC Berkeley for my undergraduate degree; the requirement there was a one-semester course, which looks comparable to UW 424 plus the Riemann integral.\r\nThe Lebesgue material in 426 is usually reserved for graduate courses, and most of 425 was covered in an optional second analysis course.\r\n\r\nAlso, be wary of the difference between semesters and quarters. Most of the EPGY courses look like semester courses, which are normally offered as one course for the subject. The real analysis class you linked to has about 40 lectures- that's standard for a semester class meeting 3 days a week. This model changes somewhat at the graduate level- courses like algebra and real analysis are usually 1-year sequences anywhere.\r\nAt a quarter school, expect 1 quarter of credit and some extra material. The EPGY course definitely contains material equivalent to UW 424, and very little from the other two quarters. At a semester school, you'll get a semester of credit if the topics line up- but you probably won't get useful credit if the course you took is missing a major topic from the course your college requires.\r\n\r\nOh- I'm a grad student and TA in math here at UW. Nice to meet a fellow student." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "$ a_1\\equal{}1$, $ a_2\\equal{}12$, $ a_3\\equal{}20$\r\n$ a_{n\\plus{}3}\\equal{}2a_{n\\plus{}2}\\plus{}2a_{n\\plus{}1}\\minus{}a_n$\r\n\r\nprove that for every natural $ n$, $ 1\\plus{}4a_{n\\plus{}1}a_{n}$ is perfect square.", "Solution_1": "$ a_n\\equal{}\\minus{}15\\plus{}\\frac{\\minus{}15\\plus{}19\\sqrt 5 }{10}(\\frac{1\\plus{}\\sqrt 5}{2})^{2n\\minus{}1}\\plus{}\\frac{\\minus{}15\\minus{}19\\sqrt 5}{10}(\\frac{1\\minus{}\\sqrt 5}{2})^{2n\\minus{}1}$.\r\n$ 1\\plus{}4a_na_{n\\plus{}1}\\equal{}b_n^2$, were\r\n$ b_n\\equal{}\\minus{}15\\plus{}(11\\plus{}\\frac{13}{\\sqrt 5})(\\frac{1\\plus{}\\sqrt 5}{2})^{2n\\minus{}2}\\plus{}(11\\minus{}\\frac{13}{\\sqrt 5})(\\frac{1\\minus{}\\sqrt 5}{2})^{2n\\minus{}2}.$" } { "Tag": [ "inequalities" ], "Problem": "What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?", "Solution_1": "[quote=\"Humpty_Dumpty\"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]\r\n\r\nIf you are going to buy it, why not form your own opinion after purchase?\r\nAnyway, it is an all-time classic.", "Solution_2": "[quote=\"Humpty_Dumpty\"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]\r\n\r\nI don`t know about this book, but I know that there it is an inequality of Polya Szego:D", "Solution_3": "[quote=\"Zamfirmihai\"]I don`t know about this book, but I know that there it is an inequality of Polya Szego:D[/quote]\r\n\r\nThey also wrote a book on inequalities in mathematical physics.", "Solution_4": "[quote=\"Humpty_Dumpty\"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]\r\n\r\nThese books are wonderfull !\r\n\r\n[color=blue][b]\n Georg P\\'olya and Gabor Szeg\"o , [/b][/color]\r\n[b] Aufgaben und Lehrs\"atze aus der Analysis (I)-(II) ,[/b] Springer-Verlag , 1970,1971.\r\nISBN 3-540-054656-1\r\nISBN 0-387-05456-1\r\nHeidelberger Taschenbuecher Band 73,Band 74 ,Vierte Auflage.\r\nEditions 1,2,3 are published in Springer Verlag [vol.20 in series \"Grundlehren der Mathematischen Wissenschaften\"]\r\nFirst edition=1925\r\n\r\n\r\nThese nice books were also translated in English, see:\r\n[b]\nG. Polya and G. Szego, \"Problems and Theorems in Analysis\", volume I,Springer Verlag 1972.\nVol I, ISBN 0-387-90224-4 or 3-540-90224-4\n\nG.Poya and G. Szego,\"Problems and Theorems in Analysis,\" Volume II,Springer-Verlag, New York, 1976. \nVol II, ISBN 0-387-90291-0 or 3-540-90291-0 \n[2 Volumes in Springer Study Edition, (c) 1972, 1976]\n[/b]" } { "Tag": [ "algebra", "polynomial", "geometry", "3D geometry", "algebra unsolved" ], "Problem": "Let $a,b,c\\in\\mathbb Q$ and suppose $f(x) =x^{3}+ax^{2}+bx+c$ has three real but irrational roots. \r\nProve or disprove : the roots of $f$ cannot be written in [i]closed form[/i] (i.e. using only rational numbers and \u2013 maybe nested \u2013 $\\sqrt$ & $\\sqrt[3]{}$ )", "Solution_1": "There [b]is[/b] a closed form: Cardano's formula.", "Solution_2": ":D I think Cardano gives you only a closed form if two roots are complex. As far as I know, if all roots are real, you need to divide a certain angle by 3... it gives a [i]method [/i]but for that case no closed formula. :maybe:", "Solution_3": "Any ideas?", "Solution_4": "The roots ARE in closed form with Cardano's formula, but the things being cube-rooted are not necessarily real. In the end, however, if all 3 roots are real, two things are added together that are conjugates to give the second and third roots, so they are real.\n\nIn short the roots are in closed form, but the closed form involves doing arithmetic with (possibly) non-real numbers.", "Solution_5": "Yes, I should have been more precise. Of course I mean [b]a closed form involving only real numbers[/b]. It is not about giving a formula (because there is nothing other than Cardano's formula), but the following:\n\n[i]is there a polynomial $x^3+ax^2+bx+c$ with its three roots in $\\mathbb{R}\\backslash\\mathbb{Q}$ where the roots can be written with radicals and everything happens inside $\\mathbb{R}$?[/i] \nNote that $ a,b,c\\in\\mathbb{Q} $ is essential, otherwise it is trivial to find one, e.g. $(x^2-x-1)(x-\\sqrt{5})$." } { "Tag": [ "geometry", "search", "MATHCOUNTS" ], "Problem": "I need help with 2 of last year's national team problems (#9 & 10 to be specific)! All help is much appreciated, but what I really need is an explanation... :)\r\n\r\n1. Five different rockets, A, B, C, D and E, are to be launched from two separate launch pads labeled 1 and 2. Each pad can accomodate only one rocket at a time. The rockets can launch from either pad, in any order and at any time (sequentially or simultaneously). One example of a launch pattern is (C1, A1d2, E1B2) where two commas separate three different launching times. Including the example given, what is the total number of different possible launch patterns?\r\n\r\n2. In the figure, what is the area of triangle ABD? Express your answer as a common fraction.\r\n[geogebra]ef20c87a46a183af117bb81f326761c59ca65b2f[/geogebra]\r\n\r\nThanks! :)", "Solution_1": "see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=606166542&t=268847]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=606166542&t=268977]this[/url].\r\n\r\nnumber 9 can be confusing, if you still don't get it, i'll do it using the method that i used when solving this problem.", "Solution_2": ":idea: Those are very helpful links! (But I still don't under stand the recursion method on number nine)", "Solution_3": "Hint: If you don't wish to use recursion for #9, use simple casework. Divide up the cases where there are certain numbers of pairs of rockets launched simultaneously, and work from there.", "Solution_4": "Does anyone know where you can get national mathcounts problems?", "Solution_5": "okay, here's my solution:\r\n\r\n[hide=\"no recursion.\"]\n\n[b]CASE ONE: all on different launch pads:[/b]\n$ 5!\\cdot2^5\\equal{}32\\cdot120\\equal{}3840$\n\n[b]CASE TWO: one set of two, rest on their own[/b]\nfour sets of two consecutive, so $ \\binom52\\cdot3\\cdot2\\cdot4\\cdot2^4\\equal{}3840$\n\n[b]CASE THREE: two sets of two[/b]\nthree ways to do that, so $ 5\\cdot\\binom42\\cdot2^3\\cdot3\\equal{}720$\n\n$ 3840\\cdot2\\plus{}720\\equal{}\\boxed{8400}$\n\nif you don't understand where any of those numbers are coming from, feel free to ask. \n[/hide]" } { "Tag": [ "linear algebra", "matrix", "LaTeX" ], "Problem": "In my Algebra 2 textbook, it wasn't clear just WHY you have to multiply two matrices the way ou do, like, say: a 3x2 matrix by a 2x2 one. Is there a simple way to remember this? \r\n\r\nI don't know how to Latex them. :blush:", "Solution_1": "For typing matrices in Latex go here:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=151274", "Solution_2": "well, if i understand what u mean, if u have a 3x[b]2[/b] matrix and u wanna multiply it by a [b]2[/b]x2 , the reason u can do this is because the two numbers in bold are the same otherwise u can not multiply the matrices. for example u can not multiply 2x2 by 3x2. in other words the number of columns of the left matrix has to be equal to the number of rows of the right matrix. remember matrix multiplication is not commutative! and ur resulting matrix of the afore-mentioned example will have dimensions 3x2 because u take the rows of the first matrix and the columns of the second. \r\ni hope this helps!" } { "Tag": [ "inequalities" ], "Problem": "If $ a,b,c$ are the sidelenghts of a triangle prove that :\r\n\r\n$ 9a^2\\plus{}9b^2\\plus{}17c^2\\minus{}14ac\\minus{}14bc\\minus{}6ab>0$", "Solution_1": "No one ? :(", "Solution_2": "[quote=\"alex2008\"]If $ a,b,c$ are the sidelenghts of a triangle prove that :\n\n$ 9a^2 \\plus{} 9b^2 \\plus{} 17c^2 \\minus{} 14ac \\minus{} 14bc \\minus{} 6ab > 0$[/quote]\r\nIt very weak,not use $ a,b,c$ be the sidelengths of a triangle :wink: ,we can easyli prove that:\r\n$ 9a^2\\plus{}9b^2\\plus{}17c^2 \\ge 6ab\\plus{}\\sqrt{204}c(a\\plus{}b)$" } { "Tag": [ "vector", "linear algebra" ], "Problem": "Let $ F: W \\to V$ be a linear map. Show that $ F$ is surjective iff transforms a spanning set of $ W$ in a spanning set of $ V$.", "Solution_1": "Suppose $ F$ transforms spanning set of $ W,: k_{1}, k_{2} ... k_{t}$ into a spanning set of $ V, : l_{1}, l_{2} ... l_{t}$ - i.e. $ F(k_{i}) \\equal{} l_{i}$. For any element $ x$ of $ V$, we can choose reals $ a_{i}$ so that $ x \\equal{} \\sum_{1 \\leq i \\leq t} a_{i} * l_{i}$, since $ l_{1} ... l_{t}$ spans $ W$. Then we have $ x \\equal{} F ( \\sum_{1 \\leq i \\leq t} a_{i} * k_{i})$, so for any $ x$ in $ V$ we can find $ y$ in $ W$ with $ F(y) \\equal{} x$, so $ F$ is surjective as required. \r\nFor the other direction, suppose $ F$ transforms spanning set of $ W: k_{1} ... k_{t}$ into some list in $ V : l_{1} ... l_{t}$. Suffices to prove $ l_{i}$ is a spanning set. For any $ x$ in $ W$, we can find $ a_{1}, a_{2} ... a_{t}$ so that $ x \\equal{} \\sum_{1 \\leq i \\leq t} a_{i} * k_{i}$, so $ F(x) \\equal{} \\sum_{1 \\leq i \\leq t} a_{i} * l_{i}$. Thus the range of $ F$ is equal to vectors of the form $ \\sum_{1 \\leq i \\leq t} a_{i} * l_{i}$, which is just $ span (l_{1} ... l_{t})$. But since $ F$ is surjective, the range of $ F$ is $ V$, so thus $ span (l_{1} ... l_{t}) \\equal{} V$, and $ (l_{1} ... l_{t})$, is a spanning set, as required." } { "Tag": [], "Problem": "Okay, I'm going to try to actually run a signups this time. Everyone's starting HP may depend on the number of people who sign up.\r\n\r\n\r\nHurt and Heal. Each post you hurt someone and you heal someone (this decreases and increases their health by 1 point, respectively). When someone reaches 0 HP they die.\r\n\r\nAlso, I will be here as the ever-loving god to keep rule-breakers in control.\r\n\r\nYou cannot heal yourself.\r\nYou cannot hurt yourself.\r\nYou cannot double-post in this thread.\r\nOnce you hurt and heal, you must recharge (wait) for [b]1 hour[/b] (at least) before hurting & healing again.", "Solution_1": "I'm signing up...", "Solution_2": "sign me up.", "Solution_3": "I'm in for this!", "Solution_4": "I'll sign up, too.", "Solution_5": "I'm in[i][/i]", "Solution_6": "I'll play.", "Solution_7": "Nice, I'll get it started when ~10 people join.", "Solution_8": "/in", "Solution_9": "Amendments to rules:\r\n\r\nAlliances: This rule is going to be impossible to enforce, so I'm not going to bother with it.\r\n\r\nRecharge time: 30 minutes is too short, I have made it 1 hour instead.\r\n\r\nAlso, no more random events. They will actually be very constructed to keep the game from lasting forever (~200 posts, everyone loses 1)", "Solution_10": "im in\r\nfo sho", "Solution_11": "i'm in", "Solution_12": "Oi, that's 10. Game shall commence.\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=770965#770965\r\n\r\nIf people continue signing up another one might commence as well.", "Solution_13": "I'll sign up :D", "Solution_14": "I'm de-signing up.", "Solution_15": "yeah im signing up too", "Solution_16": "sign me in, set me free, insert metalcore lyrics.", "Solution_17": "ill sign up", "Solution_18": "I'll play if everyone doesn't gang up on me.", "Solution_19": "That might be tough.\r\n\r\nI'll do it.", "Solution_20": "I'll sign up", "Solution_21": "That is 7 ... should I wait for more people or randomly add people in?\r\n\r\nHint: Threats of endless poking from the afterlife should stop people from hurting you at least in the beginning.", "Solution_22": "I'll join.", "Solution_23": "I would liek to join", "Solution_24": "http://www.artofproblemsolving.com/Forum/viewtopic.php?p=786695\r\n\r\ngame #2 has started with the next batch of signups.\r\n\r\ngood luck everyone" } { "Tag": [ "trigonometry" ], "Problem": "$ \\tan^2{2x}.\\tan^2{3x}.\\tan{5x} \\equal{} \\tan^2{2x} \\minus{} \\tan^2{3x} \\minus{} \\tan{5x}$", "Solution_1": "What's with the $ .$? WHat do those mean? And it looks slightly nicer if you use \\tan{x} as opposed to tanx.\r\n\r\n$ \\tan{x}$\r\n$ tanx$", "Solution_2": "[quote=\"thanhnam2902\"]$ \\tan^2 2x \\cdot \\tan^23x\\cdot \\tan 5x \\equal{} \\tan^22x \\minus{} \\tan^23x \\minus{} \\tan 5x$\n\n[/quote]\r\n\r\nThis is what OP meant (dots were there for multiplying).", "Solution_3": "Ok thanks! :) Now to solve the problem.", "Solution_4": "[hide=\"Hint\"]Let $ a \\equal{} \\tan ^2 2x$, $ b \\equal{} \\minus{} \\tan ^2 3x$, and $ c \\equal{} \\minus{} \\tan 5x$, so $ abc \\equal{} a \\plus{} b \\plus{} c$. Solve for $ c$.[/hide]\n[hide=\"More\"]$ c \\equal{} \\minus{} \\frac{a\\plus{}b}{1\\minus{}ab} \\equal{} \\minus{} \\frac{\\tan ^2 2x \\minus{} \\tan ^2 3x}{1 \\minus{} \\tan ^2 2x \\tan ^2 3x} \\equal{} \\minus{}\\frac{\\tan 2x \\plus{} \\tan 3x}{1 \\minus{} \\tan 2x \\tan 3x} \\cdot \\frac{\\tan 2x \\minus{} \\tan 3x}{1 \\plus{} \\tan 2x \\tan 3x}$.[/hide]", "Solution_5": "How do you go on from there though?", "Solution_6": "Use the sum of tangents formula..." } { "Tag": [ "geometry", "email", "MATHCOUNTS", "\\/closed" ], "Problem": "I heard somewhere that AoPS is planning on releasing an Into to Geometry book? If so, does anyone know when it will be available for purchase?", "Solution_1": "Probably in March or April.", "Solution_2": "If you join the email list [url=http://www.artofproblemsolving.com/StayInformed/AoPS_Stay_Informed.php]here[/url], you'll get an email shortly after it is available.", "Solution_3": "Is there any chance of an Intermediate Geometry book? I don't know if I would want an intro book because I've never had trouble with amc and mathcounts geometry, but it would be great to learn some of the AIME/olympiad level geometry.", "Solution_4": "[quote=\"xxazurewrathxx\"]Is there any chance of an Intermediate Geometry book? I don't know if I would want an intro book because I've never had trouble with amc and mathcounts geometry, but it would be great to learn some of the AIME/olympiad level geometry.[/quote]\r\nGet the \"Geometry Revisited\" by Coxeter and Greitzer.", "Solution_5": "[quote=\"xxazurewrathxx\"]Is there any chance of an Intermediate Geometry book? I don't know if I would want an intro book because I've never had trouble with amc and mathcounts geometry, but it would be great to learn some of the AIME/olympiad level geometry.[/quote]\r\n\r\nThat's a year or two away, unfortunately. The problems in Challenging Problems in Geometry are good training at that level." } { "Tag": [ "inequalities", "floor function", "function", "number theory proposed", "number theory" ], "Problem": "Let $ n$ be a positive integer, $ q_n$ the product of the primes less than or equal to $ n$. \r\n\r\n[b]a[/b] Prove that \r\n\r\n\\[ \\sum_{d|q_n}{2\\mu(n)\\left\\lfloor\\frac{n}{d}\\right\\rfloor}\\le\\sum_{d|q_n}{\\mu(n)\\left\\lfloor\\frac{2n}{d}\\right\\rfloor}\\]\r\n\r\n[b]b[/b] For fixed $ n$ determine the greatest integer $ k$ that can substitue number $ 2$ in both hands of the above expression. \r\n\r\n :)", "Solution_1": "[quote=\"ElChapin\"]Let $ n$ be a positive integer, $ q_n$ the product of the primes less than or equal to $ n$. \n\n[b]a[/b] Prove that\n\\[ \\sum_{d|q_n}{2\\mu(n)\\left\\lfloor\\frac {n}{d}\\right\\rfloor}\\le\\sum_{d|q_n}{\\mu(n)\\left\\lfloor\\frac {2n}{d}\\right\\rfloor}\\]\n[b]b[/b] For fixed $ n$ determine the greatest integer $ k$ that can substitue number $ 2$ in both hands of the above expression. \n\n :)[/quote]\r\nWhat is $ \\mu(n)$?", "Solution_2": "$ \\mu$ is the M\u00f4bius function http://en.wikipedia.org/wiki/M%C3%B6bius_function. \r\n\r\nAlso, the sums consider positive divisors only, as it is usual whit that notation. \r\n\r\n:)" } { "Tag": [ "function", "Alcumus", "group theory" ], "Problem": "How many permutations of $ \\{1,2,3,4,5,6,7,8\\}$ have multiple cycles?\n\nNote: Such a permutation acts as a function, taking $ k$ to $ \\sigma(k)$ ($ k\\equal{}1,2,\\ldots,8$), where \\[ \\{\\sigma(1),\\sigma(2),\\ldots,\\sigma(8)\\}\\equal{}\\{1,2,\\ldots,8\\}.\\] A cycle is a list of elements $ a_1,a_2,\\ldots,a_k$ where $ \\sigma(a_1)\\equal{}a_2$, $ \\sigma(a_2)\\equal{}a_3$, $ \\ldots$, $ \\sigma(a_k)\\equal{}a_1$.", "Solution_1": "Seems advanced, but the answer might be $ 8!$ divided by $ 8$ for the cycles, which is $ 5040$.", "Solution_2": "I think it is $ 8!\\minus{}7!\\equal{}\\boxed{35280}$.\r\n\r\nI sort of guessed, and I never really understood what cycles were...", "Solution_3": "I don't really understand this problem (so I just gave up :( ), but when is it possible to have only one cycle, and when is it possible to have multiple cycles? Also what exactly is a cycle (I think it is like for example when we have $ (1,2,3,4)$, a cycle of this would be $ (2,3,4,1)$, when is the cycle unique?)?", "Solution_4": "I seriously think that the people at AoPS should take this problem down because it is tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo difficult for those who haven't had any experience in group theory (I know this because I researched cycles a billion times). And cycles are made of permutations like this: (1, 2, 3, 6, 5, 4, 8, 7) and then this permutation can be broken into many different cycles I believe. i was an inch away from getting the problem correct. I wasn't sure how to rid of permutations that had only 1 cycle.", "Solution_5": "A cycle is a property of a permutation. When you permute something, you are rearranging the objects. Cycles show how the objects are rearranged.\nFor example, consider permuting the set $\\{1,2,3,4\\}$ into $\\{3,1,4,2\\}$. What you are doing is putting 1 into the second spot, putting 2 into the fourth spot, putting 4 into the third spot, and putting 3 back into the first spot. If you were to draw this out, making and arrow from 1 to 2 and from 2 to 4 and from 4 to 3 and from 3 to 1, you would get one \"loop\" with one beginning and one end. Every \"loop\" is a cycle so this example had one cycle.\nSome permutations are made with more than one \"loop\". If you permute $\\{1,2,3,4\\}$ into $\\{3,4,1,2\\}$, drawing it out, you have an arrow from 1 to 3 and from 3 to 1, and another arrow from 2 to 4 and from 4 to 2. That has two beginnings and ends or two \"loops\" so it has 2 cycles. Also, if a number stays in the same place, it is a separate cycle. So changing $\\{1,2,3,4\\}$ into $\\{1,2,4,3\\}$ takes 3 cycles.\nApplying it to the problem, there are $8!$ total ways to permute the set. We need to find the number of ways to make a single cycle. Define \"1\" as an object we are permuting and define 1 as the first spot. \"1\" has $7$ places it can go to. It can't stay at the same spot since staying in the same spot creates another cycle. Without loss of generality, let \"1\" land on \"3\". \"3\" now has $6$ options, it can to to every place but 1 and where \"3\" is currently at. By the same logic, whatever \"3\" lands on has $5$ options and so on. That makes $7!$ permutations with only one cycle. $8!-7!=\\boxed{35280}$", "Solution_6": "By the way, I think this should be for Holy Grail even thought it's not.", "Solution_7": "I am surprised how simple the alcumus solution was!", "Solution_8": "The hardest part is figuring out what a cycle is..." } { "Tag": [ "trigonometry", "calculus", "integration" ], "Problem": "edit . . . thanks PiDeltaPhi", "Solution_1": "Why did you post in the Solved Problems section.\r\n\r\nAnyways, the answer is 1, but I think it is a bit simple even for beginner's section. :D", "Solution_2": "[quote=\"Fierytycoon\"]Why did you post in the Solved Problems section.\n\nAnyways, the answer is 1, but I think it is a bit simple even for beginner's section. :D[/quote]\r\ni agree..\r\nit's a bit simple\r\nbut u hv made a mistake\r\nit should be undefined....\r\nsin pi =0 !", "Solution_3": "[quote]but u hv made a mistake \nit should be undefined.... \nsin pi =0 ![/quote]\r\n\r\nhaha, yes, you are right. I should double-check my answer next time before saying its too simple. :D", "Solution_4": "90% Kaplan's fault, 10% my fault", "Solution_5": "[quote]No mistake. The answer is 1. \n[/quote]\n\n\nYou can't divide by 0...\n\n[quote]I thought the solved problems section was for \"already solved problems\" . . . but now I guess it is for problems solved in the other sections? The other moderators can move this if they want.[/quote]\r\n\r\nThe solved problems section is for problems that are posted in the unsolved section which have then been solved by other forum posters. Usually moderators move threads to the solved problems sections. This is my understanding, at least.", "Solution_6": "90% Kaplan's fault, 10% my fault", "Solution_7": "[quote]First of all, multiply both sides by sin(pi) to get rid of the fraction in the denominator. [/quote]\r\n\r\nYou can't do that, since your original expression is blah/sin(pi)=blah/0, which is clearly undefined right from the start.", "Solution_8": "Integral0:\r\n\r\nFind the mistake:\r\n\r\n\r\nStep 1: a and b > 0\r\nStep 2: a = b\r\nStep 3: a^2 = a^b\r\nStep 4: a^2 - b^2 = ab - b^2\r\nStep 5: (a + b)(a - b) = b(a - b)\r\nStep 6: (a + b) = b\r\nStep 7: b + b = b\r\nStep 8: 2b = b\r\nStep 9: 2 = 1\r\n\r\n\r\nYou're doing the same mistake :D" } { "Tag": [ "trigonometry", "geometry", "LaTeX" ], "Problem": "I don't see how to derive $\\sin\\angle{B}=\\frac{b}{2R}$ and $\\sin\\angle{C}=\\frac{c}{2R}$ from this diagram (page 3 of Geometry Revisited). :huh:", "Solution_1": "Well,\r\n$\\frac{a}{2R}= \\sin{J}= \\sin{\\frac{\\widehat{BAC}}{2}}= \\sin{\\frac{360^\\circ-\\widehat{BJC}}{2}}= \\sin{(180^\\circ-A)}= \\sin{A}$.\r\n\r\nWe can similarly derive $\\sin{B}= \\frac{b}{2R}$ and $\\sin{C}= \\frac{c}{2R}$ in almost exactly the same way, but it's not necessary to show this step in a proof because it follows by symmetry. :)\r\n\r\nEDIT: By the way, if you want to see how someone typed something up in LaTeX, you can just hover your mouse cursor over it and you'll see it, or you can quote the post and take a look at the LaTeX. :)", "Solution_2": "by the way, how did you write an unitalicized \"sin\" in math mode? :maybe:", "Solution_3": "[code]sin x[/code] produces $sin x$\n\n[code]\\sin x[/code] produces $\\sin x$" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "It takes 1 quart of paint to paint the six faces of one unit cube. How many quarts of paint are needed to paint the exterior faces of this structure (including the bottom) made from 10 unit cubes?\n\n[asy]size(101);\nimport three;\ncurrentprojection=orthographic(3/8,-1,1/4);\ndraw((0,0,0)--(5,0,0)); draw((0,0,1)--(5,0,1)); draw((1,0,2)--(5,0,2)); draw((2,0,3)--(3,0,3));\ndraw((0,0,0)--(0,0,1)); draw((1,0,0)--(1,0,2)); draw((2,0,0)--(2,0,3)); draw((3,0,0)--(3,0,3)); draw((4,0,0)--(4,0,2)); draw((5,0,0)--(5,0,2));\ndraw((0,0,1)--(0,1,1)); draw((0,1,1)--(.63,1,1)); draw((1,0,2)--(1,1,2)); draw((1,1,2)--(1.63,1,2)); draw((2,0,3)--(2,1,3)--(3,1,3)--(3,0,3));\ndraw((3,1,3)--(3,1,2)--(3,0,2)); draw((3,1,2)--(4,1,2)--(4,0,2)); draw((4,1,2)--(5,1,2)--(5,0,2));\ndraw((5,1,2)--(5,1,0)--(5,0,0)); draw((5,1,1)--(5,0,1));[/asy]", "Solution_1": "Okay, the SA of this figure is $ 10(2) \\plus{} 5(2) \\plus{} 3(2)$ or $ 36$. Since it takes $ 1$ quart to paint $ 6$ square units, it takes $ \\boxed{6}$ quarts to paint $ 36$ square units." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "let $ A \\equal{} \\{a_1,a_2,...,a_n\\}$ and $ B \\equal{} \\{b_1,b_2,...,b_n\\}$ be two sets of $ n$ integers, where $ A \\neq B$,\r\nif the following two multisets (where the same element can occur twice or more, and in this problem, $ |X|\\equal{}|Y|\\equal{}{n \\choose 2}$) $ X \\equal{} \\{a_i \\plus{} a_j | a_i, a_j \\in A, a_i \\neq a_j\\}$, $ Y \\equal{} \\{b_i \\plus{} b_j | b_i, b_j \\in B, b_i \\neq b_j\\}$ are the same ($ X \\equal{} Y$),\r\nprove that $ n \\equal{} 2^k$ for some positive integer $ k$", "Solution_1": "Famous and hard:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=150844\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=13258\r\n\r\n darij" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "AMC 10", "AMC 12", "AIME I" ], "Problem": "I think this year's cutoff will be lower than last year's (assuming this year's AIME will be about the same in difficulty level as last year's AIME), because of the expansion this year and harder AMC12's this year. \r\n\r\nCutoff: 210 to 215?", "Solution_1": "211.5 .", "Solution_2": "I'm guessing that it's going to be a 213. Just a guess. It's pretty hard for it to be above a 215, though, since the AMC was hard, as $\\beta$ said.", "Solution_3": "I was thinking 205-210, but what do I know?", "Solution_4": "All of these predictions are premature. The single most important factor will be the difficulty of the AIME, and none of you has seen that.", "Solution_5": "yeah none have seen that.", "Solution_6": "I would like a number ending in a \"2\", since that's the same number my AMC 12 scores ends in, and I'm less confident in my AIME than I would like to be.", "Solution_7": "I would like a number ending in 2^2, lol, since that is mine.", "Solution_8": "I'm thinking... 215.5.", "Solution_9": "I'm going to be bold and say the AIME is going to be hard this year in addition to the AMCs, and the cutoff will be 207.", "Solution_10": "I am thinking it will be in the 210-215 range.", "Solution_11": "What about the AIME floor value for AMC10 people?\r\nSince this is my first year taking AMC stuff, i think that it's will be $1\\le \\text{value}\\le 15$ :D", "Solution_12": "The AIME floor will be 8.\r\n\r\n[hide=\" :ninja:\"]Disclaimer: I do not back up any of my statements in this thread![/hide]", "Solution_13": "I'll take a shot with:\r\n\r\nIndex: 211\r\nFloor: 7", "Solution_14": "cutoff: 214\r\nfloor: 7", "Solution_15": "Oh great, this means I need at least a 10 on AIME, if not an 11...yup, I've got no chance. :dry:", "Solution_16": "Ah, the joys of being a freshman and not even knowing how to calculate index.\r\n\r\nOn a serious note, I think the AMCs were not actually much harder than last year, but the blank-value difference made it seem that way.", "Solution_17": "[quote=\"worthawholebean\"]]\nOn a serious note, I think the AMCs were not actually much harder than last year, but the blank-value difference made it seem that way.[/quote]\r\n\r\nI'm not sure if they made it seem harder or harder to qualify with. Instead of having to answer 11 correctly, you have to answer 14. For those teetering on the edge of qualifying, 3 problems makes a lot of difference because they start getting \"difficult\" around 12/13.", "Solution_18": "I am not sure how the whole thing would go, but I hope it is lower than 217.5, so I can get 9 in AIME and pass.", "Solution_19": "I predict a 211, because I suspect there will be many people with 210s.", "Solution_20": "[quote=\"[b]Andrew Kim[/b]\"]I am not sure how the whole thing would go, but I hope it is lower than 217.5, so I can get 9 in AIME and pass.[/quote]\r\nHey I knew an Andrew Kim. He was in this GT thing called PACE in third grade. I don't think its you though.", "Solution_21": "I fee like a contestant on Deal or No Deal...\r\n\r\n[size=200]Keep it loooow!!![/size]", "Solution_22": "[quote=\"nat mc\"][quote=\"[b]Andrew Kim[/b]\"]I am not sure how the whole thing would go, but I hope it is lower than 217.5, so I can get 9 in AIME and pass.[/quote]\nHey I knew an Andrew Kim. He was in this GT thing called PACE in third grade. I don't think its you though.[/quote]\r\n\r\nI think that it is one of the most common asian-american names, at least from my experience... (I know many)", "Solution_23": "I just checked my 7-year record of students taking our local math contest in Southern California, about 700 names. There were 16 people named Kim and 5 named Andrew. There wasn't an Andrew Kim, but there were both an Andrew Lee and an Andrew Park.", "Solution_24": "Due to the USAMO expansion, I feel the index cutoff will be no greater than 210 and the floor no greater than 8", "Solution_25": "[quote=\"Kent Merryfield\"]All of these predictions are premature. The single most important factor will be the difficulty of the AIME, and none of you has seen that.[/quote]\r\n\r\nGood point.\r\n\r\nI based my guess upon the assumption that the AIME will be of the same higher difficulty increase than the AMC increase was.", "Solution_26": "Well these are pretty approximate but here's my idea:\r\n\r\nFloor: 8, maybe 7, doubt it'll be anything other than that, but 6 or 9 is possible I guess.\r\n\r\nIndex: Hmm I highly doubt it will be any higher than last year seeing as (a) there's [i]more[/i] of an expansion and (b) the AMC scoring system lowers scores.", "Solution_27": "I think there might be a curve. Maybe 200 for the amc12 b and 210 for the A (most people including I thought the B was harder). But lets say someone gets a 150 on the B and 5 on the AIME. Would the floor only be 5 then?", "Solution_28": "If he/she got in then yeah. But an index of 200 would be most probably not enough to make USAMO.", "Solution_29": "They could just make it 200.5 and thus force a floor of 6...\r\n\r\nRealistically, I am making a random prediction that the cutoffs will be 218 and 8 for the A and 212 and 8 for the B, for index and floor, respectively.", "Solution_30": "I hope it is 208!\r\n\r\nLast year's one was in the 210's I thought?", "Solution_31": "Would there be four different index floors, since there is the AMC A, AMC B, and AIME and alternate AIME? (assuming that the two AIMEs have enough of a difficulty difference). How about tie breakers for people qualifying by the floor value, since the AMC A and B were different difficulty, will they separate the tie breakers as well?", "Solution_32": "There might be two different qualifying scores for the USAMO from the AIME I and II depending on difficulty, but I'm pretty sure that they won't make a difference between the AMC 12A and the B, since the qualifying scores were only 2.5 points apart.", "Solution_33": "[quote=\"neelnal\"]I hope it is 208!\n\nLast year's one was in the 210's I thought?[/quote]\r\nLast year's was 217.5 I think.", "Solution_34": "If it's <= 217, then I only need 10 on AIME.\r\n\r\nMy goal is still 14 though.\r\n(wanna break 250 index)", "Solution_35": "Is \"index\" sum of AMC12 scores or sum of AMC10 and AMC12?", "Solution_36": "[quote=\"ra5249\"]\t\nIs \"index\" sum of AMC12 scores or sum of AMC10 and AMC12?[/quote]\r\n\r\nIndex=AMC 12 +10*AIME score." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[color=darkblue]Let $ f(x) \\equal{} a_6x^6\\plus{}a_5x^5\\plus{}a_4x^4\\plus{}a_3x^3\\plus{}a_2x^2\\plus{}a_1x\\plus{}a_0$; $ a_6 \\not\\equal{} 0$. Know that $ f(1)\\equal{}f(\\minus{}1)$,$ f(2)\\equal{}f(\\minus{}2)$,$ f(3)\\equal{}f(\\minus{}3)$. \n\nProve that $ f(x)\\equal{}f(\\minus{}x)$, $ \\forall x\\in R$.[/color]", "Solution_1": "consider $ g(x)\\equal{}(f(x)\\minus{}f(\\minus{}x))/2\\equal{}a_5x^5\\plus{}a_3x^3\\plus{}a_1x$\r\n\r\nIt has the 6 roots x=-3,-2,-1,1,2,3. But it has degree 5. So $ g(x)$ is constant. So f(x)=f(-x) for all x in R.\r\n\r\nQ.E.D.", "Solution_2": "It's true. Thank! :P" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c \\ge 0$, prove that\r\n\\[(a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2})^{2}\\ge 3abc(a^{4}b^{3}+b^{4}c^{3}+c^{4}a^{3}) \\]\r\n:)", "Solution_1": "am i welcome?\r\n[hide=\"not nice\"]\n$a^{6}b^{4}+b^{6}c^{4}+c^{6}a^{4}\\ge a^{4}bc^{5}+b^{4}ca^{5}+c^{4}ab^{5}$\n$2(a^{3}b^{5}c^{2}+b^{3}c^{5}a^{2}+c^{3}a^{5}b^{2})\\ge 2(ab^{5}c^{4}+bc^{5}a^{4}+ca^{5}b^{4})$\n[/hide]", "Solution_2": "Yes, you are welcome. :) This inequality is not hard at all, but it is related to the hard one\r\n\\[\\sum\\frac{a^{4}}{a^{3}+abc+b^{3}}\\ge \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}\\]\r\n:)", "Solution_3": "[quote=\"toanhocmuonmau\"]Yes, you are welcome. :) This inequality is not hard at all, but it is related to the hard one\n\\[\\sum\\frac{a^{4}}{a^{3}+abc+b^{3}}\\ge \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}\\]\n:)[/quote]\r\n\r\nHow are they related, Toanhocmuonmau? :oops: \r\n\r\nThank you very much." } { "Tag": [ "inequalities", "calculus", "induction", "inequalities solved" ], "Problem": "Let a1,a2, ... an be real numbers. Prove that\r\n

a1+ 2/(1/a1+1/a2) + 3/(1/a1+ 1/a2 + 1/a3) + ... + n/(1/a1+ 1/a2 + 1/a3 +... \r\n+1/an) <= 2(a1+ a2+ ...+ an)", "Solution_1": "well, I think you ommited something I think that a1,a2,..,an are not just real numbers but \r\npositive real numbers <>0. well I am just going to post a solution for n=3 just to show on what my solution is based in case I am right about a1,.. an!\r\n\r\na1+2/(1/a1+1/a2)+3/(1/a1+1/a2+1/a3)<2(a1+a2+a3)\r\n\r\nwe just apply the fact that 2/(1/a+1/b)<=(a+b)/2 \r\nypu apply it twice and then you get that\r\n\r\na1+2/(1/a1+1/a2)+3/(1/a1+1/a2+1/a3)1,a2, ... an be positive real numbers.", "Solution_3": "By applying your technique , Lagrangia , the coefficient of a1 you end up with is :\r\n\r\n1 + 1/2 + .. + 1/n , which is unbounded and will be greater than 2 from some value of n on . \r\n\r\nActually , 1 + 1/2 + .. + 1/n - ln(n) is convergent ! So you can see it grows quite fast ..", "Solution_4": "[b]Solution[/b]:\r\n\r\ndenote x1=1/a1, ... , xn=1/xn and the ineq is equivalent to:\r\n1/x1+2/(x1+x2)+..+n/(x1+..+xn)<2(1/x1+..+1/xn).\r\n\r\nS=sum_(k=1,n) (k/(x1+..+xk))=sum_(k=1,n) 2/(k+1)* (k(k+1)/2)/(x1+2*x2/2+...+k*ak/k))<= sum_(k=1,n) 2/(k+1)*2/(k*(k+1))(1/x1+2*2/x2+..+k*k/xk)=sum_(k=1,n) 4/(k*(k+1)^2) sum_(h=1,k) (h^2/xh)=sum_(1<=h<=k<=n) 4/(k*(k+1)^2)*h^2/xh=4*sum_(h=1,n) h^2/xh*sum(k=h,n) 1/(k(k+1)^2) where in the last ineq I used the mean ineqs: MH and MA.\r\n\r\nand we also have:\r\nsum_(k=h,n)1/(k(k+1)^2)=1/2*sum_(k=h,n) 2k/(k^2*(k+1)^2) < 1/2 sum_(k=h,n) ((k+1)^2-k^2)/(k^2(k+1)^2)=1/2*sum_(k=h,n) (1/k^2-1/(k+1)^2)=1/2(1/h^2-1/(n+1)^2)<1/2h^2\r\nso we have that:\r\nS<4*sum(h=1,n) h^2/xh*1/2h^2=2*sum(h=1,n) 1/xh.\r\n\r\nnow I hope this solution is okay! :D :D \r\n\r\ncheers! :D :D", "Solution_5": "That inequality were presented at All-Russia olympiad in 1986 with const 4 instead of 2, but solution was more elegant.", "Solution_6": "do you know that solution myth and can you post it?\r\n\r\ncheers! :D :D", "Solution_7": "Suppose WLOG x1 \\leq x2 \\leq ... \\leq xn.\r\n\r\nWe have\r\n\r\n2/(x1+x2) \\leq 1/x1\r\n\r\n(2k-1)/(x1+x2+...+x2k-1) \\leq (2k-1)/(xk+...+x2k-1) \\leq (2k-1)/(k.xk-1) \\leq 2/xk-1, k=2,3,...,(n+1)/2\r\n\r\n2k/(x1+x2+...+x2k) \\leq (2k)/(xk+1+...+x2k) \\leq 2/xk, k=2,3,...,n/2", "Solution_8": "[quote=\"hxtung\"]Let a1,a2, ... an be real numbers. Prove that\n

a1+ 2/(1/a1+1/a2) + 3/(1/a1+ 1/a2 + 1/a3) + ... + n/(1/a1+ 1/a2 + 1/a3 +... \n+1/an) <= 2(a1+ a2+ ...+ an)[/quote]\nLet $ a_1,a_2, \\cdots,a_n $ be [b]positive[/b] real numbers. Prove that\\[a_1+ \\frac{2}{\\frac{1}{a_1}+\\frac{1}{a_2}}+ \\frac{3}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\frac{1}{a_3}}+ \\cdots+ \\frac{n}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_n}}\\le 2( a_1+a_2+\\cdots+a_n )\\]\n(All-Russia olympiad in 1986 )", "Solution_9": "[quote=\"sqing\"]\nLet $ a_1,a_2, \\cdots,a_n $ be [b]positive[/b] real numbers. Prove that\\[a_1+ \\frac{2}{\\frac{1}{a_1}+\\frac{1}{a_2}}+ \\frac{3}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\frac{1}{a_3}}+ \\cdots+ \\frac{n}{\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_n}}\\le 2( a_1+a_2+\\cdots+a_n )\\]\n(All-Russia olympiad in 1986 )[/quote]\n\nWe have the following:\n\n\\begin{align*}\nLHS & \\leq \\sum_{1 \\leq j \\leq n} \\sum_{1 \\leq i \\leq j} \\frac{a_i}{j} \\\\\n& = \\sum_{1 \\leq i \\leq n}H_{n-i+1} a_i \\\\ \n\\end{align*}\n\nHere, $H_n$ is the $n^{\\text{th}}$ harmonic number,\n\n\\[ H_n = \\sum_{1 \\leq j \\leq n} \\frac{1}{j} \\]\n\nbut we have a problem since this series diverges.\n\nAlso, another site has the problem statement differently @sqing from the '86 All-Soviet Olympiad:\n\n\\[ \\sum_{1 \\leq j \\leq n} \\frac{j}{\\sum_{1 \\leq i \\leq n} a_i} \\]\n\nSee problem 427 from [url=http://olympiads.win.tue.nl/imo/soviet/RusMath.html]here.[/url]", "Solution_10": "Thank SZRoberson." } { "Tag": [ "geometry", "geometric transformation", "homothety", "geometry proposed" ], "Problem": "ABCD has an inscribed circle center O. The lines AB and CD meet at X. The incircle of XAD touches AD at L. The excircle of XBC opposite X touches BC at K. X, K, L are collinear. Show that O lies on the line joining the midpoints of AD and BC.", "Solution_1": "It looks strange. Aren't the two circles you mention one and the same with the incircle of $ABCD$? Second of all, I must be misunderstanding it at a more profound level, because I drew a sketch on my computer which suggests it's not true. Could you please restate it or something? I picture would help..", "Solution_2": "[quote=\"grobber\"]It looks strange. Aren't the two circles you mention one and the same with the incircle of $ABCD$?[/quote]\r\n\r\nYou probably drew a sketch where the point X lies on the rays AB and DC. Try it with another sketch, where the point X lies on the rays BA and CD. With this sketch, the problem makes much more sense. Anyway, I have not tried it since I have no time, but it doesn't look wrong...\r\n\r\n Darij", "Solution_3": "[quote=\"grobber\"]It looks strange. ...\nI picture would help..[/quote]\r\n\r\nHere is a picture, on how I interpreted the problem. About the text the only think I made was cut and paste from the page\r\n\r\n[url]http://www.kalva.demon.co.uk/russian/rus00.html[/url].", "Solution_4": "It looks like the line connecting the midpoints of $AD,BC$ is parallel to $KL$, but I have to give it some more thought..\r\n\r\nEdit: I do have a solution now, but I'd like to make it simpler. The idea is proving that, if $M,N$ are the midpoints of $AD,BC$, then $MO\\|KL,NO\\|KL$.", "Solution_5": "Let $K'$ the point closer to $L$ where $LK$ cuts the circle and let $U$ be the poine where $CB$ touches the circle. It's easy to see that $K'$ is the antipode of $U$ (because the tangent in $K'$ to the circle is obtained from $CB$ through the homothety taking $K$ to $K'$ and centered at $X$, so this tangent is parallel to $CB$). From here the conclusion $NO\\|KK'=KL$ follows easily. $MO\\|KL$ is proven analogously.", "Solution_6": "sprmnt21,\r\nmay I ask, with what software did you produce this nice diagram?", "Solution_7": "[quote=\"grobber\"]Let $K'$ the point closer to $L$ where $LK$ cuts the circle and let $U$ be the poine where $CD$ touches the circle. It's easy to see that $K'$ is the antipode of $U$ (because the tangent in $K'$ to the circle is obtained from $CD$ through the homothety taking $K$ to $K'$ and centered at $X$, so this tangent is parallel to $CD$). From here the conclusion $NO\\|KK'=KL$ follows easily. $MO\\|KL$ is proven analogously.[/quote]\r\n\r\nmmh ... ok! I can understand the idea. But, could be interesting to make some comment to that. First of all remark a typo your CD should be, I believe, CB. \r\n\r\nBut most important aspect, IMHO, to make clearer the argumentation, yuo should explain because if N is the midpoint of CB then NO//KL. For instance I can see more immediately that if N is the mid point of KU then NO//KL and from this, using well known proprierties, justify the claim. \r\n\r\nDo you agree, or am I too pernickety?", "Solution_8": "Indeed, every occurence of $CD$ in my message should be replaced by $CB$. Terribly sorry about that, I'll edit it.", "Solution_9": "[quote=\"jhaussmann5\"]sprmnt21,\nmay I ask, with what software did you produce this nice diagram?[/quote]\r\n\r\nWINGEOM which is a FANTASTIC little (in size) tool for drawing dinamic sketchs.\r\n\r\nYou can dowload, FOR FREE, the program from [url=http://math.exeter.edu/rparris/wingeom.html]here[/url]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Does there exist an infinite sequence of positive integers, containing every positive integer exactly once , such that the sum of the first n terms is divisible by n for every n?", "Solution_1": "Yes, it does.\r\nConstruction is just simple.\r\nLet we choose $a_1,a_2,\\ldots$ consequently.\r\nSuppose that we chose $a_1,\\ldots,a_n$.\r\nNow suppose that the least number that does not appear in $a_1,\\ldots,a_n$ is $k$.\r\nNow we choose $a_{n+1}$ such that\r\n\\[ n|a_{n+1}+a_1+\\ldots+a_n \\]\r\n\\[ n+1|a_{n+1}+k+a_1+\\ldots+a_n \\]\r\nSee that by chinese remainder theorem and the fact that we already only chose finite numbers we can choose such $a_{n+1}$.\r\nThen we just make put $a_{n+2}=k$.\r\nIt's easy to check that this sequence has the requested property and also the fact that every number appears exactly once in the sequence.", "Solution_2": "There exists another such sequence, namely the one with $a_1=1$ and such that $a_i$ is the smallest positive integer that has not yet appeared in the sequence satisfying the condition $i|\\sum_{k=1}^{i}a_i$. It can be shown that $a_{a_n}=n, \\forall n \\in \\mathbb{N}$ so every positive integer is a term of such a sequence.", "Solution_3": "To complete Xixas' post:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=30381&highlight=" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "This is perhaps quite famous problem, but nonetheless i consider it very nice:\r\n\r\nThere are 100 wizards in a dungeon and there are 100 possible colored hats. Every wizard gets a hat, which he doesn't see, but he sees the colors of all the rest 99 hats.\r\nAt the same moment every wizard tries to guess the color of his own hat and says the answer, in case he doesn't guess he dies.\r\n\r\nIt is possible that there are several hats of the same color and all the wizards say the answer at the same time, so that they cannot make decision based on the analysis of the other answers, but they all can invent some strategy the day before.\r\n\r\nProve that there is a strategy so that at least one of the wizard will be alive.", "Solution_1": "You wrote that one dies if he does not guess. So, if he guess, although it is wrong, he is alive, right?\r\n\r\nIf you mean that one dies if he does not guess correctly, you didn't mention that they cannot communicate each other. So it is very easy.\r\n\r\nIf they cannot communicate each other, I think it is very difficult.", "Solution_2": "[quote=\"Johan Gunardi\"]You wrote that one dies if he does not guess. So, if he guess, although it is wrong, he is alive, right?\n[/quote]\n\nNo, the wizard stays alive only in case the color he said indeed coincides with the color of his hat. \nThe purpose is to have at least one wizard, who tells the right color. \n\n[quote=\"Johan Gunardi\"]\nIf you mean that one dies if he does not guess correctly, you didn't mention that they cannot communicate each other. So it is very easy.\n\nIf they cannot communicate each other, I think it is very difficult.[/quote]\r\n\r\nNo, they cannot communicate each other, i mentioned that, when i said that they don't know the answers of other wizards, they only see the colors of the rest 99 hats.\r\n\r\nThe point is that there must be a strategy for all of them, doesn't matter which colors they would have so that at least one guesses correct color.", "Solution_3": "[quote=\"eugene\"]This is perhaps quite famous problem, but nonetheless i consider it very nice:\n\nThere are 100 wizards in a dungeon and there are 100 possible colored hats. Every wizard gets a hat, which he doesn't see, but he sees the colors of all the rest 99 hats.\nAt the same moment every wizard tries to guess the color of his own hat and says the answer, in case he doesn't guess he dies.\n\nIt is possible that there are several hats of the same color and all the wizards say the answer at the same time, so that they cannot make decision based on the analysis of the other answers, but they all can invent some strategy the day before.\n\nProve that there is a strategy so that at least one of the wizard will be alive.[/quote]\r\n\r\nIs this formulation really complete and correct?\r\nThere seem to be so many details missing:\r\n\r\nIf the colors of the 100 hats are NOT known a priori, the problem is clearly unsolvable. (Give every wizard\r\na hat with a different, unique phantasy-color; there is no way of deducing one's own phantasy color\r\nfrom looking at the other ones.)\r\n\r\nIf the colors of the 100 hats are known a priori, the problem is trivial. (Every wizard looks at the\r\n99 other hats, and deduces his own hat color as the missing hat.)\r\n\r\nIf there are only two colors, but the number of hats with either color is unknown:\r\nConsider three wizards A,B,C among the 100 wizards.\r\nA looks at B, and guesses B's hat color.\r\nB looks at C, and guesses C's hat color.\r\nC looks at A, and guesses A's hat color.", "Solution_4": "[quote=\"test20\"][quote=\"eugene\"]This is perhaps quite famous problem, but nonetheless i consider it very nice:\n\nThere are 100 wizards in a dungeon and there are 100 possible colored hats. Every wizard gets a hat, which he doesn't see, but he sees the colors of all the rest 99 hats.\nAt the same moment every wizard tries to guess the color of his own hat and says the answer, in case he doesn't guess he dies.\n\nIt is possible that there are several hats of the same color and all the wizards say the answer at the same time, so that they cannot make decision based on the analysis of the other answers, but they all can invent some strategy the day before.\n\nProve that there is a strategy so that at least one of the wizard will be alive.[/quote]\n\nIs this formulation really complete and correct?\nThere seem to be so many details missing:\n\nIf the colors of the 100 hats are NOT known a priori, the problem is clearly unsolvable. (Give every wizard\na hat with a different, unique phantasy-color; there is no way of deducing one's own phantasy color\nfrom looking at the other ones.)\n\nIf the colors of the 100 hats are known a priori, the problem is trivial. (Every wizard looks at the\n99 other hats, and deduces his own hat color as the missing hat.)\n\n[/quote]\r\n\r\nIt is known that every hat is colored in one of the a priori known 100 fixed colors, say enumerated by 1,2,...,100.\r\n\r\nFor example here is the solution for the sample problem with two wizards and two colors: 1'st wizard says the color of second wizard and the second wizard says the color, different to the color of the 1'st. wizard. In this way at least one of them will say his own color.", "Solution_5": "[quote=\"eugene\"]It is known that every hat is colored in one of the a priori known 100 fixed colors, say enumerated by 1,2,...,100.\n\nFor example here is the solution for the sample problem with two wizards and two colors: 1'st wizard says the color of second wizard and the second wizard says the color, different to the color of the 1'st. wizard. In this way at least one of them will say his own color.[/quote]\r\n\r\nAha!\r\nThis changes everything.\r\n\r\nWizard number $ k$ ($ k\\equal{}1,2,3,\\ldots,100$) guesses that the sum $ S$ of the numbers of all 100 hats equals $ k\\bmod100$. Then he deduces his own hat color from $ S$.\r\nClearly, exactly one of the 100 wizards guesses the correct congruence class for $ S$, and hence will guess his correct hat color." } { "Tag": [ "probability", "geometry", "3D geometry", "sphere" ], "Problem": "I stand exactly 10 meters away from a revolving ball of radius 1. I throw three darts at the ball, and my shot is equally likely to hit any part of the ball. What is the probability that all three darts hit the same hemisphere of the ball given that all darts hit the ball?", "Solution_1": "It's not hard...... :)\r\n[hide=\"Hint\"]What are the possible ways for the points to form a hemisphere?[/hide]", "Solution_2": "OK..... no one seems to want to do this, so:\r\n\r\n[hide=\"Solution\"]$p=\\fbox{1}$. Any 3 points on a sphere determine a hemisphere. Wasn't that bad, was it?[/hide]" } { "Tag": [ "ratio", "geometry unsolved", "geometry" ], "Problem": "Two circles $O_{1},O_{2}$ intersect at $A,B$. The two tangents to $O_{1}$ at $A,B$ meet at $K$. $M$ is a point on $O_{1}$, and $KM$ meets $O_{1}$ again at $C$. $CA$ meets $O_{2}$ again at $Q$, and $AM$ meets $O_{2}$ again at $P$. \r\n\r\nProve that $\\frac{CA}{AM}=\\frac{QB}{BP}$.", "Solution_1": "[color=indigo]We have: $CA\\cdot BM=CB\\cdot AM\\;\\textrm{(well-known)}\\Longrightarrow\\frac{CA}{AM}=\\frac{CB}{BM}\\quad (1)$\n\nOn the otherhand, ${\\begin{cases}\\angle QBP=\\angle QAP=\\angle CBM\\\\ \\angle QPB=\\angle QAB=\\angle CMB}\\end{cases}\\Longrightarrow\\triangle BQP\\sim\\triangle BCM\\Longrightarrow\\frac{QB}{BP}=\\frac{CB}{BM}\\quad (2)$\n\nFrom $(1)$ and $(2)$ we have: $\\frac{CA}{AM}=\\frac{QB}{BP}\\quad\\mathbb{QED}$[/color]" } { "Tag": [ "geometry", "calculus", "integration" ], "Problem": "If the radius of a circle is a rational number, its area is given by a number which is:\r\n\r\n$ \\textbf{(A)}\\ \\text{rational} \\qquad\\textbf{(B)}\\ \\text{irrational} \\qquad\\textbf{(C)}\\ \\text{integral} \\qquad\\textbf{(D)}\\ \\text{a perfect square}$\r\n$ \\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide=\"Click for solution\"]\nSince $ A\\equal{}\\pi r^2$ and $ \\pi$ is irrational and $ r$ is rational, $ A$ is irrational, or $ \\boxed{\\textbf{(B)}}$.\n[/hide]" } { "Tag": [ "real analysis", "real analysis theorems" ], "Problem": "The set $ A\\subset l^p(p\\ge 1)$ has $ \\overline{A}$ is compact iff $ A$ is bounded and $ \\forall \\epsilon>0$, we find $ N(\\epsilon)>0$ such that:\r\n$ \\sum_{k\\equal{}n}^\\infty |x_k|^p<\\epsilon$, $ \\forall x\\equal{}(x_1,x_2,...)\\in A$, $ \\forall n>N(\\epsilon)$\r\n\r\nHow's about the proof?", "Solution_1": "Which implication are you having trouble with? One is easier than the other, as I recall." } { "Tag": [ "function", "induction", "calculus", "derivative", "algebra", "binomial theorem", "real analysis" ], "Problem": "Prove the following:\r\n\r\nIf $f(t)$ and $g(t)$ are two functions of t, then \r\n\r\n$\\frac{d^n}{dt^n} fg = \\sum_{k=0}^{n} {n \\choose k} \\left [ \\frac{d^{n-k}f}{dt^{n-k}} \\right ] \\left [ \\frac{d^{k}g}{dt^{k}} \\right ]$", "Solution_1": "That is Leibniz formula(rule), i think you can find it in every textbook on Calculus. Alternatively, try induction on $n$.", "Solution_2": "I have never done induction using derivatives before, so can anyone tell me how should I go about proving that $p_n \\Rightarrow p_{n+1}$ after doing the base case?", "Solution_3": "You differentiate $\\frac{d^n}{dt^n} fg = \\sum_{k=0}^{n} {n \\choose k} \\left [ \\frac{d^{n-k}f}{dt^{n-k}} \\right ] \\left [ \\frac{d^{k}g}{dt^{k}} \\right ]$ :)", "Solution_4": "Hmmm...\r\n\r\nTaking the derivative of both sides be get:\r\n\r\n$\\frac{d}{dt} \\frac{d^n}{dt^n} fg = \\frac{d}{dt}\\sum_{k=0}^{n} {n \\choose k} \\left [ \\frac{d^{n-k}f}{dt^{n-k}} \\right ] \\left [ \\frac{d^{k}g}{dt^{k}} \\right ]$\r\n\r\n$\\Rightarrow \\frac{d^{n+1}}{dt^{n+1}} fg = \\sum_{k=0}^{n} {n \\choose k} \\frac{d}{dt} \\left [ \\frac{d^{n-k}f}{dt^{n-k}} \\frac{d^{k}g}{dt^{k}} \\right ]$\r\n\r\nNow using the product rule on the R.H.S, we get:\r\n\r\n$\\Rightarrow \\frac{d^{n+1}}{dt^{n+1}} fg = \\sum_{k=0}^{n} {n \\choose k} \\left [\\frac{d^{n-k+1}f}{dt^{n-k+1}} \\frac{d^{k}g}{dt^{k}} + \\frac{d^{n-k}f}{dt^{n-k}} \\frac{d^{k+1}g}{dt^{k+1}} \\right ]$\r\n\r\nNow what?", "Solution_5": "What now? Split it into two sums and re-index. At some point, you'll need to use Pascal's recursion:\r\n\r\n${n \\choose k-1}+{n \\choose k}={{n+1} \\choose k}.$\r\n\r\nOne thing to note: the algebraic technicalities of the inductive step are nearly identical to the algebraic technicalities involved in proving the Binomial Theorem.", "Solution_6": "[quote=\"Kent Merryfield\"]One thing to note: the algebraic technicalities of the inductive step are nearly identical to the algebraic technicalities involved in proving the Binomial Theorem.[/quote]I seem to recall that there is a proof of the Leibniz formula (due to Peter Lax, via the Fourier transfrom,) that reduces it to the binomial theorem.", "Solution_7": "[quote]What now? Split it into two sums and re-index. [/quote]\r\nI am not quite sure how would you do that. I tried but I don't think its getting me anywhere. :(", "Solution_8": "swapnillium: try backing away from this problem for a moment and see if you can prove the Binomial Theorem - that is, for all $n\\in\\mathbb{N},$\r\n\r\n$(x+y)^n=\\sum_{k=0}^n{n\\choose k}x^{n-k}y^k.$", "Solution_9": "Oh WoW! I think I know how to do this! OK, here it goes:\r\n\r\n$(fg)^{(n)} = \\sum_{k=0}^{n} {n \\choose k} f^{(n-k)}g^{(k)}$\r\n\r\nThe base case of $n = 0$ is obvious: $fg = fg$\r\n\r\nTaking the derivative of both sides and using the product rule on the R.H.S we get:\r\n\r\n$(fg)^{(n+1)} = \\sum_{k=0}^{n} {n \\choose k} \\left ( f^{(n-k+1)}g^{(k)} + f^{(n-k)}g^{(k+1)} \\right )$\r\n\r\nNow we distribute the sum and create a new dummy variable $j$:\r\n\r\n$(fg)^{(n+1)} = \\sum_{k=0}^{n} {n \\choose k} f^{(n-k+1)}g^{(k)} + \\sum_{j=0}^{n} {n \\choose j} f^{(n-j)}g^{(j+1)}$\r\n\r\nNow we set $j = k - 1$ and we get:\r\n\r\n$(fg)^{(n+1)} = \\sum_{k=0}^{n} {n \\choose k} f^{(n-k+1)}g^{(k)} + \\sum_{k=1}^{n+1} {n \\choose {k-1}} f^{(n+1-k)}g^{(k)}$\r\n\r\nNow we take the $k = 0$ term out from the left sum and we take the $k = n + 1$ term out from the right sum. So,\r\n\r\n$(fg)^{(n+1)} = f^{n+1}g + \\sum_{k=1}^{n} {n \\choose k} f^{(n-k+1)}g^{(k)} + \\sum_{k=1}^{n} {n \\choose {k-1}} f^{(n+1-k)}g^{(k)} + fg^{n+1}$\r\n\r\nCombining the sums together we have:\r\n\r\n$(fg)^{(n+1)} = f^{n+1}g + \\sum_{k=1}^{n} \\left [ {n \\choose k} + {n \\choose {k-1}} \\right ] f^{(n+1-k)}g^{(k)} + fg^{n+1}$\r\n\r\nFrom Pascal's identity, we get:\r\n\r\n$(fg)^{(n+1)} = f^{n+1}g + \\sum_{k=1}^{n} {{n+1} \\choose k} f^{(n+1-k)}g^{(k)} + fg^{n+1}$\r\n\r\nNow we notice that the right and left terms are exactly the $k = 0$ and $k= n + 1$ terms of the sum, respectively. Thus,\r\n\r\n$(fg)^{(n+1)} = \\sum_{k=0}^{n+1} {{n+1} \\choose k} f^{(n+1-k)}g^{(k)}$\r\n\r\nOur Induction is Complete! :D :lol: ;)" } { "Tag": [ "geometry", "trapezoid", "Pythagorean Theorem" ], "Problem": "Let $ ABCD$ be a trapezoid for which $ AD\\parallel BC$ and $ AB\\perp AD$ . Denote $ I\\in AC\\cap BD$ and the midpoint \r\n\r\n$ M$ of $ [AB]$ . Prove the the point $ I$ is the orthocenter of the triangle $ MCD$ if and only if $ AD^2\\plus{}BC^2\\equal{}CD^2$ .", "Solution_1": "[hide=\"Solution\"]$ CA \\perp MD \\iff \\triangle ABC \\sim \\triangle AMD \\iff BC \\cdot AD = 2 AM^2$\n$ BD \\perp MD \\iff \\triangle ABD \\sim \\triangle BMC \\iff BC \\cdot AD = 2 AM^2$\nSo I is the orthocenter of $ \\triangle CMD \\iff BC \\cdot AD = 2AM^2$.\nBy the Pythagorean theorem,\n$ CD^2 = 4AM^2 + (AD - BC)^2 = AD^2 + BC^2 - 2 \\cdot AD \\cdot BC + 4AM^2 = AD^2 + BC^2$[/hide] \r\n:)" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Prove or give a counterexample: if $ T \\in L(V)$, then\r\n\r\n$ V \\equal{} \\text{null }T \\oplus \\text{range }T$.\r\n\r\n\r\n\r\n\r\nFurthermore, prove that if $ T \\in L(V)$, then\r\n\r\n$ V \\equal{} \\text{null } T^{n} \\oplus \\text{range }T^{n}$\r\n\r\nwhere $ n \\equal{} \\text{dim} V$.", "Solution_1": "For the first one, consider $ T\\equal{}\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix}.$\r\n\r\nThen $ \\text{null }T\\equal{}\\text{range }T\\equal{}\\text{span }\\begin{pmatrix}1\\\\0\\end{pmatrix}.$" } { "Tag": [ "function", "linear algebra", "linear algebra theorems" ], "Problem": "Hello all.\r\n\r\nI'm thinking about automorphisms - I don't really know much about them though. However, if I have some mathematical object is it true that it is always isomorphic to itself? If so, then if I take a mapping that leaves everything unchanged then is this an automorphism? For instance, if I take a graph G and want to preserve edges and non-edges, i claim the graph is isomorphic to itself I will define functions of vertex set V(G) and edge set E(G) as a:V(G)->V(G) and b: E(G) ->E(G) as a(i)=i, b(i)=i, for all i. So this is an automorphism of my graph, right?", "Solution_1": "A mathematical object is always isomorphic to itself. The mapping that leaves everything unchanged is the identity function and it's always an automorphism. \r\n\r\nThe identity function:\r\n$ \\mathsf{id}_A: A \\to A$\r\n$ a \\mapsto a$\r\n\r\nIt's an isomorphism because it's a biiective function and if $ \\varphi(a)$ is true in A, where $ \\varphi$ is a quantifier-free A-formula, then $ \\varphi(\\mathsf{id}_A(a)) \\equal{} \\varphi(a)$ is true in $ \\mathsf{id}_A(A) \\equal{} A$. I used the model theory definition.", "Solution_2": "In other words, this is some kind of permutation of the elements of a group." } { "Tag": [ "inequalities", "superior algebra", "superior algebra solved" ], "Problem": "Let $A,B \\in M_{n}(R)$ with the property $AB=BA$. Prove that: $ det[(8n^{2}+4n+1)(A^{2}+B^{2})+4n(A+B)+2B+I_{n}] \\geq 0$ $\\forall n \\geq 3$.\r\n\r\nI hope this is original and even more: correct! :)\r\n\r\nenjoy", "Solution_1": "Hi Lagrangia, it is... \r\n[quote=\"Lagrangia\"]... corect! :) \n[/quote]\r\ncorrect. :P", "Solution_2": "Did you solve it?\r\n\r\ncheers! :D", "Solution_3": "[quote=\"Lagrangia\"]Let $A,B \\in M_{n}(R)$ with the property $AB=BA$. Prove that: $\\det[(8n^{2}+4n+1)(A^{2}+B^{2})+4n(A+B)+2B+I_{n}] \\geq 0$ $\\forall n \\geq 3$.\n\nenjoy[/quote]\r\n\r\nLet \r\n$A'=A+\\frac{2n}{8n^2+4n+1}$ and let \r\n$B'=B+\\frac{2n+1}{8n^2+4n+1}$.\r\n\r\n$(8n^2+4n+1)(A^2+B^2)+4n(A+B)+2B+I_n=(8n^2+4n+1)A'^2-\\frac{4n^2}{8n^2+4n+1}I_n+(8n^2+4n+1)B'^2-\\frac{(2n+1)^2}{8n^2+4n+1}I_n+I_n=(8n^2+4n+1)(A'^2+B'^2)$\r\n\r\n$\\det[(8n^{2}+4n+1)(A^{2}+B^{2})+4n(A+B)+2B+I_{n}]=(8n^2+4+1)^n\\det[A'^2+B'^2]\\geq 0$ \r\n(because A' and B' commute)." } { "Tag": [], "Problem": "An escalator in pune central is moving downwards.\r\nsam starts at the bottom of the escalator and takes 40\r\nsteps to reach the top of the escalator taking him to\r\nfood bazaar. however, while coming down from the top\r\nof the escalator he takes just 10 steps to reach the\r\nbottom. what is the total number of steps on the\r\nescalator when it is not moving? given that sam walks\r\nat a constant speed while going up and coming down.", "Solution_1": "[hide]uhhh... 25? \n\nthis can be solved using a 2-variable equation system. [/hide]", "Solution_2": "[quote=\"236factorial\"][hide]uhhh... 25? \n\nthis can be solved using a 2-variable equation system. [/hide][/quote]\r\n\r\nHow exactly????", "Solution_3": "I think 20.", "Solution_4": "Ignore my answer, I undercomplicated the problem.", "Solution_5": "post the solution?", "Solution_6": "Ah, but some steps are always below the others and rotating.", "Solution_7": "[hide=\"my solution\"]The effect of the escalator on his speed when he is going up in the opposite of when he is going down, so the number of steps would have to be multplied by one number to get his number of steps on the way up, and its recipricol to get the number of steps he took on the way down. Thus, $20*2=40$ and $20*\\frac{1}{2}=10$, and the answer is $\\boxed{20}$. This is the geometric mean of the two numbers, which is implied by what I said before, but I can't remember the formula for that. I just found it with logic once I saw that $r=2$.\n\n[hide=\"geometric mean\"]Is this what you do for geometric mean? Something like $Ar=\\frac{B}{r}$ for two numbers $A$ and $B$, so $Ar^2=B\\Rightarrow r^2=\\frac{B}{A}\\Rightarrow r=\\sqrt{\\frac{B}{A}}$?[/hide][/hide]", "Solution_8": "[hide]Geometric Mean=$(a_1,a_2...a_n)^{1/n}\\Rightarrow(40\\cdot20)^{1/2}=20$\n\nOh, will it always be two and 0.5? I was trying to apply opposite rates.[/hide]", "Solution_9": "[quote=\"surge\"]\nOh, will it always be two and 0.5? I was trying to apply opposite rates.[/quote]\r\n\r\nYes.", "Solution_10": "I also came up with 20, by applying pretty simple thinking: let's just say that when he's going down, for every step he actually takes the escalator brings him down the equivalent of two steps. Since he and the escalator are each going at constant rates, when he is going up one of his steps only brings him half a step up. Therefore, 40 equals twice the number of actual steps, and 10 equals half the number of actual steps." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "If you have n squares with total area 4 show that they can cover a square with side lenght 1.\r\n\r\nThanks.", "Solution_1": "[b]cover[/b] means overlap allowed or not ?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "hi guys! solve it if u can. here goes:\r\n\r\n2n friends namely A0, A1, A2, \u2026, A2n-1 sat around a circular table in this order to have dinner. The waiter served them the dishes and they started eating. Suddenly Aj fainted. On the medical report it was found that his food was poisoned. So the police tried to arrest the waiter but he escaped. Instead they found a letter. It said, \"Serve the dishes in the order you see them with the first one to me and the next one after some persons on my right. The next one the same number of persons on his right etc. This number must be such that each of my friends get one of the 2n dishes.\"\r\n\r\nFinding this the police came to the conclusion that one of the 2n dishes was already poisoned and arranged in such an order that if \r\n1. the serving was started with the killer,\r\n2. the interval number was such that each of them got a dish.\r\nThen whatever the number was only Aj would get the poisoned one. With the above information, can you help the police finding the killer?", "Solution_1": "Unless we consider the possibilty that murderer and murdered are the same person, they could find the killer.\r\n\r\nIf $j\\geq n$ the murderer is $A_{(}j-n)$\r\n\r\nIf $j0, and \r\n\r\n p+q+r>0, p*(q+r)>=q^2,0<=q*(r+p)-p*r<=q^2+r^2 (*)\r\n\r\nthen\r\n\r\n x^p*y^q*(y^r-z^r)+y^p*z^q*(z^r-x^r)+z^p*x^q*(x^r-y^r)>=0 (**) \r\n\r\n \r\n Proof\uff1aLet\r\n\r\n a = (p+q+r)*(p*q-r*p+q*r)/(r^2+p*q+r*p)\uff0c\r\n\r\n b = (p+q+r)*(-p*q+r*p-q*r+r^2+q^2)/(r^2+p*q+r*p), \r\n\r\n c = (p+q+r)*(p*q+r*p-q^2)/(r^2+p*q+r*p),\r\n\r\nthen \r\n\r\n a+b+c=p+q+r>0, a>=0,b>=0,c>=0,\r\n\r\n and by weighted AM-GM we have\r\n\r\n\r\n f(x,y,z)=1/(p+q+r)*[a*x^p*y^(q+r)+b*y^p*z^(q+r)+c*z^p*x^(q+r)]-x^p*y^q*z^r>=0,\r\n\r\n\r\n thus f(x,y,z)+f(y,z,x)+f(z,x,y)>=0 \uff1d>(**).", "Solution_1": "Therem1 .Let x,y,z>0, and \r\n\r\n p+q+r>0, p*(q+r)>=q^2,0<=q*(r+p)-p*r<=q^2+r^2 (1-1)\r\n\r\nthen\r\n\r\n x^p*y^q*(y^r-z^r)+y^p*z^q*(z^r-x^r)+z^p*x^q*(x^r-y^r)>=0 (1-2)\r\n\r\n Therem2 .Let x,y,z>0, and \r\n\r\n p+q+r>=0, r*(r-p)>=0,r*(p-q)>=0,p^2+q^2>=p*(q+r) (2-1) \r\n\r\nthen\r\n\r\n x^p*y^q*(y^r-z^r)+y^p*z^q*(z^r-x^r)+z^p*x^q*(x^r-y^r)<=0 (2-2) \r\n\r\n \r\n Proof\uff1aLet\r\n\r\n a = (p+q+r)*(p^2+q^2-p*(q+r))/(p^2+q^2+r^2-p*q-q*r-r*p),\r\n\r\n b = (p+q+r)*r*(p\uff0dq)/(p^2+q^2+r^2-p*q-q*r-r*p),\r\n\r\n c = (p+q+r)*r(r-p)/(p^2+q^2+r^2-p*q-q*r-r*p), \r\n\r\nthen \r\n\r\n a+b+c=p+q+r>0, a>=0,b>=0,c>=0,\r\n\r\nand by weighted AM-GM we have \r\n\r\n f(x,y,z)=1/(a+b+c)*[a*x^p*y^q*z^r+b*y^p*z^q*x^r+c*z^p*x^q*y^r]\r\n \r\n -x^p*y^(q+r)>=0;\r\n\r\nthus f(x,y,z)+f(y,z,x)+f(z,x,y)>=0 \uff1d>(2-2)." } { "Tag": [], "Problem": "The numbers 2, 4, 6, 8 are a set of four consecutive even numbers. Suppose the sum of five consecutive even numbers is 320. What is the smallest of the five numbers?", "Solution_1": "[hide]let x be the smallest number. \nx + (x+2) + (x+4) + (x+6) + (x+8) = 320. \n5x + 20 = 320\nx = 60. [/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $F=\\{f|f: \\mathbb{R}\\rightarrow\\mathbb{R}\\}$ and $C(f)=\\{g|g\\in F;\\;g\\circ f=f\\circ g\\}$. Prove that:\r\na) if $C(f)=F$ then $f=1_{\\mathbb{R}}$;\r\nb) there exist $f,g\\in F$, $f\\neq g$, so that $C(f)=C(g);$\r\nc) for any $f,g\\in F$: $C(f)\\cap C(g)\\subset C(f\\circ g)$.", "Solution_1": "a) for any $y\\in\\mathbb{R}$, choose $g(x)=y\\forall x\\in\\mathbb{R}$ , then $f(y)=y$.\r\nc)Trivial from defined of $C(f)$.", "Solution_2": "b) Take $f\\in F$ with $f\\not =id_\\mathbb{R}=f^{3}$.\r\nThen $C(f)\\subseteq C(f^{2})\\subseteq C(f^{4})=C(f)$, but $f\\not =f^{2}$.\r\n\r\nExample $f(x)=\\{\\begin{array}{ll}x+{1\\over 3}& \\text{for }\\{x\\}\\in [0,{2\\over 3})\\\\ &\\\\ x-{2\\over 3}& \\text{for }\\{x\\}\\in [{2\\over 3},1) \\end{array}$" } { "Tag": [ "function" ], "Problem": "$f(f(x))=-5x$\r\n\r\nFind $f(x)$", "Solution_1": "Is f defined to be reals to reals?", "Solution_2": "Yes, $f:R->R$" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "The centres of the exscribed circles of triangle ABC are O1, O2, O3. Prove that the area of triangle (O1 O2 O3 )is at least four times the area of triangle ABC.", "Solution_1": "[quote=\"ashrafmod\"][color=darkred]The exincentres of the triangle ABC are $I_{a}$ , $I_{b}$ , $I_{c}\\ .$ Prove that the area $[I_{a}I_{b}I_{c}]$ is at least four times the area $[ABC]\\ .$[/color][/quote]\r\n[color=darkblue][b]Proof.[/b] Denote the $A$- exincircle $w_{a}=C(I_{a},r_{a})$ a.s.o. From the well-known relations $S\\equiv [ABC]=pr=(p-a)r_{a}$ obtain $ar_{a}=p(r_{a}-r)$ a.s.o. \nAlso remark the relations $r_{a}+r_{b}+r_{c}=4R+r$ and $2\\cdot [BI_{a}C]=ar_{a}$ a.s.o. Therefore, $[I_{a}I_{b}I_{c}]\\ge 4S$ $\\Longleftrightarrow$ $\\sum [BI_{a}C]\\ge 3S$ $\\Longleftrightarrow$ \n$\\sum ar_{a}\\ge 6pr$ $\\Longleftrightarrow$ $p\\sum (r_{a}-r)\\ge 6pr$ $\\Longleftrightarrow$ $\\sum r_{a}\\ge 9r$ $\\Longleftrightarrow$ $4R+r\\ge 9r$ $\\Longleftrightarrow$ $R\\ge 2r\\ ,$ what is truly.[/color]" } { "Tag": [ "irrational number" ], "Problem": "Is $\\sqrt2+\\sqrt3$ rational?", "Solution_1": "No,[hide=\"Hint\"] Its square is $5+2\\sqrt 6$ and it is easy to show that $\\sqrt 6$ is not rational.[/hide]", "Solution_2": "Using Gyan's hint, wouldn't we also have to prove that the square root of an irrational number is also irrational?", "Solution_3": "Yeah, but I don't think it's hard to prove that the square root of an irrational number is also irrational... Just let the square root of an irrational number be a rational number $\\dfrac{m}{n}$ and look for a contradiction.", "Solution_4": "Yes, to add to above, easiest to see that $m^2 = 6n^2$ where $m$ and $n$ are natiural numbers is not possible because\r\nif you factor both sides into its prime factors, on LHS number of factors of $2$ is even while on right had side its odd which is a contradiction... (For any square, all prime number in its factors have to appear even times)" } { "Tag": [ "Asymptote", "geometry", "incenter", "\\/closed" ], "Problem": "Long asymptote code, such as that posted in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149405]Now incenter[/url] in the Test forum, is cut off at under 750 characters in the popup window in IE7. Firefox is fine though.", "Solution_1": "There's nothing that I can do about it. It is created by the limitation in IE's url length. Unfortunately IE is still miles back compared to Firefox, Opera, or other browsers. Right now though almost 50% of the users use Firefox :)" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Construct a set $A \\subset \\mathbb{R}$ and a function $f: A \\to \\mathbb{R}$ such that for every $a_{1}, a_{2}\\in A$ holds $|{f(a_{1})-f(a_{2})}|\\leq|{a_{1}-a_{2}}|^{3}$ and the range of $f$ is uncountable. \r\n(V. Brayman)", "Solution_1": "[hide=\"Hint\"]$A$ = standard Cantor set.[/hide]\n[hide=\"Solution\"]Every element of $A$ has base 3 expansion with only $0$ and $2$ in it. Therefore, for every $a_{1},a_{2}\\in A$ the difference $|a_{1}-a_{2}|$ is at least $1/3^{n}$, where $n$ is the first digit (after the dot) where the expansions of $a_{1}$ and $a_{2}$ disagree. Define $g(0.xyzw\\dots)=0.00x00y00z00w\\dots$. Since $|g(a_{1})-g(a_{2})|$ is at most the $3/3^{3n}$ with $n$ as above, it follows that $|g(a_{1})-g(a_{2})|\\le 3|a_{1}-a_{2}|^{3}$. So, $f(a)=g(a)/3$ works.[/hide]" } { "Tag": [ "pigeonhole principle", "number theory", "relatively prime" ], "Problem": "Prove or disprove the following statement: Every positive integer has a multiple whose decimal representation consists only of the digits 0, 1, and 2 repeated as many times as necessary.", "Solution_1": "You don't need any 2s.", "Solution_2": "If it isn't clear already, make 11...100....0, where the 0's make the number divisible by any power of 2 or 5 necessary, and the right amount of 1's to make the number divisible by any number relatively prime to 10.", "Solution_3": "Let our number be $ N$. Consider the sequence $ 1,11,111,...,111...111$ where the last term consists of $ N\\plus{}1$ ones. Then by pigeonhole, two have the same remainder on division by $ N$. Subtracting the smaller from the larger, we have a number of the form $ 111...000...$ Which is clearly divisible by $ N$.\r\n\r\n[b]Remark[/b]: If $ N$ is not divisible by $ 5$, we can simply divide out the trailing $ 0$s to give a number divisible by $ N$ consisting entirely of ones. Also, we can get numbers with only $ 2,0$ or $ 3,0$ etc by simply multiplying this number by $ 2,3$ etc.", "Solution_4": "Why can't 1 be the multiple?", "Solution_5": "$ 1$ is not a multiple in general, but $ 1$ is always a divisor. e.g. multiples of $ 12$ are $ 0,12,24,36...$ and divisors are $ 1,2,3,4,6,12$ but $ 1$ is certainly not always a multiple of any number." } { "Tag": [ "function", "calculus", "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c>0$ such that $a+b+c=\\frac{3}{2}$ And\r\n$F=27abc+\\frac{8}{ab+bc+ca}$\r\nHere is my question:$minF=?$ :wink: :wink: \r\nWhen the equality occur? :wink:", "Solution_1": "How do you think? :blush:", "Solution_2": "It is easy to find minimum of this function if you replace 27 by 256/27 but for 27???", "Solution_3": "[quote=\"delegat\"]It is easy to find minimum of this function if you replace 27 by 256/27 but for 27???[/quote]\r\nYes, :wink: but with 27, I had a nice pro.", "Solution_4": "q: a,b,c > 0 and a+b+c = 1, find minimum value of f(a,b,c) = 27abc + 8/(ab+bc+ca).\r\na: not yet a full solution but some thought about it leads me to say that the minimum value is m = f(1/2,1/2,1/2). as you can see if you let a --> 0 then f(0,b,c) attains a minimum at b = c = 3/4. but m <= f(0,3/4,3/4) as you can verify it yourself. the strategy is perhaps the AC method. \r\n-------------------------------------------------------------------------------------", "Solution_5": "[quote=\"Minh Can\"]q: a,b,c > 0 and a+b+c = 1, find minimum value of f(a,b,c) = 27abc + 8/(ab+bc+ca).\na: not yet a full solution but some thought about it leads me to say that the minimum value is m = f(1/2,1/2,1/2). as you can see if you let a --> 0 then f(0,b,c) attains a minimum at b = c = 3/4. but m <= f(0,3/4,3/4) as you can verify it yourself. the strategy is perhaps the AC method. \n-------------------------------------------------------------------------------------[/quote]\r\nThe equality don't occur when $a=b=c$ or $a=0$ :wink:", "Solution_6": "$MinF=14$ when $(a,b,c)=(\\frac{1}{6},\\frac{2}{3},\\frac{2}{3}) or (\\frac{2}{3},\\frac{1}{6},\\frac{2}{3}) or (\\frac{2}{3},\\frac{2}{3},\\frac{1}{6})$\r\n :lol:", "Solution_7": "[quote=\"Minh Can\"] The strategy is perhaps the AC method. \n-------------------------------------------------------------------------------------[/quote]No AC method, but certainly EV Method.", "Solution_8": "Can we see a solution? :)", "Solution_9": "Mixing variables works, assume c is the smallest, then we can check that $f(a,b,c) \\ge f(t,t,c)$ where $t = \\frac{a+b}{2}$ After that we are to prove $\\frac{81}{2}t^{2}-54t^{3}+\\frac{8}{3t-3t^{2}}\\ge 14$ for $t \\in [\\frac{1}{2}, \\frac{3}{4}]$. Expanding we get\r\n$324t^{5}-567t^{4}+243t^{3}+84t^{2}-84t+16 \\ge 0$ which is equivalent to\r\n$(3t-2)^{2}(36t^{3}-15t^{2}-9t+4)\\ge 0$ which is true.", "Solution_10": "[quote=\"Soarer\"]After that we are to prove $\\frac{81}{2}t^{2}-54t^{3}+\\frac{8}{3t-3t^{2}}\\ge 14$ for $t \\in [\\frac{1}{2}, \\frac{3}{4}]$. Expanding we get\n$324t^{5}-567t^{4}+243t^{3}+84t^{2}-84t+16 \\ge 0$ which is equivalent to\n$(3t-2)^{2}(36t^{3}-15t^{2}-9t+4)\\ge 0$ which is true.[/quote]\r\n\r\nI see, and it is nice, but you only verify that 14 is right after he tells us 14. How would we find the value?", "Solution_11": "After mixing variables, you are left with an expression of one variable. At least calculus works.", "Solution_12": "By EV-Theorem, for $a+b+c=3/2$ and $abc=const$, the sum $a^{2}+b^{2}+c^{2}$ (hence E) is minimal when $0 b > or equal to 10 with square root of a - b = sqrt2\r\n\r\nA. 13\r\nB. 14\r\nC. 15\r\nD. 16\r\nE. None of them\r\n\r\n7. In how many distinguishable ways can the letters in word acacia can be written?\r\n\r\nA. 20\r\nB. 30\r\nC. 40\r\nD. 50\r\nE. None of them\r\n\r\n8. How many numbers are there that satisfy the following three conditions? \r\nA) No digit contains digit 2\r\nB) Greater than 100, less than 1000\r\nC) Such that in abc where a,b,and c each represent different number a > b > c.\r\n\r\nA. 52\r\nB. 56\r\nC. 60\r\nD. 64\r\nE. None of them\r\n\r\n9. Three points that are not on a line determine three points. How many lines are determined by ten points, no three of which are on a line?\r\n\r\nA. 42\r\nB. 45\r\nC. 70\r\nD. 100\r\nE. None of them\r\n\r\n10. The product of three consecutive multiple of 2 is 26880. Then whats the sum of the three numbers?\r\n\r\nA. 84\r\nB. 90\r\nC. 92\r\nD. 94\r\nE. None of them\r\n\r\n11. The triangle ABC is scalene in which the length of sides are 5,9, and k. If altitude is 8, what is the sum of largest and smallest possible area? (round to whole number)\r\n\r\nA. 26\r\nB. 28\r\nC. 30\r\nD. 32\r\nE. None of them\r\n\r\n12. Find the distance between points (1,-2,7) and (6,5,-9).\r\n\r\nA. 18\r\nB. sqrt330\r\nC. 19\r\nD. sqrt411\r\n\r\n13. There are eight bridges that connects city D,E,F, and G. The bridges are paired like this: 1 from D to F, 2 from F to E, 2 from E to G, 2 from G to D, and 1 from D to E. If you must begin and return to the same point, which vertex should you start on? (You must cross all bridges)\r\n\r\nA. D\r\nB. E\r\nC. F\r\nD. G\r\nE. Impossible\r\n\r\n14. (a + b)^7 s second term and third term is?\r\n\r\nA. 7a^6b^1 + 21a^5b^2\r\nB. 13a^6+b^2 + 21a^5b^1\r\nC. 21a^6b^2 + 7a^5b^1\r\nD. 13a^6b^1 + 7a^5b^2 \r\nE. None of them\r\n\r\n15. The lengths of the sides of triangle RST are ST = 12, TR = 13, and RS = 14. If J is the midpoint of TR, and K is the point where TR is cut by the bisector of angle B, find JK.\r\n\r\nA. 0.5\r\nB. 1\r\nC. 1.5\r\nD. sqrt2\r\nE. None of them\r\n\r\n16. The triangle ABC is a right triangle with angle B of a right triangle and AC as hypotenuse in length of 10. An altitude from right angle B is met with AC at P, resulting CP = 3 and PA = 7. If altitude BP has length of h and other two legs of right triangle ABC has length of a and b, calculate a*b*h.\r\n\r\nA. 120\r\nB. 150\r\nC. 180\r\nD. 210\r\nE. None of them\r\n\r\n17. F (x) = a^5 - a^4 - 8a^3 - 24a^2 - 153a - 135 s one of solutions is 3i. Find the sum of other real solutions.\r\n\r\nA. 1\r\nB. 2\r\nC. 3\r\nD. 4\r\nE. None of them\r\n\r\n18. Find the area of a regular hexagon with apothem, a line thats perpendicularly distanced to the center, 12.\r\n\r\nA. 264 sqrt3\r\nB. 272 sqrt3\r\nC. 280 sqrt3\r\nD. 288 sqrt3\r\nE. None of them\r\n\r\n* Proof *\r\nProve that a^5 + a^4 is divisible by any 2a such a > 0.", "Solution_1": "I don't get #2- why are there three equations and two variables- when you solve for a and b they don't work ini all of the equations...", "Solution_2": "For #5, shouldn't it be \"find the probability that if a dart hits the board, it will also hit a balloon\"?", "Solution_3": "Shouldn't #9 read: three non-collinear points determine 3 lines. How many do 10 non-collinear points determine?", "Solution_4": "For #15, shouldn't it be the bisector of S?", "Solution_5": "Rep123max wrote:Ya, that last thing is useless/doesn't work with what a and b are.\n\n\n\nWell...the way the problem was designed (poorly) it did have a use. Sorry to ruin the problem, but it's no good to begin with:\n\n\n\n[hide]a+2b=14, b+2a=16, a+b=10, a^2+2ab+b^2=100\n\nab=35\n\na^2+b^2=30\n\n(a+b)(a^2+b^2-ab)=a^3+b^3=(10)*(30-35)=-50.[/hide]\n\n\n\nToo bad that doesn't really work with real values of a and b.", "Solution_6": "For #18, just to clarify, an apothem is the segment with endpoints on the center and a side so that it is perpendicualr to that side.", "Solution_7": "Ya, I saw the way you were \"supposed\" to do it, but then I checked and it didn't work.. oh well..." } { "Tag": [ "inequalities", "calculus", "calculus computations" ], "Problem": "Two metrics $ d$ and $ \\rho$ on a set $ M$ are [i]equivalent[/i] if they generate the same convergent sequences; that is, $ d(x_{n},x)\\rightarrow 0$ iff $ \\rho(x_{n},x)\\rightarrow 0$.\r\n\r\nShow that the metrics induced by $ \\left\\|\\cdot\\right\\|_{1}$, $ \\left\\|\\cdot\\right\\|_{2}$, and $ \\left\\|\\cdot\\right\\|_{\\infty}$ on $ \\mathbb{R}^{n}$ are equivalent.\r\n\r\nI know that $ \\left\\|x\\right\\|_{\\infty}\\leq\\left\\|x\\right\\|_{2}\\leq\\left\\|x\\right\\|_{1}$, and that $ \\left\\|x\\right\\|_{1}\\leq n\\left\\|x\\right\\|_{\\infty}$ and $ \\left\\|x\\right\\|_{1}\\leq\\sqrt{n}\\left\\|x\\right\\|_{2}$. I've already proved these. I'm just not seeing how to connect them to the current problem.", "Solution_1": "Rearrange your inequalities a bit into this slightly more illuminating form:\r\n\\[ \\frac {1}{\\sqrt {n}}\\left \\Vert x \\right \\Vert _1\\leq \\left \\Vert x \\right \\Vert_2 \\leq \\left \\Vert x \\right \\Vert _1 \\\\\r\n \\\\\r\n\\frac {1}{n}\\left \\Vert x \\right \\Vert _{2} \\leq \\left \\Vert x \\right \\Vert _{\\infty} \\leq \\left \\Vert x \\right \\Vert _{2} \\\\\r\n \\\\\r\n\\left \\Vert x \\right \\Vert _{\\infty} \\leq \\left \\Vert x \\right \\Vert _1 \\leq n \\left \\Vert x \\right \\Vert _{\\infty}\r\n\\]\r\nNow if $ \\{a_k\\}$ is a sequence of real numbers that converges to $ 0$, what can you say about the convergence of the sequence $ \\{b_k\\}$ if there are constants $ m,M$ such that $ m a_k \\leq b_k \\leq M a_k$ for all $ k$? \r\n\r\nUse this reasoning to show that the first inequality proves $ (\\text{convergence in }p \\equal{} 1)\\Rightarrow (\\text{convergence in }p \\equal{} 2)$, the second proves $ (\\text{convergence in }p \\equal{} 2)\\Rightarrow (\\text{convergence in }p \\equal{} \\infty)$, and the third proves $ (\\text{convergence in }p \\equal{} \\infty )\\Rightarrow (\\text{convergence in }p \\equal{} 1)$, so all three are equivalent.", "Solution_2": "Let me see if I've got this: From what you wrote, since $ \\{a_{k}\\}\\rightarrow 0$, then $ c\\{a_{k}\\}\\rightarrow 0$ as well for some constant $ c$. So if $ ca_{k}\\leq b_{k}\\leq Ca_{k}$, then $ b_{k}\\rightarrow 0$ as well since it is bounded by two sequences that converge to $ 0$.\r\n\r\nThen from the first inequality, $ \\left\\|x\\right\\|_{2}$ converges to whatever $ \\left\\|x\\right\\|_{1}$ converges to, so they are equal. From the second, $ \\left\\|x\\right\\|_{\\infty}$ converges to whatever $ \\left\\|x\\right\\|_{2}$ converges to so they are equal, and from the third, $ \\left\\|x\\right\\|_{1}$ converges to whatever $ \\left\\|x\\right\\|_{\\infty}$ converges to, so they are equal. Hence all three are equal." } { "Tag": [ "search" ], "Problem": "I'm asked to find some images or films about reaction between Phosphorus and other substances, such as $ Ca$, $ O_{2}$, $ Cl_{2}$, $ KClO_{3}$, $ KNO_{3}$, $ K_{2}Cr_{2}O_{7}$, condensed $ HNO_{3}$, and also, the preparation of $ P$ from $ Ca_{3}(PO_{4})_{2}$.\r\nCan anyone post them here? Or give me link to download them?\r\nThank you very much.", "Solution_1": "For a start, what are the reactions that might happen between phosphorus and those substances?\r\n\r\nRegarding the images/films, have you tried a google search?", "Solution_2": "I tried to google them, but the quality is very bad, not good enough for my presentation in my class (I need them to put in my PowerPoint)\r\nHere are the reactions\r\n$ 2P\\plus{}3Ca\\longrightarrow^{t^{o}}Ca_{3}P^{2}$\r\n$ 4P\\plus{}5O_{2}\\longrightarrow2P_{2}O_{5}$\r\n$ 2P\\plus{}5Cl_{2}\\longrightarrow2PCl_{5}$\r\n$ 6P\\plus{}5KClO_{3}\\longrightarrow^{t^{o}}3P_{2}O_{5}\\plus{}5KCl$\r\nand so on ...\r\n$ Ca_{3}(PO_{4})_{2}\\plus{}3SiO_{2}\\plus{}5C\\longrightarrow^{t^{o}}3CaSiO_{3}\\plus{}2P\\plus{}5CO$\r\n\r\nThanks again for any help.", "Solution_3": "The third and fifth reactions you have written are wrong. You have also to specify in the reactions if you are using (or not) white phosphorus or other allotrope: because if you are using the white allotrope then all your reactions are wrong.\r\n\r\nCheck also [url=http://boyles.sdsmt.edu/pwithcl/reaction_of_white_phosphorus_and.htm]this site[/url].", "Solution_4": "Huh? What's wrong with my reactions? You mean I have to right $ P_{4}$ instead of $ P$ only?\r\nBtw, thank you for your link. :)", "Solution_5": "Yes, normally white phosphorus is used, so $ P_{4}$ should be written." } { "Tag": [], "Problem": "this is my first challenge (even though it is easy) \r\n\r\na theater has infinity seats. in the first row there is one seat. in the second row there is 3 seats and on the third there are there are 5 seats. if the seating order coninues like this, what seat number is behind [b]seat[/b] #144", "Solution_1": "the anwser is seat #168", "Solution_2": "[quote=\"Programmer_Xia\"]the anwser is seat #168[/quote]\r\nTry to show your work when answering a problem and welcome to AoPS. :)", "Solution_3": "Try not to post problem solving problems like this in the Classroom Math forum. If a problem takes some ingenuity beyond middle school curriculum, it belongs in a different forum.", "Solution_4": "it is really 168 and you [b]have[/b] to draw a picture. If you do you can notice the numbers on the left gining up all are \"that row number\"^2 so you know that 144 is 12^2 and the number on the right is 13^2=169. Then go back one and you get 168" } { "Tag": [ "geometry", "circumcircle", "parallelogram", "symmetry", "perpendicular bisector", "geometry solved" ], "Problem": "Given an isosceles triangle $ABC$ (with $AB=BC$). A point $X$ is chosen on a side $AC$. Some circle passes through $X$, touches the side $AC$ and intersects the circumcircle of triangle $ABC$ in points $M$ and $N$ such that the segment $MN$ bisects $BX$ and intersects sides $AB$ and $BC$ in points $P$ and $Q$. Prove that the circumcircle of triangle $PBQ$ passes through the circumcentre of triangle $ABC$.\r\n\r\n[b]proposed by Sergej Berlov[/b]", "Solution_1": "Let's look at it somewhat backwards. \r\n\r\nIf $X$ is fixed, there is clearly only one circle satisfying the conditions, so let's show that if $BX$ bisects the chord $MN$ of the circumcircle of $ABC$, then $M,N$ are the points we are looking for.\r\n\r\n$BNXM$ is a parallelogram, so the entire figure is symmetric wrt the midpoint of $BX$. Since the parallel through $B$ to $AC$ is tangent to $(ABC)$, it means that $AC$ is tangent to $(MXN)$, so $M,N$ are the points mentioned in the problem.\r\n\r\nWe have shown that, if $T$ is the midpt of $BX$, then $P,Q$ are the intersections of $BA,BC$ with the perpendicular through $T$ to $OT$ (call this property $(*)$). On the other hand, if we take $P'\\in BA,Q'\\in BC$ s.t. $XP'\\|BC,XQ'\\|BA$, then it's easy to see that they satisfy $(*)$ and $\\angle P'OQ'=\\pi-\\angle ABC$, so $P=P',Q=Q'$ and $\\angle POQ=\\pi-\\angle ABC$, which is what we wanted.", "Solution_2": "Can you please explain to me why : \"If $ X $ is fixed, there is clearly only one circle satisfying the conditions \" ?", "Solution_3": "I have just solved a generalized version of this problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4. In fact, in that post, I proved the following:\r\n\r\n[color=blue]Construct the circle passing through the vertices A and B of triangle ABC and touching the side BC at the point B, and construct the circle passing through the vertices B and C and touching the side AB at the point B. These two circles intersect at B; let their second point of intersection be called F. Then, the circumcircle of triangle PBQ always passes through the point F.[/color]\r\n\r\nThis actually holds for arbitrary triangles ABC. Now, for isosceles triangles ABC such that AB = BC, the point F can be characterized in a simpler way: It is the circumcenter of triangle ABC. In order to prove this, note that, at first, the point F was defined in a way symmetric with respect to the vertices C and A; thus, since the triangle ABC is isosceles with base CA, the point F must lie on the symmetry axis of this isosceles triangle, i. e. on the perpendicular bisector of the segment CA. Hence, FC = FA. Now, as I showed in the above-mentioned post, the triangles FAB and FBC are directly similar; hence, FA : FB = FB : FC, so that $FA\\cdot FC=FB^2$. Since FC = FA, this becomes $FA\\cdot FA=FB^2$, so that $FA^2=FB^2$ and thus FA = FB. Combining this with FC = FA, we obtain FA = FB = FC, and thus the point F is the circumcenter of triangle ABC. So it follows that the circumcircle of triangle PBQ always passes through the circumcenter of triangle ABC. And the problem is solved.\r\n\r\n[quote=\"nttu\"]Can you please explain to me why : \"If $ X $ is fixed, there is clearly only one circle satisfying the conditions \" ?[/quote]\n\nWell, I don't understand why Grobber finds this so obvious, but it follows from one of my observations in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4, actually from the following one:\n\n[quote=\"darij grinberg\"]First, it's time to involve the assumption that the radical axis of the circle k [b][note: this is our circle through the point X touching the side AC][/b] and the circumcircle of triangle ABC bisects the segment BX. In other words, if M is the midpoint of the segment BX, then this point M has equal powers with respect to the circle k and to the circumcircle of triangle ABC. Again, let's compute these powers: If the line BX meets the circle k at a point U (apart from X) and the circumcircle of triangle ABC at a point T (apart from B), then the power of the point M with respect to the circle k is $MX\\cdot MU$, and the power of the point M with respect to the circumcircle of triangle ABC is $MB\\cdot MT$. Since the two powers of M are equal, we thus have $MX\\cdot MU=MB\\cdot MT$. In other words, $MX\\cdot MU=BM\\cdot TM$. But since M is the midpoint of the segment BX, we have MX = BM, and thus MU = TM. Hence, BU = BM + MU = MX + TM = TX,[/quote]\r\n\r\nIn fact, this yields an easy construction of the circle k: Construct the point T as the point where the line BX meets the circumcircle of triangle ABC (apart from B), and construct the point U as the point on the line BX such that BU = TX (there is exactly one such point U, since we are working with directed segments). Then, the circle k must pass through the points X and U and touch the line AC at the point X; this is already enough for uniquely determining the circle k (you can construct its center as the point where the perpendicular to AC at X meets the perpendicular bisector of the segment XU). So the circle k is unique.\r\n\r\n Darij" } { "Tag": [ "inequalities", "calculus", "integration" ], "Problem": "What is the sum of the integers x that satisfy $-5\\leq\\frac{x}{\\pi}\\leq10$?", "Solution_1": "[hide]\nMultiply everything by $\\pi$. Since $x \\geq -5\\pi$, $x \\geq -15$, since $5\\pi < 16$ and $x$ is integral.\n\nFrom the second part of the inequality, $x \\leq 10\\pi$, so $x \\leq 31.4$, and since $x$ has to be integral, $x \\leq 31$.\n\nThe numbers from $-1$ to $-15$ cancel the numbers from $1$ to $15$, so we are left with summing $16$ through $31$, which is $376$.[/hide]" } { "Tag": [ "real analysis", "calculus", "calculus open" ], "Problem": "If $ m(E) > 0$ then $ E \\minus{} E \\equal{} \\{x \\minus{} y: x,y\\in E\\}$ contains an interval centered around 0. The only other result that I know that could possibly be useful is that for any $ E$ such that $ m(E) > 0$ we have an interval $ I$ such that $ m(E\\cap I) > \\alpha m(I)$. I don't really see how this is going to be useful in proving the statement but here's what I have so far. If $ E \\minus{} E$ doesn't contain an interval around 0 then for any neighborhood $ N$ of 0 there exists an $ a\\in N$ such that $ a\\not\\in E \\minus{} E$ which is equivalent to saying that $ E \\plus{} a$ and $ E$ are disjoint. Since they are disjoint so are $ E\\cap I \\plus{} a$ and $ E\\cap I$ where $ I$ is an interval such that $ m(E\\cap I) > \\alpha m(I)$. This means $ m(E\\cap I \\plus{} a\\bigcup E\\cap I) \\equal{} m(E\\cap I) \\plus{} m(E\\cap I) \\equal{} 2m(E\\cap I)$ which by assumption leads to $ 2m(E\\cap I) > 2\\alpha m(I) > m(I)$ if $ \\alpha > \\dfrac{3}{4}$. This is where I get stuck since I don't see where I can take it from here. Forgot to mention that $ \\alpha<1$ otherwise the statement that $ m(E\\cap I)>\\alpha m(I)$ doesn't make sense.", "Solution_1": "Never mind. Apparently this is a theorem and the proof is at http://en.wikipedia.org/wiki/Steinhaus_theorem.", "Solution_2": "It has also been posted on this forum- see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113431]here[/url]." } { "Tag": [ "vector", "inequalities", "geometry proposed", "geometry" ], "Problem": "Here are three brothers-problems (they are solved by the very same method):\r\n1. Let $m_1,\\cdots,m_n$ be positive reals and points $A_1,\\cdots,A_n$\r\nLet $M_0$ be a point s,t, $\\sum_{i=1}^n m_i \\overrightarrow{e_i}=\\overrightarrow{0}$ \r\nwhere $\\overrightarrow{e_i}=\\frac{\\overrightarrow{M_0A_i}}{M_0A_i}$\r\nShow the inequality \r\n$\\sum_{i=1}^n m_iMA_i\\geq \\sum_{i=1}^n m_i M_0A_i$\r\nwhere $M$ is an arbitrary point.\r\n2. Given are positive reals $m_1,\\cdots,m_n$ and points $A_1,\\cdots,A_n$ with $\\sum_{i=1}^{n-1} m_i\\overrightarrow{e_i}|\\leq m_n$ \r\nwhere $\\overrightarrow{e_i}\\frac{\\overrightarrow{M_0A_i}}{M_0A_i}$\r\nShow that $\\sum_{i=1}^n m_i MA_i\\geq \\sum_{i=1}^{n-1} m_i A_nA_i$\r\nwhere $M$ is an arbitrary point.\r\n3. Given are positive reals $m_1,\\cdots,m_n$ and convex polygon $B_1\\cdots B_n$ inscribed in convex polygon $A_1\\cdots A_n$ ($B_i \\in [A_iA_{i+1}]$, $A_{n+1}=A_1$) and\r\n$\\overrightarrow{A_iA_{i+1}}(m_{i-1}\\overrightarrow{e_{i-1}}-m_i\\overrightarrow{e_i})=\\overrightarrow{0}$\r\n($A_{-1}=A_n$).\r\nwhere $\\overrightarrow{e_i}=\\frac {\\overrightarrow{B_iB_{i+1}}}{B_iB_{i+1}}$, $B_{n+1}=B_1$.\r\nShow the inequality $\\sum_{i=1}^n m_i C_iC_{i+1}\\geq \\sum_{i=1}^n m_i B_iB_{i+1}$ where \r\n$C_1\\cdots C_n$ is convex polygon inscribed in $A_1\\cdots A_n$ the same way as $B_1\\cdots B_n$\r\n(i.e. $B_i \\in [A_iA_{i+1}]$) and $C_{n+1}=C_1$.", "Solution_1": "Solution for the first problem:\r\n\r\n$\\displaystyle MA_i*M_0A_i\\geq\\overrightarrow{MA_i}*\\overrightarrow{M_0A_i}$\r\n\r\n$\\displaystyle MA_i*M_0A_i\\geq (\\overrightarrow{MM_0}+\\overrightarrow{M_0A_i})*\\overrightarrow{M_0A_i}$\r\n\r\n$\\displaystyle MA_i-M_0A_i \\geq\\frac{\\overrightarrow{MM_0}*\\overrightarrow{M_0A_i}}{M_0A_i}$\r\n\r\nSumming all terms multiplied $m_i$ times:\r\n$\\displaystyle \\sum m_i(MA_i-M_0A_i)\\geq 0$" } { "Tag": [ "trigonometry" ], "Problem": "Sin(x) = 1/7\r\nSin(y) = 7/8\r\n\r\nWhat is sin (x+y) \r\n\r\nDo I use one of the sum and difference formulas?\r\n\r\nIf so, which one?", "Solution_1": "[hide=\"hint\"]\njust use the formula $\\sin(x+y)=\\sin x\\cos y+\\sin y\\cos x$\n\n$\\cos(x)$ and $\\cos(y)$ can be evaluated using $sin^{2}\\theta+\\cos^{2}\\theta=1$\n[/hide]", "Solution_2": "Wouldn't there be four different values then?", "Solution_3": "yes, i suppose there would be, due to the fact that there are multiple values of x such that sin(x)=1/7, etc." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ If a, b, c \\ge 0such that abc \\equal{} 1, then\r\n(a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\plus{} 7\\ge 5(a \\plus{} b \\plus{} c).$", "Solution_1": "See here\r\nhttp://www.mathlinks.ro/viewtopic.php?t=205620" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Hi everyone ! \r\n I have a question. Do they have a stable number of participants in the USAMO every year ? Is that number around 250 or 160 ? I used to think it is 250, according to the USAMO website.", "Solution_1": "It is approximately 250. Selection process for this is in another thread." } { "Tag": [ "email", "search", "real analysis", "real analysis theorems" ], "Problem": "Kani or Kain, I don't remember your name. I'm so sorry. I deleted your email you sent to me. If you or every one who has this book \"Problems in Mathematical Analysis I, II, III\" of Kaczor and Nowak, please send to my mailbox: antueminh@yahoo.com or gay_chay_so.\r\nThanks a lot. :)", "Solution_1": "You can search the member list (see the button above) for the name of your correspondent and contact them directly.\r\nWhatever you do, [u]don't double-post[/u]." } { "Tag": [ "function", "inequalities", "LaTeX", "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "Hey there, I realy need some guidance in the following questions:\r\n\r\n1. Let f,g be entire functions and there exists a real constant M such as\r\n Re(f(z)) <= M * Re(g(z)) for every z in C.\r\nProve that there exist complex numbers a,b such as f(z) = a*g(z) +b for every z in C.\r\n\r\n2. Let f(z) be analytic at the open unit circle and |f ' (z) | <= 1/ (1 - |z| )for every |z| < 1.\r\nProve that the coefficients in that taylor series f(z) = Sigma_ an*zn are : |an| < e for every n>=1.\r\n\r\nAbout 1- If we'll define h(z) = f(z)/g(z) we'll get a constant function by Liouville's theorem. But I can't understand how to get the \"b\" in the expression...\r\n\r\nAbout 2-I'm pretty sure we need to use Cauchy's Inequality but I can't figure out how...\r\n\r\nI will be delighted to get some guidance around here...\r\n\r\nThanks a lot", "Solution_1": "In 1, why in the world do you think $ f/g$ is entire? You have an inequality on the real parts. Since $ Re f$ determines the size of $ e^f$, try exponentiating.\r\n\r\n2. is a cute problem. Noting $ f^{(n)}(0) \\equal{} (f')^{(n\\minus{}1)}(0)$, apply Cauchy's estimates to $ f'$ on the circle of radius $ r$ centered at $ 0$, and obtain a simple estimate. Now choose the right $ r$ for each $ n$.\r\n\r\n[quote=\"WannaBE\"]\n\nI will be delighted to get some guidance around here...\n\nThanks a lot[/quote]\r\n\r\nWe'll be delighted if you take the time to learn a little LaTex.", "Solution_2": "Hey there...\r\nf/g isn't necessarily entire... I was thinking about taking f-Mg...\r\nIt's entire and we know that Re(f(z)) - M*Re(g(z)) <= 0 so by Liouville's theorem we know that there exists constant b such as: f(z)-Mg(z)=b. Hence, we can define M=a and then we'll get f(z)=ag(z)+b... Am I right in this one?\r\n\r\nAbout the second one... I'll think about your guidance and if I'll have any further questions I'll ask here...\r\n\r\nHope you'll be able to verify my response...\r\n\r\nThanks a lot", "Solution_3": "Where can I learn some Latex?\r\n\r\nAnd about the second one, if I'll take r= n / (n+1) it'll work, isn't it?\r\n\r\n\r\nI'll be delighted if you'll be able to verify my two answers...I'm pretty sceptic about the first one because a,b are suppose to be complex numbers and not real ones...\r\n\r\n\r\nThanks a lot" } { "Tag": [ "algebra", "function", "domain", "calculus", "integration", "inequalities", "superior algebra" ], "Problem": "Let $ A$ be a normal domain, $ A \\to B$ is an integral extension, $ B$ is torsion-free over $ A$. Let $ P$ be an ideal of $ B$ with $ p\\equal{}P\\cap A$. Show that the height(P)=height(p).\r\n\r\nI can show using the going down theorem that height(p) $ \\leq$ height(P). I don't know how to show the other inequality", "Solution_1": "I think only the condition integral extension is needed. If we have a chain of primes in $ B$, restrict each of them to $ A$. We are going to show that none of these restrictions coincide. This is because\r\n\r\nClaim: If $ A \\rightarrow B$ is integral, then for any $ p \\subset A$ prime, its fiber (all the primes of $ B$ that restricts to $ p$) is of dimension 0. There is no containing relation.\r\n\r\nProof: $ A_p \\rightarrow B_p$ is still integral. Moreover, $ pA_p$ is maximal, and we know that all the primes that restricts to $ A$ would be exactly the primes in $ B_p$ that restricts to $ pA_p$. By integrality, we know that only maximal ideals restrict to $ pA_p$, so there can't be any containing relation." } { "Tag": [], "Problem": "Ok, so I know everyone's going to laugh at this question, but I was just wondering...\r\n\r\nDoes anyone have any advice as to how to memorize the periodic table of the elements?", "Solution_1": "Er... don't?\r\n\r\nThe way I memorized the first few periods is by pronouncing the abbreviations out loud in chunks. Here:\r\n\r\nHHe\r\nLiBeBCNOFNe\r\nNaMgAlSiPSClAr\r\nKCaScTiVCrMnFe CoNiCuZn GaGeAsSeBrKr\r\nRbSrYZr NbMoCtRu RhPdAgCdIn SnSbTeIXe", "Solution_2": "I find that doing them in chunks helps. You can't expect to memorize it in one week but you can do 20 elements at a time. I did it two summers ago...but then, I've forgotten all of the elements past 50. =/", "Solution_3": "Why would you want to?\r\n\r\nI've been told that Albert, when being teased for not knowing his own phone number said, why would I spend time learning something I can look up when I need it?" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Prove that there doesn't exist any function $ f: (0,\\infty)\\rightarrow(0,\\infty)$ such that $ f^2(x)\\geq f(x\\plus{}y)(f(x)\\plus{}y)$.\r\n\r\nNotes:\r\n- no continuity of any sort is given\r\n- obviously if it would exist, $ f$ should be decreasing\r\n\r\nPretty thanks and enjoy", "Solution_1": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71344]here[/url]. ;)", "Solution_2": "I seee :)" } { "Tag": [], "Problem": "Playing soccer with 3 goes as follows: 2 field players try to make a goal past the goalkeeper, the one who makes the goal stands goalman for next game, etc.\r\n\r\nArne, Bart and Cauchy played this game. Later, they tell their math teacher that A stood 12 times on the field, B 21 times on the field, C 8 times in the goal. Their teacher knows who made the 6th goal.\r\n\r\nWho made it?", "Solution_1": "My attempt...\n\n\n\n[hide]\n\nLet's let A and C be represented by D\n\n\n\nClearly, B cannot stand in the goal two straight times. B also stood in the goal one time more than D. Therefore, B must have started in goal and ended in goal as such...\n\nBDBDBDBDBDBDBDB....etc.... B\n\n\n\nIf B started in goal, D scored the first goal, B scored the second, etc... meaning D (A or C) scored all the odd numbered goals and B scored all the even numbered goals...\n\n\n\nTherefore, B must've scored the 6th goal\n\n\n\n[/hide]", "Solution_2": "eum.. yes.. but how do you proof there's nowhere ...BDDB... in the list? strings ..BCAB.. and ..BACB.. are not contradicting..\r\n\r\nSo you're still missing something...", "Solution_3": "Hey lucky you were almost there, why don't you continue? You've found the hardest part yet, but you miss the easy link to make it consistent :)", "Solution_4": "No one up for completing?", "Solution_5": "Wow, this topic is old! :)", "Solution_6": "I didn't know I am that good at soccer :D", "Solution_7": "We know that A and C were in the goal 21 times. C was there 8 times, so A was goalie 13 times. Similar reasoning, B was goalie 4 times. So in total there were 25 rounds played, and C was on the field 17 times.\r\n\r\nThe only way for A to be the goalie 13 times in 25 games is if he starts out being goalie, and then is goalie as often as possible. This would mean he starts out being the goalie, then is it the third round, the fifth, the seventh.... The player who was goalie the seventh round was the one who made the sixth goal. So A made the sixth goal.", "Solution_8": "That's correct, Kalle. :)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Hi, I'm new here people . There's a problem that i'd like to solve but i can't.\r\n\r\nThere are 27 numbers. You can use three times number 1, three times number 2, three times number 3, ... , three times number 8 and three times number 9.\r\nWhen you use number 1 you must \"jump\" one space and use number one again. Ex: 1 (other number) 1 (other number) 1\r\nWhen you use number 2 you must \"jump\" two spaces and use number two again. Ex: 2 (other number) (other number) 2 (other number) (other number) 2\r\nWith number 3 three spaces, number 4 four spaces ... number 8 eight spaces and number 9 nine spaces.\r\nNow try to organize the 27 numbers :D, there's more than one solution.", "Solution_1": "I understood nothing :?", "Solution_2": "There are 27 numbers _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .\r\nYou can use the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9, three times each one.\r\nWhen you use number 1, you must skip one space and then put number one again and skip other space and put number 1 again. Exemple 1 _ 1 _ 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .\r\nWith number 2 you must skip to spaces then put another number 2 skip more two spaces and put another number two. Exemple 2 _ _ 2 _ _ 2 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .\r\nWith number 3 you must skip three spaces put aother number 3, skip 3 spaces and put another number 3. \r\nExemple : 3 _ _ _ 3 _ _ _ 3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.\r\n...\r\nWith Number 9 you must skip nine spaces the put another number 9, skip nine spaces and put another number 9.\r\nExemple : 9 _ _ _ _ _ _ _ _ _ 9 _ _ _ _ _ _ _ _ _ 9 _ _ _ _ _ _ .", "Solution_3": "I think he meant the following:\r\n\r\nFind all permutations $(a_1, a_2, ..., a_{27})$ of $(1, 1, 1, 2, 2, 2, ..., 9, 9, 9)$ such that $a_i = a_{i - a_i - 1}$ or $a_i = a_{i + a_i + 1}$ (if existent) for all $1 \\leq i \\leq 27$." } { "Tag": [], "Problem": "factorise:\r\n\r\n2x^4 + 3x^3 + x^2 + x + 4", "Solution_1": "[hide=\"answer\"]Not really sure how to explain how I got this...\n\n$ 2x^{4}\\plus{}3x^{3}\\plus{}x^{2}\\plus{}x\\plus{}4\\equal{}(2x^{2}\\plus{}5x\\plus{}4)(x^{2}\\minus{}x\\plus{}1)$\n\nI just realized that it had to be of the form $ (2x^{2}\\plus{}a_{1}x\\plus{}b_{1}x)(x^{2}\\plus{}a_{2}x\\plus{}b_{2}x)$, and $ b_{1}\\in\\{1,2,4\\}$, $ b_{2}\\equal{}\\frac{4}{b_{1}}$and you can very easily limit the possibilities even further.[/hide]" } { "Tag": [], "Problem": "How many positive odd integers with middle digit 5 and no digit repeated are there between 10,000 and 69,999?\r\n\r\nFor certain integers $ n$, the expression $ n^2 \\minus{} 3n \\minus{} 126$ is a perfect square. What is the sum of all distinct possible values of n?", "Solution_1": "[hide=\"Hint 1\"]$ n^2\\minus{}3n\\minus{}126\\equal{}m^2$\n$ \\iff \\left(n \\minus{} \\frac{3}{2}\\right)^2 \\minus{} \\frac{513}{4} \\equal{} m^2$\n$ \\iff (2n\\minus{}3)^2 \\minus{} 513\\equal{} (2m)^2$\n$ \\iff (2n\\minus{}2m\\minus{}3)(2n\\plus{}2m\\minus{}3) \\equal{} 3^2 \\cdot 19$[/hide]\n[hide=\"Hint 2\"]$ (2(\\minus{}n\\plus{}3)\\minus{}3)^2 \\equal{} (\\minus{}2n\\plus{}3)^2 \\equal{} (2n\\minus{}3)^2$[/hide]\n[hide=\"Solution\"]$ n \\equal{} \\minus{}127, \\minus{}42, \\minus{}15, \\minus{}10, 13, 18, 45, 130$[/hide]", "Solution_2": "SINCE ALL NUMBERS ARE 5 DIGIT,\r\nAND CENTER DIGIT 5 IS CONSTANT,\r\nLET US ASSUME 4 CASES:HAVING LAST DIGIT 1,3,7,9\r\n [b] CASE1[/b]\r\nFOR 1 AS LAST DIGIT:\r\nEXCLUDING 5, 0, 1 THERE ARE 7 COMBINATIONS FOR FIRST DIGIT\r\nFOR SECOND ALSO THERE ARE 7 DIGITS AS 0 CAN BE USED\r\nFOR FOURTH, 6\r\n THEREFORE. COMBINATION=7*7*6\r\nCOMBINING ALL 4 CASE WE GET,\r\n4(7*7*6)\r\n=24*49\r\n=1176\r\n EXCLUDING 69,999 THE ANS. IS 1175 :wink:", "Solution_3": "STOP SHOUTING :P \r\nthe last digit has to be 1,3,5,7,9 because the number has to be odd. \r\nCase 1(last digit 1 because the number has to be odd):\r\nyou have 10 choices for the first digit\r\nyou have 5 choices for the middle digit\r\nyou have 8 choices for the fouth digit\r\nand only one choice for the last digit\r\nso the answer for this case is $ 10*5*8$\r\nall the next cases get the same answer so the final answer would be $ 10*5*8*5$ which is $ 2000$.", "Solution_4": "[quote=\"Poincare\"]STOP SHOUTING :P \nthe last digit has to be 1,3,5,7,9 because the number has to be odd. \nCase 1(last digit 1 because the number has to be odd):\nyou have 10 choices for the first digit\nyou have 5 choices for the middle digit\nyou have 8 choices for the fouth digit\nand only one choice for the last digit\nso the answer for this case is $ 10*5*8$\nall the next cases get the same answer so the final answer would be $ 10*5*8*5$ which is $ 2000$.[/quote]\r\n?\r\n\r\nThe middle digit has to be 5 and the first digit can only be 1,2,3,4,6", "Solution_5": "thanks chemalete for pointing out\r\n\r\nSINCE ALL NUMBERS ARE 5 DIGIT, \r\nAND CENTER DIGIT 5 IS CONSTANT, \r\nLET US ASSUME 4 CASES:HAVING LAST DIGIT 1,3,7,9 \r\nCASE1 \r\nFOR 1 AS LAST DIGIT: \r\nEXCLUDING 5, 0, 1 and 7, 8, 9 THERE ARE 4 COMBINATIONS FOR FIRST DIGIT \r\nFOR SECOND THERE ARE 7 DIGITS AS 0 CAN BE USED \r\nFOR FOURTH, 6 \r\nTHEREFORE. COMBINATION=4*7*6 \r\n same as for 3\r\nbut for 7 and 9,\r\nthere are 5 choices for first digit.\r\nCOMBINING ALL 4 CASE WE GET, \r\n4*7*6+4*7*6+5*7*6+5*7*6 \r\n=42*18\r\n=756 \r\nEXCLUDING 69,999 THE ANS. IS 755", "Solution_6": "Why do you exclude 69,999 when you didn't count it? I think the answer is just 756.", "Solution_7": "oh :oops: \r\nthats right", "Solution_8": "[hide=\"thorough solution\"]The middle digit is $ 5$. The last digit can be either $ 0, 2, 4, 6, 8$ and the first digit can be any number $ 1\\minus{}6$, excluding 5. \n2 Cases-\n1) last digit $ (1,3)$\nThere are 2 ways to choose the last digit, 4 ways for the first digit, and (7)(6) ways to choose the middle two $ \\implies (2)(4)(42) \\implies 336$ ways.\n2) last digit $ (7,9)$\n2 ways for the last digit, 5 for the first, and 42 for the middle two digit for a total of $ (10)(42) \\equal{} 420$ ways to the answer is $ 420 \\plus{} 336$ = [b]756 ways[/b][/hide]" } { "Tag": [], "Problem": "Can some body give me an example from of direct variation and solve it so i can understand?", "Solution_1": "As in direct and inverse proportionality? \r\n\r\n[hide=\"Direct\"]\n[hide=\"Problem\"]\nIf $ a$ is directly proportional to $ b$ and $ a\\equal{}7$ when $ b\\equal{}13,$ find $ a$ when $ b\\equal{}27.$ [/hide]\n\n[hide=\"Solution\"]\nSince they are directly proportional, their quotient is constant. In other words, \\[ \\frac{a}{b}\\equal{}\\frac{7}{13}\\] is the constant quotient between $ a$ and $ b.$ Now we find the value of $ a$ such that \\[ \\frac{a}{27}\\equal{}\\frac{7}{13}\\Rightarrow a\\equal{}\\boxed{\\frac{189}{13}}.\\][/hide][/hide]\n\n[hide=\"Inverse\"]\n[hide=\"Problem\"]\nIf $ a$ is inversely proportional to $ b$ and $ a\\equal{}17$ when $ b\\equal{}24,$ what is $ a$ when $ b\\equal{}15?$[/hide]\n\n[hide=\"Solution\"]\nSince they are inversely proportional, their product is constant. In other words, \\[ ab\\equal{}17\\cdot24\\equal{}408\\] is constant. Now we have to find $ a$ such that \\[ 15a\\equal{}408\\Rightarrow a\\equal{}\\boxed{\\frac{136}{5}}.\\][/hide][/hide]\r\n\r\nThese are some simpler proportions. Some more complicated proportions involve more variables.", "Solution_2": "[quote=\"A.C\"]Can some body give me an example from of direct variation and solve it so i can understand?[/quote]\r\n\r\n\r\nIf one quantity increases then other quantity also increases (or) If one quantity decreases then other quantity also decreases. It is called direct variation. May be seems little confusing.\r\n\r\nLets see this example:\r\n\r\nCost of 5 pens = $ \\$$10\r\n\r\nWould the cost of 10 pens will be more than the cost of 5 pens?\r\n\r\nYes, It will be more.\r\n\r\nWhat would be the cost of 2 pens? Would it be less than cost of 5 pens?\r\n\r\nAs number of pens increases, the cost also increases, if the number of pens decreases, then the cost also decreases.\r\n\r\nI hope you understand this.\r\n\r\nThis link may be useful to work with word problems.\r\n\r\n[url]http://www.mathocean.com/[/url]" } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Ok so here I go.\r\n\r\nI'm no math genius, so bare with me.\r\n\r\nMy problem is this.\r\nI have a triangle, or at least part of it.\r\nI know one of the the angles is 30deg, and the length of A to B, but I have no other measurements available.\r\n\r\nNow, what I need to figure out is the length of B to C.\r\n\r\nCan anyone help me with this?\r\nThere must be a way.\r\n\r\nThanks in advance.\r\n\r\n", "Solution_1": "A math genius would also be unable to compute a triangle given only one sidelength and one angle.\r\n\r\nYou need another measurement, else the \"length of B to C\" is not uniquely determined.\r\n\r\n darij", "Solution_2": "Ok darij, Yes I realize that. I have it figured out anyways.\r\nBeing that all angles added up will equal 180, 30deg + 60 deg + 90deg=180.\r\n\r\nI forgot to mention that C to B, will always be 90 deg, leaving me with 60deg.\r\n\r\nThanks for the response anyways.\r\n\r\nOff to do some calcs.", "Solution_3": "Which angle is $90^\\circ$?\r\nif $\\angle C = 90^\\circ$ , then $BC$ = 5cm\r\nif $\\angle B = 90^\\circ$ , then $BC$ = $\\frac{10\\sqrt{3}}{3}$cm" } { "Tag": [ "trigonometry" ], "Problem": "A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 23-22. Show that if the incident angle theta is small, the emerging ray is displaced a distance d = t (theta) (n-1)/n from the incident ray, where t is the thickness of the glass and theta is in radians.\r\n(note: sin (theta) is approximately tan (theta) which is approximately theta for small thetas).", "Solution_1": "where is the figure?", "Solution_2": "I don't know how to draw the figure, but I'll just describe it.\r\n\r\nThe figure consists of a rectangular piece of glass with the index n.\r\nThe surrounding medium is air, with index approximately 1.\r\n\r\nThe light ray enters the glass at angle.\r\nThe light ray is then refracted inside the glass.\r\nWhen the light ray leaves the glass, it is refracted again.\r\nWhen the light ray leaves the glass, the refraction angle is equal to the incident angle( the angle at which the light ray approaches the glass)", "Solution_3": "Refraction (Snell) Law:\r\n\\[ \\sin\\theta = n . \\sin \\theta^'\\]\r\nWith small angles you can assume : $ \\cos \\theta = \\cos \\theta^' = 1$,\r\n\r\nso $ \\sin(\\theta - \\theta^') = \\sin \\theta - \\sin \\theta^'$, and at the end, \r\n\r\n$ \\sin \\theta = \\theta.$", "Solution_4": "Okay, I don't see how that proves d=t(theta)(n-1)/n where theta is in radians.", "Solution_5": "I don't about you guys, but I got $ d \\equal{} \\frac {t\\theta}{n}$\r\n\r\nis the original equation faulty? or am I missing something.", "Solution_6": "I find that d = tn/(n-1)\r\n\r\nI have the angle of refraction, theta R, is equal to the angle of incidence divided by n (snell's law and sin theta = theta for small thetas)\r\n\r\nIn my drawing I have 2 triangles.\r\nOne with base d and height x and an angle theta R.\r\n\r\nThe other with base (t-d) and height x with angle theta I.\r\n\r\nso then tan (theta R) = x/d \r\ntheta R = x/d\r\ntheta I / n = x/d\r\nx=theta I * d / n\r\n\r\nThe other triangle gives me \r\ntan (theta I) = x /( t-d)\r\ntheta I = x / (t-d)\r\nx = theta I * ( t-d)\r\n\r\ntheta I * d / n = theta I * (t-d)\r\nd/n = t-d\r\nd=tn-dn\r\ndn-d = tn\r\nd(n-1) = t n\r\nd = t n /(n-1)\r\nObviously my equation is wrong, but I'm not sure what I I did wrong.\r\n\r\nI wish there was some way to draw my figure.", "Solution_7": "May be this will help:\r\n\r\n[url=http://img195.imageshack.us/i/mathlinksdifraction.jpg/][img]http://img195.imageshack.us/img195/402/mathlinksdifraction.th.jpg[/img][/url]\r\n\r\n\r\nFrom the rectangular triangle NCA , $ |AC| = {d \\over {\\cos \\theta^'}}$.\r\n\r\nFrom the rectangular triangle ABC , $ t = |AC|.\\sin (\\theta - \\theta^')$\r\n\r\nSorry permuting t and d...", "Solution_8": "The main mistake that we both made was what our assumption of what the displacement. As Immanuel has it, displacement is the perpendicular distance between the refracted light right at the point of refraction. I think we thought it was the distance between the imaginary light and actual refracted light striking the other side of the glass.", "Solution_9": "That's what I thought it would be too." } { "Tag": [ "probability" ], "Problem": "Given that m+n people are waiting in a line at a box office and: n people have 5-dollar bills, other m people don't have less than 10-dollar bill. Now the ticket costs 5 dollars each.When the box office opens there is no money in the till.If each customer just buys one ticket,what is the probability that none of them have to wait for change?\r\nPS: I clearly don't know how we arrange the line in case there are 6 people having 5-dollar bills and 4 people having 10-dollar bills.If we arrange the line in oder to have at least as many people with 10-dollar bills as people with 5-dollar bills ahead of the point, we don't get the result, do we? Because if we have at least 1 people with 10-dollar bill among others, that person will force the person right behind him to be waited. But if we arrange all of 6 people with 5-dollar bills in a row and then arrange 4 people in the rest of 4 positions, it won't work. So I don't think we can have any arrangements in this case?? I read the solution from the book but it's vague to me, not because of its solution but because of my argument I use to break their solutions :huh:.Am I wrong anywhere in the argument above?", "Solution_1": "By any chance, is the answer\r\n[hide=\"?\"]$ \\frac{n\\minus{}m\\plus{}1}{n\\plus{}1}$[/hide]\r\n\r\nI'm not sure that I follow your logic; I think that 6 people paying with $ 5$ dollar bills followed by four paying with $ 10$ dollar bills will work.", "Solution_2": "Yes, your answer is right, azips(in the case m4 \r\nNote : 2n,2n-1,2n+1,2n+2 they are subscript from a.\r\nThanks and best regards!", "Solution_1": "I'm going to latex this for clarity, and come back if I solve it:\r\n\r\n$ a_1\\equal{}2, a_2\\equal{}8$\r\n$ a_n$ is defined by $ a_{2n\\plus{}1}\\equal{}\\frac{a_{2n}\\plus{}a_{2n\\minus{}1}}{2}, a_{2n\\plus{}2}\\equal{}\\frac{a_{2n}a_{2n\\minus{}1}}{2a_{2n\\plus{}1}}$.\r\n\r\nProve that $ a_n$ is convergent and converges to 4.", "Solution_2": "Thank you very much :)I dont know how can i will use latex :oops:", "Solution_3": "[hide=\"Are you sure this sequence converges to four?\"]\nEmpirically, Excel has shown that this sequence does not approach 4.\n\nThis can also be seen since taking the product of any odd term and the term after it yields:\n\n\\[ a_{2n\\plus{}1}a_{2n\\plus{}2}\\equal{}\\frac{a_{2n}\\plus{}a_{2n\\minus{}1}}{2}\\cdot \\frac{a_{2n}a_{2n\\minus{}1}}{2a_{2n\\plus{}1}}\\]\n\\[ a_{2n\\plus{}1}a_{2n\\plus{}2}\\equal{}\\frac{a_{2n}\\plus{}a_{2n\\minus{}1}}{2}\\cdot \\frac{a_{2n}a_{2n\\minus{}1}}{a_{2n}\\plus{}a_{2n\\minus{}1}}\\]\n\\[ a_{2n\\plus{}1}a_{2n\\plus{}2}\\equal{}\\frac{a_{2n}a_{2n\\minus{}1}}{2}\\]\n\nThat is, the product of an odd term and the term after it is half of the product of the previous odd term and the term after it. If the sequence did converge, the product of the terms would tend towards some constant product as well, which is not the case.\n[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "how do you do it?", "Solution_1": "Only moderators may delete posts. Other questions about the forum should be posted here:\r\n[url]http://www.artofproblemsolving.com/Forum/index.php?f=144[/url]" } { "Tag": [ "national olympiad" ], "Problem": "where are they?", "Solution_1": "wait, does nobody have apmo 2006 results...?", "Solution_2": "http://www.amt.canberra.edu.au/apmo2006.html\r\nThe overall results can be found by scrolling down the link. This page is primarily for Australian results, though.", "Solution_3": "the link does not have the full results though(ie who got which award)\r\nThe results are not up on the official site. :(", "Solution_4": "[quote=\"rem\"]\nThe results are not up on the official site. :([/quote]\r\n\r\nThey are now\r\n\r\nhttp://www.kms.or.kr/competitions/apmo/data/APMO_Results_2006.pdf" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let a_1 1* (a_2)<= 2*(a_1).", "Solution_1": "[quote=\"shoki\"]Let $ a_1 1* (a_2)<= 2*(a_1).[/quote]\r\nIt seems not very hard problem . From the condition we have :\r\n$ 1<\\frac{a_j}{a_j\\minus{}a_1}<\\frac{a_j}{a_j\\minus{}a_2}<...<\\frac{a_j}{a_j\\minus{}a_{j\\minus{}1}}$ . \r\nNotice that all numbers are integers and $ \\frac{a_j}{a_j\\minus{}1}>1$ , it is easy to prove that $ \\frac{a_j}{a_j\\minus{}a_i}\\geq i\\plus{}1$ or $ (i\\plus{}1)a_i\\geq ia_j$ ,because $ i\\plus{}1\\leq j$ so $ ja_i\\geq ia_j$ for all $ j>i$ . That is what we must prove .\r\nAdded problem :Show that for any n ,exist a sequence that satisfies condition :$ a_j\\minus{}a_i|a_j$ for all $ i,j\\in \\{1,..,n\\}$" } { "Tag": [], "Problem": "In how many ways can a convex n-gon be divided into triangles by nonintersecting\r\ndiagonals?\r\n\r\nany help would be appreciated.\r\nthank you so much!!", "Solution_1": "Is it gonna be n-2? I am not sure my answer,though.\r\nCould somebody help me to take a look?\r\nThanks.", "Solution_2": "No, it's not $n - 2$. \r\n\r\nHint\r\n\r\n[hide]Catalan numbers[/hide]" } { "Tag": [ "geometry", "geometric transformation", "homothety" ], "Problem": "Let $ABC$ and $DEF$ be triangles such that $DEF$ is inside $ABC$. It is known that $AB\\parallel DE$, $BC\\parallel EF$, $CA\\parallel FD$. Show that $AD,BE,CF$ are concurrent.", "Solution_1": "The pairs of parallel lines make the triangles homothetic, and the lines AF, etc, are through corresponding points, so all 3 lines go through the center of homothecy, so they are concurrent", "Solution_2": "[quote=\"indybar\"]Let $ABC$ and $DEF$ be triangles such that $DEF$ is inside $ABC$. It is known that $AB\\parallel DE$, $BC\\parallel EF$, $CA\\parallel FD$. Show that $AD,BE,CF$ are concurrent.[/quote]\r\n\r\nI'm not sure, but is this problem from the book \"Geometry Revisited\"???" } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "$ \\int{\\ln(\\cos(e^{\\arctan(x)})}$", "Solution_1": "[quote=\"Kent Merryfield\"]Do you have any reason for thinking that there's anything interesting about this? You realize that if you just slap random functions together, the most likely result is a function whose antiderivative is not an elementary function.[/quote]\r\n\r\nAs Kent said in [url=http://www.mathlinks.ro/viewtopic.php?t=200744]here[/url], what's the point of this kind of \"problem-making\"?" } { "Tag": [ "puzzles" ], "Problem": "Folks,\r\n\r\n\r\nCheck this out:\r\nhttp://www.highiqsociety.org\r\n\r\nInteresting part for people here is World Smartest Persons Puzzle Contest 2006.\r\nDeadline for submission of the answers is 31.12.2006.", "Solution_1": "I can't find World Smartest Persons Puzzle Contest 2006.", "Solution_2": "me neither :blush: we must be awefully idiotic not to find that contest :o or it doesn't exist", "Solution_3": "http://www.highiqsociety.org/wsp_challenge.php\r\n\r\nwith the actual questions [url=http://www.highiqsociety.org/wsp_challenge.php]here[/url]", "Solution_4": "cule that kid lied to us about the deadline :mad:", "Solution_5": "Didn't lied on purpose.\r\n :mad: \r\nI misread it in a hurry and thought the deadline was December 31st (not October 31st).Sorry,I don't have a time machine to fix that.\r\nCool puzzles I think anyway.", "Solution_6": "I did the IQ test and didn't pass :(", "Solution_7": "I failed too...by 1 point!! :mad: :(", "Solution_8": "[quote=\"Eros88\"]Didn't lied on purpose.\n :mad: \nI misread it in a hurry and thought the deadline was December 31st (not October 31st).Sorry,I don't have a time machine to fix that.\nCool puzzles I think anyway.[/quote]\r\noh i forgive you i have the worst problems with my eyes i miss problems on the IOWA test because my eyes are so bad :)", "Solution_9": "New IHIQS contest starts September $ 1^{st}$,2007.\r\n\r\nFirst prize 1000 bucks :)" } { "Tag": [ "analytic geometry", "trigonometry" ], "Problem": "Let $ O$ be the center of a quarter of circle. the radius that delimet this region are $ OA$ and $ OB$. A circle with center $ O'$ is inscribed in this region, Being tangent to $ OA,OB$ and the arc $ AB$ . The chord $ AB$ cuts the circle $ (O')$ in $ M,N$ . Find the value of $ cos\\angle MO'N$\r\n\r\nProve also that this angle is close to $ 135^0$", "Solution_1": "[quote=\"Tiger100\"]delimet[/quote]\r\n\r\nWhat does this word mean?", "Solution_2": "He probably meant delimit.", "Solution_3": "Yes, sorry i wanted to mean [b]Delimit[/b]\r\n\r\nAny ideas to solve this problem? it is nice in my opinion", "Solution_4": "Any ideas? :maybe:", "Solution_5": "I'm working on it.\r\nOf course there is coordinate bash, I could ask pythag011 to do that... :P\r\n\r\nEDIT: I have a solution:\r\n\r\nSince OAB is a quarter sector, the midpoint of MN is the intersection of the perpendicular bisectors of OA and OB. WLOG let the radius of circle $ O'$ be 1. The radius of $ OB$ is therefore $ 1\\plus{}1\\sqrt{2}\\equal{}\\sqrt{2}\\plus{}1$. The midpoint of $ MN$ (let's call it $ P$) is therefore $ \\frac{\\sqrt{2}\\plus{}1}{2}$. Hence $ OP\\equal{}\\frac{2\\plus{}\\sqrt{2}}{2}$. Since $ OO'\\equal{}\\sqrt{2}$, we have $ O'P\\equal{}\\frac{2\\minus{}\\sqrt{2}}{2}$. Since $ MO'\\equal{}1$, we have $ \\cos{MO'P}\\equal{}\\frac{2\\minus{}\\sqrt{2}}{2}$, and $ \\angle MO'N\\equal{}2\\angle MO'P$. We can now find $ \\cos{MO'N}$, and it should be roughly $ \\minus{}\\frac{\\sqrt{2}}{2}$, since $ \\cos{135^{\\circ}}\\equal{}\\minus{}\\frac{\\sqrt{2}}{2}$." } { "Tag": [ "function", "calculus", "derivative", "absolute value", "calculus computations" ], "Problem": "SO here's the problem... seems simple, but I just can't get the jist of it because of the absolute value:\r\n\r\nFind the local max/min, as well as the absolute max/min values of the function:\r\n\r\nf(x) = |x^2 - 4x + 3|\r\n\r\nin the interval (1,4].\r\n\r\n\r\nAny help would be greatly appreciated!", "Solution_1": "Either $|x^2-4x+3|=x^2-4x+3$, or $|x^2-4x+3|=-(x^2-4x+3)$. You have critical points where the derivative is zero, and also where you switch formulas.", "Solution_2": "alright... I think that I've just been overcomplicating this.... is the derivative simply f'(x) = |2x-4| ? And if so, how do I prove where I switch formulas? I'm aware that it is at x=1 and x=3, so do I just solve for x-intercepts of the original function to prove this?", "Solution_3": "No, that's not the derivative. It has the same zeros, but its sign is only right some of the time.\r\n\r\nThe derivative is $2x-4$ where $x^2-4x+3$ is positive, and $4-2x$ where $x^2-4x+3$ is negative.\r\nNever try to do calculus with a formula that still has an absolute value sign in it- you're always better off looking at it as the piecewise defined function it is.", "Solution_4": "If you write $|x|$ as $\\sqrt{x^2}$, you get\r\n\\[ \\frac{d}{dx}|x|=\\frac{x}{|x|} \\]\r\nfor $x\\not=0$, which is a reasonably nice way of looking at things." } { "Tag": [], "Problem": "The sum of the number of sides of two convex polygons is 10,\r\n\r\nand the sum of the numbers of their diagonals is 11.\r\n\r\n \r\n\r\nWhat type of polygons are described? \r\n\r\nthank you~", "Solution_1": "Square and hexagon.", "Solution_2": "If s is the number of sides, the number of diagonals is $\\frac{n(n-3)}{2}$.\r\n\r\nTherefore, $x+y=10$, and $x^{2}-3x+y^{2}-3y=22$.\r\n\r\nThis can be solved easily using substitution; \r\n\r\n\\begin{eqnarray*}22&=&(10-y)^{2}-30+3y+y^{2}-3y\\\\ &=&100-20y+2y^{2}-30\\\\ 0&=&2y^{2}-20y+48\\\\ &=&y^{2}-10y+24\\\\ 0&=&(y-6)(y-4) \\end{eqnarray*}\r\n\r\nThe two polygons have 6 and 4 sides, respectively, so you have a square and a circle.", "Solution_3": "Square and circle? :P", "Solution_4": "[quote=\"360_Fan\"]Square and circle? :P[/quote]\r\n\r\nI wonder if there was a joke or an accident." } { "Tag": [ "probability", "function", "real analysis", "search", "probability and stats" ], "Problem": "Find the distribution of $ \\sum_{n\\equal{}1}^{\\infty}\\frac{X_n}{3^n}$\r\n$ X_n$ are iid with $ P(X_n\\equal{}0)\\equal{}P(X_n\\equal{}2)\\equal{}\\frac{1}{2}$", "Solution_1": "This produces a \"random\" element of the usual Cantor set. Its cumulative distribution function $ F(x)$ is the Cantor-Lebesgue \"devil's staircase\" function. This distribution is continuous but not absolutely continuous. That is, it does not have a density function.\r\n\r\n\r\nThe distribution of $ \\sum_{n\\equal{}1}^{\\infty}\\frac{Y_n}{2^n}$ where $ Y_n$ takes on the values $ 0$ and $ 1$ each with probability $ \\frac12$ is simply the uniform distribution on $ [0,1].$", "Solution_2": "and what if $ X_n$ are bernoulli random variable $ P(X_n\\equal{}1)\\equal{}1\\minus{}P(X_n\\equal{}0)\\equal{}p \\neq \\frac{1}{2}$ ?", "Solution_3": "The values of the sum would still be confined to the Cantor set. The cumulative distribution function $ F(x)$ would have the following properties. (I'll use the notation $ q \\equal{} 1 \\minus{} p.$)\r\n\r\n$ F(x) \\equal{} q$ for $ \\frac13\\le x\\le\\frac23.$\r\n\r\n$ F(x) \\equal{} q^2$ for $ \\frac19\\le x\\le\\frac29.$\r\n\r\n$ F(x) \\equal{} q \\plus{} qp \\equal{} 1 \\minus{} p^2$ for $ \\frac79\\le x\\le\\frac89.$\r\n\r\nAnd so on. We will still have $ F$ as the uniform limit of a sequence of continuous functions, and hence continuous. The distribution will still be continuous but not absolutely continuous.\r\n\r\nIf we do that with my alternative of $ Y_n,$ the result will be one of my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=101950603&t=118219]favorite examples.[/url]" } { "Tag": [ "geometry", "circumcircle", "modular arithmetic", "geometric transformation", "homothety", "ratio", "rotation" ], "Problem": "[u][b]Problem:[/b][/u]The vertices of the triangle ABC are lattice points and there is no smaller triangle similar to ABC with its vertices at lattice points. Show that the circumcenter of ABC is not a lattice point. \r\n---------------------------------------------------------------", "Solution_1": "I think it's a number theory problem :?", "Solution_2": "Suppose the contrary.\r\n\r\nThere are some cases:\r\n1. $R^2=0\\pmod{4}$. Make homotety with ratio $1/2$.\r\n2. $R^2=2\\pmod{4}$. Make homotety with ratio $1/2$.\r\n3. $R^2=1\\pmod{4}$. Make homotety with ratio $1/\\sqrt{2}$ and rotate the triangle by $\\pi/4$." } { "Tag": [ "Support", "function" ], "Problem": "Why is our country going at length to ensure the rights of unborn fetuses? Now a woman can't even have an abortion without having law complications. It is kind of sad - just today Senate passed H.R. 1997 - The Unborn Victims of Violence Act. Isn't it enough that fetuses are constitutionally given the right to life no matter what the pregnant mom thinks? Now, if the mom is hurt during the pregnancy the \"living\" fetus can sue too. WHAT??!! That was exactly my reaction. The bill doesn't even specify the requirements for life, but it is assumed that fetuses are living. Why go to lengths to protect the unborn when those that are born are suffering even more. We look too far intot he future and obliviously ignore what is happening now sometimes.", "Solution_1": "I would like to pre-emptively remind people that, although this can be a contentious issue and can raise a lot of passions, you should still be polite and respectful in your discussions with your fellow forum-members. Thank you.", "Solution_2": "I am religious, I don't believe in abortions. However, it is perfectly legal to abort if its danger the mother's health (which happened in my relatives). Recently in Utah a woman who was a looser (drug taker, child abuser, divorced, etc.). was convicted of murdering a stillborn child.", "Solution_3": "wow churchill, what a contrast coming from you :).", "Solution_4": "I believe that women should be allowed to choose whether or not they want an abortion without having to face legal consequences. \r\n\r\nHowever, if a woman is going to have an abortion, she should have it done as soon as possible. (Say, during the first six weeks.) The longer she waits, the more taxing the baby will be on her body, the more she will become attached to the baby as a result of hormones, and the more developed the baby will be.\r\n\r\nSome may argue that abortion should never, or almost never, be allowed. But here are some objections to that viewpoint:\r\n\r\nFirst, what if being pregnant puts the potential mother at risk of injury or death? For instance, pregnancy in women with diabetes can result in hard-to-control blood sugar levels that require constant monitoring. Even with constant attention, this can result in the loss of an eye, kidney, leg, etc. Would you have a mother risk her life to deliver a baby? What if this were someone you [i]knew[/i], like your sister or your friend?\r\n\r\nIn the absence of special, life-threatening circumstances, some argue that a woman should not be allowed to have an abortion because ``she was irresponsible\". But if this is the case, then forcing her to deliver the baby won't remedy this problem of her irresponsibility; it'll just result in a mom who can't care for her baby. So you're really punishing the baby by placing it with a mother who cannot care for it. \r\n\r\nAnd the suggestion that the ``irresponsible'' mother deliver the baby, then put her new baby up for adoption is no better. First of all, no mother will give up her baby easily, perhaps because hormones she produces during the pregnancy make her feel attached to the baby. Second of all, there are only a limited number of families that want to adopt a baby, so adding another baby to that pool of adoptees will displace some other baby. Third, each extra baby will contribute to the overpopulation of the world.\r\n\r\nThird, many abortions are given to teenagers. Suppose a teenage girl is pregnant and wants to have an abortion but cannot, either because the law mandates that she cannot or because her friends and family cry out \"you would kill your own baby?\" Then, she's a high-school age girl who has to worry about her own school life, her kid, and her source of income. Chances are, if she was in high school, she'll start falling severely behind in school. She'll probably drop out to go to work at Wal-Mart so that she can have a source of income to support her child. This girl may have had a bright future ahead of her - maybe she was gonna go to college, get her degree, go to work somewhere, get married, raise a child together with her husband, etc. But because she had to deliver this baby, her future went ka-put.\r\n\r\nFinally, sometimes accidents happen (contraceptive measures are breached, so to speak) and a baby results unintentionally. If a mother doesn't feel she's ready to have a baby yet (or ever) - maybe she want to focus on her job, maybe she hates kids, maybe she wants to devote 8 hours a day to math - then she shouldn't be forced to.\r\n\r\nP.S. For those of you who are wondering, yes, I am religious. Roman Catholic. Does that change my view of abortion? No.", "Solution_5": "Actually, white-horse's characterization of what the Bill says is incorrect. The Bill, as it stands right now, doesn't say anything explicitly about abortion. It says that if a pregnant woman is killed then the perpetrator can be charged with two crimes - one for the murder of the woman and one for the murder of the child.\r\n\r\nOf course, it naturally brings up questions of the implications it will have for abortion; but that's not what [i]this[/i] Bill is addressing. This Bill is talking about the killing of an unborn child [i]without[/i] the woman's consent.\r\n\r\nThe Bill is very complicated, but it doesn't in itself say anything about abortion. It just opens the door to questions about what rights the fetus has. If the fetus is said to have stringent rights to life; just like any other person has - then pretty much all of mathfanatic's arguments (except for maybe the first one) become void. Because those stringent rights to life would superceed any \"utilitarian\" arguments - just like you can't kill any regular person, simply to make your life better.\r\n\r\nWhat's always intersting to me though, is why do we value the lives of infants and children more so than adults? When we hear that a young child is killed, it seems to carry so much more weight than if we hear that an adult is killed. Yet, just a few months before they are born - we consider those [i]same[/i] lives worth so much less than the value of an adult's.", "Solution_6": "[quote=\"gauss202\"]What's always intersting to me though, is why do we value the lives of infants and children more so than adults?[/quote]\r\n\r\nBecause children still have the prospect to do something impressive whereas adults have already blown it?\r\nBecause children are cute?\r\nBecause children are defenseless whereas adults theoretically have a chance to defend themselves or not enter themselves into dangerous situations in the first place?\r\n\r\n\r\nAs for the later question, there are certainly groups of people who value fetuses just as strongly. That's why we have an anti-abortion movement -- it's not moderate religious Christians who maybe think abortion is unfortunate, it's the hardcore evangelicals who really do believe that the fetus is a person. Most other people don't share that belief, so an abortion isn't such a tradgedy.", "Solution_7": "Again, another political issue I'm not entirely familiar with-but me being the gender that this most directly applies to I think I'll share what I believe. First off, I'd like to say that I am a little bit offended at mathfanatic's remarks about a girl delivering a baby and her future going \"kaput\". Yes, it's true that a girl who has a baby in her teenage years may not have all the advantages as a girl who doesn't-but it's not final. A successful future is relative, just because a girl doesn't win all the math contests after she's delivered a baby doesn't mean she's failed. Her own idea of a successful future might be just to be at peace with herself, and there's nothing wrong with that. I am pro-choice because I believe a woman has the right to do what she wants with her own body. I don't understand this \"unborn child rights\". Apparently it has to do maybe with the fact that biologically they're considered alive. Well, practically every month something biologically alive is killed in a woman's body, what about their rights? Just what I think, I'm definitely no expert.", "Solution_8": "I haven't read any of the above so I apologize if there are any duplicate opinions: Abortions are ok with me, after all I don't see any laws protecting cows and pigs from getting slaughtered and their carcass eaten, so why not babies as well? (even though very few cultures eat babies now-a-days)", "Solution_9": "[quote=\"Tare\"]I haven't read any of the above so I apologize if there are any duplicate opinions: Abortions are ok with me, after all I don't see any laws protecting cows and pigs from getting slaughtered and their carcass eaten, so why not babies as well? (even though very few cultures eat babies now-a-days)[/quote]\r\n'tis [url=http://www.english.upenn.edu/~jlynch/Courses/95c/Texts/modest.html]A Modest Proposal[/url], to say the least.", "Solution_10": "A Modest Proposal is one of the funniest pieces of political writing I've ever seen. I highly recommend you all go read it, although it's helpful to know a little bit of British-Irish history first.", "Solution_11": "I was just pointing out that laws against abortion are already, in my opinion, high strung. There are complications, but mothers shouldn't be completely deprived of the right to abortion. I was just saying as an allusion to those developing laws how I felt about giving fetuses the \"power\" to sue. I predict that this bill will be abused. Mothers may intentionally become hurt and the charge will be against the mother are the unborn fetus. Who takes the money for the damages? This is completely incredulous.\r\n\r\nMy madness isn't totally unjustified - babies are now given that right mentioned in the bill because they are considered to be living humans. Babies don't even have motor functions at that stage - how can they \"\"\"\"\"sue\"\"\"\"\"? Another thing is, I'm concerned that the major reason this bill was proposed is because of not only sympathyzers but also people greedy for a quick buck. Also, money seems to be the best compensation for anything. \r\n\r\nThe argument for the bill seems based on the fact that a fetus is being deprived. Deprived of what? It isn't even really alive yet - no brain functions, motor functions, or any functions. \r\n\r\nI'm concerned - what if people deliberately seek money through this method. People already overabuse sueing powers - people sue for everything. \r\n\r\nI would like someone to weigh out the benefits vs. the negative potential effects and give a proper justification of the bill.", "Solution_12": "white_horse: one more mis-conception. This law has nothing to do with sueing. To sue is a matter for civil court. In many jurisdictions, you can't do it until you are 18, although you might be able to get someone to sue on your behalf. (An example of this is the case currently at the Supreme Court about the Pledge of Allegiance -- the man who brought the case brought it on behalf of his daughter, since she isn't old enough to sue and he doesn't have grounds for sueing.) Anyhow, the law applies only to criminal cases. I, personally, am worried very much about the math -- 1 mother + 1 fetus = 2 crimes; 1 mother = 1 crime. Hopefully, none of our politicians will learn to subtract any time soon." } { "Tag": [], "Problem": "Let \\[A=1, \\ B=2, \\ C=3, \\ D=4, \\ldots X=24, \\ Y=25, \\ Z=26\\] Let the value of a word be the product of it's corresponding variables (i.e ALL=(1)(12)(12)=144). Take your name. Find the value of it. Multiply it by 200. Add 50 to it. Add 15 to it. Add 35 to it. Divide it by 2, subtract 27 from it and then divide it by 3. Subtract half the value of your name from it, then subtract 32(value of your name), and finally, subtract 5/6 of the value of your name from it. What are you left with? Express your answer AS A MIXED NUMBER.\r\n\r\n\r\n\r\nSorry for the uncombined sentences, but it takes all the fun outta it. :D", "Solution_1": "[quote=\"ragnarok23\"]Let \\[A=1, \\ B=2, \\ C=3, \\ D=4, \\ldots X=24, \\ Y=25, \\ Z=26\\] Let the value of a word be the product of it's corresponding variables (i.e ALL=(1)(12)(12)=144). Take your name. Find the value of it. Multiply it by 200. Add 50 to it. Add 15 to it. Add 35 to it. Divide it by 2, subtract 27 from it and then divide it by 3. Subtract half the value of your name from it, then subtract 32(value of your name), and finally, subtract 5/6 of the value of your name from it. What are you left with? Express your answer AS A MIXED NUMBER.\n\n\n\nSorry for the uncombined sentences, but it takes all the fun outta it. :D[/quote]\r\n[hide]Lets just say that the value of my name is x. $\\frac{\\frac{200x+50+15+35}{2}-27}{3}-.5x-32x-\\frac{5x}{6}=\\frac{100x+23-1.5x-96x-2.5x}{3}=\\frac{23}{3}=7\\frac{2}{3}$[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let a,b,c>0\r\nprove:\r\n\r\ncyclic sum [ ((4a+b-c)^2)/(2(a^2) + (b+c)^2) ] >=8", "Solution_1": "It is equivalent to\r\n\r\n4*H61+16*G63+16*G64+72*G65+15*G66+16*PP61+32*PP62+16*P63>=0,\r\n\r\nin which\r\n\r\nH61=sum(a^2*(a-b)^2*(a-c)^2)>=0;\r\n\r\nG63=sum(a^2*b*c*(a-b)*(a-c))>=0;\r\n\r\nG64=sum(a*(b+c)*b*c*(a-b)*(a-c))>=0;\r\n\r\nG65=sum(b^2*c^2*(a-b)*(a-c))>=0;\r\n\r\nG66=sum(a^2*(b-c)^2*(a-b)*(a-c))>=0;\r\n\r\nPP61=sum(c*(c^2*b+a^3)*(a-b)*(a-c))>=0;\r\n\r\nPP62=sum(c^2*(a^2+b^2)*(a-b)*(a-c))>=0;\r\n\r\nP63=sum(a*b*c*(2*b+c)*(a-b)*(a-c))>=0." } { "Tag": [ "analytic geometry" ], "Problem": "What is the distance from (2,3) to (3,6)?", "Solution_1": "[hide]the distance is 10^0.5. the difference in x-coordinates is 1 and 3 in y-coordinates. therefore, applying the pyhagorean theorem, we get 1^2+3^2=10, so, the answer is 10^0.5[/hide]", "Solution_2": "That would be correct. $\\sqrt{10}$ is the answer.", "Solution_3": "[hide]using the distance forumla:\ni get $\\sqrt{10}$ as my answer[/hide]" } { "Tag": [ "geometry", "3D geometry", "number theory proposed", "number theory" ], "Problem": "a)Prove that there are infinitely many possitive integers \u03b1 so \u03c6(\u03b1) be perfect square.\r\nb)Are there infinitely many possitive integers \u03b1 so \u03c6(\u03b1) be perfect square and gcd(\u03b1,(\u03c6(\u03b1))=1?\r\nc)Are there infinitely many primes \u03c0 so \u03c6(\u03c0) is rerfect cube?\r\nThanks.", "Solution_1": "What's \u03c6(\u03c0) exactly ???", "Solution_2": "It's Euler's function.", "Solution_3": "Oh, I thought it's $\\phi (n)$... As I've read in many books...", "Solution_4": "a) Let $a=5^{2m-1}$ for any positive integer $m$. $\\phi(a) = 5^{2m-2}(5-1) = (2\\cdot 5^{m-1})^{2}$\r\n\r\nc) Since $a$ is prime, then $\\phi(a)=a-1$\r\nIf $a=2$, $\\phi(a)=1^{3}$, If $a>2$, assume $\\phi(a)=m^{3}$ for positive integer $m$, then $a=m^{3}+1>2 \\Rightarrow m>1$.\r\n$a=m^{3}+1 = (m+1)(m^{2}-m+1)$\r\nSince $10$ and from $ u^p\\equal{}u^m\\equal{}1$ we conclude $ u\\equal{}1$ and $ P(x)\\equal{}(x\\minus{}1)^n$.\r\n\r\nP.S. Two nice articles on inequalities of symmetric polynomials:\r\n\r\n[url]http://www.math.helsinki.fi/EMIS/journals/JIPAM/v4n2/116_02_www.pdf[/url]\r\n[url]http://www.emis.de/journals/JIPAM/images/182_05_JIPAM/182_05.pdf[/url]" } { "Tag": [ "vector", "calculus", "derivative" ], "Problem": "So I found a stack of old Descartes contests lying around my school and decided to open them up and start solving some problems, I stumbled across these two that I thought at first were going to be really straightforward but for some reason I can't seem to solve them. I've posted them here for your solving pleasure :):\r\n\r\n6.\r\nLet $\\vec{a}$, $\\vec{b}$, and $\\vec{c}$ be linearly dependent vectors in $R^{3}$ and let:\r\n$\\vec{u}= 3\\vec{a}+2\\vec{b}-\\vec{c}$\r\n$\\vec{v}=-2\\vec{a}+4\\vec{c}$\r\n$\\vec{w}=-\\vec{a}+3\\vec{b}+k\\vec{c}$\r\nDetermine $k$ so that $\\vec{u}$, $\\vec{v}$, and $\\vec{w}$ are coplanar.\r\n\r\n10 b).\r\nIn triangle $ABC$, sides $AB$ and $AC$ have fixed lengths $c$ and $b$ respectively, with $c>b$, and $BC$ has variable length $a$. Also, $\\theta$ is the measure of $\\angle CAB$ and $h$ is the length of the altitude from $A$ to $BC$.\r\nProve that: \r\n$\\frac{da}{d\\theta}= h$", "Solution_1": "What grade level is this? It seems like it is 13th Grade.", "Solution_2": "Yeah, Descartes was the OAC contest back when OAC existed.", "Solution_3": "10 b). is actually pretty simple\r\n[hide]\n$b^{2}+c^{2}-2bccos\\theta=a^{2}$\nThen take the derivative of $\\theta$ on both sides\nAnd it's obvious that $bcsin\\theta=ah$\n[/hide]\n\n\nfOR 6, I guess you just need to find the right combination.\n[hide]$k=-1*\\frac{3}{2}+4*\\frac{11}{4}=9.5$[/hide]", "Solution_4": "10 b)... touche, probably should have persevered more.\r\n\r\n6. Sorry but could you elaborate?? I have no idea what you did :P", "Solution_5": "well, vectors u v w are coplanar if and only if they are linearly dependent which means w would be a linear combination of finitely many u and v.", "Solution_6": "!!! I should definitely review my linear algerba basics (or... everything, it's not like I know much more than the basics :P) I was totally only thinking of the definition \r\n$\\vec{u}\\times \\vec{v}\\bullet \\vec{w}$ $= 0$" } { "Tag": [ "function", "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Distance(x)(cm)\r\n0,1,5,6,8\r\nTemp t(x) (celsius)\r\n100, 93, 70, 62, 55\r\n\r\nMetal wire is 8 cm. and heated at one end. distance x is how far from the heated end you are. The function ti s decreasing and twice differentiable.\r\na) estimate t'(7)\r\nb) write an integral expression of T(x) for avg temp of wire. estimate avg temp of wire using trapezoidal sum with 4 subintervals indicated by data\r\nc) Find int up limit 8 lower limit 0. explain the meaning of this in the context of the problem\r\nd) Is that data consistent with teh assertion that T''(x)>0 for every x from 0 to 8. explain", "Solution_1": "bump...", "Solution_2": "bump", "Solution_3": "bump", "Solution_4": "That's just obnoxious. There's no reason to expect a response within a few hours- this forum moves more slowly than that.", "Solution_5": "sorry, that third bump was a double post by accident. i am just bumping every 2 hours or so, not a big deal.", "Solution_6": "Between your first post and now, there has been only one other post in this subforum; even one bump was excessive." } { "Tag": [ "algebra", "polynomial", "induction", "algebra unsolved" ], "Problem": "Let $f(X)$ be polynomial with rational coefficients. Prove that to have $f$ taking integer values for integer arguments, it is necessary and enough to be able to write $f$ as:\r\n\r\n$f(X) = \\sum _{j=0}^{N} a_j C_{X}^j$\r\n\r\nwith integer $a_1,...,a_N$", "Solution_1": "Define $\\Delta f (x) := f(x+1) - f(x)$ and observe that when $f(x) = \\sum_{i=0}^n a_i \\left( x \\atop i \\right)$ you get $\\Delta f(x) = \\sum_{i=0}^{n-1} a_{i+1} \\left( x \\atop i \\right)$. So you can do it by induction on $n$." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "find all function with reel numbers and wich stisfy\r\n$f(x+y+z)=f(x)+f(y+z)+(y+z)(x+y+z)$. :lol:", "Solution_1": "$z=0\\Rightarrow f(x+y)-f(x)-f(y)=y(x+y)$\r\nLHS is symmetric in $x,y$, RHS isn't.\r\nSo there is no such function.", "Solution_2": "oops i did have a mistake \r\nthinks olorin" } { "Tag": [ "inequalities", "trigonometry", "geometry", "function", "inequalities unsolved" ], "Problem": "Let [tex]\\alpha, \\beta, \\gamma[/tex] be angles in the interval [tex](0, \\pi/2)[/tex] whose sum is [tex]{\\pi}/2[/tex]. Then, prove that\n\n[tex]{\\sin \\alpha \\sin \\beta \\sin \\gamma}(\\sec^2 \\alpha + \\sec^2 \\beta + \\sec^2 \\gamma}) \\geq {\\cos (2 \\alpha) + \\cos (2 \\beta) + \\cos (2 \\gamma) - 1[/tex]", "Solution_1": "[color=blue][b]Problem.[/b] Let $ \\alpha$, $ \\beta$, $ \\gamma$ be three angles in the interval $ \\left]0,\\frac{\\pi}2\\right[$ such that $ \\alpha\\plus{}\\beta\\plus{}\\gamma\\equal{}\\frac{\\pi}2$. Then, show the inequality\n\n$ \\sin \\alpha \\sin \\beta \\sin \\gamma\\left(\\sec^2 \\alpha \\plus{} \\sec^2 \\beta \\plus{} \\sec^2 \\gamma\\right) \\geq \\cos \\left(2 \\alpha\\right) \\plus{} \\cos \\left(2 \\beta\\right) \\plus{} \\cos \\left(2 \\gamma\\right) \\minus{} 1$.[/color]\r\n\r\nAfter three years of triangle geometry, one doesn't find such problems tough, but rather one finds them straightforward. At first, we can easily show\r\n\r\n$ \\cos \\left( 2\\alpha \\right) \\plus{} \\cos \\left( 2\\beta \\right) \\plus{} \\cos \\left( 2\\gamma \\right) \\minus{} 1 \\equal{} 4\\sin \\alpha \\sin \\beta \\sin \\gamma$,\r\n\r\nsince\r\n\r\n$ \\cos \\left( 2\\alpha \\right) \\plus{} \\cos \\left( 2\\beta \\right) \\plus{} \\cos \\left( 2\\gamma \\right) \\minus{} 1$\r\n$ \\equal{} \\cos \\left( 2\\alpha \\right) \\plus{} \\cos \\left( 2\\beta \\right) \\minus{} 2\\sin ^{2}\\gamma \\equal{} 2\\cos \\left( \\alpha \\plus{} \\beta \\right) \\cos \\left( \\alpha \\minus{} \\beta \\right) \\minus{} 2\\sin ^{2}\\gamma$\r\n$ \\equal{} 2\\sin \\gamma \\cos \\left( \\alpha \\minus{} \\beta \\right) \\minus{} 2\\sin ^{2}\\gamma \\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (since }\\alpha \\plus{} \\beta \\plus{} \\gamma \\equal{} \\frac {\\pi }{2}\\text{)}$\r\n$ \\equal{} 2\\sin \\gamma \\left( \\cos \\left( \\alpha \\minus{} \\beta \\right) \\minus{} \\sin \\gamma \\right)$\r\n$ \\equal{} 2\\sin \\gamma \\left( \\cos \\left( \\alpha \\minus{} \\beta \\right) \\minus{} \\cos \\left( \\alpha \\plus{} \\beta \\right) \\right) \\text{\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (since }\\alpha \\plus{} \\beta \\plus{} \\gamma \\equal{} \\frac {\\pi }{2}\\text{) }$\r\n$ \\equal{} 2\\sin \\gamma \\cdot 2\\sin \\alpha \\sin \\beta \\equal{} 4\\sin \\alpha \\sin \\beta \\sin \\gamma$.\r\n\r\nTherefore, your inequality simplifies to\r\n\r\n$ \\sec ^{2}\\alpha \\plus{} \\sec ^{2}\\beta \\plus{} \\sec ^{2}\\gamma \\geq 4$.\r\n\r\nNoting that\r\n\r\n$ \\sec ^{2}\\frac {\\alpha \\plus{} \\beta \\plus{} \\gamma }{3} \\equal{} \\sec ^{2}\\frac {\\pi }{6} \\equal{} \\left( \\frac {2}{\\sqrt {3}}\\right) ^{2} \\equal{} \\frac {4}{3}$,\r\n\r\nwe see that the above inequality follows from the convexity of the function $ f: x\\rightarrow \\sec ^{2}x$ in the interval $ \\left( 0;\\;\\frac {\\pi }{2}\\right)$. In fact, here is, for the sake of completeness, a proof that this function is convex in this interval: If x and y lie in the interval $ \\left( 0;\\;\\frac {\\pi }{2}\\right)$, then we must show\r\n\r\n$ \\sec ^{2}x\\plus{}\\sec ^{2}y\\geq 2\\sec ^{2}\\frac{x\\plus{}y}{2}$.\r\n\r\nBy multiplying with $ \\cos ^{2}x\\cos ^{2}y$, we \"simplify\" this to\r\n\r\n$ \\cos ^{2}y\\plus{}\\cos ^{2}x\\geq \\frac{2\\cos ^{2}x\\cos ^{2}y}{\\cos ^{2}\\frac{x\\plus{}y}{2}}$.\r\n\r\nBut remembering\r\n\r\n$ 2\\cos \\frac{x\\plus{}y}{2}\\cos \\frac{x\\minus{}y}{2}\\equal{}\\cos x\\plus{}\\cos y$,\r\n\r\nour inequality becomes\r\n\\[ \\cos ^{2}y \\plus{} \\cos ^{2}x\\geq \\frac {2\\cos ^{2}x\\cos ^{2}y}{\\left( \\cos x \\plus{} \\cos y\\right) ^{2}/\\left( 4\\cos ^{2}\\frac {x \\minus{} y}{2}\\right) },\r\n\\]\r\nor\r\n\\[ \\left( \\cos ^{2}y \\plus{} \\cos ^{2}x\\right) \\left( \\cos x \\plus{} \\cos y\\right) ^{2}\\geq 8\\cos ^{2}x\\cos ^{2}y\\cos ^{2}\\frac {x \\minus{} y}{2}.\r\n\\]\r\nDenoting $ X \\equal{} \\cos x$ and $ Y \\equal{} \\cos y$, with $ 0\\leq X\\leq 1$ and $ 0\\leq Y\\leq 1$, and $ T \\equal{} \\cos ^{2}\\frac {x \\minus{} y}{2}$, with $ 0\\leq T\\leq 1$, the inequality we must prove becomes\r\n\r\n$ \\left( X^{2} \\plus{} Y^{2}\\right) \\left( X \\plus{} Y\\right) ^{2}\\geq 8X^{2}Y^{2}T$.\r\n\r\nNow, we know $ X^{2} \\plus{} Y^{2}\\geq 2XY$ and $ \\left( X \\plus{} Y\\right) ^{2}\\geq 4XY$; thus, $ \\left( X^{2} \\plus{} Y^{2}\\right) \\left( X \\plus{} Y\\right) ^{2}\\geq 8X^{2}Y^{2}\\geq 8X^{2}Y^{2}T$. And we are done.\r\n\r\nOf course, rather than proving the convexity of $ f: x\\rightarrow \\sec ^{2}x$ as I did, one could also bash it by calculus.\r\n\r\n Darij" } { "Tag": [ "combinatorial geometry", "geometry", "combinatorics", "length", "polygon", "IMO", "IMO 1982" ], "Problem": "Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.", "Solution_1": "There is sth wrong ith my idea:\n\nThe length of $X,Y>4*99-1$ is quite trivial in one direction, but if they ean the shortest distance, we just take points very close to the boundary and with angles close to $180$ degrees except in the corners, what is the mistake in those ideas?", "Solution_2": "See John Scholes' solution at [url]http://mks.mff.cuni.cz/kalva/[/url].", "Solution_3": "The condition should be $A_0\\neq A_n$. That's the statement appearing on the official IMO website.", "Solution_4": "The solution is difficult to understand because it involves compact sets. Can anyone explain the properties of compact sets" } { "Tag": [ "function", "quadratics", "algebra", "domain", "induction", "linear algebra", "matrix" ], "Problem": "Let\n$f(x)=\\frac{ax+b}{cx+d}$\n$F_n(x)=f(f(f...f(x)...))$ (with $n\\ f's$)\n\nSuppose that $f(0) \\not =0$, $f(f(0)) \\not = 0$, and for some $n$ we have $F_n(0)=0$,\nshow that $F_n(x)=x$ (for any valid x).", "Solution_1": "Consider the equation $f(x) = x$, which gives the fixed points of $f$:\r\n\\begin{eqnarray*} f(x) = \\frac{ax + b}{cx + d} &=& x \\\\ \\Leftrightarrow \\ ax + b &=& x(cx + d) = cx^2 + dx. \\end{eqnarray*}\r\n\r\nIf $c = 0$, then $f(x) = (ax + b)/d$, a linear function, so every iterate $F_n$ is also linear, and the above equation becomes\r\n\\[ ax + b = dx. \\]\r\nSuppose $a = d$. Then $f(x) = x + b/a$. If $b = 0$, then $f(0) = 0$, contradiction, so $b \\neq 0$. But then $F_n(x) = x + nb/a$, so $F_n(0) = nb/a$ can never be zero for $n \\ge 2$, contradiction, so $a \\neq d$.\r\n\r\nThat means the equation $ax + b = dx$ has a solution; in other words, $f$ has a fixed point, which means that $F_n$ has a fixed point for all $n$. We know this fixed point is not 0, since $f(0) \\neq 0$. But we also know that $F_n(0) = 0$ for some $n$. For this $n$, $F_n$ has at least two fixed points, and is linear, which means that $F_n(x) = x$ for all $x$.\r\n\r\nNow we can handle the case $c \\neq 0$. In this case, the above quadratic\r\n\\[ ax + b = cx^2 + dx \\]\r\nhas two roots (we can extend the domain of $f$ to the complex numbers), so $f$ has at least two fixed points. We know that 0 is not one of these fixed points, since $f(0) \\neq 0$.\r\n\r\nAgain as before, we know that $F_n(0) = 0$ for some $n$. For this $n$, $F_n$ has at least three fixed points, and by a straight-forward induction argument,\r\n\\[ F_n(x) = \\frac{a_n x + b_n}{c_n x + d_n} \\]\r\nfor some constants $a_n$, $b_n$, $c_n$, and $d_n$. Note that\r\n\\[ F_n(x) = x \\ \\Leftrightarrow \\ a_n x + b_n = x(c_n x + d_n). \\]\r\nBut the quadratic can only be satisfied by three different values if and only if it is identically 0, i.e. $c_n = b_n = 0$ and $a_n = d_n$. Hence, $F_n(x) = x$ for all $x$.\r\n\r\nNote that the condition $f(f(0)) \\neq 0$ is not used anywhere. This doesn't seem useful anyway, since you can check that $f(f(0)) = 0$ if and only if $b(a + d) = 0$, and $a + d = 0$ implies that $F_2(x) = x$ for all $x$.", "Solution_2": "Nice problem and solution :) .\r\nOnly one question :How do we know that $f(f(f(...f(x)...))$ is defined for any $x \\neq -\\frac{d}{c}$ ?\r\nMaybe its a stupid question cant see it :?", "Solution_3": "Alternatively, one could use the following result:\r\n\\[ f^{n}\\left( x\\right) =\\frac{a_{n}x+b_{n}}{c_{n}x+d_{n}}, \\]\r\nwhere $\\left( \\begin{array} [c]{cc} a_{n} & b_{n}\\\\ c_{n} & d_{n} \\end{array} \\right) =A^{n},$ the matrix being $A=\\left( \\begin{array} [c]{cc} a & b\\\\ c & d \\end{array} \\right) .$ This is easy to prove by induction.\r\n\r\nNow, if $f^{n}\\left( 0\\right) =0$ for some $n,$ then $A^{n}$ has the form $\\left( \\begin{array} [c]{cc} \\alpha & 0\\\\ \\beta & \\gamma \\end{array} \\right) .$ Since $A^{n}\\cdot A=A\\cdot A^{n},$ we obtain $a\\alpha +b\\beta=a\\alpha$ and $b\\gamma=b\\alpha.$ Since $f\\left( 0\\right) =b\\neq0,$ it follows that $\\beta=0$ and $\\alpha=\\gamma,$ so $f^{n}\\left( x\\right) =\\frac{\\alpha x}{\\alpha}=x$ for all $x.$", "Solution_4": "Well, in the day of this exam, the professor Edmilson Motta discovered that the condition $f(f(0)) \\not = 0$ is superfluous. The demo is along the last we post here...", "Solution_5": "Nicely done, enescu! I tried the matrix approach, but I couldn't get it to work myself." } { "Tag": [ "HCSSiM" ], "Problem": "I'm having some trouble writing my summer science program essay, when they ask \"Tell us something unique about yourself to help us understand you better as a person.\" what would be considered unique? Should I tell an anecdote about myself or state facts? would being bilingual consider unique, not anymore I guess. I like to solve problems and figure how things work, but I'll bet everyone whoes applying to SSP and on AoSP likes to do that. being the only speech club member on math team? They also ask me why math is important to me, the first answer that comes to mind is that I'm happy when I do math and that I like figuring how to solve interesting problems with twists. Its also important to me because I want to become an enginneer, not to mention math and my other favorite subject, physics, are intertwined. But my answer seems so bland and cliche. \r\n\r\nany advice?\r\nthanks", "Solution_1": "While your answer may seem bland and cliche to you, it's all in how you word it. I read all of the friendly letters submitted with HCSSiM applications, and most of them have the same basic themes---but they're all expressed in different ways. Your personal passion should come through (just let it out) and that's what I think the purpose of these questions is.", "Solution_2": "Anyone else applying to SSP?\r\nWhen do the results come?\r\nAccording to the CC guys it looks like a week or so.... :huh:" } { "Tag": [ "geometry", "inequalities", "inequalities unsolved" ], "Problem": "Let $ABCD$ be an inscribable quadrilateral of sides $a,b,c,d$ , diagonals $p,q$, semiperimeter $s$ and area $F$. Prove the two inequalities\r\n\r\n\r\n$ \\frac{1}{p^2} + \\frac{1}{q^2} \\leq (\\frac{s}{2F})^2 $\r\n\r\n\r\n\r\n$ 2pq > (\\frac{2F}{s})^2 $", "Solution_1": "[quote=\"manlio\"]Let $ABCD$ be an inscribable quadrilateral of sides $a,b,c,d$ , diagonals $p,q$, semiperimeter $s$ and area $F$. Prove the two inequalities\n\n\n$\\frac{1}{p^2} + \\frac{1}{q^2} \\leq (\\frac{s}{2F})^2$\n\n\n\n$2pq > (\\frac{2F}{s})^2$[/quote]\r\n\r\n :D :D" } { "Tag": [ "induction", "strong induction" ], "Problem": "I know the basic method of induction is to show that it's true for $n=1$, and then showing that if by assuming it's true for $n=k$, then it's true for $n=k+1$, thus showing it's true for all $n$.\r\n\r\nHowever, I have a question on a stronger form of induction: Is it still true you can still show it is true for $n=1$ and $n=2$, then assuming it's true for $n=k$ and $n=k+1$ you prove it's true for $n=k+2$? I believe it's a valid form of induction, but I'm unsure of whether it's actually true or not.", "Solution_1": "Yes! In fact, this stronger form of induction is called Strong Induction :D You can go even further, of course. You can assume it is true for $n=k, n=k+1,\\ldots, n=k+(j-1)$ and prove it is true for $n=k+j$. You have to show that the first $j$ base cases work as well.\r\n\r\nIf you understand why regular induction works, this should seem like a logical extension. There are even more ways you can use induction as well such as reverse induction and so forth (well, reverse is all I can think of right now...but I'm sure there are other well-known ways to induct).", "Solution_2": "Ah, cool, thanks.", "Solution_3": "[quote=\"joml88\"]Yes! In fact, this stronger form of induction is called Strong Induction.[/quote]\n\nNot quite true. Suppose we want to prove the statements P(1), P(2), P(3), ... . In strong induction, we assume that [b]all[/b] statements P(1), P(2), ..., P(k) are true to establish P(k + 1).\n\n[quote=\"Elemennop\"]However, I have a question on a stronger form of induction: Is it still true you can still show it is true for $n=1$ and $n=2$, then assuming it's true for $n=k$ and $n=k+1$ you prove it's true for $n=k+2$? I believe it's a valid form of induction, but I'm unsure of whether it's actually true or not.[/quote]\r\n\r\nThis is valid, because this argument shows that P(1) and P(2) imply P(3), and P(2) and P(3) imply P(4), and so on, and it is then \"clear\" that P(n) is true for all positive integers n. This is a fairly simple variation of the standard induction argument, so in a proof I would just call it induction." } { "Tag": [ "limit", "algebra", "binomial theorem", "calculus", "calculus computations" ], "Problem": "Calculate\r\n\r\n(a) $\\lim_{n\\rightarrow \\infty}(\\sqrt{4n^{2}+1}+\\sqrt{n^{2}+3n+1}-3n)$\r\n\r\n(b) $\\lim_{n\\rightarrow \\infty}(\\sqrt{x^{2}+x+1}-\\sqrt{x^{2}-2x+4})$\r\n\r\n(c) $\\lim_{n\\rightarrow \\infty}\\frac{\\sqrt{3x^{5}+1}}{\\sqrt[4]{7x^{10}+x-5}}$", "Solution_1": "Please reedit your text, Jumbler.", "Solution_2": "For (a) I'll use the binomial theorem:\r\n\r\n$\\sqrt{4n^{2}+1}=2n\\left(1+\\frac1{4n^{2}}\\right)^{\\frac12}= 2n\\left(1+\\frac12\\cdot\\frac1{4n^{2}}+O(n^{-2})\\right)$\r\n\r\n$=2n+\\frac14+O(n^{-1})$\r\n\r\n$\\sqrt{n^{2}+3n+1}=n\\left(1+\\frac3n+O(n^{-2})\\right)^{\\frac12}$\r\n\r\n$=n\\left(1+\\frac12\\cdot\\frac3n+O(n^{-2})\\right) =n+\\frac32+O(n^{-1})$\r\n\r\nSo, $\\sqrt{4n^{2}+1}+\\sqrt{n^{2}+3n+1}-3n$\r\n\r\n$=2n+n+\\frac14+frac32-3n+O(n^{-1})=\\frac74+O(n^{-1}).$\r\n\r\nThe limit is $\\frac74.$\r\n\r\nThe same methods will get us that the limit in (b) is $\\frac32.$", "Solution_3": "[quote=\"Kent Merryfield\"]For (a) I'll use the binomial theorem:\n\n$\\sqrt{4n^{2}+1}=2n\\left(1+\\frac1{4n^{2}}\\right)^{\\frac12}= 2n\\left(1+\\frac12\\cdot\\frac1{4n^{2}}+O(n^{-2})\\right)$\n\n$=2n+\\frac14+O(n^{-1})$\n\n$\\sqrt{n^{2}+3n+1}=n\\left(1+\\frac3n+O(n^{-2})\\right)^{\\frac12}$\n\n$=n\\left(1+\\frac12\\cdot\\frac3n+O(n^{-2})\\right) =n+\\frac32+O(n^{-1})$\n\nSo, $\\sqrt{4n^{2}+1}+\\sqrt{n^{2}+3n+1}-3n$\n\n$=2n+n+\\frac14+frac32-3n+O(n^{-1})=\\frac74+O(n^{-1}).$\n\nThe limit is $\\frac74.$\n[/quote]\r\nI maybe wrong here but I think the answer to [b](a)[/b] is $\\frac32.$\r\n\r\nFirst, I think it is possible to justify that \r\n\r\n$\\lim_{n\\rightarrow \\infty}(\\sqrt{4n^{2}+1}+\\sqrt{n^{2}+3n+1}-3n) = \\lim_{x\\rightarrow \\infty}(\\sqrt{4x^{2}+1}+\\sqrt{x^{2}+3x+1}-3x)$\r\n\r\nNow, we make the substitution $y = \\frac1{x}$, where $y > 0.$\r\n\r\nSo, $\\lim_{x\\rightarrow \\infty}(\\sqrt{4x^{2}+1}+\\sqrt{x^{2}+3x+1}-3x) = \\lim_{y \\to 0}\\frac{\\sqrt{4+y^{2}}+\\sqrt{1+3y+y^{2}}-3}{y}$\r\n\r\n$= \\lim_{y \\to 0}\\frac{\\frac{y}{\\sqrt{4+y^{2}}}+\\frac{3+2y}{2\\sqrt{1+3y+y^{2}}}}{1}$ ... (L'Hospital's rule)\r\n\r\n$= \\frac32$\r\n\r\n\r\nFor part [b](b)[/b] and [b](c)[/b], I would again use the same technique.", "Solution_4": "Oops, I see my mistake. Fixing:\r\n\r\n$\\sqrt{4n^{2}+1}=2n\\left(1+\\frac1{4n^{2}}\\right)^{\\frac12}= 2n\\left(1+\\frac12\\cdot\\frac1{4n^{2}}+O(n^{-3})\\right)$\r\n\r\n$=2n+\\frac1{4n}+O(n^{-2})$\r\n\r\n$\\sqrt{n^{2}+3n+1}=n\\left(1+\\frac3n+O(n^{-2})\\right)^{\\frac12}$\r\n\r\n$=n\\left(1+\\frac12\\cdot\\frac3n+O(n^{-2})\\right) =n+\\frac32+O(n^{-1})$\r\n\r\nSo, $\\sqrt{4n^{2}+1}+\\sqrt{n^{2}+3n+1}-3n$\r\n\r\n$=2n+n+\\frac32-3n+O(n^{-1})=\\frac32+O(n^{-1}).$\r\n\r\nThe limit is $\\frac32.$", "Solution_5": "Thanks.\r\n\r\nWhat about\r\n\r\n$\\lim_{x\\rightarrow \\infty}(\\sqrt{2x-\\sqrt{x^{2}-1}}-\\sqrt{x+\\sqrt{x+1}})$", "Solution_6": "$\\sqrt{2x-\\sqrt{x^{2}-1}}=\\sqrt{2x-(x+O(x^{-1}))}$\r\n\r\n$=\\sqrt{x+O(x^{-1})}=\\sqrt{x}(1+O(x^{-2}))^{1/2}= \\sqrt{x}+O(x^{-3/2})$\r\n\r\n----\r\n\r\n$\\sqrt{x+\\sqrt{x+1}}=\\sqrt{x+\\sqrt{x}+O(x^{-1/2})}$\r\n\r\n$=\\sqrt{x}\\left(1+x^{-1/2}+O(x^{-3/2})\\right)^{1/2}=\\sqrt{x}\\left(1+\\frac12x^{-1/2}+O(x^{-1})\\right)$\r\n\r\n$=\\sqrt{x}+\\frac12+O(x^{-1/2})$\r\n\r\n----\r\n\r\nSubtract these to get\r\n\r\n$-\\frac12+O(x^{-1/2})$\r\n\r\nSo the limit is $-\\frac12.$" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let x_0=1 and let x_(n+1)=3x_n+[x_n* \\sqrt 5]. What is x_2004?", "Solution_1": "[quote=\"harazi\"]Let $ \\ x_0\\equal{}1$ and let $ \\ x_{n\\plus{}1}\\equal{}3x_n\\plus{}\\left\\lfloor (\\sqrt {5}) x_n\\right \\rfloor.$ What is $ \\ x_{2004}?$[/quote]\r\n\r\n$ \\ x_0\\equal{}1\\implies x_1\\equal{}5,\\ x_2\\equal{}26,\\ x_3\\equal{}136,\\ x_4\\equal{}712,\\ x_5\\equal{}3728,$ etc.\r\n \r\nIt can be proved that $ \\ x_{n}\\equal{}6x_{n\\minus{}1}\\minus{}4x_{n\\minus{}2}$\r\n\r\n$ \\implies\\boxed{\\boxed{x_n\\equal{}\\left(\\frac {\\sqrt{5}\\plus{}2}{2\\sqrt 5}\\right)\\left(3\\plus{}\\sqrt 5\\right)^n\\plus{}\\left(\\frac {\\sqrt{5}\\minus{}2}{2\\sqrt 5}\\right)\\left(3\\minus{}\\sqrt 5\\right)^n}}$ or $ \\boxed{\\boxed{x_n\\equal{}\\frac {1}{2}\\left(\\left(3\\plus{}\\sqrt 5\\right)^n\\plus{}\\left(3\\minus{}\\sqrt 5\\right)^n\\right)\\plus{}\\frac {1}{\\sqrt 5}\\left(\\left(3\\plus{}\\sqrt 5\\right)^n\\minus{}\\left(3\\minus{}\\sqrt 5\\right)^n\\right)}}$" } { "Tag": [ "geometry", "calculus", "integration" ], "Problem": "$ \\triangle ABC$ has consecutive integral sides and the bisector of $ \\angle A$ ($ AB$ is the smallest and $ AC$ is the largest) meets $ BC$ in $ D$ while $ E$ & $ F$ are mid points of $ AC$ and $ BC$ respectively. If quadrilateral $ ADFE$ has an area that is the smallest non-zero, integer possible then find the sides of the triangle.", "Solution_1": "I found the numbers 3 , 4 , 5 (with Heron's formula).", "Solution_2": "[quote=\"mathbel\"]I found the numbers 3 , 4 , 5 (with Heron's formula).[/quote]\r\n\r\nIf the sides of the triangle be $ AB\\equal{}3, BC\\equal{}4 and AC\\equal{}5$ then one may easily verify the following: $ \\triangle ABD \\equal{} 2.25$, $ \\triangle EFC \\equal{} 1.5$ and $ QUAD ADFE \\equal{} 2.25$, which is not an integer.\r\n\r\nIf I may give you a hint, assume the sides to be $ (x\\minus{}1)$, $ x$ and $ (x\\plus{}1)$ and determine area of $ \\triangle ABC$. Now $ \\triangle ABD$ and $ \\triangle EFC$ can be determined in terms of $ x$ and thus one can write an equation for the area $ ADFE$ in terms of $ x$. Call this area $ n$. Now find an integral solution ($ n.x$) where $ n\\ge 0$. \r\n\r\n[$ n\\equal{}0$ actually gives $ 1, 2, 3$ as the solution but this isn't a triangle].", "Solution_3": "I thought the same way to solve this problem , but I took F as AB midpoint and ... :( .\r\n I try again and the result is AB=13 , BC=14 and AC=15 . Is it OK now??" } { "Tag": [ "induction" ], "Problem": "Let $a_1,\\cdots a_n$ be positive integers such that $a_1 \\leq \\cdots \\leq a_n$. Prove that\r\n\\[ \\frac{1}{a_1}+ \\cdots +\\frac{1}{a_n} =1 \\Rightarrow a_n<2^{2!} \\]", "Solution_1": "$2^{n!}$?", "Solution_2": "I remember this is a famous problem written in Problem-Solving Strategies from Arthur Engel. You can buy see it on http://www.amazon.com/gp/product/0387982191/102-0259262-9615320?v=glance&n=283155&n=507846&s=books&v=glance", "Solution_3": "A highly recommended book, i must add. i have it at home, and learnt basically all of my knowledge of olympiad level maths single-handedly from it. Its worth whatever price it is." } { "Tag": [ "probability" ], "Problem": "A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?\r\n\r\n$ \\textbf{(A)}\\ \\frac{1}{8} \\qquad\r\n\\textbf{(B)}\\ \\frac{5}{32} \\qquad\r\n\\textbf{(C)}\\ \\frac{9}{32} \\qquad\r\n\\textbf{(D)}\\ \\frac{3}{8} \\qquad\r\n\\textbf{(E)}\\ \\frac{7}{16}$", "Solution_1": "[hide]Following the rule of replacing any bead with a red bead, we have the possible orders RGG, GRG, or GGR. The sum of the probabilities is\n\\[ \\frac {1}{2} \\cdot \\frac {1}{2} \\cdot \\frac {1}{4} \\plus{} \\frac {1}{2} \\cdot \\frac {3}{4} \\cdot \\frac {1}{4} \\plus{} \\frac {1}{2} \\cdot \\frac {1}{4} \\cdot 1 \\equal{} \\frac {2}{32} \\plus{} \\frac {3}{32} \\plus{} \\frac {4}{32} \\equal{} \\frac {9}{32} \\rightarrow \\boxed{\\textbf{C}}.\n\\]\n[/hide]", "Solution_2": "You're allowed room for one red bead to pull out. (In fact, you must pull out at least one red bead...)\r\n\r\nCase RGG: $ \\frac 12 \\cdot \\frac 12 \\cdot \\frac 14 \\equal{} \\frac 1{16}$.\r\n\r\nCase GRG: $ \\frac 12 \\cdot \\frac 34 \\cdot \\frac 14 \\equal{} \\frac 3{32}$.\r\n\r\nCase GGR: $ \\frac 12 \\cdot \\frac 14 \\cdot 1 \\equal{} \\frac 18$.\r\n\r\nAdding up, $ \\frac 1{16} \\plus{} \\frac 3{32} \\plus{} \\frac 18 \\equal{} \\frac 9{32}$, which is choice $ \\mathbf C$.\r\n\r\nFrom this, I can tell that the 2003 AMC 10B was rather easy..." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Is it true that every positive rational number can be written in the form (a^2 + b^3)/(c^5 + d^7) where a,b,c,d are positive integers ?", "Solution_1": "Yes.", "Solution_2": "It seems really close to an ISL one.\r\n\r\nAs Myth says : 'Trust but check'.\r\nSo, come on Vess, give a proof.\r\n\r\nPierre.", "Solution_3": "Yes, thrust but check. But first, check the solved problems section, I am sure I have posted a solution for this problem." } { "Tag": [ "inequalities" ], "Problem": "An original permutable inequality of 3 positive variables\r\n\r\nConsider 3 positive real variables $ x, y, z$. Prove that:\r\n$ \\sum\\limits_{cyc} \\frac {x(y \\plus{} z)^2}{2x \\plus{} y \\plus{} z} \\geq \\sqrt {3(x \\plus{} y \\plus{} z)xyz}$", "Solution_1": "mods lock this topic this problem i suppose is from an ongoing competition and has been posted several times directly and indirectly and all have been locked\r\n[url=http://www.mathlinks.ro/viewtopic.php?t=168941]here[/url]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167965]here[/url]\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167949]here[/url]\r\n :huh:", "Solution_2": "You're false. It's from Uzebekistan MO 1998! We can solve it!", "Solution_3": "it may be from whatever but it's now a question from an ongoing c :) ompetition and if u don't beleive me contact the mods of inequalities section." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "Show that if $ f: X\\to \\mathbb{R}$ is measurable, and $ g: \\mathbb{R}\\to\\mathbb{R}$ is continuous, then $ g\\circ f$ is measurable.", "Solution_1": "$ (g(f(\\minus{}\\infty, c)))^{\\minus{}1} \\equal{} f^{\\minus{}1}(g^{\\minus{}1}(\\minus{}\\infty,c))$. Note that $ g^{\\minus{}1}(\\minus{}\\infty,c)$ is open in $ f(X) \\subset \\mathbb R$, then $ g^{\\minus{}1}(\\minus{}\\infty,c) \\equal{} f(X) \\cap G$ for some open set $ G$ in $ \\mathbb R$. Now $ f^{\\minus{}1}(g^{\\minus{}1}(\\minus{}\\infty,c)) \\equal{} X \\cap f^{\\minus{}1}(G)$." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\n\r\nI don't know how to proof the following proposition:\r\n\"Let $ p$ be a prim number and $ f \\in \\mathbb F_p[X]$ an irreducible polynomial with $ deg(f)\\equal{}m$. Then $ f$ divides the polynomial $ X^{p^m}\\minus{}X \\in \\mathbb F_p[X]$.\r\n\r\nMy first idea was, that there must exist a polynomial $ g \\in \\mathbb F_p[X]$ and $ f \\cdot g \\equal{} X^{p^m}\\minus{}X \\in \\mathbb F_p[X]$, but I think this is not the reasonable way to proof it. Then I thought, it is enough to show, that the roots of $ f$ are also the roots of $ X^{p^m}\\minus{}X \\in \\mathbb F_p[X].$ \r\nBut how to continue?\r\n\r\nBye,\r\nEdgar", "Solution_1": "Are you the one I talked to earlier today\u00bf :D \r\n\r\nTake a root $ a$ of $ f$ and consider the field $ \\mathbb F_p[a]$. Especially, what can you say about powers of $ a$ in the multiplicative group of that field(bigger hint: Fermat)\u00bf" } { "Tag": [ "inequalities", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $E={\\cal M}_n(\\mathbb R)$. We define for $A,B\\in E$ the scalar product: $(A|B)=\\textrm{tr}(^tA.B)$.\r\n\r\nProve that : $\\forall A, B\\in E,~~\\| AB\\|\\leq \\|A\\|\\ \\|B\\|$.", "Solution_1": "For any pair of $n\\times m$ matrices $A,B$ with entries $a_{ij},b_{ij}$ respectively, $\\text{tr}(A^TB)=\\sum_{i=1}^m\\left(\\sum_{j=1}^n a_{ji}b_{ji}\\right)$; the $ij$ entry of $A^T$ is $a_{ji}$, and that inner sum is the $ii$ entry of the product.\r\n\r\nIf we identify $M_{m\\times n}(\\mathbb{R})$ with $\\mathbb{R}^{mn}$ using matrices with only one nonzero entry as a basis, this inner product is the Euclidean inner product on $\\mathbb{R}^{mn}$. You should know how to prove the inequality over there.", "Solution_2": "More specifically, http://www.mathlinks.ro/Forum/viewtopic.php?t=84339 .", "Solution_3": "Thx a lot!" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Prove that: $ F_{3n}\\equal{}F_{n\\plus{}2}^3\\minus{}F_{n\\minus{}1}^3\\minus{}3F_{n}F_{n\\plus{}1}F_{n\\plus{}2}$", "Solution_1": "All of these problems are computations with binet's formula. I would be more interested in how these problems were thought of.", "Solution_2": "Hi! Use this problem we can prove Fibonacci 8 problem! \r\n[url]http://www.mathlinks.ro/viewtopic.php?t=182727[/url]" } { "Tag": [], "Problem": "In a $ 7\\times 7$ board $ 19$ cells are colored. We say that a row or column is colored if it contains at least four colored cells. How many colored rows and columns at most can there be in the board?", "Solution_1": "At most $ 4$ colored rows and at most $ 4$ colored columns gives $ 8$...", "Solution_2": "Right... what if $ 20$ cells are colored?" } { "Tag": [ "\\/closed" ], "Problem": "There are some pop ups that say there are new messages wating for you.\r\nBut sometimes mine say there are no new messages waiting for you!! :huh:", "Solution_1": "It happens when the message pops up, you navigate to a new page, and then click the back button.", "Solution_2": "sometimes i get that; dont hit the back button if u navigate to a new page or it'll mess everything up!" } { "Tag": [], "Problem": "What is the maximal number of intersection points that six circles can have?\r\n\r\nA. 15\r\nB. 24\r\nC. 28\r\nD. 30\r\nE. 36", "Solution_1": "[hide]The maximum occurs when each circle intersects (at two points) with all the others.\n2 x 6C2 = 30. D[/hide]", "Solution_2": "[quote=\"Done\"]maximal number [/quote]\r\n\r\nYou mean maximum, right? :) [/i]", "Solution_3": "well..it's from an Australian Contest :)", "Solution_4": "wouldnt it be infinity because if the circles were on top of eachother....", "Solution_5": "Well, by looking at the answer choices, you can assume they meant distinct circles.", "Solution_6": "would bezout's theorem not apply to this?", "Solution_7": "lol englighten us? bezout's theorem?\r\n\r\nalthough i agree with sky9073's method... just test a few out, and you'll see it works... \r\nactually for the intersection of n triangles, the maximum is twice the n'th triangular number... and the nth triangular number may be written as n(n+1)/2 or nC2\r\nso your answer is really 2*n(n+1)/2 = n(n+1) ORRR as sky said, 2*nC2" } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "(X,M,m) is a msr space, 1=1}0\r\nfor every set E in M (m(E) finite) as n----->inf prove that integral (f_n)(g)dm------>0 for every g in Lq, p and q conjugate. \r\nIf (X,M,m) is a sigma finite msr space, prove the same for p=1 and q=inf.", "Solution_1": "$(X,M,m)$ is a measurable space, $1 0$, one can find a simple function $e$ such that $|g-e|_q < \\epsilon$. Hence $|f_n g|_1 \\leq |f_n (g-e)|_1 + |f_n e|_1 \\leq |f_n|_p |g-e|_q + |f_n e|_1$. This is why, because of the boundedness of $f_n$ in $L_p$, $|f_n|_p 0$. This is precisely the desired conlusion." } { "Tag": [ "analytic geometry", "combinatorics open", "combinatorics" ], "Problem": "One of my friends asked me a question, actually i know the answer but just dont know how to tell it :D need some help:\r\n\r\nSuppose we have a set of persons. If we can find the similarity (as number) between each possible pairs of this set,then how can we show this numbers in graph? Is there any space in mathematics that we can use it for ploting the similarity between each pair. Based on my imagination, the possible $ X$ and $ Y$ axes are the set of persons.", "Solution_1": "What do you mean by \"similarity\"? Do you just mean some arbitrary metric on people?", "Solution_2": "[quote=\"t0rajir0u\"] Do you just mean some arbitrary metric on people?[/quote]\r\n\r\nsomethin' like that!!!", "Solution_3": "If the metric is the distance in some coordinate system (say, if we measure along various dimensions - age, height, political orientation), then plotting in that coordinate system is straightforward. Alternately, a [url=http://mathworld.wolfram.com/WeightedGraph.html]weighted graph[/url] serves the same purpose but seems to have less applications (to me).\r\n\r\nNote that measuring people using many dimensions we run into the [url=http://en.wikipedia.org/wiki/Curse_of_dimensionality]curse of dimensionality[/url].", "Solution_4": "Well, are the similarities are always computable ?\r\n\r\nWe can have lot of other constructs.\r\n\r\nSuppose a few like coffee, few like tea and a few like fruit juice. We create three sets $ S_c,S_t,S_f$ and a person can belong to any set. We have sieve methods to solve problems of this kind.\r\n\r\nSuppose we are looking for people who have crush on some one else. We define a ordered pair ($ a,b)$ if a has a crush on $ b$. For every $ b$ we can make a list of all the $ a$'s (they are similar in sense that they have crush on the same person)\r\n\r\nGeneralizing it, we can define ordered pairs of k-tuples of people. I feel that we have all these structures and more .. simple graphs, matchings, hypergraphs ...\r\n\r\nEither we have a well known structure to solve a concrete problem or we need to come up with them as the problem inspires." } { "Tag": [ "geometry", "Support", "inequalities", "ratio", "videos" ], "Problem": "In the topic about whether or not the usa empire will fall, i was wondering, what is considered the greatest non us power nowadays?\r\nNow I am not implying whether or not usa is the most powerful country, so if you think it is, give me your number two, if you don't give me number one\r\n\r\nPlease elaborate after doing the poll\r\nbecause the poll can never give all countries\r\ni just gave a few that have a high chance of including your choice\r\n\r\nSome of them have great historic influence, others have high population and have been chosen for that", "Solution_1": "By the word \"power\", do you mean which country has the greatest potential to be a superpower? Or which country is strongest militarily? Or economically?", "Solution_2": "It is common knowledge that in the next 50 years China will become the world's new super power, along with India if it ever manages to have its industrial revolution.", "Solution_3": "[quote=\"G-UNIT\"]It is common knowledge that in the next 50 years China will become the world's new super power.[/quote]I really really do not belive this :) \r\n\r\nI mean how many members does the Bush family have? Not enough to make presidency for 50 years I tell ya :D", "Solution_4": "Well i meant in general superpower, mix of economically, military..\r\n\r\nI do mean : today, nut in the near future", "Solution_5": "The question is slightly ill-phrased since the US is the only nation satisfying the conventional definition of \"super power\" at the moment.\r\n\r\nBut as for what non-US nation is most powerful [i]today[/i], I think it is Japan. They have not been using their power in aggressive ways; indeed, they barely impose any demands on other nations at all. They don't even want immigrants; [url=http://www.economist.com/world/asia/displaystory.cfm?story_id=5323427&no_na_tran=1]they're happier building robots to take their place[/url]. However, they clearly have the capacity to do more than any other non-US nation right now if they set their mind to it. For example, while Japan possesses no nuclear weapons, they could assemble one in four weeks if they needed to.\r\n\r\nIn a few decades this might change to China or India. But there are a lot of things that could go wrong with either nation (especially given the legacy of communism in China).", "Solution_6": "Why did you put Pakistan on that list? Because it's a nuclear power? Besides for that fact, I think there are several nations more powerful than Pakistan not on that list...", "Solution_7": "Well Pakistan is a nuclear power and it has got a huge population\r\nSome people, i hope they voted seriously, actually chose pakistan\r\nThere is still the other option but i admit that is not a very good alternative :blush: \r\nnothing however stops you from elaborating in a reply\r\n\r\nI always wondered how japan managed to win against china, russia ,korea,.. especially since china has got a lot more people.\r\n\r\nMy uncle keeps telling me not to underestimate the chinese, also economically\r\nwhen i ask him, much of the stuff they sell, like technology (like obscure brands of mp3 players) aren't that good (my sister had experiences with that), he tells me (dunno if it is true) japan went through the same thing : toyota at first wasn't much quality, and now it is top notch I hear\r\n\r\nAre japanese essentially different than chinese? or are chinese catching up?\r\n\r\nHere we have the clich\u00e9 of japanese that they are hard working, intelligent, serious people, who set their minds to it when they get started, and that they greet the flague of their company each morning\r\n\r\nMaybe i should have included israel and turkey , iran, thailand , vietnam and philippines as well (judging from population and nuclear power), but hey... I couldn't copy the entire list of nations on earth", "Solution_8": "[quote=\"fredbel6\"]I always wondered how japan managed to win against china, russia ,korea,.. especially since china has got a lot more people.[/quote]\n\nWhen they decided to modernize in the late 19th century, they did it more efficiently than any other nation on Earth.\n\nThis extremely impressive accomplishment is still worth studying today.\n\n[quote=\"fredbel6\"]My uncle keeps telling me not to underestimate the chinese, also economically\nwhen i ask him, much of the stuff they sell, like technology (like obscure brands of mp3 players) aren't that good (my sister had experiences with that), he tells me (dunno if it is true) japan went through the same thing : toyota at first wasn't much quality, and now it is top notch I hear[/quote]\n\nThe Toyota anecdote is accurate.\n\n[quote=\"fredbel6\"]Are japanese essentially different than chinese? or are chinese catching up?[/quote]\n\nThe genetic distance between Japanese and Chinese is roughly the same as that between Anglo-Saxons and Germans. Certainly not large enough to expect the Chinese to lack the potential to catch up, given their 10x population advantage.\n\nThe environmental/cultural difference between the nations was much greater until ~1980. Deng Xiaoping makes my personal list of top 5 individuals in human history because he did so much to change this. I hope China continues following in his footsteps.\n\n[quote=\"fredbel6\"]Maybe i should have included israel and turkey , iran, thailand , vietnam and philippines as well (judging from population and nuclear power), but hey... I couldn't copy the entire list of nations on earth[/quote]\r\n\r\nYour list was fine. I would have just listed China, Japan, Russia, and \"Other\". (India would make the list if we were speaking of the year 2056.) Maybe Saudi Arabia could be thrown in for thought provoking purposes.", "Solution_9": "Anyone who's read Michael Crichton's [i]Rising Sun[/i] will know that no nation stands a chance against Japan - that is, if the book's background information is entirely true. You never can tell with Crichton's books...", "Solution_10": "Michael Crichton does not have a reputation as a responsible provider of meaningful social analysis.", "Solution_11": "dog of justice, you certainly speak words that i recognise\r\nespecially if you consider that before their drastic revolution , they didn't have trains or anything\r\n(belhium had trains forty years earlier than japan) (i'm a rurouni kenshin fan, sometimes a bit over thte top, but it still provides a lot of information about japan and the meiji revolution)\r\n\r\nAbout Deng Xiaoping, I recently read an interview , i believe with robert kaplan, and there it was also claimed that deng xiaoping is underrated. He actually opposed mao didn't he?\r\n[img]http://en.wikipedia.org/wiki/Image:Destroy_liu-deng.jpg[/img]", "Solution_12": "Well anyway i am not surprised to see china on top in the poll, but to see only one person vote for russia i had not expected", "Solution_13": "I'll say that China and india really will take the world by storm within the next 50 years.[China's already begun]\r\n And India is very much well into it's Industrial revolution! What G-Unit just said is not quite the way things are going here.", "Solution_14": "The biggest non US super power......in my opinion...... is the European Union.", "Solution_15": "I would say that Japan [i]used to be[/i] one of the most powerful countries in the world, but we aren't as \"powerful\" as we used to be. Other Asian countries such as China and Korea are catching up with Japan; in fact, some of the recent economical/technological developments show that these countries have already started to have more power than Japan in some fields.\r\n\r\n[quote=\"fredbel6\"]Here we have the clich\u00e9 of japanese that they are hard working, intelligent, serious people, who set their minds to it when they get started, and that they greet the flague of their company each morning [/quote] It is true that the Japanese worked hard, but that might not be the only reason. Take the economical growth in the recent 50 years (especially 1950s-70s) for example. It was helped by the benefits from the special procurement (for Korean War, Vietnam War, etc.) as well as the cheap price of oil at that time. It was not just about the \"diligence\" of the Japanese :| \r\n\r\nI don't know what is going to happen in the future, but I'm sure that the power of eastern Asian countries will grow enormously anyway. China will become very powerful, and other Asian countries as well. Now the Japanese need to work hard again :roll:", "Solution_16": "[quote=\"frt\"]I would say that Japan [i]used to be[/i] one of the most powerful countries in the world, but we aren't as \"powerful\" as we used to be. Other Asian countries such as China and Korea are catching up with Japan; in fact, some of the recent economical/technological developments show that these countries have already started to have more power than Japan in some fields.[/quote]\r\n\r\nYes, in relative terms Japan was more powerful 20 years ago than it is now. Nevertheless, after spending most of my winter break this year in Beijing and Shanghai, and having been in Japan before, I don't think China has caught up to it quite yet (even taking its much greater size into account), though my answer may change in 10 years. And I think every other non-US country except for possibly Russia (because of its raw destructive power) is clearly behind Japan and/or China at the moment.\r\n\r\nClearly I am being massively outvoted, though. I'd be interested in one of the China voters providing a justification of why they think China is already ahead of Japan.", "Solution_17": "interestingly, someone from china just called the eu the greatest superpower\r\ni didn't list that because i meant nations\r\nwe aren't a single nation and we surely aren't going to become one very soon, we just tried to enforce a constitution about it\r\neven before it was fully written there was debate as to whether it should contain christian references as europe is a base for christianity( while belgium is in the stage of removing christian crosses from court houses...) , end in many countries referenda about it failed\r\n\r\nalso, right now the eu is expanding eastwards, this is going to be a setback i am afraid\r\npeople like valentin have said that themselves. Here in belgium we feel the effect already of cheaper polish, etc.. employees, like truck drivers\r\n\r\nas i said, we consider, economically, china as the super power in eu right now\r\n\r\non a military level, we don't have a european army...", "Solution_18": "[quote=\"Valentin Vornicu\"][quote=\"G-UNIT\"]It is common knowledge that in the next 50 years China will become the world's new super power.[/quote]I really really do not belive this :) [/quote]\r\n\r\nI do not believe that either. China doesn't even have free elections, a free press, or other attributes of truly powerful countries. I'd say it's rather doubtful that anything called \"China\" will even have effective territorial sovereignty over all the regions now combined under that name by fifty years from now. In particular, I expect to live to see a free Tibet and independent Eastern Turkistan, and Taiwan generally acknowledged to be a de jure independent country.", "Solution_19": "Hmm i have a couple of notes about that\r\n\r\nfirst of all what i meant when i started this poll, was biggest power on military, economical, etc level\r\nbut my main question was, after soviet union collapse, did another nation take its place\r\nyou see, when you look at democracy and freedom of speech, human rights, i am not so sure one could call the soviet union a super power (on the other hand, usa does not go along with everything about human rights, they do executions, the first bush election is still controversiel....)\r\n\r\nwhat is eastern turkistan, is there a relation to turkmenistan? guess not?\r\n\r\nHow central is the government of china? How autonomous are those provinces? Do they have governors like the united states of america have? And is it true that major politicans have roots in Beijing (Deng Xiaoping was not)\r\n\r\nThen one final question : as I noted before, there is an issue about one of China's most famous projects, the Three Gorges Dam, that Taiwan might have claimed it has the capability of destroying that dam (with rockets?) Are tensions really that high??\r\n\r\nDo taiwanese consider all chinese their enemies or just the communist party? Guess not seeing many still seek reunification", "Solution_20": "[quote=\"tokenadult\"][quote=\"Valentin Vornicu\"][quote=\"G-UNIT\"]It is common knowledge that in the next 50 years China will become the world's new super power.[/quote]I really really do not belive this :) [/quote]\n\nI do not believe that either. China doesn't even have free elections, a free press, or other attributes of truly powerful countries. I'd say it's rather doubtful that anything called \"China\" will even have effective territorial sovereignty over all the regions now combined under that name by fifty years from now. In particular, I expect to live to see a free Tibet and independent Eastern Turkistan, and Taiwan generally acknowledged to be a de jure independent country.[/quote]\r\n\r\nIt is true that China has severe limitations on civil liberties, but there is no reason why that would disqualify it as a superpower. Keep in mind that democracy in the modern sense did not come about until 300 years ago, and it was not a feature of many well-run states before then, including the Roman empire. Here in the U.S., we automatically assume that all countries should imitate our government, but this is just not always true. The best possible government depends on the current situation, and current governments will adapt to what the situation is.", "Solution_21": "But my point is that it really makes a country stronger to have good internal civil liberties. That makes the government of the country more aware of what is really going on, and it makes the citizenry more eager to fight to preserve the country. That's why I think smaller population countries with more civil liberty are generally stronger in world-politics power (as they often are economically also) than larger countries with less civil liberty.", "Solution_22": "China's future dominance is far from a foregone conclusion at this point. It may look easy for them to just \"continue on their current course\", which is in fact, broadly, all they need to do. Other East Asian nations have shown that authoritarianism is not actually incompatible with several decades of high economic growth. But every China enthusiast should listen to tokenadult when he says there are A LOT of things that could go wrong along the way.\r\n\r\nI see China as the big wild card in world history right now. It can be anything from a friendly competitor that drives the US and Japan to correct many of their inefficiencies, to the opponent in a new Cold War, to a Russia-style economic basket case.\r\n\r\nBut this is not something we have no control over. Quite the contrary -- maintaining US economic and moral leadership will go a very long way to ensuring China turns out okay.", "Solution_23": "[url=http://www.unpo.org/member.php?arg=21]Eastern Turkestan[/url] \r\n\r\n[url=http://www.uygur.org/]Eastern Turkistan Information Center[/url]", "Solution_24": "[quote=\"fredbel6\"]interestingly, someone from china just called the eu the greatest superpower\ni didn't list that because i meant nations\nwe aren't a single nation and we surely aren't going to become one very soon, we just tried to enforce a constitution about it\neven before it was fully written there was debate as to whether it should contain christian references as europe is a base for christianity( while belgium is in the stage of removing christian crosses from court houses...) , end in many countries referenda about it failed\n\nalso, right now the eu is expanding eastwards, this is going to be a setback i am afraid\npeople like valentin have said that themselves. Here in belgium we feel the effect already of cheaper polish, etc.. employees, like truck drivers\n\nas i said, we consider, economically, china as the super power in eu right now\n\non a military level, we don't have a european army...[/quote]\r\n\r\nwhat about Eurocorps? :P\r\n\r\n\r\nTo answer the question: China. It has a huge population and is moving in the right direction. It is by no means a sure thing, though. The EU is already about as powerful as the US economically, and I believe they still are moving towards more unity, even if it is going slowly. \r\n\r\nI would say that currently, world powers are:\r\n\r\nUS>EU>China>Japan>Russia>India.\r\n\r\nI don't actually know, of course.", "Solution_25": "capybaralet, eurocorps what?\r\n\r\nWell economically, as i already mentionned sometimes, what about textile and car industry moving to china? Factories are closed down and thousands of people fired as almost simultaneously several factories in china are announced.", "Solution_26": "If China didn't have such terrible leadership (In other words, their leaders are more concerned about maintaining their own power than the country's interests,) than they could easily become a super-power in a matter of years.", "Solution_27": "[quote=\"LordoftheMorons\"]If China didn't have such terrible leadership (In other words, their leaders are more concerned about maintaining their own power than the country's interests,) than they could easily become a super-power in a matter of years.[/quote]\nYou have no evidence supporting this kind of opinion, I believe. \n\n[quote]US>EU>China>Japan>Russia>India. [/quote]\r\nAs is known to all, Japan>China.", "Solution_28": "[quote=\"shobber\"][quote=\"LordoftheMorons\"]If China didn't have such terrible leadership (In other words, their leaders are more concerned about maintaining their own power than the country's interests,) than they could easily become a super-power in a matter of years.[/quote]\nYou have no evidence supporting this kind of opinion, I believe. \n\n[quote]US>EU>China>Japan>Russia>India. [/quote]\nAs is known to all, Japan>China.[/quote]\r\n\r\nPlease...About LordoftheMorons statement...this is completely false...do you know what Hu did in the years he got into power...\r\n\r\nChina like most other countries is a complex case...Its leader...esp. new leader must make a passive peaceful transition of power...ppl were actually afraid that jiang ze ming was not going to give up his seat as the commander of the army when Hu came out...\"In other words, their leaders are more concerned about maintaining their own power than the country's interests\"...this statement is meaningless...can you name one successful leader in the world that didn't view the preservation of their own power before others...since if they don't have the power they can't do no good so called \"their ppl\"...\r\n\r\nChina is improving in condition of civl rights and so call \"government cares about its ppl\"...before Hu...trapped miners, lacking bezene, hostages, and all this other stuff that relate to the ppl would never make it to TV...the national assembly would never be live...\r\n\r\nI understand this is a forum and we r free to express opinions...but please don't give out rash and unsupportabled info...\r\n\r\nNow about the real issue...about superpowers....China is definitely second...if and only if compared in all areas one by one with each of the other countries...(for example there can't be like a 5 on 1 war or an universal embargo)\r\n\r\nChina's economical reform have brought results...one of the few things haven't happened to the old communism countries of Soviet past...The economical improvements of china is increasing in double digits in GDA 3 and probably soon to be 4 years straight...something no country in the world had ever done...With the entry into the WTO and the \"full membership deal\"...have increased fabric trades in one quarter of what use to be 52 years...China is a country with high potentials of overtaking america...\r\n\r\nMilitary wise...china is not as powerful as the Russia, EU, or the USA technologies, but i believe China can get the technologies from favorable deals from the EU...EU is not second nor it is third...I would probably place it 4th in world power after Russia...explain that one later...EU on the outside is a strong union but there all more concerns with the economical reforms going on inside EU....the first constitution failed to pass..some of the wealther countries are concerned of cheap \"foreign labor\" and the poorer countries all afraid of foreign monoplies...therefore there is no true union...\r\n\r\nOn the otherhand China provides the favorable economical breeding grounds for foreign business since after all 1.3 billion ppl is a good consumer amount...China is also buying Airbus, magnet rails, military equipments, etc from booming European companies...in the billions...China have money...a huge saving....of trillions that can be thrown in paper tactics...to get favorable deals...\r\n\r\nChina also have strong support in Africa for China owns most of the oil and natural gas reserves in Sudan (China is protecting Sudan from UN sanction with that no vote)...and other african countries...China have also invested in much money in building a dry dock in Pakistan to issure oil can be transported to pakistan and transported by land pipelines to China...this provent blockades in the Strait of Malacca and the chain of island including Taiwan in the pacific used by USA to stunt China's growth...China also is using its UN seat to help Iran which is going to be an interesting case as USA embargoes Iran and China moves in to secure much needed oil supply currently under american companies...\r\n\r\nChina had also successfully found a pest in the american backyard as 2005 was a political landslide in South America...as Venezula, Bolivia,Cuba, Chile, Brazil, Argentine,etc turned socialist and blocked Bush's plan to create a American trade organization...and pushed him to form only a mini version in Cafto...Central American free trade organization. \r\n\r\nI stated earlier that Russia was strong than the EU...from the natural gas issue in the end of 2005 Russia had regained some national pride and a stricter control over ex-Soviet countries that r western leaning policies...With about half of the EU's energy source in Russia...Russia can easily crush the EU economically and a bitter winter that prevent direct attack...Russia is pretty strong...\r\n\r\nAs for Japan...it would be 5th for its strong economical achievements...yet it is still weak since its island landscape deprive it of a large population that would offset issues....but also that the island is resource deprived and without a successful import...it is doomed...as if a early blockade action against the tiny yet strong Japanese Navy...if successful...Japan is probably doomed...currently if a war does start you r not going to have 4 weeks to make that atomic bomb...you need it right away...for most likely wars r different now....\r\n\r\nAll the above are just my two cent....now to prep for AMC ....yawn...sleep...zzzzz", "Solution_29": "One thing that really scares Chinese social scientists about the future of China is the degree of regional economic inequality, and urban-rural economic inequality, which has increased greatly during the last two decades. The ratio of income inequality region to region now exceeds that found in Yugoslavia before Yugoslavia disintegrated. \r\n\r\nhttp://www.mtholyoke.edu/courses/sgabriel/economics/china-essays/6.html \r\n\r\nhttp://ideas.repec.org/a/aea/aecrev/v89y1999i2p306-310.html \r\n\r\nhttp://www.gwu.edu/~econ270/Taejoon.html \r\n\r\nhttp://ideas.repec.org/p/dgr/kubcen/200153.html \r\n\r\nhttp://repositories.cdlib.org/are_ucb/977/ \r\n\r\nhttp://www.wider.unu.edu/publications/rps/rps2004/rp2004-051.pdf \r\n\r\nhttp://www.zonaeuropa.com/20050621_2.htm", "Solution_30": "[quote=\"shobber\"]\nAs is known to all, Japan>China.[/quote]\r\n\r\n\r\n*sniff* i smell sarcasm", "Solution_31": "I personally think if US did fall (not that I am implying it would) then alot of countries would suddenly break out into civil war:\r\nI suspect that many labrorers would revolt because of mass layoffs (due to lack of demand from US for China) and would cause extreme unrest so I don ot think China is stable enough to be the next Super power.\r\n\r\nOn the other hand: I expect North Korea will be the first to seize power;\r\nIf u have not seen the 60 minutes artile on the insgiht on North Korea: please do and see the video also and [u]I think [/u]it is pretty obvious that from what 60 minutes got that North Korea is ready to jump at every opportuity.\r\n[url=http://www.cbsnews.com/stories/2006/01/13/60minutes/main1209741.shtml]www.cbsnews.com/stories/2006/01/13/60minutes/main1209741.shtml[/url] \r\n\r\nYet After this preliminary attack to seize power; I firmly believe Russia will reestablish the Soviet Union and came on top due to recent advances in the 20th century in spy sattelite technology.\r\nThough I expect some form of alliance with Saudi Arabia .\r\n\r\nThus, to sum it up I believe Russia is still the next superpower but is weak without allied countries", "Solution_32": "I think you missed an important possibility here: when the U.S. \"falls\" (as it inevitably will... someday), it's not necessarily going to be a sudden, devastating catastrophe. The U.S. isn't just going to go \"poof\" one day and disappear. More likely, it would be a gradual decline that takes several stages. Undoubtedly, it will put the world somewhat off-balance, but I doubt there will be any \"mass layoffs\" the way you describe it. I still think China is probably the primary candidate for the next \"superpower,\" but one can not discount countries like Japan and Russia. In response to tokenadult's post: China doesn't have the ethnic tensions or whatever that Yugoslavia had. While the inequality is still an issue, I wouldn't consider it a major one. We must remember that a nation doesn't have to meet all our ideals of living to be a strong one, even if meeting those ideals would be helpful. And I think China needs that economy boom to try to fix its problems in the long run.", "Solution_33": "[quote]In response to tokenadult's post: China doesn't have the ethnic tensions or whatever that Yugoslavia had.[/quote]\r\n\r\nI think that is a very interesting open question. To be sure, most Tibetans and Uygurs are distinctly cool about the remaining part of China, and while they are small in number, their traditional territory makes up a LARGE part of the territory of China. \r\n\r\nThe post-Qing Dynasty rulers of China, whether the KMT or the CPC, have tried hard to promote a sense of Han nationalism that embraces all the different Sinitic-language-speaking peoples of China, but I regularly hear comments from Chinese people about Chinese people from different regions that make me wonder how readily China might disintegrate. It's a historical fact that China has had many periods of regional separatism (some dynastic periods are named after the numerous regional kingdoms that arose in those eras), so it wouldn't be a new phenomenon if, for instance, the Cantonese-speaking regions of current China decided they would be better off under their own national administration than under an administration run from Beijing. A lot depends, as you note, on how well the common people of China make economic progress (and, I might add, personal liberty progress) during the next several years. It would be better for most people in China to make a Hungary-style transition to multiparty democracy than a Romania-style transition, but it would be worse than either of those choices, and more weakening to China's position in the world, not to make the transition at all. \r\n\r\nAFTER EDIT: My edit was just to fix a spelling error.", "Solution_34": "Tibet shouldn't be much a problem for China since during the transitional period of the past 50 years, most of tibet should have at least cooled down from the separation deal....\r\n\r\nChina have employed methods to slow or even prevent future separation radicals from happening like the Go Western program which on the outside is a wonderful program to sent just out of college people into the West, or where many so called other ethnic groups were \"forced\" into China throughout history. The program offer many benefits for ppl who goes and millions have gone since this prrgram started during the late 1990s...which is one way to melt the ethnic tension down through future cross marriages, cultural blending, etc....plus there is always the army...Saddly China had killed many of these resistance ethnic group down to a managable size during the 50s and 60s....\r\n\r\nThis is my guess since from history when Nixon visited China he told De Xiao Ping to open up more chances for the China to immigrate to other countries (trying to get radicals to absorb western idea and change stuff at home)...then De said how about i give you 100 million Chinese...The point was culture was on China's side...with 5000 years of it...and a sudden 100 million ppl whos kids will likely inter-marry and will soon cause the american culture to diffuse into the Chinese ones...\r\n\r\nChina have been conquered before - Mongols, parts by Japan....but honestly the Mongols failed to culturally conquer China in their 200 years, but was instead deeply influenced by the Chinese which is a good cause to the creation of the Inner Mongolia Province of China...And the Japanese failed...1) due to american forces in the pacific...2)even without one...China could have outlasted the Japanese...3) Japan took a time where China was weak...There was a fake puppet emperor created by the Dowager who resistanted Western influence...the betrayal of general Yuan and his own \"empire\"...Mr. Sun with his western influenced \"radicals\"...local warlords who took the chance to gain power....bandit everywhere....8 western forces just owned China in the Opium War (twice)...Portugal pirates on the open sea...etc...that is not today...so China should be the next Superpower" } { "Tag": [ "AMC", "AIME", "blogs" ], "Problem": "Given that: \r\nx1 + 4 x2 + 9 x3 + 16 x4 + 25 x5 + 36 x6 + 49 x7 = 1; \r\n4 x1 + 9 x2 + 16 x3 + 25 x4 + 36 x5 + 49 x6 + 64 x7= 12; \r\n9 x1 + 16 x2 + 25 x3 + 36 x4 + 49 x5 + 64 x6 + 81 x7= 123. \r\nFind 16 x1 + 25 x2 + 36 x3 + 49 x4 + 64 x5 + 81 x6 + 100 x7. \r\n\r\nI spent like an hour on this working and working out equations and got it and now i don't even remmember what all steps i used to do this :mad: :mad: How do you know what to start with in this equation and how to proceed?", "Solution_1": "Ooh, I just showed furious how to do this one Friday. I thought about writing the solution in my blog since I think it is a very nice problem (but got too busy).\r\n\r\n\\begin{eqnarray} x_1 + 4 x_2 + 9 x_3 + 16 x_4 + 25 x_5 + 36 x_6 + 49 x_7 &=& 1\\\\ 4 x_1 + 9 x_2 + 16 x_3 + 25 x_4 + 36 x_5 + 49 x_6 + 64 x_7 &=& 12\\\\ 9 x_1 + 16 x_2 + 25 x_3 + 36 x_4 + 49 x_5 + 64 x_6 + 81 x_7 &=& 123 \\end{eqnarray}\r\n\r\n\\begin{eqnarray*} (4)=(2)-(1) &=& 3x_1+5x_2+7x_3+9x_4+11x_5+13x_6+15x_7=11\\\\ (5)=(3)-(2) &=& 5x_1+7x_2+9x_3+11x_4+13x_5+15x_6+17x_7=111 \\end{eqnarray*}\r\n\r\n\\[ (6)=(5)-(4)=2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7=100 \\]\r\n\r\nNow we build back up.\r\n\r\n\\[ (7)=(6)+(5)=7x_1+9x_2+11x_3+13x_4+15x_5+17x_6+19x_7=211 \\]\r\n\r\n\\[ (7)+(3)=16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=\\boxed{334}. \\]", "Solution_2": "See #8 here: http://www.upal.ca/adeel/aime/1989.pdf", "Solution_3": "[hide=\"different approach\"]notice that we're trying to get the last line from some combination of the other 3.\nalso notice that all the coefficients are squares. thus, we have\n$a(n-1)^2 + b(n)^2 + c(n+1)^2 = (n+2)^2$.\n$an^2-2an+a + bn^2 + cn^2+2cn + c = n^2 + 4n + 4$\nmatch coefficients\n$n^2(a+b+c) = n^2 \\Rightarrow a+b+c = 1$\n$n(2c-2a) = 4n \\Rightarrow c-a = 2$\n$a + c = 4$\nFrom these 3 equations, we get $c = 3, a = 1, b = -3$\nso $3(123) - 3(12) + 1 = \\boxed{334}$[/hide]", "Solution_4": "This problem was also discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57745]here[/url].\r\n\r\nMy solution is fairly similar to alan's." } { "Tag": [ "function", "algebra solved", "algebra" ], "Problem": "Find all funtions F from N to N such that\r\n[tex]F(F(n))=n^2[/tex]", "Solution_1": "there are many such functions and they are the following:\r\ndivide the set of natural numbers which are not a perfect squares into two two sequences $A={a_1,a_2,...,a_n,...}$ and $B={b_1,b_2,...,b_n,...}$ such that the intersection of A and B is empty and the union of A and B is the whole set of natural numbers which are not a perfect square. then, set $F(a_n^{2^m})=b_n^{2^m}$ and $F(b_n^{2^m})=a_n^{2^{m+1}}$. it is easy to see that all these function satisfy the equation and that every function satisfying the equation is of that form.\r\n\r\nPeter", "Solution_2": "Dear Peter,\r\nSorry but I don't think that function is good.I assume that you said:\r\n[tex] F(a_{n}^{2^m})=b_{n}^{2^m} \\forall m \\in N [/tex]\r\nSo take $ m=1 $,we have:\r\n$ f(a_{n}^2)=b_{n}^2 $\r\nTake $a_{n}^2=b_{m}$ ,we have $F(b_m)=a_{m}^2$\r\nHence $b_{n}=a_{m}$ :)", "Solution_3": "[quote=\"Master\"]Take $a_{n}^2=b_{m}$[/quote]\r\n\r\nWe cannot do that since we are told that $a$ and $b$ are both sequences that do not contain any perfect squares.\r\n\r\nThe proof given is quite nice, but is missing a small indicator that F(1)=1 to provide completeness for the way F is defined.\r\n\r\nExample: $F(F(4))=F(F(2^2))$\r\nIf 2 is in $a$, then $F(2)$ is some element in $b$ such that $F(2)=b_n$ and $F(2^2)=b_n^2$ and $F(b_n^2)=2^{(2^2)}=16$", "Solution_4": "yeah, 1 is in neither of the sequences since 1 is a square but is not some power of a number which is not a square, so i didn't define it, that's true. we must have $F(1)=1$ since $F(1)=F(F(F(1)))=F(1)^2$. i assume 0 is not a natural number, anyway that case isn't hard to handle either.\r\n\r\nPeter", "Solution_5": "[quote=\"Peter Scholze\"]there are many such functions and they are the following:\ndivide the set of natural numbers which are not a perfect squares into two two sequences $A={a_1,a_2,...,a_n,...}$ and $B={b_1,b_2,...,b_n,...}$ such that the intersection of A and B is empty and the union of A and B is the whole set of natural numbers which are not a perfect square. then, set $F(a_n^{2^m})=b_n^{2^m}$ and $F(b_n^{2^m})=a_n^{2^{m+1}}$. it is easy to see that all these function satisfy the equation and that every function satisfying the equation is of that form.\n\nPeter[/quote]\r\nsorry i don't read it carefully" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "solve:\r\n$ a\\plus{}b\\plus{}c\\minus{}ab\\equal{}0$\r\n$ a,b,c\\in N$", "Solution_1": "equivalent transformation yields: $ c\\equal{}(a\\minus{}1)(b\\minus{}1)\\minus{}1$ ie. the solution space for the tripel $ (a,b,c)$\r\n\r\nover $ \\mathbb{N}\\setminus\\{0\\}$ is given by $ \\{(u,v,(u\\minus{}1)(v\\minus{}1)\\minus{}1)\\,|\\,10$ ,si $ x,y,z$ are in the interval I.", "Solution_1": "This inequality is well known as Popoviciu's...a very very nice one :D .If f is strictly convex the inequality becomes equality if $ x\\equal{}y\\equal{}z$\r\nWe supose that $ x \\geq y \\geq z$.We have $ x \\geq (x\\plus{}y\\plus{}z)/3 \\geq z$ we have to compare $ y$ with $ (x\\plus{}y\\plus{}z)/3$.We have 2 cases. [b]CASE 1[/b] We obtain the inequality by aplying Karamata to $ (x,y,\\frac{x\\plus{}y\\plus{}z}{3},\\frac{x\\plus{}y\\plus{}z}{3}, \\frac{x\\plus{}y\\plus{}z}{3},z)$ and $ (\\frac{x\\plus{}y}{2},\\frac{x\\plus{}y}{2},\\frac{x\\plus{}z}{2},\\frac{x\\plus{}z}{2},\\frac{y\\plus{}z}{2},\\frac{y\\plus{}z}{2}).$\r\nWe can aply Karamata's because we have the same sum of the components like:\r\n\\[ x \\geq \\frac{x\\plus{}y}{2} \\Longleftrightarrow x\\geq y\\]\r\n\\[ x\\plus{}y\\geq\\frac{x\\plus{}y}{2}\\plus{}\\frac{x\\plus{}y}{2} \\Longleftrightarrow x\\plus{}y \\geq x\\plus{}y\\]\r\n\\[ x\\plus{}y\\plus{}\\frac{x\\plus{}y\\plus{}z}{3} \\geq \\frac{x\\plus{}y}{2} \\plus{} \\frac{x\\plus{}y}{2} \\plus{} \\frac{x\\plus{}z}{2} \\Longleftrightarrow y\\geq \\frac{x\\plus{}z}{2}\\]\r\n\\[ x\\plus{}y\\plus{}\\frac{x\\plus{}y\\plus{}z}{3} \\plus{}\\frac{x\\plus{}y\\plus{}z}{3} \\geq \\frac{x\\plus{}y}{2} \\plus{} \\frac{x\\plus{}y}{2}\\plus{} \\frac{x\\plus{}z}{2} \\plus{}\\frac{x\\plus{}z}{2} \\Longleftrightarrow y \\geq {x\\plus{}z}{2}\\]\r\n\\[ x\\plus{}y\\plus{}\\frac{x\\plus{}y\\plus{}z}{3}\\plus{}\\frac{x\\plus{}y\\plus{}z}{3}\\plus{}\\frac{x\\plus{}y\\plus{}z}{3} \\geq \\frac{x\\plus{}y}{2} \\plus{} \\frac{x\\plus{}y}{2}\\plus{}\\frac{x\\plus{}z}{2}\\plus{}\\frac{x\\plus{}z}{2}\\plus{}\\frac{x\\plus{}z}{2} \\Longleftrightarrow y\\geq z\\]\r\n[b]CASE 2[/b] $ y<\\equal{}(x\\plus{}y\\plus{}z)/3 \\Longleftrightarrow y<\\equal{}(x\\plus{}z)/2.$We obtain Karamata's for $ (x,\\frac{x\\plus{}y\\plus{}z}{3},\\frac{x\\plus{}y\\plus{}z}{3},\\frac{x\\plus{}y\\plus{}z}{3},y,z)$ and $ (\\frac{x\\plus{}y}{2},\\frac{x\\plus{}y}{2},\\frac{x\\plus{}z}{2},\\frac{x\\plus{}z}{2},\\frac{y\\plus{}z}{2},\\frac{y\\plus{}z}{2})$ and the inequality can be aplied because the conditions are verified..." } { "Tag": [ "geometry", "circumcircle", "power of a point", "radical axis", "cyclic quadrilateral", "geometry unsolved" ], "Problem": "Points $P$ and $Q$ on side $AB$ of a convex quadrilateral $ABCD$ are given such that $AP = BQ.$ The circumcircles of triangles $APD$ and $BQD$ meet again at $K$ and those of $APC$ and $BQC$ meet again at $L$. Show that the points $D,C,K,L$ lie on a circle.", "Solution_1": "Let:\r\n$(APD)=o_{1}$\r\n$(BPD)=o_{2}$\r\n$(APC)=o_{3}$\r\n$(BQC)=o_{4}$\r\nLet $M$ be the point of the intersection of line $DK$ with line $AB$. Because $M$ lies on a radical axis of $o_{1}$ and $o_{2}$ it's easy to conclude from the given relation that $M$ is a midpoint of $AB$ (since $|MP|\\cdot|MA|=|MQ|\\cdot|MB|)$. Analogically, $M$ lies on a radical axis of $o_{3}$ and $o_{4}$, and it's obvious that it lies also on the radical axis of $o_{1}$ and $o_{3}$ and on the radical axis of $o_{2}$ and $o_{4}$. So it's clear that $M$ is a radical center of these $4$ circles and in particular we have:\r\n$|MK|\\cdot|MD|=|ML|\\cdot|MC|$\r\nwhich implies that $D, K, C, L$ is a cyclic quadrilateral (from the converse of the Power of the Point theorem).", "Solution_2": "more generally $AP=BQ$ isn't necessary.", "Solution_3": "We denote as $(O_{1},$ $(O_{2},$ $(O_{3},$ $(O_{4}),$ the circumcircles of the triangles $\\bigtriangleup APD,$ $\\bigtriangleup BQD,$ $\\bigtriangleup APC,$ $\\bigtriangleup BQC,$ respectively.\r\n\r\nSo, we have $K\\equiv (O_{1})\\cap (O_{2}),$ $L\\equiv (O_{3})\\cap (O_{4})$ and let $E,$ $F$ be, the intersection points of $(O_{2}),$ $(O_{3})$ $($ In my drawing the point $P,$ is between $A,$ $Q$ and $F,$ between $E$ and side line $AB$ of $ABCD$ $).$\r\n\r\nLet $M$ be, the intersection point of $AB,$ from the segment line $DK,$ as the radical axis of $(O_{1}),$ $(O_{2}).$\r\n\r\nSo we have that $(MP)\\cdot (MA) = (MK)\\cdot (MD) = (MQ)\\cdot (MB)$ $,(1)$\r\n\r\nFrom $(1)$ we conclude that the point $M,$ lies on the segment line $EF,$ as the radical axis of the circles $(O_{2}),$ $(O_{3}).$\r\n\r\nLet $M'$ be, the intersection point of $AB,$ from the segment line $CL,$ as the radical axis of $(O_{3}),$ $(O_{4}).$\r\n\r\nSo we have that $(M'P)\\cdot (M'A) = (M'L)\\cdot (M'C) = (M'Q)\\cdot (M'B)$ $,(2)$\r\n\r\nFrom $(2)$ we conclude that the point $M',$ lies also, on the segment line $EF.$ \r\n\r\nFrom $(1),$ $(2)$ $\\Longrightarrow$ $M'\\equiv M$ and that the segment lines $DK,$ $EF,$ $CL,$ are concurrent at one point, here the point $M,$ lies on the side line $AB.$ \r\n\r\nSo, we have that $(MK)\\cdot (MD) = (ML)\\cdot (MC)$ $,(3)$\r\n\r\nFrom $(3),$ we conclude that the points $C,$ $D,$ $K,$ $L,$ are concyclic and the proof is completed.\r\n\r\nKostas Vittas.", "Solution_4": "[quote=\"Umut Varolgunes\"]more generally $AP=BQ$ isn't necessary.[/quote]\nYes!\nDoing inversion to $A$ the radius isn't necessary for this problem hence the problem becomes like this [url=http://goo.gl/mc9cY][img]http://b1110.hizliresim.com/11/10/15/6349.jpg[/img][/url] and we know that $D',E',F',C'$ is concyclic because $DC$ becomes a circle by inversion. Then we will show that $K'\\in{B'L'}$ we know that $K'E'.K'D'=K'F'.K'C'$ because $E',D',C',F'$ are concyclic and $K'$ has equal power to circles and $BL$ is the radical axis of the circles so $K'\\in{B'L'}$ line hence $K',B',L'$ is collinear so $A,B,K,L$ are concyclic.\n[color=#FF0000][u][b]EDIT: [/b][/u][/color]\nIn the orginal problem $P$ is $E$, $Q$ is $F$, $A$ is $D$, $B$ is $C$" } { "Tag": [ "calculus", "function", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f:$ $ [0,+\\infty)\\rightarrow R$ differentiable on the $ [0,+\\infty)$. $ f(0)=0$ and also $ f''(x)>0$ $ \\forall x \\in (0,+\\infty)$.\r\nProove that $ g(x)=\\frac{f(x)}{x}$ is increasing on the $ (0,+\\infty)$.", "Solution_1": "$ g'(x)=(xf'(x)-f(x))/x^{2}$. The function $ xf'(x)-f(x)$ is equal to $ 0$ at $ x=0$, and its derivative is $ xf''(x)$.", "Solution_2": "A more geometric argument: $ f$ is strictly convex, so it lies strictly below its secant lines. If we had $ a>b>0$ with $ \\frac{f(a)}{a}\\le \\frac{f(b)}{b}$, then $ (b,f(b))$ would be above the line segment between $ (0,f(0))$ and $ (a,f(a))$, contradicting convexity.", "Solution_3": "Very Nice solutions!I will present to you and my solution;\r\n[quote]Let $ f:$ $ [0,+\\infty)\\rightarrow R$ differentiable on the $ [0,+\\infty)$. $ f(0)=0$ and also $ f''(x)>0$ $ \\forall x \\in (0,+\\infty)$.\nProove that $ g(x)=\\frac{f(x)}{x}$ is increasing on the $ (0,+\\infty)$.[/quote]\r\n\r\nAs $ f''(x)>0$ $ \\forall x \\in(0+\\infty)$ and $ f'$ is continuous on the $ [0+\\infty)$ (because $ f$ is twice differentiable) , therefore\r\n$ f'$ is increasing on the $ [0+\\infty)$.We have ;\r\n\r\n$ g'(x)=\\frac{f'(x)x-f(x)}{x^{2}}$ $ \\forall x \\in(0+\\infty)$\r\n\r\nApplying the Mean Value theorem on $ f$ on the $ [0,x]$ we take that exists $ \\xi \\in(0,x)$ such as \r\n$ f'(\\xi)=\\frac{f(x)-f(0)}{x-0}$$ \\Longleftrightarrow$ $ f(x)=xf'(\\xi)$ \r\n\r\nso $ g'(x)=\\frac{f'(x)x-xf'(\\xi)}{x^{2}}=\\frac{f'(x)-f'(\\xi)}{x}$. $ x>0$\r\nNow it's enough to proove that $ g'(x)>0$ $ \\forall x \\in(0,+\\infty)$ or $ f'(x)-f'(\\xi)>0$ $ \\Longleftrightarrow$ $ f'(x)>f'(\\xi)$ $ \\Longleftrightarrow$ $ x>\\xi$ (because $ f'$ is increasing) which is true :P", "Solution_4": "[quote=\"giannis\"] therefore\n$ f'$ is increasing on the $ [0+\\infty)$\n[/quote]\r\n\r\nI don't understand how you realized that, function as it is convex but for the you can't easily conclude that is increasing.\r\nCan you explain please?", "Solution_5": "the problem says that $ f''(x)>0$ $ \\forall x \\in (0,+\\infty)$ $ \\Longrightarrow$ $ f'$ is increasing. \r\n\r\nIt's the same as $ f'(x)>0$ $ \\Longrightarrow$ $ f$ is increasing. :)" } { "Tag": [], "Problem": "(1)\r\nHow many ways are there to put $ 2n\\minus{}2$ bishops on a $ n\\times n$ chess field so no two can capture each other (that is, no two are on the same diagonal)?\r\n\r\n(2)\r\nLet $ n\\ge 2$ and suppose you have $ n$ positive integers written on a blackboard. In one move, you may wipe out two integers and replace them both by their sum. For which $ n$ can this result in $ n$ equal positive integers?", "Solution_1": "1) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225452\r\n\r\n2) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225457", "Solution_2": "Ah, I see. :oops:" } { "Tag": [], "Problem": "How can the ketone $ RCOR'$ be converted into the ketone $ RR'CHCOCH_3$?", "Solution_1": "Maybe:\r\n\r\n1. Ylide $ \\ce{(Ph)3P\\plus{}C^{\\minus{}}HCH3}$\r\n\r\n2. Hydroboration $ \\ce{BH3}$ followed by $ \\ce{H2O2/OH\\minus{}}$\r\n\r\n3. Oxidation $ \\ce{CrO3}$ and $ \\ce{H3O\\plus{}}$", "Solution_2": "Yes, very good, that's one possible way. Can you propose other ways?", "Solution_3": "Is the other way shorter?", "Solution_4": "Yes it is." } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $f$ be a real-valued function of bounded variation over $[0,1]$. Show that \\[\\sup_{00$ now we need to check if there are solutions for $x<0$\r\n\r\nnow multiply -1 to both sides of this $x<0$ and you get $-x>0$\r\n\r\nok now back to the big original inequality:\r\n\r\n\\[\\frac1{x}+\\frac1{x^3}+\\frac1{x^5}+\\frac1{x^7}+\\frac1{x^9}+\\frac1{x^{11}}+\\frac1{x^{13}} \\leq \\frac7{x^7}\\]\r\n\r\nwe are going to do the same thing and multiply -1 to both sides of original inequality and get\r\n\r\n\\[\\frac1{-x}+\\frac1{-x^3}+\\frac1{-x^5}+\\frac1{-x^7}+\\frac1{-x^9}+\\frac1{-x^{11}}+\\frac1{-x^{13}} \\geq \\frac7{-x^7}\\]\r\nname this inequality (*)\r\n\r\ni hope you are following me here\r\n\r\nok, now we can rewrite (*) in this form\r\n\r\n\\[\\frac1{(-x)}+\\frac1{(-x)^3}+\\frac1{(-x)^5}+\\frac1{(-x)^7}+\\frac1{(-x)^9}+\\frac1{(-x)^{11}}+\\frac1{(-x)^{13}} \\geq \\frac7{(-x)^7}\\]\r\n\r\nbecause $-x^\\text{odd power}=(-x)^\\text{odd power}$. call this new inequality (**)\r\n\r\nbut, now we have all the terms in (**) [b]positive[/b]\r\n\r\nnow here is where we bring in AM-GM. ok you know that for a seven variables AM-GM $a,b,c,d,e,f,g$ all [b]positive numbers[/b], then\r\n\r\n\\[a+b+c+d+e+f+g\\geq 7\\sqrt[7]{abcdefg}\\]\r\n\r\nthis inequality is satisfied by all positive numbers that you can select for $a,b,c,d,e,f,g$\r\n\r\nso if you let \r\n\\[a=\\frac1{(-x)},b=\\frac1{(-x)^3},c=\\frac1{(-x)^5},d=\\frac1{(-x)^7},\\]\r\n\\[e=\\frac1{(-x)^9},f=\\frac1{(-x)^{11}},g=\\frac1{(-x)^{13}}\\]\r\n then\r\n\\[7\\sqrt[7]{abcdefg}=7\\sqrt[7]{\\frac1{(-x)^{1+3+5+7+9+11+13}}}\\]\r\n\\[=7\\sqrt[7]{\\frac1{(-x)^{49}}}=\\frac7{(-x)^7}\\]\r\nso inequality (**) is equivalent to AM-GM!!! \r\n\r\ntime to make a comparison:\r\n(**)\r\n\\[\\frac1{(-x)}+\\frac1{(-x)^3}+\\frac1{(-x)^5}+\\frac1{(-x)^7}+\\frac1{(-x)^9}+\\frac1{(-x)^{11}}+\\frac1{(-x)^{13}} \\geq \\frac7{(-x)^7}\\]\r\nAM-GM 7 variables\r\n\\[a+b+c+d+e+f+g\\geq 7\\sqrt[7]{abcdefg}\\]\r\n\r\nthey have the same form! \r\nsince all the conditions for AM-GM are satisfied and since there are no restrictions on $x$ other than $x<0$ then inequality is satisfied for all $x<0$. hope this helps (what a long post! im going to be yetti's apprentice for long posts :D ) :thumbup: :pilot:", "Solution_14": "Ah, great post!\r\n\r\nI completely understand now. Thanks for your time.", "Solution_15": "[quote=\"Palytoxin\"]but, now we have all the terms in (**) [b]positive[/b][/quote]\r\n\r\n:huuh: they look negative to me :?", "Solution_16": "[quote=\"chess64\"][quote=\"Palytoxin\"]but, now we have all the terms in (**) [b]positive[/b][/quote]\n\n:huuh: they look negative to me :?[/quote]\n\nthats because you are skimming thru my post, you have to pay attention to details\n\nlet me quote myself:\n[quote=\"Palytoxin\"]now multiply -1 to both sides of this $x<0$ and you get $-x>0$[/quote]", "Solution_17": "ahh... I didn't think carefully about $x$ negative :wallbash:", "Solution_18": "Let $\\frac{1}{x}=y\\ (x\\neq 0)$, we can rewrite the given inequality as follows.\r\n\r\n$y+y^3+y^5+y^7+y^9+y^{11}+y^{13}\\leq 7y^7$\r\n\r\n$\\Longleftrightarrow (y-y^7)+(y^3-y^7)+(y^5-y^7)+(y^9-y^7)+(y^7-y^7)+(y^{11}-y^7)+(y^{13}-y^7)\\leq 0$\r\n\r\n$\\Longleftrightarrow y(1-y^6)+y^3(1-y^4)+y^5(1-y^2)+y^7(y^2-1)+y^7(y^4-1)+y^7(y^6-1)\\leq 0$\r\n\r\n$\\Longleftrightarrow y(1-y^6)^2+y^3(1-y^4)^2+y^5(1-y^2)^2\\leq 0$\r\n\r\nSolving this for $y$, we obtain $1-y^6=1-y^4=1-y^2=0\\ or\\ y<0\\ and\\ y^3<0\\ and\\ y^5<0$\r\n\r\n$\\Longleftrightarrow y=\\pm 1\\ or\\ y<0$, the desired answer is $\\boxed{x=\\pm 1, x<0}$.\r\n\r\nkunny", "Solution_19": "haha i like kunny's proof.. very original", "Solution_20": ":)" } { "Tag": [ "function", "logarithms", "complex analysis" ], "Problem": "Hi, \r\nThis question was on a comprehensive exam last year.\r\nShow that the meromorphic function [e^(1/z) + (1/(1- e^(1/z))] on C\\{o} does not omit any value at z=0.", "Solution_1": "$w=e^{1/z}$ attains all nonzero values in any neighborhood of $z=0$. It remains to analyze the map $g(w)=w+1/(1-w)$. Being rational, $g$ is onto $\\mathbb C\\to \\mathbb C$. So we only need to check that $g(0)=1$ is covered. And it is: $g(2)=1$.", "Solution_2": "The only way to prove the first statement (without working from elementary principles--which could take a while) is to invoke Big Picard, correct? Is there a way to prove this without big Picard?", "Solution_3": "Big Picard may not be necessary when you have an explicit function. Any nonzero complex number can be written as $re^{i t}$ with $r>0$ and $t\\in \\mathbb R$. We have $e^{1/z_{k}}=re^{i t}$, where $z_{k}=\\frac{1}{\\log r+i(t+2\\pi k)}$, where $k\\in\\mathbb Z$ is arbitrary. As $k\\to\\infty$, $z_{k}\\to 0$." } { "Tag": [ "trigonometry", "invariant", "complex numbers", "number theory proposed", "number theory" ], "Problem": "prove that if $n\\in N$ isn't a power of a prime number then there is permutation $(i_1,i_2,...,i_n)$ of 1,...,n s.t :$\\sum_{k=1}^{n}k.\\cos (\\frac{2\\pi i_k}{n})=0$", "Solution_1": "dear sam-n \r\ni remember trying to solve this problem quite sum time back and dint make any headway\r\nmind giving me a small hint (as a private messges probably ) and let the others give it a try\r\nthanx", "Solution_2": "have you idea?", "Solution_3": "I will prove more: let $\\zeta= e^{\\frac{2 \\pi i}{n}}$, then we want to show that $\\sum_{k=1}^n \\zeta^{i_k}=0$ (by just looking at the real part, this solves the problem).\r\n\r\nLet $n=pq$ with coprime $pq>1$ (exist since $n$ is not power of a prime).\r\nLets set $\\xi_{s,t} = \\sum_{k=1}^q (pk+t) \\zeta^{pk+s} = \\sum_{k=1}^q pk \\zeta^{pk+s}$, where the last equality comes from $\\sum_{k=1}^q t \\zeta^{pk+s} = t \\cdot \\zeta^{s-p} \\cdot \\sum_{k=0}^{q-1} \\zeta^{pk} = t \\cdot \\zeta^{s-p} \\cdot \\frac{\\zeta^{pq}-1}{\\zeta^{p}-1}=0$ (here we need that $p>1$).\r\nAs we have seen, $\\xi_s : = \\xi_{s,t}$ is invariant under changing of $t$. Now $\\zeta^{lq} \\xi_s = \\xi_{s+lq}$. Thus $\\sum_{l=1}^p \\xi_{s+lq} = \\xi_s \\cdot \\sum_{l=1}^p \\zeta^{lq} = 0$ like above (now $q>1$ is needed).\r\nNow we know enough about these sums to conclude:\r\n$0= \\sum_{l=1}^p \\xi_{lq} = \\sum_{l=1}^p \\sum_{k=1}^q (kp+(l-p)) \\zeta^{lq+kp}$.\r\nNote that every value from $1$ to $n$ is archieved exactly once by $kp+(l-p)= (k-1)p+l$. Additionally, the value of $lq+kp$ gets every value $\\mod n$ exactly once by the chinese remainder theorem. Thus we get that \r\n\\[ \\sum_{l=1}^p \\sum_{k=1}^q (kp+(l-p)) \\zeta^{lq+kp \\mod n}=0 \\]\r\nis a sum of the desired type." } { "Tag": [ "function", "limit", "geometry", "rectangle", "analytic geometry", "real analysis", "real analysis unsolved" ], "Problem": "Prove or disprove that each function $f:\\mathbb{R}\\to\\mathbb{R}$ has at most countable number of discontinuos points of first kind ($x_0$ is a point of discontinuos of first kind iff there exist $\\lim_{x\\to x_0-0}$ and $\\lim_{x\\to x_0+}$).", "Solution_1": "It's true.\r\n\r\nFor each $n$, let $A_n$ be the set of points of discontinuity of the first kind in which the oscillation of $f$ is $\\ge\\frac 1n$. Let $x\\in A_n$. Then there is an $\\varepsilon>0$ s.t. both conditions $x0$ s.t. both conditions $x0$ s.t. $(x-\\varepsilon_x,x)$ contains no points from $A_n$, and if to each $x\\in A_n$ we associate a rational $q_x\\in(x-\\varepsilon_x,x)$, it's clear that this is an injection from $A_n$ to $\\mathbb Q$, meaning that $A_n$ is countable.", "Solution_11": "Can one find a function $f : \\mathbb R \\to \\mathbb R$ such that all $x \\in \\mathbb Q$ are points of discontinuity of the first kind?", "Solution_12": "Let $(r_n)_n$ be a enumeration of rationals form $[0,1]$\r\n\r\n$f(x)=\\sum_{r_n1$ so that \\[ \\int_{1}^{a} \\frac{e^{t}}{t} dt = \\frac{e^{a}}{a}.\\]", "Solution_1": "b) has already been discussed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40293[/url].\r\n\r\nAs to a), just diffferentiate and use the Fundamental Theorem of Calculus; you'll get $f'(x)=\\int_1^x \\frac{e^t}t\\,dt+x\\frac {e^x}x-e^x=\\int_1^x \\frac{e^t}t\\,dt>0$ if $x>1$. As to c), note that $f$ is continuous, $f(1)=-e<0$, $f\\to+\\infty$ as $x\\to+\\infty$, so, by the Intermediate Value Theorem, there must exist $a$ such that $f(a)=0$. Since $f$ is strictly increasing, there may be only one such $a$." } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let G be a non-abelian simple group and P a Sylow p-subgroup of G. Is normalizer N_G(P) of P in G a maximal subgroup?", "Solution_1": "Why should it be?\r\nLet $ P$ be a Sylow 2-subgroup of $ A_6$.\r\nThen $ P$ is also a Sylow 2-subgroup of $ A_7$, as $ 2\\nmid|A_7: A_6|\\equal{}7$.\r\nBut then $ 7$ is a fixed point of $ P$ in $ \\{1,2,3,4,5,6,7\\}$, and for any $ g\\in N_{A_7}(P)$ \r\nwe have $ g(7)$ also fixed point of $ gPg^{\\minus{}1}\\equal{}P$.\r\nSo if $ g(7)\\not\\equal{} 7$, then $ P$ would also have a fixed point in $ \\{1,2,3,4,5,6\\}$, \r\nwhich gives $ P$ conjugate to a subgroup of $ A_5$. (the stabilizer of $ 6$ in $ A_6$)\r\nThis contradicts $ 2\\mid |A_6: A_5|\\equal{}6$, and therefore $ |P|\\nmid |A_5|$.\r\n\r\nSo we must have $ 7$ a fixed point of $ N_{A_7}(P)$, and therefore $ N_{A_7}(P)\\le A_6$.\r\nBut $ A_6$ is simple, so $ P$ is not normal in $ A_6$, which gives $ N_{A_7}(P)\\equal{}N_{A_6}(P)\\lneq A_6\\lneq A_7$.\r\nSo $ N_{A_7}(P)$ is not maximal in $ A_7$." } { "Tag": [ "inequalities" ], "Problem": "$ 3|x| \\plus{} 2|x\\minus{}1| \\leq 4|2\\minus{}4x|$. is the way to solve this is by categorizing them in cases? like $ |x|$= $ x$ when $ x\\geq 0$ and so forth? or is there an easier solution? thanks! im not asking for the answer btw, im just checking for a simpler solution...", "Solution_1": "maybe squaring first of all may help you in many inequalitites with abolute values.", "Solution_2": "Note that squaring only works if both sides are positive, as $ f(x)\\equal{}x^2$ is only increasing on $ (0,\\infty)$." } { "Tag": [ "summer program", "PROMYS", "pigeonhole principle", "quadratics" ], "Problem": "Prove that for $p\\in\\mathbb{P}$ with $p>2$, there exists $x,y\\in\\mathbb{N}$ such that \\[x^{2}+y^{2}\\equiv-1\\bmod p.\\]", "Solution_1": "[hide=\"an attempt\"]\n\nCase 1: if $-1$ is a square in $\\mathbb Z_{p}$ -> obvious.\n\nCase 2: if $-1$ is not a square, we first will proove \n\n[b]Lemma:[/b] $\\exists x, y\\in\\mathbb Z_{p}^\\ast$ such that $x^{2}+y^{2}$ is not a square.\n\n[b]Proof:[/b]\nAssume the opposite : $\\forall x, y\\in\\mathbb Z_{p}$, $x^{2}+y^{2}$ is a square. Choose $x\\in\\mathbb Z_{p}^\\ast$, then there exist $x_{1}$, $x_{2}$, $...$ $\\in\\mathbb Z_{p}$ such that:\n\n$2x^{2}=x^{2}+x^{2}=x_{1}^{2}$,\n$3x^{2}=x^{2}+x_{1}^{2}=x_{2}^{2}$,\n$...$\n\nHence all multiples of $x^{2}$ are squares. Since multiplication by $x^{2}$ is an automorphism of $\\mathbb Z_{p}$, this implies that all elements of $\\mathbb Z_{p}$ are squares, which is false. (/lemma)\n\nNow fix $x$ and $y\\in\\mathbb Z_{p}^\\ast$ such that $x^{2}+y^{2}$ is not a square. Consider the set $E=\\{(zx)^{2}+(zy)^{2}|z\\in\\mathbb Z_{p}^\\ast\\}\\subset\\mathbb Z_{p}^\\ast$.\n\nI claim that $-1\\in E$. Indeed, $\\text{card}(E)=\\frac{p-1}2$. Furthermore, the set of squares $S=\\{x^{2}|x\\in\\mathbb Z_{p}^\\ast\\}$ has same cardinality. Hence $\\mathbb Z_{p}^\\ast$ is the disjoint union of $E$ and $S$.\n\nSince, by assumption, $-1\\not\\in S$, we conclude that $-1\\in E$, i.e. $\\exists z|(zx)^{2}+(zy)^{2}=-1$. QED\n[/hide]\r\n---\r\nzpz", "Solution_2": "This was in a PROMYS problem set, but I'm too lazy to remember the proof. I remember it uses pigeonhole on the fact that there are exactly $\\frac{p-1}{2}$ quadratic residues. The $p \\equiv 1\\mod{4}$ case is easy, because then -1 is a quadratic residue, and the $p \\equiv 3\\mod{4}$ case is easy if we assume the opposite, because we then get fewer than $\\frac{p-1}{2}$ quadratic residues.", "Solution_3": "[hide]If $p \\equiv 1 \\bmod 4$ just select $y=0$,\nif $p \\equiv 3 \\bmod 4$ Let $z$ be the smallest number which is not a square (obviously $z \\ge 2$ so $z-1$ can be written as $x^{2}$ \nBut it's easy to see that $-z$ is quadratic residue (-1 is non residue and (-1)(-1)=+1) . \nwe have $x^{2}+(-z) =-1$ (and -z is a square)\nQED [/hide]", "Solution_4": "Problem:\r\n\r\nGiven a prime $p$ and an integer $k$, consider the equation $x^{2}+y^{2}\\equiv k \\mod p$.\r\na) show that it has always a solution (thus all residues $\\mod p$ are sum of two squares).\r\nb) Find a necessary and sufficient condition such that there is a solution with $x,y \\not\\equiv 0 \\mod p$.\r\nc) Find the number of solutions in terms of $k$ and $p$ (also allowing $x, y \\equiv 0 \\mod p$; but that doesn't make a big difference)." } { "Tag": [], "Problem": "What does \"QED\" mean? Some people write this after the end of proofs...", "Solution_1": "It comes from Latin: \"quod erat demonstrandum\", which literally means \"that which was to be demonstrated\". So you can write QED at the end of a proof to show that the result required for the proof to be complete has been obtained. In other words, you can write QED so that you sound cool :P", "Solution_2": "QED stands for Quod Erat Demonstratum, or \"that which has been demonstrated\" (or something like that). It's, um, Latin.", "Solution_3": "Gah. TripleM beat me to it by a few seconds. Well, if you want to be fancy, you can also do [tex]\\blacksquare[/tex].", "Solution_4": "Or you can do W5, which stands for Which Was What We Wanted.", "Solution_5": "In some class handouts, I used to put a smiley-face at the end of proofs. It seemed like the right emotion. :)", "Solution_6": "One of my friends told me that he put a :) at the end of correct proofs, a :( at the end of incorrect proofs, and a :? at the end of proofs he wasn't sure about.", "Solution_7": "Our math class has only ten students in it, and one week we decided we would all use our own creative end of proof symbols -- one person used a stick figure with a voice bubble, and inside the voice bubble was a black square. I modified my black square into a television that was showing \"Q.E.D.\"", "Solution_8": "Finally! I've always wanted to know what that meant. Thanks.", "Solution_9": "Ahh.. I love how latin lives on in today's world. It makes me think that there is a point to taking it in high school... jk... i like latin a lot even though it is \"dead\"" } { "Tag": [ "function", "inequalities unsolved", "inequalities" ], "Problem": "let $ a;b;c>0$ and $ a\\plus{}b\\plus{}c\\equal{}3$; prove that: \r\n$ (3a\\plus{}\\sqrt{9a^2\\plus{}16})(3b\\plus{}\\sqrt{9b^2\\plus{}16})(3c\\plus{}\\sqrt{9c^2\\plus{}16})\\leq 512\\equal{}8^3$", "Solution_1": "I think Karamata works for funtion $ F(x)\\equal{}ln(3x\\plus{}\\sqrt{9x^2\\plus{}16})$ because $ (a,b,c)\\succ(1,1,1)$ so $ F(a)\\plus{}F(b)\\plus{}F(c)\\le 3F(1)\\equal{}3ln(8)\\equal{}ln(512)$" } { "Tag": [ "LaTeX" ], "Problem": "Hi\r\n\r\nI would like to make some tables in LaTeX. But I do not really know how to do so. Therefore I am requesting for you help to do so. And I would like tables with more than two horizontal boxes.\r\n\r\nAll the best", "Solution_1": "Visit this site's LaTeX guide at http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#tables which tells you all about tables.\r\n\r\nIf you'd like a visual table editor for Windows then have a look at [url=http://www.g32.org/latable/]LaTable[/url]. It makes designing LaTeX tables very easy indeed.", "Solution_2": "Ok thank you very much, your help is mostly appreciated." } { "Tag": [ "inequalities", "algebra proposed", "algebra" ], "Problem": "let $ z_i\\in \\mathbb{C}\\ ,\\ 1\\le i\\le n$ and $ \\sum_{i \\equal{} 1}^nz_i \\equal{} 0$ prove that for all$ z\\in \\mathbb{C}$:\r\n$ n\\le \\sum_{i \\equal{} 1}^n|z \\minus{} z_i|$", "Solution_1": "there is probably sth missing in the problem statement, set $ z_i\\equal{}0$, then $ n\\leq \\sum_{i\\equal{}1}^n|z\\minus{}z_i|$ iff $ |z|\\geq 1$, ie. it is not fulfilled $ \\forall z\\in\\mathbb{C}$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $\\large a;b;c > 0$. Prove that:\r\n$\\large \\sum \\frac{a(3b+3c-2a)}{2a^2+3bc} \\geq\\ 1$", "Solution_1": "The ineq is equivalent to\r\n\r\n$3(ab+bc+ca)(\\frac{1}{2a^2+3bc}+\\frac{1}{2b^2+3ac}+\\frac{1}{2c^2+3ab})\\geq 4$\r\n\r\nwhich is a well-known ineq posted by Vasc I think", "Solution_2": "[quote=\"SuperMath\"]Let $\\large a;b;c > 0$. Prove that:\n$\\large \\sum \\frac{a(3b+3c-2a)}{2a^2+3bc} \\geq\\ 1$[/quote]\r\nAt last, it's first your true and correct inequality." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Given any three triangles $ \\Delta_1,\\Delta_2,\\Delta_3$. Let $ H_1,H_2,H_3$ be the orthocenters;$ G_1,G_2,G_3$ the centroids;$ O_1,O_2,O_3$ the circumcenters. Prove that the centroids of the triangles $ H_1H_2H_3,G_1G_2G_3$ and $ O_1O_2O_3$ are collinear and their circumcenters are likewise collinear.", "Solution_1": "It is easy to prove the collinearity of the centroids of the triangles $ \\bigtriangleup H_{1}H_{2}H_{3},$ $ \\bigtriangleup G_{1}G_{2}G_{3},$ $ \\bigtriangleup O_{1}O_{2}O_{3},$ as an application on a well known theorem, already has been posted before in this forum.\r\n\r\nPlease see at the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url].\r\n\r\nAs for the collinearity of the circumcenters of the above triangles you mentioned, I think that it is not true, as I can see by the drawing and I will check more carefully before a definite answer.\r\n\r\nKostas Vittas." } { "Tag": [], "Problem": "In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?", "Solution_1": "We can use complementary counting. Let's find the number of ways they can be arranged if Wilma and Paul were sitting together. If we view them as one person, there are $ 6!$ ways to arrange those 6 people, times $ 2!$ ways to arrange Wilma and Paul. This gives $ 720 \\cdot 2 \\equal{} 1440$ ways with Wilma and Paul sitting together.\r\n\r\nWith no restrictions, there are $ 7!$ ways to arrange all the people. Thus our answer is $ 5040 \\minus{} 1440 \\equal{} \\boxed{3600}$.", "Solution_2": "[hide=My Solution]Without restrictions there are $7!=5040$, and the amount of cases where Wilma and Paul sit next to each other are: $$W, P,3,4,5,6,7$$ \n$$1,W,P,4,5,6,7$$\n$$1,2,W,P,5,6,7$$\n$$1,2,3,W,P,6,7$$\n$$1,2,3,4,W,P,7$$\n$$1,2,3,4,5,W,P$$\nAnd since we know each seat (1,2,3,4,5,6,7) has $5!=120$ different options so $120(5)$, and since W, P is can also be represented as P, W, (which are distinct cases), so we have $5040-120\\times 12 = 3600$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": ":?: an alphabet consists of 3 letters.some sequence of letters\r\nof length 2 0r more are prohibited,and all of the prohibited seq. have\r\ndifferent lengths.a word is a seq.of letters of any length.a correct\r\nword does not contain any prohibited seq.prove-there ara correct words\r\nof any length.(kvant)", "Solution_1": "Let $A(n)$ be the number of correct words of length $n$. We want to show that $A(n)>0$, but a much stronger statement is true: $A(n+1)>2A(n)$. Since $A(1)=3, A(2)=8$ this implies $A(n)>2^{n}$.\r\n\r\nNow we will prove $A(n+1)>2A(n)$. To do this, consider words of the form $x_{1}x_{2}\\ldots x_{n}l$, where $x_{1}\\ldots x_{n}$ is a correct word of length $n$ and $l$ is an arbitrary letter. It is easy to see that any correct word of length $n+1$ must be of this form, since its first $n$ letters must themselves form a correct word. Besides, there are $3A(n)$ words of this type ($A(n)$ ways of choosing $x_{1}\\ldots x_{n}$ and $3$ ways of choosing $l$). So we will be done if we can prove that, from these $3A(n)$ words, less than $A(n)$ are incorrect.\r\n\r\nTo do this, we can assume inductively that $A(n)>2A(n-1), A(n-1)>2A(n-2),$ etc. Now, why would a word of the form $x_{1}\\ldots x_{n}l$ be incorrect? In order to be incorrect, it must contain a prohibited sequence, but since the first $n$ letters form a correct word, no prohibited sequences can be found among them; so the prohibited sequence must be of the form $x_{i+1}\\ldots x_{n}l$ for some $i$.\r\n\r\nThus, an incorrect word of this type would be of the form $x_{1}\\ldots x_{i}B$, where B is a bad sub-word $n-i+1$ letters long. How many words like this can exist? For a given $i$, $n-i+1$ is fixed, so there's only one choice for $B$ (because there's only one prohibited sequence of each length); and there are $A(i)$ choices for the first $i$ letters, so at most $A(i)$ words of this type can exist. Since $i$ can take the values $1, 2, \\ldots, n-1$, the number of incorrect words is at most $A(n-1)+A(n-2)+\\ldots+A(1)+1$ (the $+1$ at the end is for the case when B is $n+1$ letters long).\r\n\r\nNow, using that $A(j+1)>2A(j)$, the above is less than $(0.5+0.5^{2}+...)A(n) q=0 and x=0", "Solution_2": "The solution is 0", "Solution_3": "[quote=\"Jan\"]10 MOD 5 = x\nthat means that 10=5q+x with q a natural number.\n=> q=0 and x=0[/quote]\r\nActually q=2, but that's probably what you meant anyways.\r\n\r\nsufyan, do you know what modular congruence is? It's a little like taking the remainder after division, instead fractions or decimals. You would say $a\\equiv b\\bmod{c}$ ($a,b,c\\in\\mathbb{Z}$) if $a=b+cx$ for some integer $x$.\r\n\r\nFor more information, go here:\r\nhttp://mathworld.wolfram.com/Modulus.html\r\nhttp://mathworld.wolfram.com/Congruence.html" } { "Tag": [ "modular arithmetic" ], "Problem": "The integers 1,...,2n are arranged in any order on 2n places numbered 1,...,2n. Now we add its place number to each integer. Prove that there are two among the sums which have the same remainder mod 2n", "Solution_1": "[hide]Let the permutation be $ a_{1}, a_{2}, ...a_{2n}$. Assume to the contrary; then the new sums form a complete residue class mod $ 2n$, and we have $ (a_{1}+1)+(a_{2}+2)+...+(a_{2n}+2n) \\equiv 1+2+...+2n = n(n+1) \\neq 0 \\pmod{2n}$, but $ a_{1}+a_{2}+...+a_{2n}+1+2+...+2n \\equiv 2n(n+1) \\equiv 0 \\pmod{2n}$. [/hide]", "Solution_2": "Minor correction: $ 1+2+\\ldots+2n = n(2n+1)$; you need the fact that $ 2n+1$ is odd to make that work." } { "Tag": [ "modular arithmetic" ], "Problem": "Use congruence theory to show that for $ n>\\equal{}1$\r\n(1) $ 7|(5^{2n} \\plus{} 3)(2^{5n \\minus{} 2})$.\r\n(2) $ 27|2^{5n \\plus{} 1} \\plus{} 5^{n \\plus{} 2}$.", "Solution_1": "1 should be $ 7 \\mid (5^{2n} \\plus{} 3)(2^{5n}\\minus{}2)$\r\n[hide=\"1\"]Mod 7, $ 5^{2n}\\plus{}3 \\equal{} 25^n \\plus{} 3 \\equiv 4^n \\plus{} 3$, which is equivalent to $ 0$ for odd $ n$. $ 2^{5n} \\minus{} 2 \\equal{} 32^n \\minus{} 2 \\equal{} 5^n \\minus{} 2$, which is equivalent to $ 0$ for even $ n$.[/hide]\n\n[hide=\"2\"]Mod 27, $ 2^{5n\\plus{}1} \\plus{} 5^{n\\plus{}2} \\equal{} 2 \\cdot 32^n \\plus{} 25 \\cdot 5^n \\equiv 2 \\cdot 5^n \\plus{} (\\minus{}2) \\cdot 5^n \\equal{} 0$[/hide]", "Solution_2": "[quote=\"MellowMelon\"]1 should be $ 7 \\mid (5^{2n} \\plus{} 3)(2^{5n} \\minus{} 2)$\r\n[hide=\"1\"]Mod 7, $ 5^{2n} \\plus{} 3 \\equal{} 25^n \\plus{} 3 \\equiv 4^n \\plus{} 3$, which is equivalent to $ 0$ for odd $ n$. $ 2^{5n} \\minus{} 2 \\equal{} 32^n \\minus{} 2 \\equal{} 5^n \\minus{} 2$, which is equivalent to $ 0$ for even $ n$.[/hide]\r\n\r\n I don't know if i understand you correctly, but how come that $ 4^n \\plus{} 3$, is 0 mod 7 if n = 3?", "Solution_3": "Okay, so my correction is wrong, not to mention both lines of my solution. It's true only when $ n \\equiv 1,4 \\pmod{6}$. Do you have the right version of the problem?", "Solution_4": "[quote=\"MellowMelon\"]Okay, so my correction is wrong, not to mention both lines of my solution. It's true only when $ n \\equiv 1,4 \\pmod{6}$. Do you have the right version of the problem?[/quote]\r\n\r\nThe $ \\minus{}2$ in the second binomial should be at the exponent, you have misread the problem. :D" } { "Tag": [ "search", "real analysis", "real analysis theorems" ], "Problem": "please, send me any e-book you have related to Limits. [looking for a link or a book] \r\n\r\nthanks anyway\r\ndl", "Solution_1": "[url=http://www.googlesyndicatedsearch.com/u/lamarmathnotes?q=Limit&sa=Search+The+Site]How about this?[/url]", "Solution_2": "Thanks! Thats ok.\r\n\r\ndl" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Find a polynomial such that $xP(x-1)\\equiv (x-3)P(x)$.", "Solution_1": "[quote=\"Inspired By Nature\"]Find a polynomial such that $xP(x-1)\\equiv (x-3)P(x)$.[/quote][hide] \\[x=3\\qquad\\Rightarrow\\qquad 3P(2)=0P(3)\\qquad\\Rightarrow\\qquad P(2)=0.\\] \\[x=0\\qquad\\Rightarrow\\qquad 0P(-1)=-3P(0)\\qquad\\Rightarrow\\qquad P(0)=0.\\] Let $P(x)=(x-2)(x)Q(x)$. Then \\[x(x-3)(x-1)Q(x-1)=(x-3)(x-2)xQ(x),\\] \\[(x-1)Q(x-1)=(x-2)Q(x).\\] \\[x=1\\qquad\\Rightarrow\\qquad 0Q(0)=-1Q(1)\\qquad\\Rightarrow\\qquad Q(1)=0.\\] Let $Q(x)=(x-1)R(x)$ or $P(x)=(x)(x-1)(x-2)R(x)$. Then \\[x(x-1)(x-2)(x-3)R(x)=(x-3)(x)(x-1)(x-2)R(x),\\] which is always true. Hence, \\[P(x)=x(x-1)(x-2)R(x),\\] where $R(x)$ is some polynomial.[/hide]", "Solution_2": "Does the $\\equiv$ means something special here?", "Solution_3": "[quote=\"rzsolt\"]Does the $\\equiv$ means something special here?[/quote]\r\n\r\nit means congruent", "Solution_4": "Yeah, like in modular arithmetic, but here we treated it just like an equal sighn, right? So why was it used here?", "Solution_5": "Does this imply that $P(x)$ has to be a polynomial of degree higher than $3$?\r\n\r\nBy the way, the $\\equiv$ sign was used to mean that the two polynomials are congruent.", "Solution_6": "Saying \"the $\\equiv$ sign means congruent\" is not helpful unless you explain what you think it means for two polynomials to be congruent.\r\n\r\n\r\nMost likely, in this case it meant to emphasize that they are equal as polynomials, i.e. not that they are equal for some value of $x$ but rather that they are equal for all values of $x$ (that is, they are the same polynomial).", "Solution_7": "[quote=\"JBL\"]Saying \"the $\\equiv$ sign means congruent\" is not helpful unless you explain what you think it means for two polynomials to be congruent.\nMost likely, in this case it meant to emphasize that they are equal as polynomials, i.e. not that they are equal for some value of $x$ but rather that they are equal for all values of $x$ (that is, they are the same polynomial).[/quote]\nThats exactly what the question means, thank you JBL. :)\n[hide=\"hint\"]Just find the roots of the given polynomial and you will find $p(x)$.[/hide]\n[quote=\"towersfreak2006\"] \nP(x)=x(x-1)(x-2)R(x),\nwhere $R(x)$ is some polynomial.[/quote]\r\nhmm, you might want to check that over. :read:", "Solution_8": "[quote=\"Inspired By Nature\"][quote=\"JBL\"]Saying \"the $\\equiv$ sign means congruent\" is not helpful unless you explain what you think it means for two polynomials to be congruent.\nMost likely, in this case it meant to emphasize that they are equal as polynomials, i.e. not that they are equal for some value of $x$ but rather that they are equal for all values of $x$ (that is, they are the same polynomial).[/quote]\nThats exactly what the question means, thank you JBL. :)\n[/quote]\r\n\r\nOr put in the simplest manner: $=$ denotes an equation, $\\equiv$ denotes an identity.", "Solution_9": "thats true too. :)", "Solution_10": "[quote=\"towersfreak2006\"] Hence, \\[P(x)=x(x-1)(x-2)R(x),\\] where $R(x)$ is some polynomial.[/quote]\r\n\r\nThis works when $R(x)=1$, but if $R(x)=x$, we get $x-1=x$. :maybe:", "Solution_11": "yeah, thats why I asked towersfreak2006 to check it over.", "Solution_12": "[hide=\"solution, and some other stuff\"]\n\n$P(x) = C*x(x-1)(x-2)$\n\nwhere $C$ is a constant,.... which is what towersfreak put, without the extra polynomial.\n\nIf could multiply $P(x)$ by a polynomial $R(x)$, then $R(x-1)=R(x)$ for all $x$, but then if $R(x)$ had any roots, it must have an infinite number of roots, which is a contradiction. Therefore the only polynomail you could multiply $P(x)$ by is just a constant term.\n[/hide]", "Solution_13": "[quote=\"calculuslover800\"][hide=\"solution, and some other stuff\"]\n$P(x) = C*x(x-1)(x-2)$\nwhere $C$ is a constant,.... which is what towersfreak put, without the extra polynomial.\nIf could multiply $P(x)$ by a polynomial $R(x)$, then $R(x-1)=R(x)$ for all $x$, but then if $R(x)$ had any roots, it must have an infinite number of roots, which is a contradiction. Therefore the only polynomail you could multiply $P(x)$ by is just a constant term.\n[/hide][/quote]\nI got $P(x)=x(x-1)(x-2)$. I didn't find the need to add a constant. All I did was simply find all the [hide=\"roots\"]0, 1, and 2[/hide] and multiply them together to get the value of $P(x)$. I don't think the $C$ is needed.", "Solution_14": "I'm not sure about this, but here's my solution:\r\n\r\n[hide]For $x=0$, we have $0=-3P(0)$, so $P(0)=0$.\nFor $x=1$, we have $P(0)=-2P(1)$, so $P(1)=0$.\nFor $x=2$, we have $2P(1)=-P(2)$, so $P(2)=0$.\n\nLet $P(x)=x(x-1)(x-2)Q(x)$, where $Q$ is a polynomial. Substituting, we obtain $x(x-1)(x-2)(x-3)Q(x-1)=(x-3)x(x-1)(x-2)Q(x)$. Therefore, for all $x\\ne 0,1,2,3$, we have $Q(x-1)=Q(x)$. Since $Q$ is a polynomial, $Q(x)$ is constant. Let $Q(x)=c$, so that $P(x)=cx(x-1)(x-2)$.\n\nPlugging this back into the original equation, we get $xc(x-1)(x-2)(x-3)=(x-3)cx(x-1)(x-2)$, which is true for all $x$, as desired.[/hide]", "Solution_15": "[quote=\"matt276eagles\"]$Q(x)$ is constant $\\implies Q(x)=0$[/quote]\r\n\r\nPlease explain your reasoning here. Is zero only existing constant? :roll:", "Solution_16": "[quote=\"Farenhajt\"][quote=\"matt276eagles\"]$Q(x)$ is constant $\\implies Q(x)=0$[/quote]\n\nPlease explain your reasoning here. Is zero only existing constant? :roll:[/quote]\r\n\r\nOops... I forgot that it was $R$ that had roots at $0,1,2$, not $Q$. I'll edit. :blush:", "Solution_17": "[quote=\"matt276eagles\"]I'm not sure about this, but here's my solution:\n[hide]For $x=0$, we have $0=-3P(0)$, so $P(0)=0$.\nFor $x=1$, we have $P(0)=-2P(1)$, so $P(1)=0$.\nFor $x=2$, we have $2P(1)=-P(2)$, so $P(2)=0$.\nLet $P(x)=x(x-1)(x-2)Q(x)$, where $Q$ is a polynomial. Substituting, we obtain $x(x-1)(x-2)(x-3)Q(x-1)=(x-3)x(x-1)(x-2)Q(x)$. Therefore, for all $x\\ne 0,1,2,3$, we have $Q(x-1)=Q(x)$. Since $Q$ is a polynomial, $Q(x)$ is constant. Let $Q(x)=c$, so that $P(x)=cx(x-1)(x-2)$.\nPlugging this back into the original equation, we get $xc(x-1)(x-2)(x-3)=(x-3)cx(x-1)(x-2)$, which is true for all $x$, as desired.[/hide][/quote]\r\nhmm, makes sense. Its a more specific solution than mine. Thanks everybody. :)" } { "Tag": [ "calculus", "derivative" ], "Problem": "A nuclide $X$ of decay constant $\\lambda_{1}$ decays to a nuclide $Y$ of decay constant $\\lambda_{2}$.Show that the instantaneous rate of decay of nuclide $Y$ in a sample is given by $\\frac{dN}{dt}=\\lambda_{1}N_{X}-\\lambda_{2}N_{Y}$ where $N_{X}$ and $N_{Y}$ are the instantaneous numbers of nuclei of $X$ and $Y$ respectively in the sample .\r\n\r\n\r\nI dont quite know the way to show .Anyone help?", "Solution_1": "First, define $N = N_{Y}-N_{X}$. (The net increase of $Y$ particles.)\r\n\r\nSecond, take the derivative, $\\frac{d N}{dt}= \\frac{dN_{Y}}{dt}-\\frac{dN_{X}}{dt}$\r\n\r\nThen it's simply a matter of substitution as:\r\n\r\n$\\frac{dN_{Y}}{dt}\\equiv-\\lambda_{2}N_{Y}\\\\ \\frac{dN_{X}}{dt}\\equiv-\\lambda_{1}N_{X}$\r\n\r\n$\\frac{dN}{dt}= \\lambda_{1}N_{X}-\\lambda_{2}N_{Y}\\,\\,\\, QED$", "Solution_2": "[quote=\"Dr. No\"]First, define $N = N_{Y}-N_{X}$. (The net increase of $Y$ particles.)\n\nSecond, take the derivative, $\\frac{d N}{dt}= \\frac{dN_{Y}}{dt}-\\frac{dN_{X}}{dt}$\n\nThen it's simply a matter of substitution as:\n\n$\\frac{dN_{Y}}{dt}\\equiv-\\lambda_{2}N_{Y}\\\\ \\frac{dN_{X}}{dt}\\equiv-\\lambda_{1}N_{X}$\n\n$\\frac{dN}{dt}= \\lambda_{1}N_{X}-\\lambda_{2}N_{Y}\\,\\,\\, QED$[/quote]\r\n\r\nisnt that $N_{Y} 3 . \\]", "Solution_1": "I don' think so.\r\nIf we take $b=c\\to 0,a\\to 1$? The RHS is $1$.", "Solution_2": "[quote=\"hungkhtn\"]I don' think so.\nIf we take $b=c\\to 0,a\\to 1$? The RHS is $1$.[/quote]\r\nno, I think three is fine. Anyway, this has been posted already.", "Solution_3": "Hmm... interesting. I took it from tuymaada post http://www.mathlinks.ro/Forum/viewtopic.php?t=44430, which seems all solutions are wrong..", "Solution_4": "Yeah... I'm wrong in computing.", "Solution_5": "[quote=\"indybar\"]Hmm... interesting. I took it from tuymaada post http://www.mathlinks.ro/Forum/viewtopic.php?t=44430, which seems all solutions are wrong..[/quote] \r\n I posted a solution on ML and I think Iura's solution is right ." } { "Tag": [ "topology", "graph theory" ], "Problem": "Let $ Y$ be the complement of the following subset of the plane $ \\mathbb{R}^2:$\r\n$ \\{(x,0) \\in \\mathbb{R}^2: x \\in \\mathbb{Z} \\}$\r\nProve that $ \\pi_1(Y)$ is a free group on a countable set of generators.\r\n\r\nI don't know how to start this problem. I just need a few helpful hints. Thanks in advance.", "Solution_1": "The generators can, of course, be identified with loops around each missing point.\r\n\r\nOne standard tool for identifying a fundamental group is the Seifert-Van Kampen theorem. Can you use this to show that the fundamental group of the slice $ \\minus{}n\\minus{}\\frac12< x< n\\plus{}\\frac12$ is free on $ 2n\\plus{}1$ generators? Then, can you show that the fundamental group of $ Y$ is the limit of these?", "Solution_2": "The Van Kampen theorem (as stated in May's \"A Concise Course in Algebraic Topology\", p. 17) works on covers by infinitely many open subsets; you don't need to make\r\na limiting argument after using it. Either way, there's some machinery needed (either in the proof of Van Kampen or on how $ \\pi_1$ behaves with\r\nrespect to limits and colimits.)\r\n\r\nEDIT: May have misread May a little, but the version in his text definitely at least shows that $ \\pi_1$ takes colimits of open sets \r\nto the colimit of the $ \\pi_1$s.", "Solution_3": "Actually, I take that back; $ Y$ deformation retracts to an infinite sequence of adjacent circles\r\n(I will call this space $ Z$). Try to find a covering space $ Z$. (Hint: the real line is a universal covering\r\nspace for a circle; the cayley graph for $ F_2$ is a covering for two adjacent circles, etc.\r\nTry looking at the cayley graph for $ F_\\infty$ ... and show this is a universal covering space for $ Z$.\r\nI haven't worked out all the details but this doesn't look too hard.)" } { "Tag": [ "floor function" ], "Problem": "Compute the sum of the frist 2006 numbers of the sequence 1,2,2,3,3,3,4,4,4,4.....", "Solution_1": "[hide]\n$1+2+2+3+3+3+\\ldots+n(n) = 1^2+2^2+3^2+\\ldots+n^2 = \\frac{(n)(n+1)(2n+1)}{6}$\nWe are looking for $1+2+3+\\ldots+(n-1) = \\frac{(n-1)(n)}{2} \\leq 2006$\nSolving, we find that $n=\\left\\lfloor\\frac{1+\\sqrt{16049}}{2}\\right\\rfloor = 63$\nThere are $\\frac{(63-1)(63)}{2}=1953$ terms in the original series up to $(n-1)=62$, leaving $2006-1953=53$ remaining terms when $n=63$. Thus, our sum is\n$\\frac{(62)(63)(125)}{6}+(53)(63)=\\boxed{84714}$\n\n[/hide]" } { "Tag": [ "LaTeX", "inequalities", "inequalities unsolved" ], "Problem": "In a very old post it is proven that if abc=1, then\r\n1/(1+a+b)+1/(1+b+c)+1/(1+c+a)<=1\r\nAuh Cuong had a nice proof by substituting a=x^3,b=y^3,c=z^3\r\nWhat about this, if xyz=1 ,\r\n1/(1+x^3+y^3)+1/(1+y^3+z^3)+1/(1+z^3+x^3)<=(xy+yz+zx)/(x^2+y^2+z^2)\r\nis this true?", "Solution_1": "Sorry Siuhochung If I say your ineq not true.\r\nYou can Take \r\n$a=k^5,b=k,c=k^{-6}$\r\nand You have:\r\n$\\frac{1}{1+b^3+c^3} \\ge \\frac{ab+bc+ca}{a^2+b^2+c^2}$\r\nBecause:\r\n$\\frac{1}{1+k^3+k^{-18}} \\ge \\frac{k^6+k^{-1}+k^{-5}}{k^{10}+k^2+k^{-12}}$\r\nIt will satisfy if $k \\rightarrow +infinitive$", "Solution_2": "ic, thanks :)", "Solution_3": "I think I have a nice solution for the problem.\r\nwe can prove 1/(1+a+b) <= a k/(a k+b k +c k )\r\nk=1/3\r\n\r\n\r\nIn addition,I have downloaded Latex installated it.But I can't find the mian pile.Who can help me? :)", "Solution_4": "Let a,b,c>0, abc=1.We have\r\n1/(1+a+b)+1/(1+b+c)+1/(1+a+c)<=1/(a+2)+1/(b+2)+1/(c+2)<=1<=1/(2a+1)+1/(2b+1)+1/(2c+1)", "Solution_5": "Let a,b,c>0, abc=1.We have \r\n1/(1+a+b)+1/(1+b+c)+1/(1+a+c)1/(a+2)+1/(b+2)+1/(c+2)\r\nthe ineq should come from 02-03 china team training.", "Solution_6": "[quote=\"zhaobin\"]In addition,I have downloaded Latex installated it.But I can't find the mian pile.Who can help me? :)[/quote]\r\nWhat do you mean \"main file\"? There are many files and all they are main :)\r\nYou need a LaTex Editor, like Texniccenter, to write and compile latex documents.", "Solution_7": "Can anyone prove 1/(1+a+b)+1/(1+b+c)+1/(1+a+c) <= 1/(a+2)+1/(b+2)+1/(c+2).", "Solution_8": "Are you assuming $abc=1$?", "Solution_9": "Yes!\r\nActually yesterday we had an exam in a math training camp and that was the hardest problem.", "Solution_10": "[quote=\"hardsoul\"]Can anyone prove 1/(1+a+b)+1/(1+b+c)+1/(1+a+c) <= 1/(a+2)+1/(b+2)+1/(c+2).[/quote]\r\n\r\nThis inequality is generally wrong (e. g. take a = b = c = 0.5). But if you assume that abc = 1, then this is the problem from http://www.mathlinks.ro/Forum/viewtopic.php?p=115579 . Maybe it is also true for all positive a, b, c such that $abc \\geq 1$ ?\r\n\r\n darij", "Solution_11": "what about this problem: abc=1 (a,b,c>0), prove that \r\n1/(a+b+1) + 1/(b+c+1) + 1/(c+a+1) > 1/2", "Solution_12": "Obviously, it is wrong.", "Solution_13": "i'm sure that it's not wrong ;)", "Solution_14": "What about $a=b=100$, $c=1/10000$?", "Solution_15": "sory, Myth!!\r\ni was stupid when let c=1 then take b close to 0 :blush: :blush: :blush:", "Solution_16": "[quote=Soarer]In a very old post it is proven that if abc=1, then\n1/(1+a+b)+1/(1+b+c)+1/(1+c+a)<=1\nAuh Cuong had a nice proof by substituting a=x^3,b=y^3,c=z^3\nWhat about this, if xyz=1 ,\n1/(1+x^3+y^3)+1/(1+y^3+z^3)+1/(1+z^3+x^3)<=(xy+yz+zx)/(x^2+y^2+z^2)\nis this true?[/quote]\nIn a very old post it is proven that if $abc=1$, then\n$$\\frac{1}{1+a+b}+\\frac{1}{1+b+c}+\\frac{1}{1+c+a}\\leq1$$\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=409496&hilit=Moscow+1997]Moscow Olympiads 1997[/url]" } { "Tag": [ "algebra", "function", "domain", "integration", "limit", "logarithms", "complex analysis" ], "Problem": "Find the domain and range of the two functions f(x) = x^x and g(x) = x^-x. Justify your answer.\r\n\r\nI could not find any way to solve this problem completely, given that I was unable to evaluate this function for irrational x with the method below as a definition for irrational exponents, as I am unfamiliar with integration. \r\n\r\n$ \\int_{1}^{y} \\frac {dt}{t} \\equal{} p\\int_{1}^{a} \\frac {dt}{t}$\r\n\r\nThis is how far I got for the domain of f(x) and g(x)\r\n\r\nSubsets of its domain that I found with overlaps:\r\n1) All real numbers greater than zero\r\n2) All x = a/b, a does not equal zero, where a and b have no common factors except 1, and b is odd. *\r\n\r\n*This is what throws me off. \r\n3) All x = a/b, where a and b have no common factors except 1, b does not equal zero, and b is even is undefined. \r\n\r\n2) and 3) are both infinite sets. Given two rational numbers on a number line, there is an infinite number of rational numbers in between those two numbers. So, f(x) is defined at an infinite number of points, and undefined at another infinite number of points...\r\n\r\nFor the range, f(x) appears to cover (-inf,inf), however, there appears to be holes in the range, but I am unable to determine their locations. \r\n\r\ng(x) has one noticeable gap in the range at [0, -0.698...), but I could not find others. \r\n\r\nIf anyone here is able to help, I would deeply appreciate it. :)", "Solution_1": "[quote=\"yanglunj\"]Find the domain and range of the two functions f(x) = x^x and g(x) = x^-x. Justify your answer.\n\nI could not find any way to solve this problem completely, given that I was unable to evaluate this function for irrational x with the method below as a definition for irrational exponents, as I am unfamiliar with integration. \n\n$ \\int_{1}^{y} \\frac {dt}{t} \\equal{} p\\int_{1}^{a} \\frac {dt}{t}$\n\nThis is how far I got for the domain of f(x) and g(x)\n\nSubsets of its domain that I found with overlaps:\n1) All real numbers greater than zero\n2) All x = a/b, a does not equal zero, where a and b have no common factors except 1, and b is odd. *\n\n*This is what throws me off. \n3) All x = a/b, where a and b have no common factors except 1, b does not equal zero, and b is even is undefined. \n\n2) and 3) are both infinite sets. Given two rational numbers on a number line, there is an infinite number of rational numbers in between those two numbers. So, f(x) is defined at an infinite number of points, and undefined at another infinite number of points...\n\nFor the range, f(x) appears to cover (-inf,inf), however, there appears to be holes in the range, but I am unable to determine their locations. \n\ng(x) has one noticeable gap in the range at [0, -0.698...), but I could not find others. \n\nIf anyone here is able to help, I would deeply appreciate it. :)[/quote]\r\n\r\nThe domain of $ f(x)$ and $ g(x)$ is $ \\mathbb{R}^ \\plus{} .$ (Otherwise, the definitions don't really make sense...)\r\n\r\nFirst, clearly $ f(x) > 0$ and $ \\lim_{x\\to\\infty} f(x) \\equal{} \\infty.$ Also $ f'(x) \\equal{} \\left(e^{x\\ln x}\\right)' \\equal{} \\left(x\\cdot\\frac {1}{x} \\plus{} 1\\cdot\\ln x\\right)x^x \\equal{} \\left(1 \\plus{} \\ln x\\right)x^x$ and $ f''(x) \\equal{} \\left(\\left(1 \\plus{} \\ln x\\right)^2 \\plus{} \\frac {1}{x}\\right)x^x > 0$ so $ f$ has a minimum if $ 1 \\plus{} \\ln x \\equal{} 0$ or $ x \\equal{} e^{ \\minus{} 1}$ which gives the range $ \\left[e^{ \\minus{} \\frac {1}{e}},\\infty\\right).$\r\n\r\nWe also have $ g(x) \\equal{} \\frac {1}{f(x)}$ which gives a range of $ \\left(0, e^{\\frac {1}{e}}\\right].$", "Solution_2": "[b]bertram[/b]'s solution is correct, but I would not want to take the second derivative. The formula $ f'(x)\\equal{}(1\\plus{}\\ln x)x^x$ already shows that $ f$ is decreasing for $ x 0$ and $ \\lim_{x\\to\\infty} f(x) \\equal{} \\infty.$ Also $ f'(x) \\equal{} \\left(e^{x\\ln x}\\right)' \\equal{} \\left(x\\cdot\\frac {1}{x} \\plus{} 1\\cdot\\ln x\\right)x^x \\equal{} \\left(1 \\plus{} \\ln x\\right)x^x$ and $ f''(x) \\equal{} \\left(\\left(1 \\plus{} \\ln x\\right)^2 \\plus{} \\frac {1}{x}\\right)x^x > 0$ so $ f$ has a minimum if $ 1 \\plus{} \\ln x \\equal{} 0$ or $ x \\equal{} e^{ \\minus{} 1}$ which gives the range $ \\left[e^{ \\minus{} \\frac {1}{e}},\\infty\\right).$\n\nWe also have $ g(x) \\equal{} \\frac {1}{f(x)}$ which gives a range of $ \\left(0, e^{\\frac {1}{e}}\\right].$[/quote]\r\n\r\nThose ranges seem to only apply to the positive portion of the graph, as f(-1) = -1 which lies below e^(-1/e), although I have not thought about using natural logarithms to solve it. Thank you =)\r\n\r\nSomething I found out while playing around with my calculator:\r\nThe graph of f(x) for x < 0 seems to be the union of g(-x)and -g(-x)\r\n\r\nThe graph of g(x) for x < 0 seems to be the union of f(-x)and -f(-x)\r\n\r\nFor the TI-83+: Put the viewing window as \r\nx: -3.76,0\r\ny: -2, 2\r\n\r\nand plot f(x) , g(-x), and -g(-x) on sequential. (set f(x) as dot)\r\n\r\nYou can repeat for g(x), f(-x), and -f(-x)", "Solution_4": "[quote=\"yanglunj\"][quote=\"bertram\"]\nThe domain of $ f(x)$ and $ g(x)$ is $ \\mathbb{R}^ \\plus{} .$ (Otherwise, the definitions don't really make sense...)\n\nFirst, clearly $ f(x) > 0$ and $ \\lim_{x\\to\\infty} f(x) \\equal{} \\infty.$ Also $ f'(x) \\equal{} \\left(e^{x\\ln x}\\right)' \\equal{} \\left(x\\cdot\\frac {1}{x} \\plus{} 1\\cdot\\ln x\\right)x^x \\equal{} \\left(1 \\plus{} \\ln x\\right)x^x$ and $ f''(x) \\equal{} \\left(\\left(1 \\plus{} \\ln x\\right)^2 \\plus{} \\frac {1}{x}\\right)x^x > 0$ so $ f$ has a minimum if $ 1 \\plus{} \\ln x \\equal{} 0$ or $ x \\equal{} e^{ \\minus{} 1}$ which gives the range $ \\left[e^{ \\minus{} \\frac {1}{e}},\\infty\\right).$\n\nWe also have $ g(x) \\equal{} \\frac {1}{f(x)}$ which gives a range of $ \\left(0, e^{\\frac {1}{e}}\\right].$[/quote]\n\nThose ranges seem to only apply to the positive portion of the graph, as f(-1) = -1 which lies below e^(-1/e), although I have not thought about using natural logarithms to solve it. Thank you =)\n\nSomething I found out while playing around with my calculator:\nThe graph of f(x) for x < 0 seems to be the union of g(-x)and -g(-x)\n\nThe graph of g(x) for x < 0 seems to be the union of f(-x)and -f(-x)\n\nFor the TI-83+: Put the viewing window as \nx: -3.76,0\ny: -2, 2\n\nand plot f(x) , g(-x), and -g(-x) on sequential. (set f(x) as dot)\n\nYou can repeat for g(x), f(-x), and -f(-x)[/quote]\r\n\r\nThe real pain with negative numbers is that you can't really expand it there, since it won't be defined for any irrational number and it will be discontinouus about everywhere. So it really makes sense to restrict it to $ \\mathbb{R}^\\plus{}$ (One could use De l'Hospital to include 0 with $ f(0)\\equal{}g(0) \\equal{} 1$).\r\n\r\nConclusion: Negative bases are baaaaaaaaad.", "Solution_5": "You can leave the negative integers in the domain", "Solution_6": "You can, but do you want to? Do you also want to say that $ \\minus{}\\frac13$ is in the domain, because $ (\\minus{}1/3)^{\\minus{}1/3}\\equal{}\\minus{}3^{1/3}$? I'd rather not to. For me, the natural domain of $ f(x)\\equal{}x^x$ is $ (0,\\infty)$.", "Solution_7": "Well its not an interesting problem if we dont try find these points!", "Solution_8": "[quote=\"mlok\"]You can, but do you want to? Do you also want to say that $ \\minus{} \\frac13$ is in the domain, because $ ( \\minus{} 1/3)^{ \\minus{} 1/3} \\equal{} \\minus{} 3^{1/3}$? I'd rather not to. For me, the natural domain of $ f(x) \\equal{} x^x$ is $ (0,\\infty)$.[/quote]\n\nThat's common sense =)\n\nThey are apparently discontinuous everywhere to the left of 0, as bertram pointed out. I went into it a bit today and thought\n\nFor all a and b satisfying above conditions\nf(a/b) is positive for all even a, and negative for all odd a\n\n[quote=\"bertram\"]\n\nThe real pain with negative numbers is that you can't really expand it there, since it won't be defined for any irrational number and it will be discontinouus about everywhere. So it really makes sense to restrict it to $ \\mathbb{R}^ \\plus{}$ (One could use De l'Hospital to include 0 with $ f(0) \\equal{} g(0) \\equal{} 1$).\n\nConclusion: Negative bases are baaaaaaaaad.[/quote]\r\n\r\nIt's not defined at irrationals? Thanks, I guess that sums up all numbers in R now and whether it's defined or not =)", "Solution_9": "$ x^x\\equal{}e^{x \\ln{x}}$ and thus is holomorphic except at zero, and probably along, say, the negative real axis. Basically, in Complex Numbers its continuous. So we can put any number in and get out a complex answer. Even for irrationals, you can take a limit of rational points that converge to it", "Solution_10": "[quote=\"seamusoboyle\"]$ x^x \\equal{} e^{x \\ln{x}}$ and thus is holomorphic except at zero, and probably along, say, the negative real axis. Basically, in Complex Numbers its continuous. So we can put any number in and get out a complex answer. Even for irrationals, you can take a limit of rational points that converge to it[/quote]\r\n\r\nOK, but if we use complex numbers we might just as well say that the domain is $ \\mathbb{C}.$ And then there's also the problem of multi-valued logarithm etc...", "Solution_11": "thats why we take out the negative real axis :lol: so were only working on one branch. I was just pointing out that it can still be continous, and defined for irrational values.", "Solution_12": "Bertram: when you're responding to the post directly preceding yours, it's bad form to quote it. That's what you are responding to by default, so all it does is makes the thread more difficult to read.\r\n\r\nIt's continuous in the complex numbers only as a multivalued function. I've discussed related questions several times before on the forum, so you might try a search.* The only negative points where it's defined as a real function are rational numbers with odd denominators (in particular, not for any irrational number). However, if we use the given definition of exponentiation (via integration), it's not defined anywhere left of the origin.\r\n\r\n\r\nIn somewhat related topics, I found http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=983130192&t=149758\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=983130192&t=77380\r\netc. (I was searching for my posts with the word \"complex\" in them.)" } { "Tag": [ "calculus", "integration", "inequalities", "function", "calculus computations" ], "Problem": "the function f(x) is continuos on $ [0,\\pi]$ satifying $ xf(y) \\plus{} yf(x) \\leq 1$ for all $ x \\in [0,1]$. prove that:\r\n$ \\int_0^1 f(x)dx \\leq \\frac {\\pi}4$\r\ni'm sorry if it was posted :blush:", "Solution_1": "i am sorry if i am wrong but donet we need any limits on `y` :( ?", "Solution_2": "If we suppose inequality holds for each x, y in [0,1] then put $ x \\equal{} sin(t), y \\equal{} cos(t)$, t in $ [0,\\frac{\\pi}{2}]$ and function $ G(t) \\equal{} F(sin(t)) \\minus{} F(cos(t)) \\minus{} t$ is decreasing on $ [0,\\frac{\\pi}{2}]$, having $ G'(t) \\leq 0$, where $ F(x) \\equal{} \\int_0^x f(u) du$, so $ G(0) \\geq G(\\frac{\\pi}{2})$. From here, $ F(1) \\leq \\frac{\\pi}{4}$\r\n\r\nOne interesting question is: can we find a function f with initial properties, such that $ F(1) \\equal{} \\frac{\\pi}{4}$ ?" } { "Tag": [ "calculus", "vector", "\\/closed" ], "Problem": "I am interested in buying these books/textbooks due to the fact that I feel like I am missing bits and pieces of information from algebra/geometry/problem solving in general. My friend says that they are a waste due to the fact that he does not understand any of the material inside the book (vol. 1). I've finished precalc with a little knowledge in calc, what books do you suggest I should purchase? I don't want to buy the expensive ($ \\$$50+) textbooks that I have easily mastered 90%+ of the material. But I do not want to purchase (vol 1.) and then not have a clue what they are talking about... So I'm just wondering what do I need to know in order to successfully understand (vol 1.)? What should I buy?\r\n\r\nThanks", "Solution_1": "If you do not understand something, how is it a waste of money? I believe you should buy volume 2 as volume is basic material (in fact, you probably learned the majority in middle school). Volume 2 covers introductory calculus, as well as vectors and advanced trig. \r\n\r\nTake a look at the table of contents for [url=http://www.artofproblemsolving.com/Books/Vol1/toc.pdf]volume 1[/url] and [url=http://www.artofproblemsolving.com/Books/Vol2/toc.pdf]volume 2[/url] to decide :) .", "Solution_2": "Are you getting these books to prepare for the AMC? If so, how do you do on the AMC?", "Solution_3": "I got Volume 1 over this past school year, when I was taking Algebra, and some parts were a bit difficult, all you have to do is reread and reread and try all the exercises. I'm going into 7th grade this year, and it's a challenge for me, but rewarding. I really recommend it.\r\n\r\nAlso, if you're still not sure about it, ask your friend if you can borrow it and take a look." } { "Tag": [ "geometry", "circumcircle", "conics", "projective geometry", "geometry unsolved" ], "Problem": "One circle meets BC at $ D_1, D_2$, meets CA at $ E_1, E_2$, meets AB at $ F_1, F_2$.\r\n\r\nAlso, $ D_1 E_1$ and $ D_2 F_2$ meets at L\r\n \r\n $ E_1 F_1$ and $ E_2 D_2$ meets at M\r\n\r\n $ F_1 D_1$ and $ F_2 E_2$ meets at N\r\n\r\nShow that AL, BM, CN meets at one point.", "Solution_1": "By Pascal theorem, $ A \\equiv E_1E_2 \\cap F_1F_2,$ $ L \\equiv D_1E_1 \\cap F_2D_2,$ $ X \\equiv D_2E_2 \\cap F_1D_1$ are collinear, let $ p_a \\equiv XAL.$ Similarly, if $ Y \\equiv E_2F_2 \\cap D_1E_1,$ $ Z \\equiv F_2D_2 \\cap E_1F_1,$ then $ B, M, Y$ are collinear and $ C, N, Z$ are collinear. Let $ p_b \\equiv YBM, p_c \\equiv ZCN.$ By Steiner theorem, the 3 Pascal lines $ p_a, p_b, p_c$ are concurrent.\r\n\r\nProof of Steiner theorem: Consider two arbitrary triangles $ \\triangle D_1E_1F_1, \\triangle D_2E_2F_2$ with a common circumcircle, their six sidelines $ F_1D_1 \\equiv XN,$ $ E_2F_2 \\equiv NY,$ $ D_1E_1 \\equiv YL,$ $ F_2D_2 \\equiv LZ,$ $ E_1F_1 \\equiv ZM,$ $ D_2E_2 \\equiv MX$ are tangent to a common conic (see [url]http://www.mathlinks.ro/viewtopic.php?t=233430[/url]). Therefore, the main diagonals $ XL, YM, ZN$ of the (non-convex) hexagon $ XNYLZM$ are concurrent by Brianchon theorem.", "Solution_2": "Dear Yetti and Mathlinkers,\r\nthank you for your nice proof inolving three Pascal's lines. I think that the point of concurs is a Steiner's points...\r\nDo you have a technik for chasing all the Steiner's points from starting with a cyclic hexagon?\r\nSincerely\r\nJean-Louis", "Solution_3": "Dear Jayme,\r\nsorry to disappoint you. I have never heard of this Steiner theorem about concurrent Pascal lines, until I tried this problem. After finding the 3 Pascal lines, I looked it up on [url]http://www.cut-the-knot.org/Curriculum/Geometry/PascalLines.shtml[/url], but I did not like the proof there. I think that given a cyclic hexagon, there are $ \\frac {1}{2} \\cdot \\left(\\begin{array}{c}6 \\\\\r\n3\\end{array}\\right) \\equal{} 10$ different combinations to pick the 2 triangles with common circumcircle, which would result in 10 Steiner points. But I know that there are 20 Steiner points. So, there should be 2 Steiner points per triangle pair. \r\n\r\nLet $ P \\equiv E_1F_1 \\cap E_2F_2,\\ \\ \\ Q \\equiv F_1D_1 \\cap F_2D_2,\\ \\ \\ R \\equiv D_1E_1 \\cap D_2E_1$\r\nand $ P^* \\equiv D_1E_2 \\cap E_1F_2,\\ Q^* \\equiv E_1F_2 \\cap F_1D_2,\\ R^* \\equiv D_1E_2 \\cap E_1F_2$\r\n(running out of alphabet).\r\n\r\n$ PP^*L,\\ QQ^*M,\\ RR^*N$ would be the other concurrent triple of Pascal lines per triangle pair ?", "Solution_4": "Dear Mathlinkers,\nfor an answer to my question, see\nhttp://perso.orange.fr/jl.ayme vol. 12 Hexagramma mysticum p. 54\nSincerely\nJean-Louis", "Solution_5": "This problem from China-2005 TST or NMO.\nEasy by cheva's theorem(trigonametric).", "Solution_6": "Dear Mathlinkers,\nnow xe can say that the point of concur is a Kirkman's point.\nSincerely\nJean-Louis", "Solution_7": "We have that by Pascal's theorem $ ABC $ and $ LMN $ are perspective triangles, so $ AL,BM,CL $ are concurrent by Desargues's theorem." } { "Tag": [ "number theory", "relatively prime" ], "Problem": "What is the largest score that is [u]not[/u] attainable in an (American) football game? (The different types of scoring are, in points, $2, \\ 3, \\ 6, \\ 7, \\ 8$)", "Solution_1": "So basically, we just have to consider 2,3,7. Then we can make a 3-dimensional mod chart.", "Solution_2": "[quote=\"cincodemayo5590\"]What is the largest score that is [u]not[/u] attainable in an (American) football game? (The different types of scoring are, in points, $2, \\ 3, \\ 6, \\ 7, \\ 8$)[/quote]\r\nHow do you get 2 and 8?", "Solution_3": "[quote=\"i_like_pie\"][quote=\"cincodemayo5590\"]What is the largest score that is [u]not[/u] attainable in an (American) football game? (The different types of scoring are, in points, $2, \\ 3, \\ 6, \\ 7, \\ 8$)[/quote]\nHow do you get 2 and 8?[/quote]\r\n\r\n2 is from a safety (the opposing team gets pushed back into their own endzone) and 8 is from when you get a touchdown, and then get the ball into the endzone instead of kicking it.\r\n\r\nHow do you do a 3-dimensional mod chart?", "Solution_4": "Isn't it 1? anything above 8 you get by adding 2s to 7 or 8, 5=2+3 and 4=2+2", "Solution_5": "Anything even can obviously be made. Anything odd, except 1 can too. Since anything odd can be expressed as 2k + 1 = 2(k-1) + 3. Except when k = 0.", "Solution_6": "Oh oops, I didnt see how trivial this problem was when i was writing it :oops: \r\n\r\nHere's a better problem: Get rid of the $2$-point scores, so you've got points in $3, \\ 6, \\ 7, \\ \\text{and }8$ points. What's the highest score not attainable then?", "Solution_7": "The point is to find a general formula like the Chicken McNugget formula for these sorts of problems.", "Solution_8": "Anything other than $1,2,4,5$ can be made be adding the correct number of $3$'s to one of $6,7,8$- which cover between them all residues $\\mod 3$.", "Solution_9": "Tim, you're right. \r\nPhelpedo, you're right also.\r\n\r\n(LOL i'm sorry, this problem has been pretty trivial.)\r\n\r\n[u]Ok, here's the actual problem:[/u] :D\r\n\r\nWhat is the largest number that cannot be expressed as a linear combination of relatively prime positive integers x, y, and z?", "Solution_10": "[quote=\"cincodemayo5590\"]Tim, you're right. \nPhelpedo, you're right also.\n\n(LOL i'm sorry, this problem has been pretty trivial.)\n\n[u]Ok, here's the actual problem:[/u] :D\n\n$\\text{What is the largest number that cannot be expressed as a linear combination of positive integers x, y, and z?}$[/quote]\r\n\r\nwell, if x,y,z are not relatively prime, then the answer is infinitely large", "Solution_11": "[quote=\"cincodemayo5590\"]Tim, you're right. \nPhelpedo, you're right also.\n\n(LOL i'm sorry, this problem has been pretty trivial.)\n\n[u]Ok, here's the actual problem:[/u] :D\n\n$\\text{What is the largest number that cannot be expressed as a linear combination of positive integers x, y, and z?}$[/quote]\r\ni think I heard somewhere(stanford math circle, i think) that this problem was unsolved(meaning there's no nice closed form or w/e) when x,y,z are pairwise coprime(or something like that)", "Solution_12": "[quote=\"junggi\"][quote=\"cincodemayo5590\"]Tim, you're right. \nPhelpedo, you're right also.\n\n(LOL i'm sorry, this problem has been pretty trivial.)\n\n[u]Ok, here's the actual problem:[/u] :D\n\n$\\text{What is the largest number that cannot be expressed as a linear combination of positive integers x, y, and z?}$[/quote]\ni think I heard somewhere(stanford math circle, i think) that this problem was unsolved(meaning there's no nice closed form or w/e) when x,y,z are pairwise coprime(or something like that)[/quote]\r\n\r\nok. thanks!" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Does there exist $ f: N\\rightarrow N$ such that\r\n$ f(f(n))\\equal{}n\\plus{}2[\\frac{n}{2}]\\plus{}1$ for all $ n\\leqslant 2006$", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=185272]Yes[/url]." } { "Tag": [ "function", "limit", "algebra", "domain", "real analysis", "real analysis unsolved" ], "Problem": "Find all $f : ] 0, \\infty [ \\mapsto \\mathbb{R}$ s.t. $f(x) - x \\to 0$ as $x \\to \\infty$ and $f(f(f(x)))+2x=f(3x)$.", "Solution_1": "[quote=\"perfect_radio\"]Find all $f : ] 0, \\infty [ \\mapsto \\mathbb{R}$ s.t. $f(x) - x \\to 0$ as $x \\to \\infty$ and $f(f(f(x)))+2x=f(3x)$.[/quote]\r\n\r\n :blush:", "Solution_2": "It's a nice one ;). First considr $\\alpha(x)=f(x)-x$ so your condition can be reformulated as:\r\n\\[ \\lim_{x\\longrightarrow \\infty}\\alpha(x)=0\\quad and\\quad \\alpha(3x)=3\\alpha(x) \\] \r\nNow for $x\\in]0,\\infty[$, we have $\\alpha(3^nx)=3^n\\alpha(x)\\longrightarrow 0$ as $n\\longrightarrow\\infty$, that means that $\\alpha(x)=0$, then $f$ must be the identity.", "Solution_3": "A problem which is solved in the same way:\r\n Let $f: (0,\\infty)\\to\\Re$ such that :\r\n $1)f(x)+2x=f(3x),\\forall x>0$\r\n $2)\\lim_{x\\to\\infty}(f(x)-x)=0$\r\n Determine $f$.", "Solution_4": "No, you solved only that one from the shortlist, which unfortunately had a typo. The original problem (that I submitted and which is posted by Perfect Radio) is trickier. ;) Needless to say, the solution is wrong (for the original problem, the other one being of course trivial).", "Solution_5": "You are right, I couldn't sleep until finished it :P . The correct reformulation is:\r\n\\[ \\lim_{x\\longrightarrow \\infty}\\alpha(x)=0\\quad and\\quad \\alpha(3x)=\\alpha(x)+\\alpha(f(x))+\\alpha(f(f(x))) \\] \r\nSince $f$ takes positives values, $f(3x)=f(f(f(x)))+2x>2x$ or $f(x)>(2/3)x$.\r\nNow if $f(x)>\\beta x$, $f(f(f(x)))>\\beta^3x$ then $f(3x)>\\beta^3x+2x$ or \\[ f(x)>\\beta x\\Longrightarrow f(x)>\\frac{\\beta^3+2}{3}x \\] Consider $h(\\beta)=\\frac{\\beta^3+2}{3}$, then $f(x)>(2/3)x\\Longrightarrow f(x)>h(2/3)x\\Longrightarrow f(x)>h(h(2/3))x\\ldots$ that is \\[ \\forall\\,n\\in\\mathbb{N},\\,f(x)>h^n(2/3)x \\] \r\nSince $\\{h^n(2/3)\\}\\subseteq [0,1]$ is monotone it must converges to a fix point of $h$ so it must be 1, then $f(x)\\geq x$ or $\\alpha(x)\\geq 0$. Now the reformulation of the second condition implies \r\n\\[ \\alpha(3x)\\geq \\alpha(x) \\] \r\nSo the same argument will work :lol: (for each $x$ we have $0\\leq\\alpha(x)\\leq \\alpha(3^nx)\\longrightarrow 0$ as $n\\longrightarrow\\infty$)", "Solution_6": "[quote=\"hcast\"]Since $f$ takes positives values, $f(3x)=f(f(f(x)))+2x>2x$ or $f(x)>(2/3)x$.[/quote]\r\n\r\nThe problem doesn't state this.", "Solution_7": "mmm.. it must, if not how can you iterate it?, in other words if $f(f(f(x)))+2x=f(3x)$ have sence for all $x>0$ as I tought, $f(x)$ must be in the domain of $f$, $]0,\\infty[$, in other case there must be a condition about the $x$ st $f(f(f(x)))+2x=f(3x)$." } { "Tag": [], "Problem": "List of threads pertaining to High School Chemistry or General Chemistry", "Solution_1": "Discussion of chemistry books that are good for independent study: click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=254832]here[/url].", "Solution_2": "General Discussion of Unanswered Problems in Chemistry.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=181194" } { "Tag": [ "inequalities" ], "Problem": "$ a^{8} \\plus{} b^{8} \\ge \\frac{c^{4}}{128}$\r\nIf $ a \\plus{} b \\ge c \\ge 0$", "Solution_1": "[quote=\"anurags92\"]$ a^{8} \\plus{} b^{8} \\ge \\frac {c^{4}}{128}$\nIf $ a \\plus{} b \\ge c \\ge 0$[/quote]\r\ntry a=b=1/3 c= 2/3\r\nLHS = $ \\frac{2}{3^8}$\r\nRHS = $ \\frac{2^4}{3^4 . 2^7} \\equal{} \\frac{1}{3^4.2^3}$\r\nLHS < RHS !" } { "Tag": [], "Problem": "\u0393\u03b9\u03b1 \u03c3\u03b1\u03c2 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03bf \u0393\u03b9\u03ce\u03c1\u03b3\u03bf\u03c2", "Solution_1": "\u0393\u03b5\u03b9\u03b1 \u03c3\u03bf\u03c5 \u03b3\u03b9\u03ce\u03c1\u03b3\u03bf \u03c0\u03bf\u03bb\u03cd \u03c3\u03c9\u03c3\u03c4\u03ae \u03b7 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae\r\n :wink:", "Solution_2": "\u039a\u03b1\u03bb\u03ce\u03c2 \u03ae\u03c1\u03b8\u03b5\u03c2 \u03c3\u03c4\u03bf\u03bd ''\u03c4\u03b5\u03ba\u03ad'' \u03bc\u03b1\u03c2 \u0393\u03b9\u03ce\u03c1\u03b3\u03bf. \u0395\u03b4\u03ce \u03cc\u03bb\u03bf\u03b9 \u03c4\u03b7 ''\u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5'' \u03bc\u03b5 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac :)", "Solution_3": ":welcomeani: \u0393\u03b9\u03ce\u03c1\u03b3\u03bf \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03b5\u03bc\u03ad\u03bd\u03b1.\r\n\r\n\u039a\u03b1\u03b9 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf topic \u03bf\u03bd\u03bf\u03bc\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9, \u03b2\u03c1\u03ae\u03ba\u03b1 \u03ad\u03bd\u03b1 \u03b1\u03bd\u03ad\u03ba\u03b4\u03bf\u03c4\u03bf \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bc\u03bf\u03c5 \u03ac\u03c1\u03b5\u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1 \u03c0\u03bf\u03bb\u03cd, \u03b3\u03b9\u03b1\u03c4\u03af \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b1\u03c0\u03ad\u03c7\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc\u03c4\u03b7\u03c4\u03b1\r\n\r\n\"So how's your boyfriend doing, the math student?\" \r\n\"Don't mention that crazy pervert to me anymore! We broke up.\" \r\n\"How can you say such a nasty thing about him? He seemed to be such a nice boy.\" \r\n\"Imagine! He was restless during the days and couldn't sleep at night - always trying to solve his math problems. When he had finally done it, he wasn't happy: he would call himself a complete idiot and throw all his notes into the garbage. One day, I couldn't take it anymore, and I told him to drop math. You know what he told me?\" \r\n\"No.\" \r\n\"He said, he enjoyed it!!!\"\r\n\r\n\u03a0\u03b1\u03c1\u03b5\u03bc\u03c0\u03b9\u03c0\u03c4\u03cc\u03bd\u03c4\u03c9\u03c2, \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03ac\u03bb\u03bb\u03bf \u03b1\u03bd\u03ad\u03ba\u03b4\u03bf\u03c4\u03bf, \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b1\u03c1\u03c7\u03af\u03b6\u03b5\u03b9\r\n\r\nQ: How can you tell that a mathematician is extroverted? \r\n\r\n\u0394\u03b5\u03bd \u03b8\u03b1 \u03c0\u03c9 \u03c4\u03ce\u03c1\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7 (\u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5 \u03bb\u03af\u03b3\u03bf \u03c8\u03ac\u03be\u03b9\u03bc\u03bf \u03c3\u03c4\u03bf Internet), \u03b1\u03bb\u03bb\u03ac \u03b1\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03ce\u03c3\u03b5\u03b9 \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03c9\u03c4\u03cc\u03c4\u03c5\u03c0\u03b7 \u03b4\u03b9\u03ba\u03ae \u03c4\u03bf\u03c5 \u03b1\u03c0\u03ac\u03bd\u03c4\u03b7\u03c3\u03b7, \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c7\u03b9\u03bf\u03c5\u03bc\u03bf\u03c1\u03b9\u03c3\u03c4\u03b9\u03ba\u03ae." } { "Tag": [ "geometry", "incenter", "Pythagorean Theorem" ], "Problem": "Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3,4,12, and 13, respectively, and $\\measuredangle CBA$ is a right angle. The area of the quadrilateral is\n\n[asy]\nsize(200);\ndefaultpen(linewidth(0.7)+fontsize(10));\nreal r=degrees((12,5)), s=degrees((3,4));\npair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s));\ndraw(A--B--C--D--cycle);\nmarkscalefactor=0.05;\ndraw(rightanglemark(A,B,C));\npair point=incenter(A,C,D);\nlabel(\"$A$\", A, dir(point--A));\nlabel(\"$B$\", B, dir(point--B));\nlabel(\"$C$\", C, dir(point--C));\nlabel(\"$D$\", D, dir(point--D));\nlabel(\"$3$\", A--B, dir(A--B)*dir(-90));\nlabel(\"$4$\", B--C, dir(B--C)*dir(-90));\nlabel(\"$12$\", C--D, dir(C--D)*dir(-90));\nlabel(\"$13$\", D--A, dir(D--A)*dir(-90));[/asy]\n\n$\\text{(A)} \\ 32 \\qquad \\text{(B)} \\ 36 \\qquad \\text{(C)} \\ 39 \\qquad \\text{(D)} \\ 42 \\qquad \\text{(E)} \\ 48$", "Solution_1": "[hide]Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\\frac{3\\cdot4}{2}+\\frac{5\\cdot12}{2}=36\\Rightarrow\\boxed{B}$.[/hide]", "Solution_2": "That, and none of the other answers seem remotely possible. :P\r\n\r\nEDIT: Looked at the LaTeX. The answer choices should look like this:\r\n\r\n$\\text{(A)} \\ 32 \\qquad \\text{(B)} \\ 36 \\qquad \\text{(C)} \\ 39 \\qquad \\text{(D)} \\ 42 \\qquad \\text{(E)} \\ 48$\r\n\r\n-BLt", "Solution_3": "How do you know that the larger triangle is a right triangle?", "Solution_4": "[quote=\"breez\"]How do you know that the larger triangle is a right triangle?[/quote]\r\n\r\nWhen you join C and A, you know that that length is 5, because triangle ABC is a right triangle. Now the larger triangle has side lengths 5-12-13, which is a Pythagorean triple (or, they satisfy $5^2+12^2=13^2$, making it a right triangle).", "Solution_5": "Ahh, ok so thats just like how 3 sides can only make 1 unique triangle, thus if it is pythagorean triplet, the resulting triangle must be a right triangle.", "Solution_6": "[hide=\"Solution\"]$\\text{Answer: (B)}$\n\nFirst, we use the right angle. Applying the Pythagorean theorem, $AC=5$. Next, we recognize the lengths 5-12-13 to be in the form of a Pythagorean Triple, which means that $\\triangle ACD$ is right.\n\nOur answer is simply the sum of the areas of these two right triangles, which is $\\frac{5\\cdot 12}{2}+\\frac{3\\cdot 4}{2}=36$[/hide]" } { "Tag": [], "Problem": "Consider the real numbers $a,b,c$ such that the equation $ax^2+bx+c=0$ has two roots in the segment [0,1]. Find the greatest value of the expression $P=\\frac{(a-b)(2a-b)}{a(a-b+c)}.$", "Solution_1": "Hint: using Viete's formula and...\r\nCase 1: $a=0$\r\nCase 2: $a\\neq 0$\r\n$P=\\frac{(a-b)(2a-b)}{a(a-b+c)}= \\frac{(1-\\frac{b}{a})(2-\\frac{b}{a})}{1-\\frac{b}{a}+\\frac{c}{a}}$\r\nGood luck. :D", "Solution_2": "$MaxP=3$", "Solution_3": "Lovasz,why don't you just post the entire solution?\r\n\r\n[hide]Considering that $m$ and $n$ are the roots of the equation,by Viete's formula: $-a(m+n)=b$ and $a(mn)=c$ substituing in $P$ we get:\n\n$P=\\frac{(m+n+1)(m+n+2)}{m+n+1+mn}=\\frac{m}{n+1}+\\frac{n}{m+1}+2$\n\nnotice that when $m$ and $n$ tends to zero the two fractions tends to zero too,consequently the maximum value for $P$ will occur when $m,n=1$:\n\n$P_{max}=\\frac{1}{2}.2+2=3$[/hide]" } { "Tag": [ "inequalities" ], "Problem": "Solve for real numbers $ x$:\r\n\r\na) $ \\frac{x\\minus{}2}{x^2\\minus{}3x\\minus{}4} \\ge 0$.\r\n\r\nb) $ \\frac{5}{x\\plus{}4} \\le 1$.", "Solution_1": "[hide=\"a.\"]\nFirst start by factoring the denominator and finding the restrictions.\n$ x^2-3x-4 = (x-4)(x+1)$. $ x$ cannot be 4 or -1 because that would make the denominator equal 0.\nMultiplying both sides by (x-4)(x+1),\n$ x-2\\ge 0\\\\\nx\\ge 2$\nIf you plot the restrictions and solution on a number line, you can test all the intervals to see which make the original equation true. The intervals are $ (-\\infty,-1), (-1,2], [2,4), (4,\\infty)$.\nOf these, the intervals $ (-1,2]$ and $ (4,\\infty)$ satisfy the original equation.\nSo the values of x that work are $ \\boxed{(-1,2]\\cup (4,\\infty)}$.\n[/hide]\n[hide=\"b.\"]\nAgain start by finding the restriction.\nx cannot be -4 because that would make the denominator equal 0.\nMultiplying both sides by (x+4),\n$ 5\\le x+4\\\\\n1\\le x\\\\\nx\\ge 1$\nPlot the restriction and solution and then test all the intervals, which are: $ (-\\infty,-4), (-4,1], [1,\\infty)$\nOf these, $ (-\\infty,-4)$ and $ [1,\\infty)$ satisfy the original equation.\nSo the values of x that work are $ \\boxed{(-\\infty,-4)\\cup [1,\\infty)}$.\n[/hide]", "Solution_2": "You cannot just multiply both sides by a quantity, you must make sure that a) it's nonnegative or b) you flip the sign when it's negative.\r\n\r\nHint: \r\n\r\nBreak into cases: when the denominator is nonnegative and when it is negative.", "Solution_3": "okay...I think this would be how you would do the cases (but I'm not completely sure):\r\n[hide=\"a.\"]\nThe denominator is positive when x is in $ (-\\infty,-1)\\cup (4,\\infty)$.\nIt is negative when x is in $ (-1,4)$, and undefined when $ x=-1$ or $ x=4$.\n\nStarting with the case when the denominator is positive, multiplying both sides by the denominator (x-4)(x+1) will give:\n$ x-2\\ge 0 \\\\\nx \\ge 2$\nThen the only solutions when the denominator is positive are $ (4,\\infty)$, since x must be in $ (-\\infty,-1)\\cup (4,\\infty)$ but also must be $ \\ge 2$.\n\nWhen the denominator is negative, multiplying both sides by the denominator (x-4)(x+1) will switch the inequality sign and give:\n$ x-2\\le 0 \\\\\nx \\le 2$\nThen the only solutions when the denominator is negative are $ (-1,2]$, because x must be in $ (-1,4)$ but also must be $ \\le 2$.\n\nCombining these, the solutions are $ \\boxed{(-1,2]\\cup (4,\\infty)}$.\n[/hide]\n[hide=\"b.\"]\nThe denominator is positive when x is in $ (-4,\\infty)$.\nIt is negative when x is in $ (-\\infty,-4)$, and undefined when $ x=-4$.\n\nWhen the denominator is positive, multiplying both sides by the denominator (x+4) will give:\n$ 5\\le x+4\\\\\nx\\ge 1$\nThen the solutions when the denominator is positive are $ [1,\\infty)$, because x must be in $ (-4,\\infty)$, but also $ \\ge 1$. \n\nWhen the denominator is negative, multiplying both sides by the denominator (x+4) will switch the inequality sign and give:\n$ 5\\ge x+4\\\\\nx\\le 1$\nThen the solutions when the denominator is negative are $ (-\\infty,-4)$, because x must be in $ (-\\infty,-4)$ and also $ \\le 1$, and all values of x in this range are less than 1.\n\nCombining these, the solutions are $ \\boxed{(-\\infty,4)\\cup [1,\\infty)}$.\n[/hide]\r\n\r\nI came up with the same answers, though making the cases is probably a better solution. \r\nI think the first time I should have said to solve the inequality with equal signs instead, and then test the intervals (especially because I didn't use the inequality at the end anyway).", "Solution_4": "a.\r\n\r\n$ \\frac{x\\minus{}2}{x^{2}\\minus{}3x\\minus{}4}\\ge 0$\r\n\r\n$ \\frac{x\\minus{}2}{(x\\minus{}4)(x\\plus{}1)} \\ge 0$\r\n\r\nBy looking at either side of the critical values $ \\minus{}1, 2$ and $ 4$, we see that the required range is \r\n\r\n$ \\minus{}1\\le x \\le 2$ and $ x\\ge 4$.\r\n\r\nb. \r\n\r\n$ \\frac{5}{x\\plus{}4} \\minus{} 1 \\le 0$\r\n\r\n$ \\frac{5\\minus{}(x\\plus{}4)}{x\\plus{}4} \\le 0$\r\n\r\n$ \\frac{1\\minus{}x}{4\\plus{}x} \\le 0$\r\n\r\n$ x\\le \\minus{}4$ or $ x\\ge1$." } { "Tag": [ "geometry", "3D geometry", "analytic geometry", "vector" ], "Problem": "How can it be proven that the line [b]A`C [/b]is perpendicularly on the plane [b] AB`D` [/b] in the cube[b] ABCDA`B`C`D`[/b]?\r\nHas anyone an idea for the above problem?", "Solution_1": "Please pay attention to the posting guidelines of each forum where available. This problem is far too difficult for the forum in which you posted it. Thanks.", "Solution_2": "3D coordinate system and vectors would help.", "Solution_3": "It is a problem from the Romanian 8th class book :)" } { "Tag": [ "algorithm" ], "Problem": "Specifically,what kind of algorithm[b] is use in Robotics[/b] artificial[b] intelligence[/b] programming? Is there any available free[b] information that deals specifically with Artificial Intelligence in the internet?[/b]", "Solution_1": "All I know is that you need QBASIC and it's very simple to program them...don't know how...", "Solution_2": "The simplest algorithm would be to list all rational available choices to the program powering the AI, and after what happens, if it was successful, to put more *weight* on the option (i.e. to make it more likely to choose that option once the original branching is encountered again), or if it was a failure, to remove weight.\r\n\r\nYour use of bold is very distracting, please stop." } { "Tag": [ "blogs" ], "Problem": "Any AoPSer do Myspace?\r\n\r\nmine is myspace.com/blysh3r", "Solution_1": "I think most AoPS-MathLink-ers have their own blog right here on the site :P http://www.mathlinks.ro/Forum/apsjour.php ;)", "Solution_2": "[quote=\"Valentin Vornicu\"]I think most AoPS-MathLink-ers have their own blog right here on the site :P http://www.mathlinks.ro/Forum/apsjour.php ;)[/quote]\r\n\r\nYes, you are right. But there are probably many AoPSers with myspace as well.\r\n\r\nYou can find mine [url=http://www.myspace.com/index.cfm?fuseaction=user.viewProfile&friendID=7810069&Mytoken=20050530061907]here[/url].\r\n\r\nI started my account not too long ago...so I don't have all that many friends compared to a lot of ppl I know :(" } { "Tag": [ "calculus" ], "Problem": "What would be some good classes to take in high school if I am planning to have a career in politics? I know history would be good, but what else? Psychology?", "Solution_1": "just wondering...do you want to know for curiosity, or are you actually considering a career in politics?", "Solution_2": "I'm not quite sure there is really good class you can prepare for politics in high school. Just get good grades on main subjects..\r\n\r\nWhy don't you ask this question on College Forum for the detailsed courses in college?\r\n\r\nIt sure would be better than asking about high school.", "Solution_3": "[quote=\"peter\"]just wondering...do you want to know for curiosity, or are you actually considering a career in politics?[/quote]\r\n\r\nHehe, good question.\r\n\r\nWell, with everything politics have become today, you might want to take Public Speaking. But if they're offered at your high school, try Bending the Truth, Digging up Dirt, and Criticism 101.", "Solution_4": "[quote]But if they're offered at your high school, try Bending the Truth, Digging up Dirt, and Criticism 101.[/quote]\r\n\r\nDebate.", "Solution_5": "[quote=\"tetrahedr0n\"][quote]But if they're offered at your high school, try Bending the Truth, Digging up Dirt, and Criticism 101.[/quote]\n\nDebate.[/quote]\r\n\r\n :blush: :blush: \r\n\r\nKind of forgot about that...", "Solution_6": "Statistics, so you can manipulate the truth while claiming that you're doing it in a mathematically rigorous way.", "Solution_7": "This is not really a class but if your school has it, join Model UN. Also, take classes like Constitutional Law (if your school has it of course)", "Solution_8": "dont concentrate too much on classes. i dont believe there are any good politics-related high school classes. instead id suggest reading books on it.", "Solution_9": "most of your political knowledge will be gained in college. for now, just take as many advanced courses as possible (AP/IB) and maintain good grades. also, join forensics (speech and debate) or Model United Nations (it rules all) to develop your public speaking skills. if your school has a JSA chapter (jsa.org), join it and consider attending some of the conferences. if not, hey, start your own chapter! that's what i did, and the organization makes it very easy for you. oh yes, and AP Government, but most take that course anyway. perhaps look into some summer programs or evening classes at a local university? volunteer for a campaign? intern for a law firm? politics is an insanely competitive field; let me just tell you that.\r\n\r\ni suppose, for now, just focus on satisfying and developing your interest in politics, as well as gaining admission to a good college. colleges love the passionate type, so if politics is really your passion, pursue it! taking initiative is key.", "Solution_10": "Okay...the most obvious one here is probably Speech and Debate.\r\n\r\nMany public schools run Model UN programs but let me tell you right now that such programs don't come close to portraying how politics realli is, unless you become an ambassador to the UN...\r\n\r\nYMCA runs a program known as Youth and Government in 36 states. Check to see if your state has it. Each state has some freedom with their own program, but generally this is actually an extremely good simulation of real politics. In Texas, we have all three branches of government and extensive offices in each plus some groups that are not actually part of the government (media, executive government planners, lobbyists, etc). The organization is almost completely identical to that of the actual state government. (In fact we actually use the Capitol chambers and other government buildings to compete in.) There's officers, elections, campaigns, political parties, etc. You use parliamentary and/or courtroom procedures depending on where you are. But all-in-all it's just an awesome thing to pursue. Each state also sends representatives to the Conference on National Affairs every July, which is a huge honor. This program is actually not all that advertised because you participate through your YMCA instead of your school.\r\n\r\nIf you need more information, let me know. But I'll tell you straight out that this is the single best way you can actually experience what it's like to be in public service. No other program can come close to giving you the experiences and memories that this does. And in case you're wondering, I am not unfamiliar with the \"normal\" speech and debate programs.", "Solution_11": "Bible Study :D", "Solution_12": "psycology would be one to consider. Also speech and debate. :) \r\nThere is also political science.", "Solution_13": "How about....Pre-Calculus!!", "Solution_14": "what???????? jk\r\n\r\nI recomment AP Government and Politics.\r\n\r\nIt has Politics in the title" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "If the function $ f(x) \\equal{} 3x^2 \\plus{} 2ax \\plus{} b$ satisfies $ \\int_{ \\minus{} 1}^1 |f(x)|\\ dx < 2,$ then show that $ f(x) \\equal{} 0$ has distinct real roots.", "Solution_1": "$ f(x) \\equal{} 3x^2 \\plus{} 2ax \\plus{} b\\equal{}3(x\\plus{}\\frac{a}{3})^2\\plus{}b\\minus{}\\frac{a^2}{3}$ \r\n\r\nSuppose $ f$ has not distinct roots then $ b\\minus{}\\frac{a^2}{3}>\\equal{}0$ and $ |f(x)|\\equal{}f(x)$\r\n\r\nWe have $ \\int_{ \\minus{} 1}^1 |f(x)|\\ dx \\equal{} \\int_{ \\minus{} 1}^1 f(x)\\ dx \\equal{}2\\plus{}2b< 2$ then $ b<0$ but $ b\\minus{}\\frac{a^2}{3}>\\equal{}0$=>$ 0>a^2$ contradiction.", "Solution_2": "That's right, AYMANE. :)" } { "Tag": [], "Problem": "How many ways are there to choose four cards of different suits and different values from a deck of 52 cards?\r\n\r\nI have three different ways of reasoning this:\r\n\r\n1. For suit 1, we have 13 choices for values.\r\n\r\nFor suit 2, we have 12 choices for values.\r\n\r\nFor suit 3, we have 11 choices for values.\r\n\r\nFor suit 4, we have 10 choices for values.\r\n\r\nThus, we have 13 x 12 x 11 x 10 choices.\r\n\r\n2. For value 1, we have 4 choices for suits.\r\n\r\nFor value 2, we have 3 choices for suits.\r\n\r\nFor value 3, we have 2 choices for suits.\r\n\r\nFor value 4, we have 1 choices for suits.\r\n\r\nThus, we have 4 x 3 x 2 x 1 choices.\r\n\r\n3. For card 1, we have 52 choices for cards.\r\n\r\nFor card 2, we have (52 - 13 - 1) choices for cards.\r\n\r\nFor card 3, we have (52 - 13 - 1) - 13 - 1 choices for cards.\r\n\r\nFor card 4, we have (52 - 13 - 1) - 13 - 1) - 13 - 1 choices for cards.\r\n\r\n\r\nThus, we have 52 x (52 - 13 - 1) x ((52 - 13 - 1) - 13 - 1 ) x ( (52 - 13 - 1) - 13 - 1) - 13 - 1 ) choices for cards.\r\n\r\nEach of these reasonings seem correct to me. \r\n\r\nWhat is wrong with the reasoning in 2 and 3?\r\n\r\nPlease help because I am very confused.", "Solution_1": "is there supposed to be something wrong? because it seems correct, at least at first glance.", "Solution_2": "My solution is none of those three.\n\n\n\n[hide]Consider the values first. There are 13C4 ways of choosing 4 cards.\n\n\n\nThen consider the suits. They have to appear in some particular order to match up with the values; then there are 4P4 = 4! ways of doing that.\n\n\n\nSo, there are 13C4 * 4! = 17 160 ways of choosing 4 cards of different suits and different values from a deck of 52 cards.[/hide]\n\n\n\nEdit: Sorry, this is really identical to solution #1.", "Solution_3": "first you can choose one of the 52 cards, then 52-13-3 cards, then 52-13-3-12-2, and finally you can choose from 10 cards, so it 52*36*22*10=411840,\r\n\r\n\r\n\r\nadditionally your no. 3 is wrong because after taking away 13 and 1 (you should have taken away 13 and 3 (thirteen for suit three for number, then next time you only take away 12 for suit since 1 number is gone.... etc.)", "Solution_4": "Order doesn't matter though, so you should divide 411 840 by 24 (4!) to get 17 160.", "Solution_5": "your number 2 reasoning is mistaken because you have not chosen the actual values. when you do, you get osiris's answer (you will have to multiply by 13C4). also, osiris got the same thing as your reasoning in number 1. (13C4*4!=13P4=what you got)" } { "Tag": [ "\\/closed" ], "Problem": "If your inbox is full and someone sends you messages, what will happen? Would you still be able to receive them (That is, if you delete the older ones) or they'll go back to the sender?\r\n\r\nThanks.", "Solution_1": "sadly, it just deletes the old ones without confirmation ... [here stood some very evil trick :censored:]", "Solution_2": "[quote=\"Peter VDD\"]sadly, it just deletes the old ones without confirmation. ^^[/quote]\r\n\r\nThat's harsh but hilarious. :rotfl: :rotfl: Don't do that to me though Peter. :lol:", "Solution_3": "Yeah I'll do something about that soon ...", "Solution_4": "argh, censor! censor! :P:P j/k\r\n\r\nAnyway Valentin, I think there's something more annoying, but caused by ourselves... we have 120 inbox slots now (more than one can use actually) but the outbox is very very limited... maybe you could increase the outbox size? It is annoying when sent messages get gone because you send some other pm's, and it would not take any new disk space as the message is already stored for the recipient's inbox. :)", "Solution_5": "[quote=\"Peter VDD\"]argh, censor! censor! :P:P j/k\n[/quote]Yes, let's not give people ideas :diablo: \r\n\r\nI will see what I can do about the outbox thing.", "Solution_6": "*just for the purpose of details: I meant sentbox, which is different from outbox*", "Solution_7": "[quote=\"Peter VDD\"]*just for the purpose of details: I meant sentbox, which is different from outbox*[/quote]\r\n\r\nHmm.. I meant \"inbox\" though.. Outbox usually doesn't matter to me though. The best way is to keep quoting the text. :D That way, you can see what you said first to what other person said last (assuming both of you are keep quoting)." } { "Tag": [], "Problem": "Hey i need help making a program that can be interjected into other programs easily without going through the necessarily implications of assigning variables.... for example... is there a way to just insert a program into the calculator w/o needing a prompt or input maybe just assign the variable in a pair of parenthesis following the program name? Trying to make a program that basically goes x(abs(x)), very simple. I have a TI-84 SE. Help!", "Solution_1": "[quote=\"0mega\"]Hey i need help making a program that can be interjected into other programs easily without going through the necessarily implications of assigning variables.... for example... is there a way to just insert a program into the calculator w/o needing a prompt or input maybe just assign the variable in a pair of parenthesis following the program name? Trying to make a program that basically goes x(abs(x)), very simple. I have a TI-84 SE. Help![/quote]\r\n\r\nyou can't do that on a TI-84 SE, only on a TI-89" } { "Tag": [ "number theory", "greatest common divisor", "Divisibility Theory" ], "Problem": "Determine the greatest common divisor of the elements of the set \\[\\{n^{13}-n \\; \\vert \\; n \\in \\mathbb{Z}\\}.\\]", "Solution_1": "Since this kind or problem occurs so often, I think we should directly solve the following:\r\n\r\nDetermine the greatest common divisior $d_k$ of all the elements of $S_k = \\{n^k-n |n \\in \\mathbb{Z} \\}$.\r\n\r\nThe result is $d_k=\\prod_{\\substack{p \\in \\mathbb P \\\\ (p-1)|(k-1)}} p$ (being the denominator of the $(k-1)$-th Bernoulli number if $2|(k-1)$) and follows directly if using primitive roots.", "Solution_2": "* note to self: work out ZeTaX's solution with primitive roots and move to that chapter if appropriorate *", "Solution_3": "For the initial one:\r\n2^13-2 factors as 2*3^2*5*7*13 so the only possible factors of the gcd are 2,3(twice),5,7,13\r\nBut using Fermat's little theorem we can prove that 2,3,5,7,13|(n^13-n)\r\nWe now need to consider whether 9|n^13-n for every n, but this does not happen for n=3 so the gcd is 2*3*5*7*13=2730\r\n\r\nPostScript\r\nIn Fermat's little theorem we can only assume that n is not congruent to zero modp(where p prime).\r\nBut obviously n^13 is congruent to n for every n congruent to zero modp", "Solution_4": " there will not exist a prime with its exponential more than 1 as when we let $n=p$ then it is clear that $V_{p}p(p^{12}-1)=1$\n$ord_p(n)| p-1$ and also $ord_p(n)|12$\nso the possible values for $ord_p(n)= 1,2,3,4,6,12$\nWe must note that p cant be bigger than 12 as there must exist a primitive root so any prime bigger than 12 will not be a common divisor\nso the only possible primes are $3,5,7,13$ \ngreatest common divisor: $3\\times5\\times7\\times13$" } { "Tag": [ "inequalities", "algebra", "function", "domain", "inequalities unsolved" ], "Problem": "hi guys\r\ni was scrolling through one math jam transscript and discovered the following problem.\r\ni tried to solve but i couldnt, so im asking u to do so ...\r\n[hide]without wanting to spoil anything i tell u it was asked in the context of cauchy-schwarz[/hide]\r\n[img]http://www.artofproblemsolving.com/Images/ClassLaTeX/lx-15398132.gif[/img]\r\n\r\nthx 4 ur help\r\npeeta", "Solution_1": "Right now I have only a solution using AM-GM.\r\n\r\nWe will actually prove that if x, y, z are three positive reals satisfying $x^4+y^4+z^4=1$, then\r\n\r\n$\\frac{x^3}{1-x^8}+\\frac{y^3}{1-y^8}+\\frac{z^3}{1-z^8}\\geq\\frac{9\\sqrt[4]{3}}{8}$, with equality if and only if $x=y=z=\\frac{1}{\\sqrt[4]{3}}$.\r\n\r\nIn fact, from the AM-GM inequality,\r\n\r\n$\\frac13+\\frac13+\\frac13+\\frac13+\\frac13+\\frac13+\\frac13+\\frac13+3\\sqrt[4]{3}x^9\\geq 9\\sqrt[9]{\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot\\frac13\\cdot 3\\sqrt[4]{3}x^9}$.\r\n\r\nIn other words,\r\n\r\n$\\frac83+3\\sqrt[4]{3}x^9\\geq 9\\sqrt[9]{\\frac{1}{3^8}\\cdot 3\\sqrt[4]{3}x^9}=9\\sqrt[9]{\\frac{1}{3^7}\\cdot\\sqrt[4]{3}}x=3\\sqrt[4]{3}x$\r\n\r\n(since $9\\sqrt[9]{\\frac{1}{3^7}\\cdot\\sqrt[4]{3}}=3\\sqrt[4]{3}$). Hence,\r\n\r\n$\\frac83\\geq 3\\sqrt[4]{3}x-3\\sqrt[4]{3}x^9=3\\sqrt[4]{3}x\\left(1-x^8\\right)$.\r\n\r\nMultiplying this with $\\frac38x^3$, we get\r\n\r\n$x^3\\geq\\frac{9\\sqrt[4]{3}}{8}x^4\\left(1-x^8\\right)$.\r\n\r\nSince $x^4+y^4+z^4=1$, while the numbers x, y, z are positive, we conclude that $x^4<1$, so that x < 1, and thus $1-x^8>0$. Hence, we can divide this inequality by $1-x^8$ and obtain\r\n\r\n$\\frac{x^3}{1-x^8}\\geq\\frac{9\\sqrt[4]{3}}{8}x^4$.\r\n\r\nSimilarly,\r\n\r\n$\\frac{y^3}{1-y^8}\\geq\\frac{9\\sqrt[4]{3}}{8}y^4$;\r\n$\\frac{z^3}{1-z^8}\\geq\\frac{9\\sqrt[4]{3}}{8}z^4$.\r\n\r\nAdding up these three inequalities, we get\r\n\r\n$\\frac{x^3}{1-x^8}+\\frac{y^3}{1-y^8}+\\frac{z^3}{1-z^8}\\geq\\frac{9\\sqrt[4]{3}}{8}x^4+\\frac{9\\sqrt[4]{3}}{8}y^4+\\frac{9\\sqrt[4]{3}}{8}z^4$\r\n$=\\frac{9\\sqrt[4]{3}}{8}\\left(x^4+y^4+z^4\\right)=\\frac{9\\sqrt[4]{3}}{8}$\r\n\r\n(since $x^4+y^4+z^4=1$). This completes the solution. The equality case is an exercise for the reader. ;)\r\n\r\nProbably the Cauchy-Schwarz solution is trivial, but I don't see it (what could partly be due to my general blindness with respect to Cauchy-Schwarz solutions, but also please take into account that I'm very tired now).\r\n\r\n darij", "Solution_2": "thanx darij\r\nbut yet i dont understand, why this problem was asked in context with cauchy-schwarz.\r\n\r\nthx to all who might post a possible solution involving cauchy-schwarz", "Solution_3": "CS soln:\r\n\r\n$\\sum \\frac{x^3}{1-x^8} = \\sum \\frac{x^3}{2-2x^4} + \\frac{x^3}{2+2x^4} \\geq \\frac{(\\sum x^{\\frac{3}{2}})^2}{4} + \\frac{(\\sum x^{\\frac{3}{2}})^2}{8} = \\Bigg( \\frac{3}{8} \\Bigg) \\Bigg(\\sum x^{\\frac{3}{2}} \\Bigg)^2$\r\n\r\nIt remains to minimize $\\sum x^{\\frac{3}{2}}$. Obviously it occurs when $x=y=z$. Then $x = y = z = \\frac{1}{\\sqrt[4]{3}}$ and the minimum is $\\frac{9\\sqrt[4]{3}}{8}$ as darij said.", "Solution_4": "[quote=\"Singular\"]\nIt remains to minimize $\\sum x^{\\frac{3}{2}}$. Obviously it occurs when $x=y=z$. [/quote]\r\n\r\nI'm sorry, I think the MAXIMUM of this is achieved when the variables are equal. (Think about power mean)", "Solution_5": "thx singular\r\nalthough i wouldnt call $\\frac{x_1^2}{y_1}+\\ldots+\\frac{x_n^2}{y_n}\\ge\\frac{(x_1+\\ldots+x_n)^2}\r\n{y_1+\\ldots+y_n}$ exactly cauchy-schwarz, i guess it was reminding me of this useful inequality, which is good.\r\n\r\nsorry for being so picky, but is there anyone out there in this [size=59]web[/size]space that can find a solution with the $\\sum x_i^2\\cdot \\sum y_i^2\\ge \\left(\\sum x_iy_i\\right)^2$-inequality. u can check [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=74]the transcript[/url], but i just think the is a simple solution with cs.", "Solution_6": "[quote=\"peeta\"]thx singular\nalthough i wouldnt call $\\frac{x_1^2}{y_1}+\\ldots+\\frac{x_n^2}{y_n}\\ge\\frac{(x_1+\\ldots+x_n)^2}\n{y_1+\\ldots+y_n}$ exactly cauchy-schwarz, i guess it was reminding me of this useful inequality, which is good.\n\nsorry for being so picky, but is there anyone out there in this [size=59]web[/size]space that can find a solution with the $\\sum x_i^2\\cdot \\sum y_i^2\\ge \\left(\\sum x_iy_i\\right)^2$-inequality. u can check [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=74]the transcript[/url], but i just think the is a simple solution with cs.[/quote]\r\n\r\nI don't think you're going to get closer than that to Cauchy... in fact that IS Cauchy, just written differently.\r\n\r\n$(\\sum a_i^2)(\\sum b_i^2) \\ge (\\sum a_ib_i)^2$\r\n\r\nLet $a_i = \\frac{x_i}{\\sqrt{y_i}}$ and $b_i = \\sqrt{y_i}$\r\n\r\nThen $(\\sum \\frac{x_i^2}{y_i})(\\sum y_i) \\ge (\\sum x_i)^2 \\Rightarrow \\sum \\frac{x_i^2}{y_i} \\ge \\frac{(\\sum x_i)^2}{\\sum y_i}$.", "Solution_7": "[quote=\"siuhochung\"][quote=\"Singular\"]\nIt remains to minimize $\\sum x^{\\frac{3}{2}}$. Obviously it occurs when $x=y=z$. [/quote]\n\nI'm sorry, I think the MAXIMUM of this is achieved when the variables are equal. (Think about power mean)[/quote]\r\n\r\nIf what you said is true, then let $r = 3^{-4}$, and $f(x,y,z) = \\sum x^{\\frac{3}{2}}$. Then you claim $f(r,r,r) \\geq f(x,y,z)$ for all $x,y,z$ such that $\\sum x^4 = 1$. But clearly $f(r,r,r) \\leq f(1,0,0)$ because $3r^{3/2} \\leq 1$, contradicting your statement.", "Solution_8": "then can you prove your assertion? It's not obvious to me.", "Solution_9": "[quote=\"Singular\"]because $3r^{3/2} \\leq 1$[/quote]I'm afraid you made a mistake here : with $r=3^{-1/4}$, we have\r\n\r\n$3r^{3/2}=3 \\cdot 3^{-(1/4\\times 3/2)}=3\\cdot 3^{-3/8}= 3^{5/8} > 3^{4/8}=\\sqrt{3} > 1$.\r\n\r\nI suggest the following example to illustrate that [b]siuhochung[/b]'s statement is correct : let $f(x,y)=x+y$ and $g(x,y)=x^2+y^2$. Consider the problem of searching\r\n\r\n$\\max f()$ and $\\min f()$ under the constraint $g(x,y)=1$.\r\n\r\nIf we draw the (quarter) circle $x^2+y^2=1$ in the $(x,y)$-plane and play with the line $x+y=C$ for different values of $C$, we see that the maximum is achieved for $C=\\sqrt{2}$, at $x=y=1/\\sqrt{2}$ (inside the domain), where the line is tangent to the circle, while the minimum is attained for $C=1$, at $(x,y)=(0,1)$ or (1,0) (on the boundary).\r\n\r\nIf we consider the inverse problem, that is, to seek\r\n\r\n$\\max g()$ and $\\min g()$ under the constraint $f(x,y)=1$,\r\n\r\nwe now have to draw with the line $x+y=1$ and play with the circles $x^2+y^2=C$. We now see that, now, the minimum is achieved for $C=1/2$ at $x=y=1/2$ (inside), where the circle is tangent to the line, while the maximum is attained for $C=1$, at $(x,y)=(0,1)$ or (1,0) (boundary).\r\n\r\nConclusion (based on this and other experiments) : when the exponent of the objective function is higher than the exponent of the constraint, the maximum is inside and the minimum is on the boundary ; when the exponent of the objective function is lower than that of the constraint, the minimum is inside and the maximum is on the boundary." } { "Tag": [ "function" ], "Problem": "Find all functions $f(x)$ such that $f^{-1}(x)=\\frac{1}{f(x)}$.", "Solution_1": "[hide=\"Discussion\"] In other words, $x = f \\left( \\frac{1}{ f(x) }\\right)$. \n\nSuppose we have some point $f(a) = b$ on this graph. Repeatedly plugging in we have\n\n$a = f \\left( \\frac{1}{b}\\right)$\n$\\frac{1}{b}= f \\left( \\frac{1}{a}\\right)$\n$\\frac{1}{a}= f \\left( b \\right)$\n\nIn other words, the value of $f$ at $a$ uniquely determines the value of $f$ at three other points [b]only[/b], so we can partition the reals (or the rationals, for that matter) nearly arbitrarily for a number of really strange functions that can't be succinctly described, so we should probably restrict $f$ continuous. \n\nAlthough, compounding two appropriate pairs of these, we conclude that\n\n$f(f(x) = \\frac{1}{x}$\n\nThis might restrict solutions a little. We do have the strange complex solution $f(x) = x^{i}$, which gives $f^{-1}(x) = f(x)^{-1}= x^{-i}$. [/hide]" } { "Tag": [], "Problem": "Five real numbers are chosen and put in order from smallest to largest. The average of all five is 14 . The average of the three middle numbers is only 13 . What is the average of the largest and smallest ?", "Solution_1": "The sum of the five numbers is 70, the sum of the middle three is 39. Hence the remaining two have sum 70-39=31, with average 15.5." } { "Tag": [ "linear algebra", "matrix", "quadratics" ], "Problem": "Let $A$ be a $2\\times 2$ matrix over the real field, which is similar exactly to one matrix (i.e. itself). Prove that $A=\\alpha I, \\alpha \\in R$.", "Solution_1": "This can be generalized and is probably better known in the following form:\r\n\\[ Z(M_n(K)) = K I_n \\]\r\nThis means: Quadratic matrices, commuting with all other matrices of the same size (hence only similar to itsself), are exactly the scalar matrices, i.a. matrices of the form\r\n\\[ \\begin{pmatrix}k & 0 & \\cdots & 0 \\\\ 0 & k & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & k \\end{pmatrix} \\]\r\n \r\nSee attachment for a proof.", "Solution_2": "Thanks for your reply. But would you please clarify this part? \"commuting with all other matrices of the same size \"\r\nI suppose we are required to check this condition: \"commuting with all nonsingular matrices of the same size\"\r\nrather than that. \r\nDoes it change anything?", "Solution_3": "You're right. \"similar only to ittself\" is weaker then \" commuting with all other matrices\". So let's enjoy $n=2$ ;-)\r\n \r\nWrite\r\n\\[ A=\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\]\r\nFrom\r\n\\[ A \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix} = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix} A \\]\r\nwe get $b = c$ and $a = b + d$. And from\r\n\\[ A \\begin{pmatrix} 1 & 0 \\\\ 1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 1 & 1 \\end{pmatrix} A \\]\r\nwe get $a+b = a$, hence $b = 0$. So, we have\r\n\\[ A = \\begin{pmatrix} a & 0 \\\\ 0 & a \\end{pmatrix} \\]" } { "Tag": [ "algebra", "polynomial", "complex numbers", "imaginary numbers" ], "Problem": "$\\sqrt{x} + y = 11$\r\n\r\n$\\sqrt{y} + x = 7$\r\n\r\nproper method is required ,please.I know you can do it by hit and trial, but i would appreciate a proper soln :D\r\n\r\nP.S. This question is NOT for shadysaysurspammed as he knows the soln and answer too!", "Solution_1": "move over the y and square for the first, and mover over the x and square for the second, and any answer that is negative is bad, since it makes an imaginary number", "Solution_2": "I got a sol but it is kinda factors and all.\r\nand then you have to assume.\r\n[hide=\"Solution\"]\nsubstracting the 2 equations,\n$\\sqrt{x}-\\sqrt{y}+y-x=4$\n$\\sqrt{x}-\\sqrt{y}-(\\sqrt{x}-\\sqrt{y})(\\sqrt{x}+\\sqrt{y})=4$\n$(\\sqrt{x}-\\sqrt{y})(1-\\sqrt{x}+\\sqrt{y})=4$\nLet $\\sqrt x=a$\nand$\\sqrt y=b$\n$(a-b)(1-a-b)=2*2*1$\nIm taking into consideration that a and b are integers but they also have to be positve because root of something is always positive except $i$\nTest them out and then find the possible value\n[/hide]\nAnother try,\n[hide]\nJust substitute x in both equations and you will get a 4 degree polynomial and then you have to factorise it\n[/hide]" } { "Tag": [], "Problem": "Say I recieved a 1730 (75th percentile of college-bound seniors) on the SAT as a freshman.\r\n\r\nWhat percentile of college-bound seniors of 2011 will I be in?\r\n\r\nWhat score will I get on the SAT as a sophomore, junior, senior, assuming that I stay in the same percentile?\r\n\r\nYou might have to find the data .. I'm not sure.\r\n\r\nBut if you could find a way to figure this out, it would be great!\r\n\r\nThanks.", "Solution_1": "There is no way to predict anything. No one knows what seniors in 2011 will score, but you can assume the percentiles will be similar to every other year. If you got a 173 on the PSAT as a sophomore, that's okay but nothing amazing.", "Solution_2": "[quote=\"mj93\"]Say I recieved a 1730 (75th percentile of college-bound seniors) on the SAT as a freshman.\n\nWhat percentile of college-bound seniors of 2011 will I be in?\n\nWhat score will I get on the SAT as a sophomore, junior, senior, assuming that I stay in the same percentile?\n\nYou might have to find the data .. I'm not sure.\n\nBut if you could find a way to figure this out, it would be great!\n\nThanks.[/quote]\r\nIf all freshman score like you and do everything like you, and all freshman this year become seniors in 2011, and only seniors of 2011 are part of college-bound seniors of 2011, then on average you will be right in the middle.\r\nThat's the only thing I can predict.\r\nHowever, since you're a freshman and you were ranked among college-bound seniors, then i doubt that the college-bound seniors of 2011 will only have seniors." } { "Tag": [ "\\/closed" ], "Problem": "Well, I could've made the title a bit less revealing, but I decided not to.\r\n\r\nMy statement here is that new moderators need to be assigned to Games and Fun Factory.\r\nNo, I don't mean adding moderators.\r\nI mean revoking moderatorship of the two mods there now, and adding 2-3 competent members to mod the forum.\r\n\r\nMathfiend has not posted this year, in fact it has been almost 6 full months since her last post.\r\nMithsApprentice has 14 posts in the last 2 months+. That's 7 posts a month, which means not even 1 post every 4 days.\r\n\r\nAlso, not only is G&FF the most likely forum for people to spam uselessly, it IS the forum where people spam most uselessly.\r\nOn top of that, every once in a while we have a user like kotsba advertising for gambling sites, and I recall only 1 or 2 of his topics being locked.\r\n\r\nI can see someone arguing along the lines of moderators not really necessary, or something of the sort; however, what is the point of moderators then? \r\n\r\nAs I recall, the point of moderatorship of a forum is so that there is not too much spam, and so the users of the forum can feel comfortable posting in the forum.\r\n\r\nI don't believe this is accomplished by either, nor both, of MathFiend and MithsApprentice.", "Solution_1": "[quote=\"Klebian\"]Well, I could've made the title a bit less revealing, but I decided not to.\n\nMy statement here is that new moderators need to be assigned to Games and Fun Factory.\nNo, I don't mean adding moderators.\nI mean revoking moderatorship of the two mods there now, and adding 2-3 competent members to mod the forum.\n\nMathfiend has not posted this year, in fact it has been almost 6 full months since her last post.\nMithsApprentice has 14 posts in the last 2 months+. That's 7 posts a month, which means not even 1 post every 4 days.\n\nAlso, not only is G&FF the most likely forum for people to spam uselessly, it IS the forum where people spam most uselessly.\nOn top of that, every once in a while we have a user like kotsba advertising for gambling sites, and I recall only 1 or 2 of his topics being locked.\n\nI can see someone arguing along the lines of moderators not really necessary, or something of the sort; however, what is the point of moderators then? \n\nAs I recall, the point of moderatorship of a forum is so that there is not too much spam, and so the users of the forum can feel comfortable posting in the forum.\n\nI don't believe this is accomplished by either, nor both, of MathFiend and MithsApprentice.[/quote]\r\nI agree. There was a discussion about this a while ago, and someone said that there would be no mod changes for a while...but the G&FF forum is going to get out of control soon (If it already hasn't)", "Solution_2": "Actually,FF does not really need constant monitering but lately people have started to psam a lot,so yes,amybe keeping a mod who is regular is fine but the mod should not be \"not fun\"", "Solution_3": "How about just removing G&FF posst from people's post count like in the Test Forum. That might curb some of the spamming.", "Solution_4": "[quote=\"Centy\"]How about just removing G&FF posst from people's post count like in the Test Forum. That might curb some of the spamming.[/quote]\r\nJust my point :)\r\nThere is only one hand of pure spammers there, all others behave like they should.\r\n(and doing the moderation of it should be hard and a lot of work...)", "Solution_5": "We'll take this under consideration.\r\n\r\nAlso, feel free to report commercial spamming to an AoPS admin like me. I'll happily ban gambling site spammers.", "Solution_6": "If you still are looking for a moderator of any forum, I would like to be a candidate because i have good leadership qualities, i haven't spammed in a while, and i can control levels of spam. :)", "Solution_7": "[quote=\"math92\"]i haven't spammed in a while[/quote] :D", "Solution_8": "[quote=\"Peter VDD\"][quote=\"math92\"]i haven't spammed in a while[/quote] :D[/quote]\r\n\r\nWhat about the \"bump the thread\" topic? That's pure spam :)", "Solution_9": "To be clear, the point of having mods around is not for us to lead forum discussions...at least, not in F&G. On spam-acceptable boards, we just make sure the content is appropriate and productive, sort of. Of course, any commercial advertising should've been deleted. If anything like this occurs, please bring it to our attention via PM. Thanks.", "Solution_10": "[quote=\"MithsApprentice\"]To be clear, the point of having mods around is not for us to lead forum discussions...at least, not in F&G. On spam-acceptable boards, we just make sure the content is appropriate and productive, sort of. Of course, any commercial advertising should've been deleted. If anything like this occurs, please bring it to our attention via PM. Thanks.[/quote]\r\n\r\nHowever, if we're going to have mods, why not choose a more active user over an inactive user?\r\nI mean, why NOT have a mod lead forum discussions?\r\nAnd if a moderator isn't active enough to see a user posting gambling advertisements, should he really be a moderator?", "Solution_11": "Klebian,\r\n\r\nYou're a moderator now. Those questions are better raised at this point in the moderators community. Miths does a good job moderating. If a commercial poster goes unnoticed, it's in G&FF -- there's an enormous amount of posting to keep up with and not every link is going to be clicked through. Now that you've noted the problem, the user and IP is banned and everyone will know better what to look for in the future. Remember, this whole message board is an experiment in human resource management -- and overall the community does a remarkable job at fixing problems as they arise. Thanks for helping out.", "Solution_12": "[quote=\"robinhe\"][quote=\"Peter VDD\"][quote=\"math92\"]i haven't spammed in a while[/quote] :D[/quote]\n\nWhat about the \"bump the thread\" topic? That's pure spam :)[/quote]\r\nI thought i posted that in the test forum, where it eventually gets deleted.", "Solution_13": "Depends what you call spam. I tend to live the Games and Fun Factory forum more liberal, a more friendly place for users of young age. We will soon change part of the forum's software, so posts in that forum will not be counted anymore. Other than that I see nothing wrong in having some spam in Games and Fun (or course some limitations will apply, like no 2 consecutive posts, no more than 2 posts per minute in that forum, etc.). \r\n\r\nAnd of course all \"real spam\" (like commercial, or xxx ads) should be deleted and admins notified for user/IP banning (please send links with the posts before you delete them as well as links to the users profiles to make the job easier)." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let $ a,b,c>0$ prove that $ \\sum(\\frac{a^2}{b}\\plus{}c)^2 \\ge 12\\frac{a^3\\plus{}b^3\\plus{}c^3}{a\\plus{}b\\plus{}c}$", "Solution_1": "[quote=\"dhvdochilinh\"]let $ a,b,c > 0$ prove that $ \\sum(\\frac {a^2}{b} \\plus{} c)^2 \\ge 12\\frac {a^3 \\plus{} b^3 \\plus{} c^3}{a \\plus{} b \\plus{} c}$[/quote]\r\n[hide=\"Hint(MAYBE)\"]\nI am new to the idea of homogenisations so I apologise if this idea is not correct.... :( \n\nReplace $ a, b, c$ with $ ka, kb, kc$ in your inequality to notice that for $ k\\neq0$, your inequality does not change at all.\nSo we can (perhaps) assume $ a\\plus{}b\\plus{}c\\equal{}1$... :| \nSorry if I am incorrect :oops: .[/hide]" } { "Tag": [ "algebra", "polynomial", "Ring Theory", "superior algebra", "superior algebra theorems" ], "Problem": "Hi, I'm getting stuck trying to solve this question: what are the maximal ideals in $ Z[x]$?\r\n\r\nThanks", "Solution_1": "Grammar note: ideals are not people, so \"who\" is inappropriate. Also, there are many maximal ideals, and thus \"is\" is the wrong form. That should be \"what are the maximal ideals in $ \\mathbb{Z}[x]$\".\r\n\r\nEach maximal ideal is generated by a prime $ p$ and a polynomial $ f$ which is irreducible mod $ p$.", "Solution_2": "Thank you very much for your answer, and sorry for my english.........." } { "Tag": [ "topology" ], "Problem": "Im studying for a prelim for January and need help with the following:\r\n\r\nSuppose $ X \\subset %Error. \"field\" is a bad command.\n{R}^n$ is not homeomorphic to any proper subset of itself. Show that $ X$ has empty interior. Give an example of an infinite space $ X$ satisfying the above. \r\n\r\nAny hints would be greatly appreciated.", "Solution_1": "For the first part, just note that any open ball is homeomorphic to the entire $ \\mathbb R^n$, so you can reproduce a topological copy of $ X$ in it.\r\n\r\nThe second part is easy if $ n\\ge 2$: just take a circle. I still have to think what to do when $ n\\equal{}1$ though...", "Solution_2": "Thanks for the response fedja. I think I understand where you want to go with this-am I supposed to \r\nthink about what this homeomorphic copy of X looks like (no segments of the ball?)?" } { "Tag": [ "induction" ], "Problem": "Show that there is an infinite number of pairs $(a,b)\\in N$x$N$ such that $a$ divides $b^2+b+1$ and $b$ divides $a^2+a+1$", "Solution_1": "After some (rather hard :mad:) tries the following works :\r\n\r\nLet's define the following sequence :\r\n\r\n$u_0 = 1, u_1 = 1, u_{n+1} = ({u_i}^2+u_i+1)/u_{i-1}$\r\n\r\nThen each $(u_n, u_{n+1})$ works.", "Solution_2": "Just wondering - is it trivial that all terms of that sequence are integers? I can prove it, but it is not that trivial that I would omit it. So, am I missing something?", "Solution_3": "Hi,\r\n\r\n\r\nI think the following can be proved by induction :\r\n\r\nu_{n+3} = 6 u_{n+2} - 6 u_{n+1} + u_{n}", "Solution_4": "[quote=\"t\u00b5t\u00b5\"]\n\n$u_0 = 1, u_1 = 1, u_{n+1} = ({u_i}^2+u_i+1)/u_{i-1}$\n\n[/quote]\r\n\r\nWell, how did you come up with this?" } { "Tag": [ "induction", "ratio", "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "Prove that one can colour the $n$ vertices of a graph, each having degree at most 5, with 2 colours such that at least $\\frac 35$ of it's edges have different coloured endpoints.\r\n\r\nThe uckrainean problem was proposed for a regular $n$-gon with 2004 vertices.", "Solution_1": "We can use induction: I\u2019ll always assume that whenever we remove some vertices from the graph, a coloring with the required proeprty (call it $P$) is possible, and whenever I say that we color the vertices of a graph, I\u2019m referring to a coloring with $P$.\r\n\r\nFirst of all, we may assume that all vertices have degree $2$ or $4$: if a vertex has odd degree (we disregard isolated vertices, so no vertex has degree $0$), then it\u2019s either $3$ or $5$. Assume it\u2019s $5$. Remove the vertex, color the remaining vertices, and add the vertex we have removed. It\u2019s connected to at least $3$ vertices of one color (red) and at most $2$ vertices of the other color (blue). If we color it blue, $P$ is preserved. We do something similar if the vertex has degree $3$.\r\n\r\nAssume now we have a vertex $x$ of degree $2$, and let $y$ be one of its neighbours, which we assume to have degree $4$ (if it has degree $2$, the ratios are even better, as can easily be checked, so assuming it has degree $4$ will suffice). Remove $x$, then remove $y$, and then color the rest of the graph. When we add $y$ again, we get at least $2$ bicolor edges out of $3$, and when we add $x$ again we get at least $1$ bicolor edge out of $2$, so, in total, we have added at least $3$ bicolor edges out of $5$, so $P$ is preserved. This means that we can now assume that all the vertices have degree $4$.\r\n\r\nAssume we have a vertex $a$ connected to $a,b,c,d$, and that $a,b$ are not connected to each other. We do something very similar to the above, removing $x$, then $a$ and then $b$, coloring the rest, and, after we add the three vertices, we will have added $10$ edges ($4$ for $x$, $3$ for $a$, and $3$ for $b$), out of which at least $6$ will be bicolor ($2$ for $x$, $2$ for $a$, and $2$ for $b$), and we\u2019re done. \r\n\r\nThis shows that we may assume that whenever $x$ is connected to $a,b,c,d$, these $4$ vertices form a complete graph $K_4$. This means that there are no connecttions between the set $x,a,b,c,d$ and the rest of the vertices, so our graph consists of several connected components, each one being a $K_5$, and it\u2019s easy to color each $K_5$ so as to obtain $P$." } { "Tag": [], "Problem": "Suppose that d is a factor of $ n^4 \\plus{} 2n^2 \\plus{} 2$ such that $ d>n^2 \\plus{} 1$,where n is some natural number n>1. Prove that $ d > n^2 \\plus{} 1 \\plus{} \\sqrt {n^2 \\plus{} 1}$", "Solution_1": "$ n^{2}\\plus{}1\\equal{}x$ $ \\implies$ $ d|x^{2}\\plus{}1$ and $ d>x$\r\n$ \\implies$ $ d\\equal{}x\\plus{}a$\r\n$ \\frac{x^{2}\\plus{}1}{x\\plus{}a}\\equal{}x\\minus{}a\\plus{}\\frac{a^{2}\\plus{}1}{x\\plus{}a}$ $ \\implies$ $ x\\plus{}a|a^{2}\\plus{}1$\r\n$ a^{2}\\minus{}a\\plus{}1>x$ $ \\implies$ $ a>\\sqrt{x}$ QED :wink:", "Solution_2": "Why do you have this?\r\n $ x \\plus{} a|a^{2} \\plus{} 1$\r\n$ \\equal{}> a^{2} \\minus{} a \\plus{} 1 > x$", "Solution_3": "$ x \\plus{} a|a^2 \\plus{} 1\\implies a^2 \\plus{} 1 \\geq x \\plus{} a$, or $ a^2 \\minus{} a \\plus{} 1 \\geq x$. But $ x\\equal{}n^2\\plus{}1$, so $ a^2\\minus{}a\\plus{}1\\geq n^2\\plus{}1\\implies a^2\\minus{}a\\geq n^2$. Clearly equality cannot occur so we have $ a^2\\minus{}a>n^2$, and $ a^2\\minus{}a\\plus{}1>x$." } { "Tag": [ "probability", "calculus", "articles", "integration", "Putnam" ], "Problem": "An infinite number of lines, all parallel to each other and spaced one metre apart, cover an infinite floor. If I stand on this floor and throw a rod of negligible thickness and 2 metres long up in the air, what is the probability that the rod lands and touches a line?", "Solution_1": "This looks very, very much like Buffon's Needle.", "Solution_2": "What's Buffon's Needle?", "Solution_3": "It is a problem exactly like this one, except the needle is shorter than the distance between the lines.", "Solution_4": "Yup i just read it on Wikipedia.. is there a way to solve this without calculus though?", "Solution_5": "[quote=\"CircleSquared\"]It is a problem exactly like this one, except the needle is shorter than the distance between the lines.[/quote]\r\nThe article on Wikipedia has a formula for this case also, but it is simply a variation of the original solution and uses calculus.", "Solution_6": "I dont think this is that difficult a problem at all! It is a simple case of buffon's needle.\r\n\r\nFirst note that the only thing that makes a difference is that the length. the thing about whether its curved or straight will not make a difference logically. \r\n\r\nNow, consider the circle with radius 0.5m.\r\n\r\nObviously, it will land such that it cuts the line twice (or cuts two lines once)\r\nso each throw of the \"needle\" will result in 2 cuts of the line. \r\n\r\nnow, we know that the length of the needle is 2*Pi*0.5=Pi\r\nso we know that for a needle Pi m long, it hit 2 lines(on average). For a needle 2 meters long, It will hit 4/Pi lines(on average). \r\n\r\nHowever note that the question here is ambiguous as the question asks for the probability that it hits a line and not on average how many lines does it hit. (note the difference between these two statements I solved the question for my statement and not Aidan's case)\r\n\r\nHi aidan, its Chai here! :)", "Solution_7": "just treat the angle the rod makes to the lines and distance to the nearest line from the center of the rod as two uniform random variables and solve it", "Solution_8": "Can you elaborate? maybe offer a solution?", "Solution_9": "Hi Chai, didn't see that u'd replied =P Usually AoPS informs me when someone replies to a thread i start. Why does the curvature of the needle not matter? Actually, i asked Mr Lum (my math tutor in NJC) and he had no idea how to start. That's a good common test problem =P", "Solution_10": "that would be hard to explain. But let me try. \r\nconsider a bend as two disjoint needles. \r\n\r\nso a needle which looks like |_ has a bend. Now we consider the probability that it hits a line. \r\n\r\nnotice that we can just look at one part of the needle. i.e. we can consider it as just |+_. then we just sum the number of times each of these smaller needles will cross a line and we are done.\r\nusing that as an example for your problem, you can consider your 2m long needle to be two individual 1m long needles. On average, a buffon's needle hits 2/Pi times. So yours will hit 2(2/Pi) times. \r\n\r\nHope that helps!", "Solution_11": "Ok i get it now, thanks. :) However, how would that lead to an answer to my original problem? I suspect some integration would have to be done as the angle at which the needle lands relative to the lines has to be considered. 30 degrees relative to the lines would be the limit, but smaller the angle than 30 degrees, the more leeway the needle has to settle in between any two lines. That looks hard!\r\n\r\nAnd what if we throw the needle on a grid? Haha.. Say if the grid consisted of 2m by 2m squares", "Solution_12": "The problem can be solved by calculus too... you basically treat the angle and the position of the middle of the rod as two random variables and integrate over the possibility with all angles. :)", "Solution_13": "Ok thanks.. that implies it will never appear in any math contest (save Putnam maybe). :lol:" } { "Tag": [ "limit", "pigeonhole principle", "irrational number", "calculus", "calculus computations" ], "Problem": "Let $ f: [0,1]\\rightarrow\\mathbb{R}$ for which:\r\nif $ x\\equal{}\\frac{p}{q}$; $ (p,q)\\equal{}1$; $ \\{p,q\\}\\in\\mathbb{Z}^{\\plus{}}$, then $ f(x)\\equal{}\\frac{1}{q}$.\r\nelse, $ f(x)\\equal{}0$ [i.e. when $ x$ is irrational]\r\n\r\ni) Determine $ \\lim_{x\\rightarrow a}f(x)$ where $ a\\in[0,1]$ is irrational. \r\n\r\nProof:\r\nI will use two 'basic' facts:\r\n1) The statement: \"two nonequal real numbers are arbitrarily close\" is nonsense. \r\n2) For any irrational number, $ \\alpha$, there are other irrational numbers that are arbitrarily close to $ \\alpha$.\r\n\r\nThe limit either exists or it doesn't. If it exists, it must be $ 0$ by 2). Lets assume that the limit does not exist.\r\n\r\nThen for some $ \\epsilon_{0}>0$, there does not exist some $ \\delta$ such that \r\n$ 0<|x\\minus{}\\alpha|<\\delta\\implies |f(x)\\minus{}0|<\\epsilon_{0}$.\r\n\r\nIn other words, we can always find an $ x_{i}$ that is arbitrarily close to $ \\alpha$ for which |$ f(x_{i})|\\ge\\epsilon_{0}$. But by assumption 1), $ x_{i}$ must be a sequence because it is nonsense to say that one number is 'arbitrarily close' to another number. \r\n\r\nNoting that $ \\epsilon_{0}>0\\implies f(x_{i})>0$ which implies that $ x_{i}$ is rational. Define $ x_{i}\\equal{}\\frac{p_{i}}{q_{i}}$. If we have that $ \\lim_{n\\rightarrow\\infty}\\frac{1}{q_{n}}\\equal{}0$, then we our limit is $ 0$. Assume this is not true. Then there exists some $ M\\in\\mathbb{Z}^{\\plus{}}$ such that $ \\frac{1}{q_{i}}\\ge\\frac{1}{M}$ for all $ i$. But then we are saying that $ q_{n}\\le M$. But as $ q_{n}$ has to be an infinite sequence of positive integers, it follows by the pidgeonhole principle, that there is an infinite sequence, $ y_{i}\\equal{}\\frac{a_{i}}{q}$ that gets arbitrarily close to $ \\alpha$. We have:\r\n$ \\left|\\alpha\\minus{}\\frac{a_{i}}{q}\\right|$ or $ |q\\alpha\\minus{}a_{i}|$ gets arbitrarily small [but not $ 0$]. However, this is nonsense since $ q\\alpha$ is constant and $ a_{i}$ is an integer. Contradiction. Hence the limit is equal to $ 0$.", "Solution_1": "Comment: the end of the argument can be tightened up with the following note: We define our sequence $ x_{i}$ to get closer to $ \\alpha$. Furthermore, we can take this sequence to be the one where we have all possible $ q_{i}$. [In my proof, I took $ x_{i}$ to be an arbitrary sequence. But by the constructive nature of this proof, we can add in extra terms 'in between' so it satisfies the properties that we want it to.]\r\n\r\nHence $ |q\\alpha\\minus{}a_{i}|$ is strictly decreasing. But if we consider $ \\parallel q\\alpha\\minus{}a_{2i}|\\minus{}\\{q\\alpha\\}|$, we have an infinite decreasing sequence of positive integers that is bounded below, which is a contradiction. [the $ 2i$ accounts for the fact that we can approximate a noninteger on the left and righthand sides.]\r\n\r\nii) Determine $ \\lim_{x\\rightarrow a}f(x)$ were $ a\\in [0,1]$ is rational.\r\n\r\nProof: The statement is the same as the previous where we just need to pay attention to $ 0 < |x\\minus{}a|$.\r\n\r\n\r\n---\r\n\r\nIt trivially follows by definitions of continuity that $ f(x)$ is continuous at irrational inputs and discontinuous at rational inputs." } { "Tag": [ "AMC", "AIME" ], "Problem": "The question:\r\nhttp://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2008&p=1088465\r\nI found the quotients of the two sequences, then found that the quotient is increasing in a series of triangular numbers, so I used the (n)(n+1)/2 equation to find the number. I assume everything right up until I got 33*16 instead of 33*17, which is the right answer. Was n supposed to be 32 or 33? I'm guessing it's 33 because the original series start with n=0. If this is wrong, please correct me.", "Solution_1": "Please keep all the discussion confined to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088465#p1088465]original thread[/url].\r\n\r\nThe posted solution seems fine to me." } { "Tag": [ "algebra", "inequalities" ], "Problem": "Daca $ x,y$ sunt numere pozitive, atunci $ \\frac{1}{(1\\plus{}x)^2}\\plus{}\\frac{1}{(1\\plus{}y)^{2}} \\geq\r\n\\frac{1}{1\\plus{}xy}.$", "Solution_1": "Solutia mea este una urata , o sa ma gandesc si la alta metoda , dar pana atunci :\r\n\r\n[hide=\"Click Aici\"]\\[ \\frac{1}{(1\\plus{}x)^2}\\plus{}\\frac{1}{(1\\plus{}y)^{2}} \\geq \\frac{1}{1\\plus{}xy} \\Rightarrow \\frac{(1\\plus{}x)^2\\plus{}(1\\plus{}y)^2}{(1\\plus{}x)^2(1\\plus{}y)^2} \\geq \\frac{1}{1\\plus{}xy}\\]\n\nMergem inainte , calculand :\n\n\\[ \\Rightarrow (2\\plus{}2x \\plus{} 2y \\plus{} x^2 \\plus{} y^2)(1\\plus{}xy) \\geq (x^2\\plus{}2x\\plus{}1)(y^2\\plus{}2y\\plus{}1)\\]\n\nDesfacem parantezele , facem simplificarile de rigoare si ajungem la :\n\n\\[ x^3y\\plus{}xy^3 \\minus{}x^2y^2 \\minus{}2xy \\plus{}1 \\geq 0\\] Dar stim ca din $ AM\\minus{}GM \\Rightarrow$ ca $ x^3y \\plus{} xy^3 \\geq 2x^2y^2$ , deci ramane sa demonstram ca:\n\n\\[ x^2y^2 \\minus{}2xy \\plus{} 1 \\geq 0 \\Leftrightarrow (xy\\minus{}1)^2 \\geq 0\\] . Care este evident adevarat . Deci $ \\frac{1}{(1\\plus{}x)^2}\\plus{}\\frac{1}{(1\\plus{}y)^{2}} \\geq \\frac{1}{1\\plus{}xy}$ este adevarata .[/hide]", "Solution_2": "$ {\\frac {1}{(1 \\plus{} x)^2} \\plus{} \\frac {1}{(1 \\plus{} y)^2}}\\ge{\\frac {1}{1 \\plus{} xy}}$ $ \\Longleftrightarrow$ $ {\\frac {2 \\plus{} 2(x \\plus{} y) \\plus{} (x^2 \\plus{} y^2)}{[(1 \\plus{} x)(1 \\plus{} y)]^2}}\\ge{\\frac {1}{1 \\plus{} xy}}$ $ \\Longleftrightarrow$\r\n$ \\Longleftrightarrow$ $ {(1 \\plus{} xy)[2 \\plus{} 2(x \\plus{} y) \\plus{} (x^2 \\plus{} y^2)]}\\ge{[(1 \\plus{} xy) \\plus{} (x \\plus{} y)]^2}$ $ \\Longleftrightarrow$ \r\n$ \\Longleftrightarrow$ $ {(1 \\plus{} xy)[2 \\plus{} 2(x \\plus{} y) \\plus{} (x^2 \\plus{} y^2)]}\\ge{(1 \\plus{} xy)^2 \\plus{} 2(1 \\plus{} xy)(x \\plus{} y) \\plus{} (x \\plus{} y)^2}$ $ \\Longleftrightarrow$\r\n$ \\Longleftrightarrow$ $ {(1 \\plus{} xy)[2 \\plus{} (x^2 \\plus{} y^2]}\\ge{(1 \\plus{} xy)^2 \\plus{} (x^2 \\plus{} y^2 \\plus{} 2xy)}$ $ \\Longleftrightarrow$ \r\n$ \\Longleftrightarrow$ $ {2(1 \\plus{} xy) \\plus{} (1 \\plus{} xy)(x^2 \\plus{} y^2)}\\ge{(1 \\plus{} xy)^2 \\plus{} (x^2 \\plus{} y^2) \\plus{} 2xy}$ $ \\Longleftrightarrow$\r\n$ \\Longleftrightarrow$ $ {xy(x^2 \\plus{} y^2) \\minus{} 2xy}\\ge{(1 \\plus{} xy)^2 \\minus{} 2(1 \\plus{} xy)}$ $ \\Longleftrightarrow$ $ {xy(x^2 \\plus{} y^2) \\minus{} 2xy \\plus{} 1}\\ge{(1 \\plus{} xy)^2 \\minus{} 2(1 \\plus{} xy) \\plus{} 1}$ $ \\Longleftrightarrow$\r\n$ \\Longleftrightarrow$ $ {xy(x^2 \\plus{} y^2) \\minus{} 2xy \\plus{} 1}\\ge{(xy)^2}$ $ \\Longleftrightarrow$ $ {x^3y \\minus{} x^2y^2 \\plus{} xy^3 \\minus{} 2xy \\plus{} 1}\\ge{0}$ $ \\Longleftrightarrow$\r\n$ \\Longleftrightarrow$ $ {xy(x \\minus{} y)^2 \\plus{} (xy \\minus{} 1)^2}\\ge{0}$. $ {OK!}$.", "Solution_3": "[quote=\"Virgil Nicula\"][color=darkred]Daca $ x,y$ sunt numere pozitive, atunci $ \\frac {1}{(1 \\plus{} x)^2} \\plus{} \\frac {1}{(1 \\plus{} y)^{2}} \\geq \\frac {1}{1 \\plus{} xy}.$[/color][/quote]\r\n[b]Metoda I[/b] ([u]Darij Grinberg[/u]). Vom aplica inegalitatea [b]Cauchy-Buniakovschi-Schwartz[/b] (C.B.S.) : $ x(1 \\plus{} y)^2 \\equal{} \\left(\\sqrt x\\cdot 1 \\plus{} \\sqrt y\\cdot \\sqrt {xy}\\right)^2\\le$ $ (x \\plus{} y)(1 \\plus{} xy)$ $ \\Longleftrightarrow$ $ \\frac {1}{(y \\plus{} 1)^2}\\ge$ $ \\frac {x}{(x \\plus{} y)(1 \\plus{} xy)}\\ .$ Analog se arata $ \\frac {1}{(x \\plus{} 1)^2}\\ge$ $ \\frac {y}{(x \\plus{} y)(1 \\plus{} xy)}\\ .$ Suma ultimelor doua relatii conduce la inegalitatea propusa. Avem egalitate daca si numai daca $ \\frac {\\sqrt x}{1} \\equal{} \\frac {\\sqrt y}{\\sqrt {xy}}$ si $ \\frac {\\sqrt y}{1} \\equal{} \\frac {\\sqrt x}{\\sqrt {xy}}$ , adica $ x \\equal{} y \\equal{} 1$ .\r\n\r\nVa ofer trei demonstratii la nivelul clasei a VII - a, fara utilizarea inegalitatii celebre C.B.S.\r\n\r\n[b]Metoda II.[/b] Notam $ x \\plus{} y \\equal{} s$, $ xy \\equal{} p$ . Deoarece $ x > 0$ si $ y > 0$ avem $ s > 0$ , $ p > 0$ si $ (x \\plus{} y)^2\\ge 4xy$ , adica $ s^2\\ge 4p$ . Inegalitatea devine : $ (p \\plus{} 1)(s^2 \\plus{} 2s \\minus{} 2p \\plus{} 2)\\ge (s \\plus{} p \\plus{} 1)^2$ $ \\Longleftrightarrow$ $ ps^2\\ge 3p^2 \\plus{} 2p \\minus{} 1$ . Insa $ s^2\\ge 4p$ $ \\Longrightarrow$ $ ps^2\\ge 4p^2\\ge 3p^2 \\plus{} 2p \\minus{} 1$ deoarece $ (p \\minus{} 1)^2\\ge 0$ . Avem egalitate daca si numai daca $ p \\equal{} 1$ si $ x \\equal{} y$ , adica $ x \\equal{} y \\equal{} 1$ .\r\n\r\n[b]Metoda III.[/b] $ x^2 \\plus{} 1\\ge 2x$ $ \\Longrightarrow$ $ y(x^2 \\plus{} 1)\\ge 2xy$ $ \\Longrightarrow$ $ y(x^2 \\plus{} 1) \\plus{} x(y^2 \\plus{} 1)\\ge 2xy \\plus{} x(y^2 \\plus{} 1)$ $ \\Longrightarrow$ $ (x \\plus{} y)(1 \\plus{} xy)\\ge x(y \\plus{} 1)^2$ $ \\Longrightarrow$ $ \\frac {1}{(y \\plus{} 1)^2}\\ge \\frac {x}{(x \\plus{} y)(1 \\plus{} xy)}$ . Analog se arata ca $ \\frac {1}{(x \\plus{} 1)^2}\\ge \\frac {y}{(x \\plus{} y)(1 \\plus{} xy)}$ . Din suma ultimelor doua relatii se obtine inegalitatea propusa. Avem egalitate daca si numai daca $ x \\equal{} y \\equal{} 1$ .\r\n\r\n[b]Metoda IV.[/b] Vom folosi o identitate interesanta : $ y(x \\plus{} 1)^2 \\plus{} x(y \\minus{} 1)^2 \\equal{} (x \\plus{} y)(1 \\plus{} xy)$. Deoarece si $ x(y \\plus{} 1)^2 \\plus{} y(x \\minus{} 1)^2 \\equal{} (x \\plus{} y)(1 \\plus{} xy)$ vom obtine lantul $ 0\\le \\frac {x(y \\minus{} 1)^2}{(x \\plus{} 1)^2} \\plus{} \\frac {y(x \\minus{} 1)^2}{(y \\plus{} 1)^2} \\equal{}$ $ \\frac {(x \\plus{} y)(1 \\plus{} xy) \\minus{} y(x \\plus{} 1)^2}{(x \\plus{} 1)^2} \\plus{} \\frac {(x \\plus{} y)(1 \\plus{} xy) \\minus{} x(y \\plus{} 1)^2}{(y \\plus{} 1)^2} \\equal{}$ $ (x \\plus{} y)(1 \\plus{} xy)\\cdot\\left[\\frac {1}{(x \\plus{} 1)^2} \\plus{} \\frac {1}{(y \\plus{} 1)^2} \\minus{} \\frac {1}{1 \\plus{} xy}\\right]\\ .$ Cu alte cuvinte, inegalitatea initiala este echivalenta cu inegalitatea evidenta $ 0\\le \\frac {x(y \\minus{} 1)^2}{(x \\plus{} 1)^2} \\plus{} \\frac {y(x \\minus{} 1)^2}{(y \\plus{} 1)^2}\\ .$\r\n\r\n[b]Remarca.[/b] Omogenizare : $ \\{x,y,z\\}\\subset\\mathcal R^*_ \\plus{} \\implies$ $ \\frac {1}{(x \\plus{} y)^2} \\plus{} \\frac {1}{(x \\plus{} z)^{2}} \\geq \\frac {1}{x^2 \\plus{} yz}\\ .$\r\n\r\nO interpretare geometrica a acestei inegalitati gasiti la\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=172083", "Solution_4": "[quote=Virgil Nicula]Daca $ x,y$ sunt numere pozitive, atunci $$ \\frac{1}{(1\\plus{}x)^2}\\plus{}\\frac{1}{(1\\plus{}y)^{2}} \\geq\n\\frac{1}{1\\plus{}xy}.$$[/quote]\nLet $a,b $ be positive real number such that $ab=1.$ Prove that $$2\\left(\\frac{1}{1+7a}+\\frac{1}{1+7b}\\right)\\geq \\frac{3}{2}\\left(\\frac{1}{1+5a}+\\frac{1}{1+5b}\\right)\\geq \\frac{1}{(1+a)^2}+\\frac{1}{(1+b)^2}\\geq \\frac{a+b}{a+b+2}\\geq \\frac{1}{1+3a}+\\frac{1}{1+3b}\\geq \\frac{1}{2}$$" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Give a proof,\r\n\r\nsuppose that one function is integrable and another is increasing,\r\n\r\nprove Second Mean Value theorem for integrals, derived by Riemann-Lebesque Lemma\r\n\r\n\r\nhttp://en.wikipedia.org/wiki/Mean_value_theorem", "Solution_1": "What are $ f$ and $ g$?", "Solution_2": "Could you state this theorem? There seems to be several variants of it on the internet, so I'm not sure which you meant." } { "Tag": [], "Problem": "Hi All,\r\n\r\n3^(5/16)\r\nx^(a/b)\r\n\r\nWhat is the best way of working these out ?\r\nWhat is the best way to deal with exponents containing fractions ?\r\n\r\nThanks.", "Solution_1": "$ a^{\\frac{b}{c}}$ is defined as $ \\sqrt[c]{a}^{b}$. Or something like that...", "Solution_2": "$ \\sqrt[c]{a^b}$ I always thought.", "Solution_3": "Well, they really are the same thing, just $ b$ has been shifted a quarter of an inch.", "Solution_4": "so what would be the answer to 3^(5/16) ?\r\n\r\nThanks again.", "Solution_5": "$ \\sqrt[16]{243}$. Can't be simplified.", "Solution_6": "[quote=\"IIL\"]so what would be the answer to 3^(5/16) ?\n\nThanks again.[/quote]\r\n\r\nWell, I think you use calculator to figure that one.. \r\n\r\nIn radical form, it's $ {}^{16}\\sqrt{3}^5$", "Solution_7": "I agree with DonkeyKong and alanchou. \r\nThis is in AoPS Volume 1.", "Solution_8": "For the first one, you would need a calculator to figure it out. But in general, you can write fractional exponents in radical form, so x^(a/b)=bth root of x^a", "Solution_9": "[quote=\"IIL\"]Hi All,\n\n3^(5/16)\nx^(a/b)\n\nWhat is the best way of working these out ?\nWhat is the best way to deal with exponents containing fractions ?\n\nThanks.[/quote]\r\n\r\nHello there,\r\n\r\nI'll answer your second question because the first has been well answered already. \r\n\r\nFor algebra, I recommend writing exponents as fractions because then you can work with them easily. First of all, don't be frightened by them. Just treat them like numbers. Secondly, try to express square roots as fractions too." } { "Tag": [ "inequalities" ], "Problem": "Let $ a,$ $ b$ and $ c$ are different non-negative numbers such that $ abc \\equal{} 1$. Prove that\r\n\\[ \\frac {c(a \\minus{} b)^5 \\plus{} a(b \\minus{} c)^5 \\plus{} b(c \\minus{} a)^5}{\\sqrt [3]{(a \\minus{} b)^5(b \\minus{} c)^5(c \\minus{} a)^5}}\\geq7.5\r\n\\]", "Solution_1": "[quote=\"arqady\"]Let $ a,$ $ b$ and $ c$ are different non-negative numbers such that $ abc \\equal{} 1$. Prove that\n\\[ \\frac {c(a \\minus{} b)^5 \\plus{} a(b \\minus{} c)^5 \\plus{} b(c \\minus{} a)^5}{\\sqrt [3]{(a \\minus{} b)^5(b \\minus{} c)^5(c \\minus{} a)^5}}\\geq7.5\n\\]\n[/quote]\r\n\r\nWe have \r\n\r\n$ \\sum c(a \\minus{} b)^5\\equal{} (a\\minus{}b)(b\\minus{}c)(c\\minus{}a)[\\sum a^3\\plus{}(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)\\minus{}11abc]$\r\n\r\nso after cancellation it remains to prove\r\n\r\n$ \\sum (3c\\plus{}a\\plus{}b)(a\\minus{}b)^2 \\geq 15(\\sqrt [3]{(a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2})$\r\n\r\ntrue by AM-GM.", "Solution_2": "Yes! :lol: \r\nBut what is a minimal value of the following expression?\r\n\\[ \\frac{c(a\\minus{}b)^5\\plus{}a(b\\minus{}c)^5\\plus{}b(c\\minus{}a)^5}{\\sqrt[3]{(a\\minus{}b)^5(b\\minus{}c)^5(c\\minus{}a)^5}}\\]\r\nIt's not $ 7.5$ :wink:" } { "Tag": [ "AMC", "AMC 8", "Support" ], "Problem": "Is there any way we can find a list of schools participating in the AMCs?\r\n\r\nI want to take the AMC8 and both days for the AMC10/12. If my school doesn't do that, and I can't convince them, can I take the test in a city that is about 70 miles away?", "Solution_1": "1st: Read the AMC FAQ #4, follow the advice there.\r\n\r\n2nd: Pay attention to the listing at:\r\nhttp://unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_2": "I did that. The closest place on the list was about 5 hours away. Thanks anyway.", "Solution_3": "[quote=\"PowerOfPi\"]I did that. The closest place on the list was about 5 hours away. Thanks anyway.[/quote]\r\n\r\nWait until about a month before the AMC 10/12 A. It's possible some schools just haven't registered yet.\r\n\r\nIf not and you're really desperate you COULD try to take it at that place 5 hours away.", "Solution_4": "Out of curiosity... my school offers only the B date. What should I do if I want to take the A test too?", "Solution_5": "[quote=\"Paiev\"]Out of curiosity... my school offers only the B date. What should I do if I want to take the A test too?[/quote]\r\n\r\nThere's not much you can do besides take the A at another school.\r\n\r\nDon't worry, I have a similar problem. I'm taking A at my school, B at another.", "Solution_6": "[quote=\"Paiev\"]Out of curiosity... my school offers only the B date. What should I do if I want to take the A test too?[/quote]\r\n\r\nSomeone correct me if I'm wrong, but I don't think you can do much to take the A test. The colleges that offer the AMC only offer it on the B dates, so unless if you actually go to another high school to take the A test (which would probably be inconvenient for you, the high school, and the MAA), I don't think you will be able to.\r\n\r\nI was in the same position as you the last 2 years, and in the end, I only was able to take the B test both years.", "Solution_7": "The ideal solution is of course to convince your school to offer the AMCs, since that would allow some other students to take it as well, and hopefully continue even after you're gone. More participation is always good.\r\nStudents taking multiple tests is a loophole created in recent years. Don't expect support for it, since the two dates are really there to allow schools to pick one that is more convenient.", "Solution_8": "Worst case, just don't take both if the school won't let you. It won't kill you. Why I'm saying this is it's harder than you think sometimes to change the school's mind, remember they pay money for this.", "Solution_9": "Wow, in my school, it's all arranged by a student-founded club, and the members have to contribute to the costs of the tests. We're not even sure some years if we can even take one test, let alone 3, due to lack of school funding.", "Solution_10": "[quote=\"pascal12\"]Wow, in my school, it's all arranged by a student-founded club, and the members have to contribute to the costs of the tests. We're not even sure some years if we can even take one test, let alone 3, due to lack of school funding.[/quote]\r\n\r\nYou could ask for donations, although i'm not sure that will help much. Too many jocks at my school :P", "Solution_11": "My school always ran it with the sponsoring teacher paying the registration fee, and then having students contribute a dollar or two each for the cost of the tests. It's currently \\$40-\\$60 (per date) for AMC 10/12 registration, plus \\$16 per ten tests. If you're worried about funding, try that system- and a teacher who's invested in it that way is likely to keep it going longer than the club.", "Solution_12": "[quote=\"jmerry\"]My school always ran it with the sponsoring teacher paying the registration fee, and then having students contribute a dollar or two each for the cost of the tests. It's currently \\$40-\\$60 (per date) for AMC 10/12 registration, plus \\$16 per ten tests. If you're worried about funding, try that system- and a teacher who's invested in it that way is likely to keep it going longer than the club.[/quote]\r\n\r\nThere are very few teachers that will pay from their own bank account to some competition that only 3 people (at max) participate in..." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Let $ {a},{b},{c}$ are positive reals, such that $ {a}\\plus{}{b}\\plus{}{c}\\equal{}1$. Find the maximum value of $ \\frac{a^{2}}{b\\plus{}1}\\plus{}\\frac{b^{2}}{c\\plus{}1}\\plus{}\\frac{c^{2}}{a\\plus{}1}$", "Solution_1": "Do not double post:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278128[/url]" } { "Tag": [ "linear algebra", "matrix", "function", "algorithm" ], "Problem": "Hi,\r\n\r\n[b]Question 1:[/b] (cutting all the crap)\r\n\r\nThere is a community hall where events are going on. People can come in and go out whenever they want. At one instant only one person may come in or go out. An observant fellow keeps track of times when ppl come in and go out in the form of integers (ie: 1 signifies, the guy has entered 1 second after the event as begin). At each instant so many people may be in the hall. Your aim is to find the maximum number of people present in the hall (this will be at one particular instant).\r\n\r\nSample Input: (the values at entry and exit are the no. of seconds after which they enter and leave)\r\n\r\nPerson Entry Exit\r\nPerson1 1 7\r\nPerson2 2 4\r\nPerson3 6 9 \r\nPerson4 3 8\r\nPerson5 5 10\r\n\r\nOutput is : 4\r\n\r\nAt the interval between second 6 and second 7, four ppl ( 1,3,4,5) are present in the hall, which is the maximum. So the answer is 4.\r\n\r\nGiven: No.of People <=10^5\r\nEntry Time and Exit time are all distinct and t<=10^7\r\n\r\n[b]Question 2::[/b] \r\n\r\nThere exists a 2 x N matrix (2 rows and n columns), which can be filled by dominoes of size 2x1, horizontally or vertically. Each cell in the matrix has a value. The Value the domino is said to be the difference between the bigger value and smaller value that it covers.\r\n\r\nEx:\r\n8 6 2 3\r\n9 7 1 1\r\n\r\nAssume a domino covers like this [8 6]. Value of this domino is said to be 2 (8-6).\r\nThe task is find the maximum possible sum of values of the domino. (ie: You can cover it in so many ways, in which way will the possible sum of values of dominoes be max)\r\n\r\nIn above question:\r\n\r\nDomino 1 - 8 9 (Value 1)\r\nDomino 2 - 6 2 (value 4)\r\nDomino 3 - 7 1 (value 5)\r\nDomino 4 - 3 1 (value 2)\r\n\r\nSum is 12\r\n\r\nOutput: 12\r\n\r\nGiven: N<=10^5. Values of cells maybe from 0 --> 10^4\r\n\r\nI've done well, what about you guys. Cutoff may be higher. [/img]", "Solution_1": "Got one sum fully right i think . :D :\r\n People say one fully right sum is enough to steer you through. :maybe: \r\nBut i dont think so as almost everyone in my centre has got atleast one sum fully right \r\nAnd i am hence not expecting anything. :D :\r\nAnd i didnt even attempt the second so absolutely no chance of extra part marks \r\nSo if the cutoff is only one sum i may get through. \r\nAnd it took me so much time bcos of one part in the first sum which worked in my house but which didnt work in the exam centre (and even Shreyas says he doesnt know wat is wrong in wat i did)", "Solution_2": "congrats shreyas and madness hope u both get through :)", "Solution_3": "I have done pretty decently and got both the problems.\r\nThe first one is fine for me and works for any permutation.\r\nThe second one works for low values in the grid but fails when the values are too big(dunno why). But I have no hope and will accept any result that comes my way. I am only happy that I atleast managed to have a crack at the paper.", "Solution_4": "from whatever i heard of i complete gets u thru", "Solution_5": "where i tends to 2 i suppose :D :D :D", "Solution_6": "@rock: Is your first sum $ O(nlogn)$ or $ O(n^2)$? \r\n\r\nnlogn logic isn't so straight forward, I dunno how rohit claims it is.\r\n\r\n2nd was a std DP..\r\n\r\nOver the previous years, getting 1 fully right and atleast 30% of 2nd sum has usually got ppl through. Might by 1 & 50% this year, so let's wait for 1 month.", "Solution_7": "Mine is O(n^2)\r\nI didn't mind on the logic part because my primary aim is that the program should compile and give u the required result(Shre it worked for 10^7 I was surprised).", "Solution_8": "[quote=\"chemrock\"]Mine is O(n^2)\nI didn't mind on the logic part because my primary aim is that the program should compile and give u the required result(Shre it worked for 10^7 I was surprised).[/quote]\r\n\r\nHey, you entered 10^7 values or are you talking about int??\r\n\r\nInt is 4 bytes and accepts till $ 32767^2$ which is $ 10^9$\r\n\r\nIncase of 2nd sum int is okay compared because values range from 0 --> 10^4.\r\nThe maximum sum can be 10^5 x 10 ^4 = 10^9. which falls within ints range. \r\n\r\n@pardesi: how did your friend do info da?\r\n\r\nAbout 20 students will qualify from INOI-2008 for the training camp in June to select the Indian team to IOI-2008. ====> This worries me a lot :(", "Solution_9": "shreyas he got the first one...not the second\r\na friend of his got the first one and wrote the program for second which didn't work\r\ndon't worry da u have done well ...u will make the cut", "Solution_10": "[quote=\"pardesi\"]shreyas he got the first one...not the second\na friend of his got the first one and wrote the program for second which didn't work\ndon't worry da u have done well ...u will make the cut[/quote]\r\n\r\nThere are lot of kattans in India. The guy who got bronze medal in IOI last year wrote INOI in our centre. And when the number is only 20. Only 24 students made the cut last year//http://www.iarcs.org.in/inoi/2007/inoi2007/results_inoi2007.php But lol the guy who had my roll number (90) last year went to the camp > :D:D\r\n\r\n\r\nBut ofcourse would be nice if I could go here (http://www.tisb.org/). .campus is nice. :D", "Solution_11": "finding similarities :D", "Solution_12": "well whatever logic i used is really simple accord to me (Well bcos i wrote it :D ).\r\nAs u dont want me to discuss my logic wid u then how do u expect to find my logic simple :P (seeing u dont know wat it is)\r\nAnd did u ask Venkatesh whether $ (1.6)\\cdot 10^6$ will pass (im talking abt the memory) :maybe:", "Solution_13": "[quote=\"madness\"]well whatever logic i used is really simple accord to me (Well bcos i wrote it :D ).\nAs u dont want me to discuss my logic wid u then how do u expect to find my logic simple :P (seeing u dont know wat it is)\nAnd did u ask Venkatesh whether $ (1.6)\\cdot 10^6$ will pass (im talking abt the memory) :maybe:[/quote]\r\n\r\nI have asked him. he hasn't replied..... And I still don't think yours might be nlogn... Perhaps I will hear your logic after results come out. \r\n\r\nHmm...I thought 2nd problem was much easier than first but ppl say otherwise. :((", "Solution_14": "It was to you (and could've been to me had i tried but i didnt :D ).\r\nAnd am sure that my logic is nlogn and am almost sure that it works completely. :D", "Solution_15": "[quote=\"Cyber-Shreyas\"]\n\n\n\nAnd surely won't there be a memory limit within 40-64MB???? The region is pretty volatile. :) I've used 1.6 MB. \n\nThe inbuilt function in algorithm.h sort(a,a+n) where a is a pointer uses the most efficient sort. I am guessing quick sort but it might be merge sort as awell.\n\n@sid: I don't think the cut-offs are class dependent. :( But not sure. Who all have got both right??? Anyway many ppl can get both right but lot of other factors come into play. somebody in our centre i heard used an array of size 10^8, which obviously fails. :(\n\n@Pradeep: Well, for an input where n=10^5, n2 = 10^10. 10^10 operations on the array might not pass.[/quote]\r\n\r\n\r\nthe memory limits at the ioi last year were in the range 16MB to 150MB . but since nothing was mentioned here i dont know if they will keep a limit\r\n\r\nsort(a, a+n) is introsort, a mixture of quicksort and heapsort . for details , read this http://en.wikipedia.org/wiki/Sort_%28C%2B%2B%29\r\n\r\ni dont have my code with me , but it will look somewhat like this ( I used printf and scanf but i 'm writing cin and cout here for better understanding \r\n\r\n#include\r\n\r\nusing namespace std;\r\nint i,p,q,k=0,answer=0, a[10000002],n;\r\nint main()\r\n{\r\ncin>>n;\r\nfor( i=0;i>p>>q;\r\na[p]++;\r\na[q]--;\r\n}\r\nfor( i=0;i<10000002;i++)\r\n{\r\nk+=a[i];\r\nif( k>answer) answer = k;\r\n}\r\nreturn 0;\r\n}\r\n\r\ni've simply written it now from my mind and haven't compiled it, so there might be some small errors.", "Solution_16": "[quote=\"bruteforce\"]\n\nforget about the first question :D \n\nthe memory limits at the ioi last year were in the range 16MB to 150MB . but since nothing was mentioned here i dont know if they will keep a limit\n\nsort(a, a+n) is introsort, a mixture of quicksort and heapsort . for details , read this http://en.wikipedia.org/wiki/Sort_%28C%2B%2B%29\n\ni dont have my code with me , but it will look somewhat like this ( I used printf and scanf but i 'm writing cin and cout here for better understanding \n\n#include\n\nusing namespace std;\nint i,p,q,k=0,answer=0, a[10000002],n;\nint main()\n{\ncin>>n;\nfor( i=0;i>p>>q;\na[p]++;\na[q]--;\n}\nfor( i=0;i<10000002;i++)\n{\nk+=a[i];\nif( k>answer) answer = k;\n}\nreturn 0;\n}\n\ni've simply written it now from my mind and haven't compiled it, so there might be some small errors.[/quote]\r\n\r\nAhh..Privacy concerns. Okay, then. I've edited my post. :D\r\n\r\nYour program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...", "Solution_17": "[quote=\"Cyber-Shreyas\"]\nYour program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...[/quote]\r\n\r\ni've already said that the logic was same , but he wanted to see my code. I think the other logic mentioned here ( sorting the entry time and exit time individually ) is better than this. if a good sorting algorithm is used, that one would be O(nlogn) where n= 10^5. this method is O(n) where n is 10^7 . moreover , it uses a lot of memory.", "Solution_18": "[quote=\"bruteforce\"][quote=\"Cyber-Shreyas\"]\nYour program is more or less same as vihangs. But I guess O(n) is better than O(n^2) (mine)...[/quote]\n\ni've already said that the logic was same , but he wanted to see my code. I think the other logic mentioned here ( sorting the entry time and exit time individually ) is better than this. if a good algorithm is used, that one would be O(nlogn) where n= 10^5. this method is O(n) where n is 10^7 . moreover , it uses a lot of memory.[/quote]\r\n\r\nAnother classic case of Space was Memory. :D I guess nlogn for 10^5 and n for 10^7 is the same. Anyways lets wait for the results. we same to have saturated all discussions. \r\n\r\n@bruteforce: Loads of people claim that they got both right. What about last time?? Dunno if cut-offs might be higher.", "Solution_19": "Last time at the camp , there were rumours like u need a positive score in both , u need atleast 1 perfect score , etc. \r\n\r\nbut our sir said that u can get selected with just one bruteforce method. infact full marks in both questions are rare. \r\n\r\nbtw , almost half of the people in the camp were from chennai, i think there is something special about that place.", "Solution_20": "[quote=\"bruteforce\"]Last time at the camp , there were rumours like u need a positive score in both , u need atleast 1 perfect score , etc. \n\nbut our sir said that u can get selected with just one bruteforce method. infact full marks in both questions are rare. \n\nbtw , almost half of the people in the camp were from chennai, i think there is something special about that place.[/quote]\r\n\r\nhow much did you get last time??\r\n\r\nI hope this time too chennai rocks. Can I assume you too are from Chennai?? I can see some pride there in the post. \r\n\r\nLots of people claim that they got both right this time. :( Lets see.", "Solution_21": "I got both last time . i dont remember about the first . for the second one i got only partial mark because there was a solution with better complexity .", "Solution_22": "I've received my official certifcate. (letter) stating that I qualified in ZIO and have written INOI. What about you??", "Solution_23": "hey u got through shreyas CONGRATS DA :coool: :coool: :coool: :coool: :coolspeak: :coolspeak: :coolspeak: \r\ntreat le photo ve kude da :D", "Solution_24": "[quote=\"pardesi\"]hey u got through shreyas CONGRATS DA \ntreat le photo ve kude da :D[/quote]\r\n\r\nAvan photo va kudukamattan da :D He meant the official letter for qualifying to write the inoi :D \r\nI got it too :P", "Solution_25": "seri dan poda :D", "Solution_26": "INOI results have been declared. http://www.iarcs.org.in/inoi/2008/inoi2008/results_inoi2008.php", "Solution_27": "dei shreyas congrats da u shoudn't have edited that\r\ndei treata kude da :10: :10: :10: :first: :first: :first: :coolspeak: :coolspeak: :coolspeak:", "Solution_28": "Treat da shre\r\n[color=blue][size=200][u][i][b]CONGRATS[/b][/i][/u][/size][/color]", "Solution_29": "i am locking this for obvious reasons :D" } { "Tag": [], "Problem": "Let $ S(n)$ denote the sum of the digits of $ n$. Show that for positive integers $ a$, $ b$:\r\n\r\n$ S(a\\plus{}b)\\le S(a)\\plus{}S(b)$\r\n\r\nwith equality when there are no carry overs in the addition of $ a\\plus{}b$.", "Solution_1": "[hide=\"Maybe not the best proof, but it is a start\"]WLOG let $ a\\ge b$. Let $ A_{m}A_{m\\minus{}1}A_{m\\minus{}2}\\ldots A_{1}$ represent the digits of $ a$ (where $ m$ is the number of digits in $ a$), and let $ B_{m}B_{m\\minus{}1}B_{m\\minus{}2}\\ldots B_{1}$ the digits of $ b$. If $ a$ and $ b$ have a different number of digits, then we let $ B_{m}\\equal{}B_{m\\minus{}1}\\equal{}\\underbrace{\\ldots\\ldots\\ldots\\ldots\\ldots}_{\\text{continue as necessary}}\\equal{}0$. For example, if $ a\\equal{}1787569$ and $ b\\equal{}12345$, then $ B_{7}\\equal{}B_{6}\\equal{}0$, while $ B_{5}\\equal{}1$. Clearly, $ S(a)\\equal{}\\sum_{i\\equal{}1}^{m}A_{i}$ and $ S(b)\\equal{}\\sum_{i\\equal{}1}^{m}B_{i}$\n\nDefine $ C_{n}\\equal{}A_{n}\\plus{}B_{n}$. Therefore $ a\\plus{}b\\equal{}C_{m}C_{m\\minus{}1}C_{m\\minus{}2}\\ldots C_{1}$. However, if there exists a $ C_{n}\\ge 10$, then we carry, which is actually subtracting 10 from $ C_{n}$ and adding 1 to $ C_{n\\plus{}1}$. Therefore we subtract 9 from $ S(a\\plus{}b)$, hence $ S(a\\plus{}b)\\le S(a)\\plus{}S(b)$, with equality holding when no carrying is necessary.\n\n$ \\mathbb{QED}$[/hide]", "Solution_2": "Let's see that for one digit number a,b.\r\n[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2/d/b/2db75d491ea39d87f646cf88d5517885269a3356.gif[/img]\r\nIt's true becuse each number 9s_n+2\\sqrt{s_n}+1$, and proved by chebeshev theorem.", "Solution_2": "The following inequality gives an elementary proof of this statement:\r\n \r\nFor all natural numbers n, \r\n\r\n(1) $s_n < (p_{n+1}/2)^2$.\r\n\r\nProof: The inequality is obviously true for n=1 since $s_1 = p_1 = 2$ and $p_2 = 3$. Next assume (1) is true for $n=k-1$. Then \r\n\r\n$s_k = s_{k-1} + p_k < (p_k/2)^2 + p_k < (p_k/2 + 1)^2 \\leq (p_{k+1}/2)^2$.\r\n\r\nThus $s_k < (p_{k+1}/2)^2$, which completes the induction proof of the inequality (1).\r\n\r\nSuppose this statement of \"dieuhuynh\" is false. Then there must exits a natural number n such that the interval $[s_n,s_{n+1}]$ contains no integer squares. In other words, there is a natural number a such that\r\n\r\n$(2) \\;\\; a^2 < s_n < s_{n+1} < (a + 1)^2$\r\n\r\nwhich implies that\r\n\r\n$2a + 1 = (a + 1)^2 - a^2 \\geq (s_{n+1} + 1) - (s_n - 1) = s_{n+1} - s_n + 2 = p_{n+1} + 2$,\r\n\r\ni.e. $p_{n+1} < 2a$. Now $a < \\sqrt{s_n}$ by (2), hence $p_{n+1} < 2\\sqrt{s_n}$. Therefore $s_n > (\\frac{p_{n+1}}{2})^2$, contradicting the inequality (1)." } { "Tag": [], "Problem": "Okay . . . Lets have some fun . . . :lol: \r\n\r\nCan you arrange 6 9s to make 100. You can use Multiplication, Division, Addition, Subtraction and decimal as well.\r\n\r\nGood luck", "Solution_1": "$9*9+9+\\frac{9}{9}+9=100$", "Solution_2": "$9\\cdot 9\\cdot 9\\cdot 9=100_{9\\cdot 9}$\r\n\r\n :D", "Solution_3": "$99+\\frac{99}{99}=100$\r\n\r\n$\\frac{999}{9.99}=100$", "Solution_4": "$\\frac{999}{999}=100$%", "Solution_5": "If the solution includes 99\r\n[hide]\n$99+\\frac{9+9}{9+9}=100$\n$99+\\frac{9*9}{9*9}=100$\n$99+\\frac{9/9}{9/9}=100$\n$99+\\frac{9.9}{9.9}=100$\n$99+\\frac{9}{9}+9-9=100$\n$99+\\frac{9}{9}x\\frac{9}{9}=100$\n$99+\\frac{(\\frac{9}{9})}{(\\frac{9}{9})}=100$\n$99+\\frac{9}{(\\frac{9}{(\\frac{9}{9})})}=100$\n$99+\\frac{(\\frac{9}{(\\frac{9}{9})})}{9}=100$\n[/hide]", "Solution_6": "Try using 9 nines to make 108.", "Solution_7": "$99+9+\\frac{9\\cdot 9\\cdot 9}{9\\cdot 9\\cdot 9}=108$", "Solution_8": "No, that's 109. And I said 6 nines.", "Solution_9": "[quote=\"nutz_for2.718281828\"]No, that's 109. And I said 6 nines.[/quote]\r\nActually, you did say 9 nines. And perfectnumber's solution can be corrected by saying $99+9\\cdot\\frac{999}{999}$\r\nI'm still working on the 6 nines one", "Solution_10": "[quote=\"nutz_for2.718281828\"]No, that's 109. And I said 6 nines.[/quote]\r\nwow, my bad. yeah that's supposed to be a multiplication there at the end. and you said 9 9s.", "Solution_11": "oops... I meant 6 nines.", "Solution_12": "For six 9's to get 108: $9\\times(9+\\frac{9+9+9}{9})$", "Solution_13": "[hide]9*(9+(9+9+9)/9)=108[/hide]", "Solution_14": "[hide]$(\\frac{9}{9}+9)*9+9+9$[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "For a positive integer k, let $f_{1}(k)$ be the square of the sum of the digits of k. (For example $f_{1}(123)=(1+2+3)^{2}=36$.) Let $f_{n+1}(k)=f_{1}(f_{n}(k))$. Determine the value of the $f_{2007}(2^{2006})$. Justify your claim.", "Solution_1": "$f_{5}(2^{2006})=169,f_{6}(2^{2006})=256,...$ Therefore $f_{2k+1}(2^{2006})=169, k\\ge 2.$", "Solution_2": "How did you calculate f5(2^2006)?", "Solution_3": "$f_{1}(2^{2}006)\\le (603*9)^{2}=29452329$ and $f_{1}(2^{2006})=7(mod 9)$.\r\nIt give $f_{2}(2^{2006})\\le 61^{2}=3721$ and $f_{2}(2^{2006})=4(mod 9)$.\r\nIt give $f_{3}(2^{2006})\\le 22^{2}=484$ and $f_{3}(2^{2006})=7(mod 9)$.\r\nTherefore $S(f_{3}(2^{2006}))=16 \\ or \\ 7$. It give $f_{4}(2^{2006})=256 \\ or \\ 49.$ Therefore $f_{5}(2^{2006})=13^{2}=169$.", "Solution_4": "[quote=\"Rust\"]$f_{1}(2^{2}006)\\le (603*9)^{2}=29452329$ and $f_{1}(2^{2006})=7(mod 9)$.\nIt give $f_{2}(2^{2006})\\le 61^{2}=3721$ and $f_{2}(2^{2006})=4(mod 9)$.\nIt give $f_{3}(2^{2006})\\le 22^{2}=484$ and $f_{3}(2^{2006})=7(mod 9)$.\nTherefore $S(f_{3}(2^{2006}))=16 \\ or \\ 7$. It give $f_{4}(2^{2006})=256 \\ or \\ 49.$ Therefore $f_{5}(2^{2006})=13^{2}=169$.[/quote]\r\n\r\nYou are taking large steps :P But you're absolutely right.", "Solution_5": "I also did almost the same thing\r\n\r\n$64\\cdot 10^{600}<2^{2006}=4\\cdot 8^{668}<4\\cdot 10^{668}$\r\n\r\nso $25\\cdot 10^{6}<5410^{2}10). Then for all prime p, suth that $p=k(mod \\ q(q-1))$ we have\r\n(2) $p^{p}-(p-1)^{p-1}=k^{k}-(k-1)^{k-1}=0(mod \\ q^{2})$. \r\nTherefore if exist one k, satisfyed (1), then exist infinetely many primes p, satisfyed (2).", "Solution_2": "That's what I got also, but I want to see how you find it per computer.", "Solution_3": "$59^{2}|257^{257}-256^{256}$ :D \r\nthat together with the Dirichlet theorem on a.progressions solves the problem.", "Solution_4": "I found $p=257,487,815,897,1225,1373,2247,2895 (mod 3422)$ give $q^{2}|p^{p}-(p-1)^{p-1}$ with q=59,\r\n$p=3849,4723,5487,6677(mod 6806)$ with q=83,\r\n$p=2029,17361,18705,30433(mod31862)$ with q=179,\r\n$p=26355(mod 37056)$ with q=193, e.t.c.", "Solution_5": "zzz boring computer cheaters :P" } { "Tag": [ "algebra", "polynomial", "limit", "partial fractions", "algebra unsolved" ], "Problem": "If $Q$ is a polynomial with roots $a_1,a_2,\\ldots, a_n$, and if $P$ is a polynomial of degree $2 unit squares whose corners are the points (i,j), where 0 \\leq i,j \\leq n.\r\n\r\nNow, consider the set E of the 4(n-1) points with coordinates (i,0) or (i,n) or (0,j) or (n,j) except the 4 corners of S.\r\n\r\nLet's give a set A of squares from which we may form S.\r\nAny square from A may only have two points from E as corners (at most two. The whole square has none). From the pigeon-hole principle, we deduce that A has at least 2(n-1) elements, that is f(n) \\geq 2(n-1).\r\n\r\nTo prove the reverse inequality, it suffices to give an exemple of such a drawning with only 2(n-1) squares, which will be boring with a keyboard...\r\nlet's M(0,0) and P(n,n) be two opposite vertices of the whole square. Let N(0,n),Q(n,0) be twe two other opposite vertices. Construct each of the squares with corner M (resp. P) and side 2,3,...,n-1. Then, we may finish the drawning of S with the square with corner N and the one with corner Q and both with sides n-1. Thus, a drawning does exist with only 2(n-1) squares.\r\n\r\nPierre.", "Solution_14": "Ok, I see my problem. I interpreted it as \"find the minimum number of [b]tiles to cover [/b]an NXN checker board\" - Very similar to the 'Mrs Perkins Quilt' problem. That's a considerably more difficult question. In that case we have\r\nf(n) = 4 for n even and\r\nn f(n)\r\n3 6\r\n5 8\r\n7 9\r\n11 11\r\n13 11\r\n15 6\r\n17 12\r\n19 13\r\n\r\n And so forth.....", "Solution_15": "there doesn't seem to be any pattern in these numbers .. :?", "Solution_16": "If there was a pattern, I would have already posted a solution. Now you see why I was waiting so eagerly :D \r\n There is more to be said about my take on the problem. If you draw these out by hand you will notice a peculiar property at N=9 and 15 (and all odd composite numbers). I wont hijack this thread with the details. It's neat though.\r\nEven still, I havn't fully read the accepted solution yet but Im not sure how you can use 2(4-1)=6 squares to draw a 4X4 checkerboard. Did Pbornsztein mean instead that F(1)=1, F(2) = 2 and in general, F(n)=2*F(n-1) yielding F(n) = 2^(n-1)??? I guess I'm not sure on what drawing a NXN board really means. Is a checkerboard complete if each location either has a square or is bounded by 4 squares? The outside of the checkerboard could then also be considered a square so that the locations on the perimeter would need to either contain a square or be surrounded by 3 adjacent squares.\r\n I guess I need clarification from bugzpodder as to the details." } { "Tag": [ "inequalities theorems", "inequalities" ], "Problem": "Please explain what is CYH, PQR, SMV, EMV, SOS, and similar method that often used in this forum..\r\n\r\nThanks.. :D", "Solution_1": "[quote=\"Ronald Widjojo\"]Please explain what is CYH, PQR, SMV, EMV, SOS, and similar method that often used in this forum..\n\nThanks.. :D[/quote]\r\nAbout $ EMV$ see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=205183\r\nAbout $ SMV$ see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=113248\r\nAbout $ SOS$ see here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127" } { "Tag": [ "logarithms", "function", "real analysis", "real analysis unsolved" ], "Problem": "n>0 integer\r\n$u_n$ the smallest positive root of $e^x=x^n$ \r\n\r\nGive an asymptotic expansion of $u_n$ with two terms ?", "Solution_1": "Guess based on numerical twiddling (no proof):\r\n\r\n$u_n\\sim 1+\\frac1n+\\frac3{2n^2}$\r\n\r\n(Note that $u_n$ is only defined for $n>e.$)\r\n\r\nOK, amend that, I do have a derivation, given the assumption that an asymptotic expansion exists.\r\n\r\nWrite the equation to be solved as $x=n\\ln x.$\r\n\r\nSuppose $u_n=1+\\frac an+\\frac b{n^2}+O(n^{-3}).$\r\n\r\nThen $\\ln(u_n)=\\frac an+\\frac b{n^2}-\\frac{a^2}{2n^2}+O(n^{-3})$\r\n\r\n$1+\\frac an+\\frac b{n^2}+O(n^{-3})=a+\\frac{2b-a^2}{2n}+O(n^{-2})$\r\n\r\nEquating terms, $1=a$ and $a=\\frac{2b-a^2}2,$ from which $b=\\frac32.$", "Solution_2": "Let $z=1/n$. Then the equation to solve can be rewritten as $F(x,z)=x-e^{zx}=0$. Note that $F(x,z)$ is an analytic function of 2 complex variables and $\\frac{\\partial F}{\\partial x}(1,0)=1\\ne 0$. Thus, by the implicit function theorem, there exists an analitic function $g(z)$ in some neighborhood of the origin such that $g(0)=1$ and $F(g(z),z)=0$. This function gives our solution $u_n$ for small positive $z=\\frac 1n$. But, as any analytic function, it can be written as the sum of its Taylor series $g(z)=\\sum_{n\\ge 0}a_n z^n$. From here, it is clear that the desired asymptotic expansion of $u_n$ in inverse powers of $n$ exists and, moreover, the corresponding series converges to $u_n$." } { "Tag": [], "Problem": "1) Mixture of two liquids A and B is placed in a cylinder containing piston. Piston is pulled out isothermally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining, the mole fraction of A in vapour is $ 0.4$. Given $ P^0_A = 0.4atm$ and $ P^0_B = 1.2 atm$ at the experimental temperature. Calculate the total pressure at which the liquid has almost evaporated.(Assume ideal behaviour)\r\n\r\n2) The latent heat of fusion of ice is $ 80cal/gm$ at $ 0^0C$. What is the freezing point of a solution of KCl in water containing $ 7.45$ gm of solute in $ 500gm$ of water, assuming that the salt is dissociated to the extent of $ 95%\n$.\r\n\r\n3)The molar volume of liquid benzene(density = 0.877 g/ml) increases by a factor of 2750 as it vaporizes at $ 20^0C$ and that of liquid toluene(density = 0.867 g/ml) increases by a factor of 7720 at $ 20^0C$, at a vapour pressure of 46 torr. Find the mole fraction of benzene in the vapour phase of the above solution.\r\n\r\n4) An aqueous solution containing 288gm of a non volatile compound having stoichiometric composition $ C_xH_{2x}O_x$ in $ 90gm$ water boils at $ 101.24^0C$ at 1 atm pressure. What is the molecular formula?\r\n\r\nGiven : $ K_b(H_2O) = 0.512 K mol^{ - 1}kg$ and $ T_b(H_2O) = 100C$", "Solution_1": "[hide=\"Number 4\"]From $ \\Delta T \\equal{} K_b \\cdot \\bar{m}$ we have $ 1.24 \\,\\,K \\equal{} 0.512\\,\\,K\\,kg\\,mol^{\\minus{}1} \\cdot \\frac{288}{M(0.288 \\plus{} 0.090)} \\equal{}> M \\equal{} 314.6\\,\\,g\\,mol^{\\minus{}1}$, where M is the sought molar mass and $ \\bar{m}$ is the solutions' molality. But $ M \\equal{} 12.011x \\plus{} 2 \\cdot 1.0079x \\plus{} 15.9994x \\equal{} 30.0262x$, from which we get $ x \\equal{} 10.5 \\approx 11$.[/hide]", "Solution_2": "[quote]From $ \\Delta T \\equal{} K_b \\cdot \\bar{m}$ we have $ 1.24 \\,\\,K \\equal{} 0.512\\,\\,K\\,kg\\,mol^{ \\minus{} 1} \\cdot \\frac {288}{M(0.288 \\plus{} 0.090)} \\equal{} > M \\equal{} 314.6\\,\\,g\\,mol^{ \\minus{} 1}$, where M is the sought molar mass and $ \\bar{m}$ is the solutions' molality. [/quote]\r\n\r\nHow come this is the molality $ \\frac {288}{M(0.288 \\plus{} 0.090)}$ and not this $ \\frac {288}{M.090}$. \r\n\r\nWell actually the second gives the right answer x= 44 but I think since the solute is so much in excess , the situation might just be something else.\r\n\r\nPLease try others also.", "Solution_3": "that's cent pc rite\r\nmost books use the conventional formula..without giving a look to solute amont..even NCERT :rotfl: :P", "Solution_4": "But the question is from reliable source and we get answer using 288/M x .09 only.", "Solution_5": "Here are my answers. Please do tell if they are right. I shall post my method, if they are right.\r\n\r\n[hide=\"1\"]0.667 atm\n[/hide]\n[hide=\"2\"]$ \\approx \\minus{}0.37^oC$[/hide]\n\n[hide=\"3\"]Either I have misinterpreted the question or the question has messed up with its data. Here is what I have gathered:\nPure liquid benzene vaporises and occupies a volume 2750 times its original value.\nPure liquid toluene vaporises and occupies a volume 7720 times its original value at a VP of 46 torr.\nI checked the VP of toluene at 20C. It was 21.86 torr.\nAnd all of a sudden, there is a mention of a solution...\n :o :P [/hide]\n\n[hide=\"4\"]I agree with Carcul Sir. I did not do the calculations, but I agree with his method. Note that $ \\Delta T\\equal{}K_f\\cdot m$ is an approximation. The correct formula is $ \\Delta T\\equal{}\\frac {RT_m^2x_s}{\\Delta H_f}$\nwhere $ x_s$ is the mole fraction of solute and $ \\Delta H_f$ is the enthalpy of fusion.[/hide]\r\n\r\nA Useful link that include the derivation of the various colligative properties\r\nhttp://www.lsbu.ac.uk/water/collig.html", "Solution_6": "1 is right.\r\n2; the correct answer is -0.73\r\n3; I dont understand the question as well.\r\n\r\nPlease go on explaining your answers.", "Solution_7": "1. $ P_s$ is VP of liquid mixture. X is mole fraction in liquid phase.\r\n$ P_s\\equal{}P_A^oX_A\\plus{}P_B^oX_B$\r\nPartial pressure of A =$ P_A^oX_A\\equal{}0.4P_s$. Also $ X_A\\plus{}X_B\\equal{}1$.\r\nThis gives the values of $ X_A$ and $ X_B$. So, we know $ P_s$\r\n\r\n2. Solution can be found in the link I gave." } { "Tag": [ "inequalities", "search", "inequalities unsolved" ], "Problem": "while solving an inequality we came to have something like this\r\n\r\nx/ :sqrt: (x 2 + :lambda yz) +y/ :sqrt: (y 2 + :lambda xz) + z/ :sqrt: (z 2 + :lambda xy) >= 3/ :sqrt: (1+ :lambda ).\r\n\r\nIs this true for :lambda >= 8", "Solution_1": "You will definitely find many discussions on the forum about this inequality (from IMO 2001, I think). So, search the ssolved problems area.", "Solution_2": "Oh, my God!\r\nYet another one! A general method for such inequalities was also discussed." } { "Tag": [ "limit", "logarithms", "integration", "calculus", "calculus computations" ], "Problem": "find this limit :D\r\n\r\n$\\lim_{n\\to\\infty}\\;\\left[ \\;\\left( 1+\\frac{1}{n^{3}}\\right) \\left( 1+\\frac{8}{n^{3}}\\right) \\left( 1+\\frac{27}{n^{3}}\\right) \\cdots\\;\\;\\;\\cdots \\left( 1+\\frac{n^{3}}{n^{3}}\\right) \\;\\right]^{\\frac{1}{n}}$", "Solution_1": "${\\lim_{n\\to \\infty }\\, \\log \\left(\\left(\\prod_{i=1}^{n}\\left(\\frac{i^{3}}{n^{3}}+1\\right)\\right)^{1/n}\\right)=\\lim_{n\\to \\infty }\\, \\left(\\sum_{i=1}^{n}\\frac{\\log \\left(\\frac{i^{3}}{n^{3}}+1\\right)}{n}\\right)=\\lim_{n\\to \\infty }\\, \\int_{1}^{n}\\frac{\\log \\left(\\frac{i^{3}}{n^{3}}+1\\right)}{n}\\, di=\\log (4)+\\frac{\\pi }{\\sqrt{3}}-3}$\r\n\r\nis this allowed?\r\n\r\ni dont know, so i dont evaluate the integral, if someone say it is allowed i will evaluate", "Solution_2": "[quote=\"yagaron\"]${\\lim_{n\\to \\infty }\\, \\log \\left(\\left(\\prod_{i=1}^{n}\\left(\\frac{i^{3}}{n^{3}}+1\\right)\\right)^{1/n}\\right)=\\lim_{n\\to \\infty }\\, \\left(\\sum_{i=1}^{n}\\frac{\\log \\left(\\frac{i^{3}}{n^{3}}+1\\right)}{n}\\right)=\\lim_{n\\to \\infty }\\, \\int_{1}^{n}\\frac{\\log \\left(\\frac{i^{3}}{n^{3}}+1\\right)}{n}\\, di=\\log (4)+\\frac{\\pi }{\\sqrt{3}}-3}$\n\nis this allowed?\n\ni dont know, so i dont evaluate the integral, if someone say it is allowed i will evaluate[/quote]\r\n :lol: \r\nor you have find this sum\r\n${\\sum_{k=1}^{+oo}\\frac{(-1)^{k+1}}{k(3k+1)}=\\log (4)+\\frac{\\pi }{\\sqrt{3}}-3}$", "Solution_3": "the way you people write $\\inf = oo$ is so funny. :rotfl:" } { "Tag": [ "\\/closed" ], "Problem": "sorry but I can't post a new topic .it says; spam_title_all_caps. what does it mean?", "Solution_1": "It measn that all the letters in your title-text are uppercase. Make them lowercase and you will be fine.", "Solution_2": "Wait. I thought that (when you are not a moderator) when you post a topic, no matter what, it all gets changed to lowercase except the first letter? Has that been changed?", "Solution_3": "Yes." } { "Tag": [ "induction", "number theory proposed", "number theory" ], "Problem": "If $a=4k-1$ For $\\forall{n}\\in{N}$ Prove that $2^{n-1}\\mid{S}_{n}={C}^{0}_{n}-{C}^{2}_{n}a+{C}^{4}_{n}{a}^{2}-{C}^{6}_{n}{a}^{3}+...$", "Solution_1": "Put $u_{n}=(1+i\\sqrt{a})^{n}+(1-i\\sqrt{a})^{n}(n=1,2,...)$ , then $u_{n}=2S_{n}$. We have $u_{1}=2,u_{2}=2-2a,u_{n+2}-2u_{n+1}+(1+a)u_{n}=0(n=1,2,...)$. Now by induction on $n$ : $2^{n}|u_{n}$." } { "Tag": [ "algebra", "binomial theorem" ], "Problem": "The binomial theorem\r\n\r\n[img]http://img352.imageshack.us/img352/7360/mathwd7.png[/img]\r\n\r\nIf anyone can show the full step by step process to solving this it would be a great help!!! Please start from the beginning of the problem and work through it so I understand clearly.\r\n\r\nTo all those who do post a solutions...you guys are legends!![/img]", "Solution_1": "You're trying to find the wrong coefficient- specifically, that of $ c^{400}p^{\\minus{}200}$. Beware of the pitfalls of exponent arithmetic.", "Solution_2": "That's what they told OP the first time this was posted: \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=193318\r\nThe second time this was posted, there was a different hint:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=193470\r\nThe third time, no one has responded yet:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=193472\r\n\r\nSummary: DON'T DOUBLE POST. It's exceedingly rude." } { "Tag": [ "AMC" ], "Problem": "A circle with center $ C$ is tangent to the positive $ x$ and $ y$-axes and externally tangent to the circle centered at $ (3,0)$ with radius $ 1$. What is the sum of all possible radii of the circle with center $ C$?\r\n\r\n$ \\textbf{(A)}\\ 3 \\qquad\r\n\\textbf{(B)}\\ 4 \\qquad\r\n\\textbf{(C)}\\ 6 \\qquad\r\n\\textbf{(D)}\\ 8 \\qquad\r\n\\textbf{(E)}\\ 9$", "Solution_1": "[hide]If we draw the circle tangent to the positive axes with radius $ r$, then its center will be $ (r,r)$.\n\nIf this circle is tangent to the given circle, then the distance between its center and the center of the other circle will be $ r \\plus{} 1$.\n\nThe distance between them, in terms of $ r$, is:\n\n$ \\sqrt {(r \\minus{} 3)^2 \\plus{} r^2} \\equal{} r \\plus{} 1$\n\n$ (r \\minus{} 3)^2 \\plus{} r^2 \\equal{} r^2 \\plus{} 2r \\plus{} 1$\n\nThis simplifies to $ r^2 \\minus{} 8r \\plus{} 8 \\equal{} 0$. The sum of the roots of this will be $ 8 \\rightarrow \\boxed{\\textbf{(D)}}$\n[/hide]" } { "Tag": [ "LaTeX", "logarithms", "function", "integration" ], "Problem": "checking again the new version\r\n\r\n\\[ f(x) = \\frac 34 { n\\choose 2 } \\cdot \\sum^n_{k=1} x_k \\] \r\n\r\nor better yet \r\n\r\n\\[ f(x) - \\frac {n(n+1)}2 = g(x) \\] \r\n\r\nor just plain old simple $ f(x) = \\frac 34 $ ? :)", "Solution_1": "[quote=\"Valentin Vornicu\"]checking again the new version\n\n\\[ f(x) = \\frac 34 { n\\choose 2 } \\cdot \\sum^n_{k=1} x_k \\] \n\nor better yet \n\n\\[ f(x) - \\frac {n(n+1)}2 = g(x) \\] \n\nor just plain old simple $ f(x) = \\frac 34 $ ? :)[/quote]\r\nand quoting works now as does use of double dollar sign\r\n\\[\\frac 12\\]", "Solution_2": "[tex]f(x)+f(y)=f(x+y)[/tex]", "Solution_3": "let's try now $ f(x) = g(x) $", "Solution_4": "trying harder we obtain $ f(x) = g(x) $ ?", "Solution_5": "This is a very obscure bug - just fixing it now!", "Solution_6": "hmmm, let's test again with the fix up, $f(x)=g(x)$ ? let's also try the $ \\$ ... $ \\$ tag :)\r\n\r\nit works, so now we can use the normal latex tags to do the work. \r\n\r\nthe dolar sign is to be made like this: [code] $\\$ \\ $\\$ [/code]. \r\n\r\nhopefully you won't need it that much :)", "Solution_7": "$\\frac 12$ and \\[\\frac 12\\] but\r\n[code]$\\frac 12$[/code] and [code]$$\\frac 12$$[/code]\r\nYippee!", "Solution_8": "[quote=\"Anonymous\"]$\\frac 12$ and \\[\\frac 12\\] but\n[code]$\\frac 12$[/code] and [code]$$\\frac 12$$[/code]\nYippee![/quote]\r\nForgot to log in so testing quote\r\nEdited and still OK :)", "Solution_9": "why some numbers vanished sometimes ?\r\nl wrote 28^x=19^y+97^z and 8^x=9^y+7^z appeared\r\nl try again :\r\n$\\28^x=\\19^y+\\87^z$\r\n\r\n(you can solve it if you want)", "Solution_10": "l maybe forgot some spaces or \\... l dunno\r\nwhat can you tell me about that ?\r\n\r\n\r\n$\\\\28^x=\\\\19^y+\\\\87^z$", "Solution_11": "it's because you wrote \\28^n = \\19^n etc. \r\nyou must use the backslash \\ only for commands. \\28 is not a command :D", "Solution_12": "[quote=\"arthur\"]why some numbers vanished sometimes ?\nl wrote 28^x=19^y+97^z and 8^x=9^y+7^z appeared\nl try again :\n$\\28^x=\\19^y+\\87^z$\n\n(you can solve it if you want)[/quote]\r\nCommands in LaTeX always start with \\ It tells the program that the next character is the start of a command and is to be treated in a special way. So when LaTeX sees \\28 it thinks ... yes, 2 is the start of a command, but no command starts with that so it can be ignored. So never put in \\ unless you want LaTeX to do something.\r\nFor example, \\log 2 means that you want the word log to be the function , not the 3 variables l o and g.\r\nCompare [code]$\\log 2$[/code] which gives $\\log 2$ with [code]$log 2$[/code] which gives $log 2$ and which means $l \\times o \\times g \\times 2$\r\n\r\nDoes this makes sense?", "Solution_13": "[quote=\"arthur\"]$\\\\28^x=\\\\19^y+\\\\87^z$[/quote]\r\nValentin beat me to it! In this case you had\r\n[code]$\\\\28^x=\\\\19^y+\\\\87^z$[/code] and \\\\ means a new line so LaTeX puts a new line and is ready to deal with 28 as a number", "Solution_14": "[quote=\"stevem\"][quote=\"arthur\"]$\\\\28^x=\\\\19^y+\\\\87^z$[/quote]\nValentin beat me to it! In this case you had\n[code]$\\\\28^x=\\\\19^y+\\\\87^z$[/code] and \\\\ means a new line so LaTeX puts a new line and is ready to deal with 28 as a number[/quote]\r\n\r\njust a lucky shot :P", "Solution_15": "thank you, it's clear now.", "Solution_16": "test :\r\n$\\sum^n_{p=1}a_p=2$ and $x_1=\\frac{a}{\\sqrt{px}}$", "Solution_17": "Test\r\n1) $1+1=2$\r\n2) $f(x) = f(x-1) \\times x$\r\n3) $1/3x$\r\n4) $ \\zeta(z) = \\sum n{}^-{}^z = \\prod (1-p{}^-{}^z){}^-{}^1$\r\n5) $ \\sqrt\\pi = \\int e{}^-{}^x{}^2 dx $ \r\n6) $ n \\choose k $ \r\n7) $ (1+x)^\\mu = \\sum \\mu \\choose k x^k $" } { "Tag": [ "geometry", "angle bisector", "geometry proposed" ], "Problem": "A,B,C,D are 4 points on the line d. let a be a certain line on the plane. Show an approaching to construct the points M on that line, satisfies the following condition: angleAMB =angleCMD", "Solution_1": "We'll find all the points $M$ on the plane such that $\\angle AMB=\\angle CMD$\r\n\r\n\r\nSuppose that the points $A,B,C,D$ lie on this order on the line $d$\r\nIf their order is like $A,C,B,D$, then just rename the points $B,C$ - it is equivalent\r\nIf their order is like $A,C,D,B$, then $\\widehat{ AMB}> \\widehat{CMD}$\r\n\r\n\r\nLet $MX$ be the angle bisector of $\\angle BMC \\ (X \\in d)$. Then of course $MX$ is also the angle bisector of $\\angle AMD$\r\n\r\nLet $Y\\in d$ such that $MY \\perp MX$ (the line $AY$ is external angle bisector of both $\\angle BMC$ and $\\angle AMD$)\r\n\r\nThe points $X,Y$ are harmonic conjugates of both $A,D$ and $B,C$. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16821]Here[/url] there is a way to construct the points $X,Y$\r\n\r\nThe locus of the point $M$ is the circle with diameter $XY$" } { "Tag": [ "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "In preparation for my multi-var. calc exam next week i came across this question:\r\n\r\nLet $f(x,y) = \\frac{|x|^{a}|y|^{b}}{|x|^{c}+|y|^{d}}$ where a, b, c, d are positive.\r\ni) Prove that if $\\frac{a}{c}+\\frac{b}{d}> 1$ then $\\lim_{(x,y)\\to(0,0)}f(x,y)$ exists and equals zero\r\n\r\n..im pretty much stuck...any help would b welcome", "Solution_1": "I think you mean:$\\frac{a}{c}+\\frac{b}{d}>1$\r\nhere is the proof(Using Inequality estimating):use the Young's Inequality:\r\n(for $a+b=1,a>0,b>0$,then $x^{a}y^{b}\\le ax+by$)\r\nwe have:\r\n$\\frac{ad}{bc}|x|^{c}+|y|^{d}\\ge \\left(\\frac{ad}{bc}+1\\right) |x|^\\frac{acd}{ad+bc}|y|^\\frac{bcd}{ad+bc}$\r\nthe rest will be easy. :)", "Solution_2": "thanks!\r\n..any other solutions?", "Solution_3": "Consider $|x|^{c}$ and $|y|^{d}$ as new variables.", "Solution_4": "[quote=\"mlok\"]Consider $|x|^{c}$ and $|y|^{d}$ as new variables.[/quote]\r\n?...sorry. i dont follow", "Solution_5": "Let $u=|x|^{c}$ and $v=|y|^{d}$. Note that $(u,v)\\to (0,0)$ as $(x,y)\\to (0,0)$. It remains to prove that \\[\\lim_{(u,v)\\to (0,0)}\\frac{u^{a/c}v^{b/d}}{u+v}=0\\] where the limit is restricted to the first quadrant $u,v\\ge 0$. \r\n\r\nThe degree of homogeneity of this fraction is $\\frac{a}{c}+\\frac{b}{d}-1>0$.", "Solution_6": "thats a nice logical way to look at it. Thank you!" } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Let $x_0,x_1,x_2\\ldots$ an infinite sequence of integer numbers.\r\n$x_{n+2}=2x_n+x_{n-1}$\r\nAnd\r\n$x_0=1,x_1=0,x_2=2$\r\nProve that for all positive integer $m$ exists $m|x_k$ $m|x_{k+1}$ for some positive integer $k$", "Solution_1": "one easy induction prove that $x_n= F_{n-1} + (-1)^{n+1}$, for all $n \\ge1$ ($F$ is the fibonacci sequence, whit $F_0 = 0$and $F_1 = 1$).\r\n\r\nNow, we have to prove that for all positive integers $m$, exists $k$ such that $m| F_{2k} + 1$ \r\nand $m| F_{2k+1} - 1$.\r\n\r\n(the same that prove that exists $k$ such that $m| F_{2k}$ and $F_{2k-1} = 1 (mod. m)$).\r\n\r\nwe know that for all positive integers $m$, exists $k$ such that $m| F_{2k}$, and the rest is easy. :)" } { "Tag": [ "search", "inequalities unsolved", "inequalities" ], "Problem": "[color=blue][b]Let$ 4$ real numbers $ x_1 ,x_2 ,x_3 ,x_4$ such that:\n\n$ \\frac {1}{2} \\le (x_1 )^2 \\plus{} (x_2 )^2 \\plus{} (x_3 )^2 \\plus{} (x_4 )^2 \\le 1$\n\nFind $ Min \\minus{} Max$ of M which:\n\n$ M \\equal{} (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$ \n [/b][/color]", "Solution_1": "Please search here http://www.mathlinks.ro/resources.php?c=186&cid=41 .", "Solution_2": "[color=blue][b]Find the minimum value of $ f(x)\\equal{}x^x$ for $ x>0$[/b][/color]", "Solution_3": "[color=blue][b]Let $ {a^2} \\plus{} {b^2} \\plus{} {c^2} \\equal{} 3.$, Find the minimum value of the expression:\n$ M \\equal{} \\frac {{{a^5}}}{{{b^3} \\plus{} {c^2}}} \\plus{} \\frac {{{b^5}}}{{{c^3} \\plus{} {a^2}}} \\plus{} \\frac {{{c^5}}}{{{a^3} \\plus{} {b^2}}} \\plus{} {a^4} \\plus{} {b^4} \\plus{} {c^4}$ [/b][/color]", "Solution_4": "[quote=\"quocbao153\"][color=blue][b]Let $ {a^2} \\plus{} {b^2} \\plus{} {c^2} \\equal{} 3.$, Find the minimum value of the expression:\n$ M \\equal{} \\frac {{{a^5}}}{{{b^3} \\plus{} {c^2}}} \\plus{} \\frac {{{b^5}}}{{{c^3} \\plus{} {a^2}}} \\plus{} \\frac {{{c^5}}}{{{a^3} \\plus{} {b^2}}} \\plus{} {a^4} \\plus{} {b^4} \\plus{} {c^4}$ [/b][/color][/quote]\r\n\r\nIt's a very weak ineq. It's obviously trues with the stronger condition: $ a^4 \\plus{} b^4 \\plus{} c^4 \\equal{} 3$\r\n :wink:" } { "Tag": [ "arithmetic sequence" ], "Problem": "I am [i]really[/i] bad at problem solving and have absolutley no idea how to answer this question:\r\n\r\nHow many natural numbers less than 200 leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5?", "Solution_1": "[hide]Let the number you are looking for be $m$.\n\n$m = 7x + 3$\n\n$m = 5y + 4$\n\nbecause it's four more than a multiple of 5 and three more than a multiple of 7.\n\n$7x + 3 = 5y + 4$ so $y = \\frac {7x - 1}5$\n\nWe must find all natural number values of $x$ and $y$.\n\nQuickly we find that the values of $y$ are an arithmetic sequence in the form $7p - 3$. The maximum value of $p$ such that $7p - 3 < 200$ is $28$ so there must be $28$ values.[/hide]", "Solution_2": "My teacher skipped the sequences unit for some reason so I have no idea what they are or how to do them. Also what is p?", "Solution_3": "P is an arbitrary natural number in his case I think.", "Solution_4": "[quote=\"Philmac\"]How many natural numbers less than 200 leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5?[/quote]\r\n\r\nOk...help me out please...what did I do wrong? :? \r\n\r\n[hide]\nIf a number has a remainder of 4 when divided by 5, then it must have either a 4 or a 9 in it's ones' place.\n\nOf the first ten multiples of 7 (7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...) only 21 and 56 will have a unit's digit of 4 or 9 when 3 is added to it. So, for every 70 integers, 2 will leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5.\n\nThe smallest integer of $\\frac{200/70}$ is 2, leaving a remainder of 60. Since 56 is less than 60, there must be a total of 6 numbers that will leave a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5. :? \n[/hide]", "Solution_5": "[quote=\"tarquin\"][hide]Let the number you are looking for be $m$.\n\n$m = 7x + 3$\n\n$m = 5y + 4$\n\nbecause it's four more than a multiple of 5 and three more than a multiple of 7.\n\n$7x + 3 = 5y + 4$ so $y = \\frac {7x - 1}5$\n\nWe must find all natural number values of $x$ and $y$.\n\nQuickly we find that the values of $y$ are an arithmetic sequence in the form $7p - 3$. The maximum value of $p$ such that $7p - 3 < 200$ is $28$ so there must be $28$ values.[/hide][/quote]\r\n\r\nWait, what values are you looking for? x and y or m?\r\n\r\nBy the Chinese Remainder THeorem, there is at least one solution mod 35. If we examine all the integers less than 35, we see that 24 is the only such number. Therefore we can say that the only such m are: 24 59 94 129 164 199", "Solution_6": "[quote=\"mcalderbank\"]Therefore we can say that the only such m are: 24 59 94 129 164 199[/quote]\r\n\r\nYay! Comforting! :P", "Solution_7": "We haven't learned mod either, I only know what it is from computer programming :( Can you guys please come up with a dumber explanation that's at the eleventh grade level?", "Solution_8": "Oh I messed up.\r\n\r\nWas supposed to plug $7p-3$ for $y$ in $5y+4$\r\n\r\n$35p-11 < 200$\r\n\r\n$p<6.028$\r\n\r\nSo 6 values.", "Solution_9": "[quote=\"Philmac\"]We haven't learned mod either, I only know what it is from computer programming :( Can you guys please come up with a dumber explanation that's at the eleventh grade level?[/quote]\r\n\r\nJust scroll up and read my post. I don't think there are mods in there. (The idea behind it is still there, but it's disguised so you don't need knowledge of mods to understand my post...I think.) Let me know if you still have trouble." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Let $f: R\\to R$ an additive function: $f(x+y)=f(x)+f(y) \\forall x,y \\in R$ which is bounded on $[0,1]$ and $f(1)=1$.\r\nProve that $f(x)=x \\forall x \\in R$", "Solution_1": "you know that $f(q)=q$ far every rational number $q \\in \\mathbb Q$. Now, suppose that there exists $a \\in \\mathbb R$ such that $f(a)=a+d$ where $d > 0$ for example. Then for every rational number $f(a-q)=(a-q)+d$, which shows that there exists arbitrary small real number $0d$. You can easily find a contradiction with $f$ bounded on $[0;1]$. The case $d<0$ is similar.." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all two digit numbers $ \\overline{AB}$ such that $ \\overline{AB}$ divides $ \\overline{A0B}$.", "Solution_1": "[quote=\"Bugi\"]Find all two digit numbers $ \\overline{AB}$ such that $ \\overline{AB}$ divides $ \\overline{A0B}$.[/quote]\r\n\r\nWe have $ 100x\\plus{}y\\equal{}k(10x\\plus{}y)$ and so $ y\\equal{}10\\frac{10\\minus{}k}{k\\minus{}1}x$\r\n\r\nWe have obviously the solution $ k\\equal{}10$ and $ \\overline{A00}\\equal{}10\\cdot \\overline{A0}$\r\n\r\nIf $ y>0$, $ 100x\\plus{}y < 10(10x\\plus{}y)$ and so $ k<10$. Then we need $ 10\\minus{}k5$ So :\r\n\r\n$ k\\equal{}6$ $ \\implies$ $ y\\equal{}10\\frac{10\\minus{}6}{6\\minus{}1}x\\equal{}8x$ $ \\implies$ $ \\overline{108}\\equal{}6\\cdot \\overline{18}$\r\n\r\n$ k\\equal{}7$ $ \\implies$ $ y\\equal{}10\\frac{10\\minus{}7}{7\\minus{}1}x\\equal{}5x$ $ \\implies$ $ \\overline{105}\\equal{}7\\cdot \\overline{15}$\r\n\r\n$ k\\equal{}8$ $ \\implies$ $ y\\equal{}10\\frac{10\\minus{}8}{8\\minus{}1}x\\equal{}\\frac{20}{7}$ $ \\implies$ no solution.\r\n\r\n$ k\\equal{}9$ $ \\implies$ $ y\\equal{}10\\frac{10\\minus{}9}{9\\minus{}1}x\\equal{}\\frac 54x$ $ \\implies$ $ \\overline{405}\\equal{}9\\cdot \\overline{45}$", "Solution_2": "[quote=\"Bugi\"]Find all two digit numbers $ \\overline{AB}$ such that $ \\overline{AB}$ divides $ \\overline{A0B}$.[/quote]\n\n[u]Alternate solution:[/u]\n\nLet $\\overline{AB}=x$\n\n$10A\\equiv -B\\:mod\\:x\\Longrightarrow 100A \\equiv -10B\\:mod\\:x$ but $100A\\equiv -B\\:mod\\:x$ and so, $-B\\equiv -10B\\:mod\\:x$\n\nSo, $x=\\overline{AB}|9B$\n\n$B=0\\Longrightarrow A=1, 2, 3, \\cdots 9$\n\n$B=1$ has no solutions for $A$\n\n$B=2$ has no solutions for $A$\n\n$B=3$ has no solutions for $A$\n\n$B=4$ has no solutions for $A$\n\n$B=5\\Longrightarrow A=1, 4$\n\n$B=6$ has no solutions for $A$\n\n$B=7$ has no solutions for $A$\n\n$B=8\\Longrightarrow A=1$\n\n$B=9$ has no solutions for $A$\n\nSo, $\\overline{AB}=\\{10, 20, 30, 40, 50, 60, 70, 80, 90, 15, 45, 18\\}$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "pardon me if this problem has been discussed\r\nA and B play a game on an $ n\\times n board$. They take turn put either 0 or 1 into any of the blank squares. When all the squares have been filled, B wind if the sum of all numbers in each row is even. Otherwise A wins\r\nWhich player has a winning strategy?", "Solution_1": "hmm i'm not sure.... nevertheless \r\n\r\nIf i understand the problem correctly sum has to be even only in rows but not in colums. If that is the case then \r\n\r\nIf $ n$ is even and A starts the game B has a wining strategy. If A puts number $ i$ in some row then B only puts the same number $ i$ in the same row.\r\n\r\nIf $ n$ is odd and A starts than A has a winning strategy. Because he inserts the last number he can control the oddnes/eveness of the last row.\r\n\r\nIf $ n$ is even and B starts the game,A has a wining strategy. B begins by putting some number in some row. Now A puts a different number in the same row.Then everytime B puts a number in this row A just puts the same number in that row.\r\n\r\n\r\nIf $ n$ is odd and B starts than I don't know :) \r\n(in all cases the other player is eventualy forced to put the number in that row)" } { "Tag": [], "Problem": "Ryt u guys! Wonder if u could give me a logical explanation to this. It is really simple and some think quite stupid.....\r\n\r\n3 brothers have saved up money towards buying a cd player for their bedroom. They each save \u00a310 and go to buy this \u00a330 Cd player they found in the shops. They take the player to the cash desk and the cashier runs it through and realises that there is a special \u00a35 off bargain on the product. The sneaky cashier puts \u00a32 in his pocket and gives each of the brothers a \u00a31 back. So...........\r\n\r\nEach brother had \u00a310 but got a \u00a31 back so each of the brothers have paid \u00a39 each. \r\n\r\n\u00a39 X 3 = \u00a327\r\n\r\nand the cashier has \u00a32 in his pocket. \r\n\r\n\u00a327 + \u00a32 = \u00a329... so where has the other \u00a31 gone?????", "Solution_1": "Well that's just the bellhop problem in a different situation, and it's a very simple solution- $\\$$27 and $\\$$2 should not be added, although the problem is worded to make you think they should be. Of the 30, each brother has 1 (3), the cashier has 2, and 25 went to buying the CD player. 3+2+25=30.\r\n\r\nYou should actually subtract the 2 (that the guy took) from the 27 (that they paid) to get the price of the CD player." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "evaluate \r\n\r\n$\\int\\frac{1}{3\\cos x+4\\sin x+5}dx$\r\n\r\n\r\nAny idea how to tackle it ?", "Solution_1": "[hide=\"hint\"]$\\frac{1}{3\\cos{x}+4\\sin{x}+5}=\\frac{1}{5}(\\frac{1}{\\sin{(x+\\alpha)+1}})$[/hide]", "Solution_2": "I think it is becuase the lengths $3,4,5$ are a right triangle :wink:", "Solution_3": "Let $\\tan \\frac{x}{2}=t$", "Solution_4": "why not just $\\tan(x)=t$?", "Solution_5": "what's your answer?", "Solution_6": "ok i see. using $\\tan(x)=t$ makes the integral kind of messy.", "Solution_7": "let $\\tan{\\frac{x}{2}}=t$,\r\n\\[\\int{\\frac{1}{{3\\cos x+4\\sin x+5}}dx = \\int{\\frac{1}{{t^{2}+4t+4}}}}dt =-\\frac{1}{{t+2}}+C \\]", "Solution_8": "Your answer is incorrect.", "Solution_9": "now maybe true. :D", "Solution_10": "That's right." } { "Tag": [], "Problem": "On a drive from your ranch to Austin, you wish to average 40mph. The distance from your ranch to Austin is 70 miles. However, at 35 miles (half way), you find you have averaged only 30 mph. What average speed must you maintain in the remaining distance in order to have an overall average speed of 40mph? Answer in units of mph.", "Solution_1": "[hide]To average 40mph, you need to make the total trip in 70/40=7/4 hours. At the halfway point you have used 35/30=7/6 hours, so you must make the remaining journey in 7/4-7/6 = 7/12 hours. The speed you need is 35/(7/12)=60 mph[/hide]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "let $ \\lambda: [a,b]\\rightarrow{\\mathbb{R}^n}$ a closed path such that $ \\lambda'(t)$ is continuous, ie, is $ C^1$. Show that there is some $ t\\in(a,b)$ such that $ <\\lambda(t),\\lambda'(t)>\\equal{}0$\r\n\r\nThanks", "Solution_1": "Hint: $ t \\mapsto \\parallel{}\\lambda(t)\\parallel{}^2$ is $ C^1$ and defined on a compact interval.", "Solution_2": "Hello dihuana, thanks for your help, I do not understande wery well, can you give some details?\r\nThanks a lot\r\nHugs", "Solution_3": "as dihuana stated, look at $ f(t)\\equal{} \\parallel \\lambda(t)\\parallel ^{2}\\equal{} <\\lambda(t),\\lambda(t)>$ \r\nIt is differentiable (as a sum of products of differentiable elements), and f(a)=f(b) since its a loop. now you can use a certain theorem from basic calculus.", "Solution_4": "thanks a lot guys, i understood it very well" } { "Tag": [], "Problem": "\u0388\u03c3\u03c4\u03c9 $ p$ \u03ad\u03bd\u03b1\u03c2 \u03c0\u03ce\u03c4\u03bf\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 \u03ba\u03b1\u03b9 $ m,n$ \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf\u03b9 \u03c4\u03bf\u03c5 1 . \u03a5\u03c0\u03bf\u03b8\u03ad\u03c4\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf $ n$ \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf\u03bd \r\n\r\n$ m^{p(n \\minus{} 1)} \\minus{} 1$ . \u0394\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03bf $ n$ \u03ba\u03b1\u03b9 \u03bf $ m^{n \\minus{} 1} \\minus{} 1$ \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03bf \u03ba\u03bf\u03b9\u03bd\u03cc \u03b4\u03b9\u03b1\u03c1\u03ad\u03c4\u03b7 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03c4\u03b7\u03c2 \u03bc\u03bf\u03bd\u03ac\u03b4\u03b1\u03c2 .", "Solution_1": "Kat arxas euxaristw gia thn afierwsh!!(ektos k an uparxei allos hlias k den to 3erw! :P )\r\nProfanws ta m,n den exoun koinous diairetes megaluterous ths monadas(alliws o diareths 8a diarouse to -1)\r\nEstw q enas prwtos diaireths tou n.tote uparxei generator modq kai estw oti autos einai to g\r\nestw epishs $ m\\equiv g^x(modq)$\r\nTote exoume oti $ g^{xp(n\\minus{}1)}\\equiv 1(modq)\\equal{}>q\\minus{}1|xp(n\\minus{}1)$ logw ths epiloghs tou g\r\nalla p prwtos ara an $ q\\neq2$ exoume oti \r\n$ q\\minus{}1|x(n\\minus{}1)\\equal{}>g^{x(n\\minus{}1)}\\equiv 1(modq) \\equal{}>m^{n\\minus{}1} \\equiv 1(modq)$ \r\ndhladh $ q|(m^{n\\minus{}1} \\minus{}1)$ h alliws $ q|(n,m^{n\\minus{}1} \\minus{}1)$ dhladh $ (n,m^{n\\minus{}1} \\minus{}1)>1$\r\nan twra $ q\\equal{}2$ exoume oti $ 2|m^{p(n\\minus{}1)}\\minus{}1\\equal{}>$ m perittos $ \\equal{}>2|m^{n\\minus{}1} \\minus{}1$\r\nAn einai swsth h lush m tote thn afierwnw se esena!An oxi na 3ereis apla oti thn prospa8hsa eidika epeidh thn afierwses se emena!\r\nHlias", "Solution_2": "\u03a3\u03b5 \u03c3\u03ad\u03bd\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03c6\u03b9\u03b5\u03c1\u03c9\u03bc\u03ad\u03bd\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03be\u03b1\u03b9\u03c1\u03b5\u03c4\u03b9\u03ba\u03ae :) \u039c\u03c0\u03c1\u03ac\u03b2\u03bf .", "Solution_3": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b7 (\u03ba\u03b1\u03c4\u03ac \u03c4\u03b1 \u03ac\u03bb\u03bb\u03b1 \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 :wink: ) \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c4\u03bf\u03c5 \u0397\u03bb\u03af\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03bc\u03b9\u03ba\u03c1\u03cc \u03ba\u03b5\u03bd\u03cc. \r\n\r\n\u03a3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b1 \u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 $ q\\minus{}1|xp(n\\minus{}1)$ \u03b4\u03b5\u03bd \u03c3\u03c5\u03bd\u03b5\u03c0\u03ac\u03b3\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03b1\u03c1\u03b1\u03af\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03bd $ q\\minus{}1|x(n\\minus{}1)$. \u0393\u03b9\u03b1 \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03cc\u03c4\u03b9 (\u03bc\u03b5 $ exp(t)$ \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03c9 \u03c4\u03b7\u03bd \u03bc\u03ad\u03b3\u03b9\u03c3\u03c4\u03b7 \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c4\u03bf\u03c5 $ p$ \u03c0\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf\u03bd $ t$) $ exp(q\\minus{}1) 2 and x3 \u2265 1?", "Solution_1": "You know, I've seen a lot of things like this.\r\n\r\nTry [url]http://www.openmathtext.org/downloads.html[/url].\r\n\r\nI believe the one entitled \"difference calculus generating functions\" will be of use. It explains how these things relate to generating functions and is in fact quite useful.", "Solution_2": "[quote=\"hyderman\"]plerase explain this in details....\n\nHow many solutions are there to the equation\nx1 + x2 + x3 = 12,\nwhere x1 , x2 , x3 and x4 are integers satisfying\nx1 \u2265 0,\nx2 > 2 and x3 \u2265 1?[/quote]\r\n\r\n$x_{1}\\ge 0$\r\n$x_{2}\\ge 3$ (unless you meant \"x2 \u2265 2\" which would match the signs of the others.\r\n$x_{3}\\ge 1$\r\n\r\nIf we just let $x_{1}=0, x_{2}=3,$ and $x_{3}= 1$, then the sum would be $4$. So we need to add a total of $8$ units to the three variables.\r\n\r\nconsider each of these circles as a unit\r\n00000000\r\n\r\nWe need to figure how to put these units into the variables. So we put in some seperators. For example,\r\n000|00|00\r\nwould mean put 3 into $x_{1}$ ($x_{1}=0+3=4$), put 2 into $x_{2}$, ($x_{2}=3+2=5) and put 2 into$x_3$($x_3=1+2=3).\r\n\r\nSo there are 9 slots for the seperators, and two seperators, making $\\binom{9}{2}+9=36+9=\\boxed{45}$ (I added the 9 because the $\\binom{9}{2}$ does not include the possibilities in which the seperators are in the same spot.", "Solution_3": "Using my own advice:\r\n\r\nWe are looking for the coefficient of $x^{12}$ in the expansion of $(1+x+x^{2}+\\ldots)(x^{3}+x^{4}+x^{5}+\\ldots)(x+x^{2}+x^{3}+\\ldots)$.\r\n\r\nNow notice that this can be rewritten as:\r\n\r\n$\\left(\\frac{1}{1-x}\\right)\\left(\\frac{x^{3}}{1-x}\\right)\\left(\\frac{x}{1-x}\\right)=\\frac{x^{4}}{(1-x)^{3}}$\r\n\r\n(Recall that $\\frac{1}{1-x}= 1+x+x^{2}+x^{3}+\\ldots$. Multiply both sides by $(1-x)$ if this is not apparent.)\r\n\r\nNow, since we are looking for the coefficient of $x^{12}$, it means that we need to find the coefficient of $x^{8}$ in the denominator (when written in power series form) so that the $x^{4}$ on top will multiply to an exponent of 12. Now, we use the binomial formula:\r\n\r\n$\\binom{-3}{8}= \\binom{8+3-1}{8}= \\binom{10}{8}= 45$\r\n\r\nThis is handy for many problems like this, and the interesting thing is that you do not need to come up with a unique method for each type of problem, as this generalizes.", "Solution_4": "$y_{1}-1+y_{2}+2+y_{3}=12\\implies y_{1}+y_{2}+y_{3}=11$, where $y_{i}\\in\\mathbb{N}$. We have to choose 2 plus signs from the 10 spaces in 1_1_..._1, which can be done in $\\binom{10}{2}=45$.", "Solution_5": "BALLS AND URNS\r\n\r\nlearn this...indistinguishable balls, and distinguishable urns for this problem", "Solution_6": "[quote=\"Altheman\"]BALLS AND URNS\n\nlearn this...indistinguishable balls, and distinguishable urns for this problem[/quote]\r\n\r\nHow do you change formula to account for distinguishable urns (or distinguishable balls, for that matter)?", "Solution_7": "balls/urns \r\n\r\ni/i no closed form\r\n\r\ni/d bijection with walls and balls thing\r\n\r\nd/i see stirling number of the second kind\r\n\r\nd/d balls^urns", "Solution_8": "what are balls and urns?", "Solution_9": "[quote=\"tjhance\"]what are balls and urns?[/quote]balls are just balls, urns are just urns. Dictionary can help you if you need to look it up. If you want to know more about it mathematically, it refers the number of ways of placing $n$ ([size=150][b][color=red]d[/color][/b][/size]istinguishable/[size=150][b][color=red]i[/color][/b][/size]ndistinguishable) balls into $m$ ([size=150][b][color=red]d[/color][/b][/size]istinguishable/[size=150][b][color=red]i[/color][/b][/size]ndistinguishable) urns. That's what Altheman is referring to. For more info, look up a good book with Combinatorics in it like Fred Roberts' Applied Combinatorics." } { "Tag": [ "probability", "complementary counting" ], "Problem": "A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears. Show that the probability of getting the first ace on or before the ninth card is greater than 50%. \r\n\r\nHow would you solve this problem without having to write out all the probablities and then adding them all up?\r\n\r\nThanks!", "Solution_1": "[hide]The probability he doesn't get it in the first 9 turns is:\n\n48/52*47/51*46/50...*41/45*40/44. We don't care what happens next. If you combine this, you notice that a lot of terms are repeated in the numerator and denominator, leaving 43*42*41*40/(52*51*50*49), which happens to be less than 50%. Therefore, the probability that you DO get an ace in the first 9 turns is greater than 50% (they have to add up to 100%)[/hide]", "Solution_2": "Would this be correct? \r\n\r\n[hide]The deck is $ \\frac{1}{13}$ aces so the probabilty that an ace would pop up in the first 9 cards would be $ \\frac{9}{13}$ which is greater than $ \\frac{1}{2}$[/hide]", "Solution_3": "SnipedYou, I don't think the probability an ace turns up is $ \\frac {9}{13}$. Rather, it's easier to use complementary counting and find the probability that it [i]doesn't[/i] turn up:\r\n\r\n$ \\frac {48}{52} \\times \\frac {47}{51} \\times \\frac {46}{50} ... ... \\times \\frac{40}{44}$\r\n$ \\equal{} \\frac {3526}{7735}$\r\n\r\nSo the probability it [i]does[/i] turn up is $ \\frac {4209}{7735}$ (the complement) which is more than 50%." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let a and b denote positive integers, such that $ a^n \\plus{} n$ divides $ b^n \\plus{} n$ for all positive integers n. Prove that a=b.", "Solution_1": "Let $ n(p)\\equal{}1\\plus{}(a\\plus{}1)(p\\minus{}1)$, then $ p|a^n\\plus{}n\\equal{}1\\plus{}(a\\plus{}1)(p\\minus{}1)\\plus{}a*a^{(a\\plus{}1)(p\\minus{}1)}$, therefore must be $ p|b^n\\minus{}a^n\\equal{}b\\minus{}a\\mod p$.\r\nIt mean, for any prime p $ p|b\\minus{}a$ or $ b\\equal{}a$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove for $ a,b,c \\in R_\\plus{}$ that the if\r\n\r\n$ ab\\plus{}bc\\plus{}ac \\leq 3abc$ then\r\n\r\n$ a^3\\plus{}b^3\\plus{}c^3 \\geq a\\plus{}b\\plus{}c$", "Solution_1": "We have $ \\sum a \\le \\frac {9a^2b^2c^2\\left(\\sum a \\right)}{\\left(\\sum ab \\right )^2} \\le \\sum a^3$, which is necessarily true by Muirhead.", "Solution_2": "[quote=\"NikolayKaz\"]Prove for $ a,b,c \\in R_ \\plus{}$ that the if\n\n$ ab \\plus{} bc \\plus{} ac \\leq 3abc$ then\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\geq a \\plus{} b \\plus{} c$[/quote]\r\n\r\nWe have:\r\n$ ab\\plus{}bc\\plus{}ca\\le 3abc\\le \\frac{1}{3}(ab\\plus{}bc\\plus{}ca)(a\\plus{}b\\plus{}c)$\r\nso $ a\\plus{}b\\plus{}c\\ge 3$\r\nso that\r\n$ a^3\\plus{}b^3\\plus{}c^3\\ge \\frac{(a\\plus{}b\\plus{}c)^3}{9}\\ge \\ a\\plus{}b\\plus{}c$ (by Holder) :lol:" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "For which real numbers b does the function f (x), defined by the conditions\r\nf (0) = b and f' = 2f-x, satisfy f (x) > 0 for all x \u2265 0?", "Solution_1": "[quote=\"mdk\"]For which real numbers b does the function f (x), defined by the conditions\nf (0) = b and f' = 2f-x, satisfy f (x) > 0 for all x \u2265 0?[/quote]\r\n\r\nFirst solve the differential equation,\r\n$ f'-2f = x \\implies f= Ce^{2x}-\\frac{1}{2}x-\\frac{1}{4}$\r\n\r\nNow just find the conditions on the initial value problem." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "can somone help?? I don't really know how yo count this\r\n\r\nQuestion: \r\nI need to get the number of sequences of length 20 consisting of 8 X chars, 7 Y chars and 5 Z chars in which no two Z chars appear adjacently, and X,Y do not appear adjacently.\r\n\r\n\r\nPlease Please Be specific\r\nthanks\r\nAlisha", "Solution_1": "Do not post the same problem more than once. Continue your discussion here:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193734[/url]" } { "Tag": [], "Problem": "How do I prepare so I get at least a 50 on the local exam? :(", "Solution_1": "simple :) \r\ndo the past exams.\r\nread books.", "Solution_2": "but where does one find these past tests? and which books are highly recommended? \r\n\r\nSorry, cuz I'm sure these topics have been discussed before, but I went through a number of old topics and couldn't find much.", "Solution_3": "Try here:\r\n\r\nhttp://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education\\student\\co02.html", "Solution_4": "and for books?\r\n\r\n(or more sites, but I'm lookin to get me a book soon)", "Solution_5": "The link above dosn't seem to work! Can you plz fix it!", "Solution_6": "Just include everything up to \".html\".", "Solution_7": "Oops, sorry." } { "Tag": [ "algebra", "partial fractions" ], "Problem": "Find the values of $ A,B$ such that:\r\n\r\n$ \\frac{A}{z\\plus{}2}\\plus{}\\frac{B}{2z\\minus{}3}\\equal{}\\frac{5z\\minus{}11}{2z^2\\plus{}z\\minus{}6}$", "Solution_1": "Partial Fractions??\r\nMultiply both sides by $ 2z^2 \\plus{}z \\minus{}6$ to get\r\n$ A(2z\\minus{}3) \\plus{} B(z\\plus{}2) \\equal{} 5z \\minus{} 11$\r\nLet $ z \\equal{} \\minus{}2$ to get\r\n$ \\minus{}7A \\equal{} \\minus{}21$\r\nso \r\n$ A \\equal{} 3$\r\nLet $ z\\equal{}\\frac{3}{2}$ to get\r\n$ \\frac{7}{2}B \\equal{} \\minus{}\\frac{7}{2}$\r\nso\r\n$ B\\equal{} \\minus{}1$", "Solution_2": "Nice job, thats the solution.", "Solution_3": "! :P :lol:" } { "Tag": [ "circumcircle" ], "Problem": "Given segments labeled 1, a, and b, construct a segment of length ab. Also, explain why a segment of length 1 is needed.", "Solution_1": "Given three segments with lengths $r,a,b$ their [i]fourth proportional[/i] is a segment $x$ such that $\\frac{r}{a}= \\frac{b}{x}\\ \\ (1)$\r\n\r\n[hide=\"Construction 1\"]Construct a segment $OA=a$.\nTake a point $R\\not\\in OA$ such that $OR=r$\nOn the ray $(OR$ take a point $B$ such that $OB=b$\nThrough $B$ bring a parallel line to $AR$. This will intersect the ray $(OA$ at a point $X$. Let $x=OX$\nThen from the similarity of the triangles $OAR,OBX$ we get $\\frac{r}{a}= \\frac{b}{x}$[/hide]\n[hide=\"Construction 2\"]I usually use a circle and power of point to construct the fourth proportional\nTake $OA=a$ and on the ray $(OA$ choose a point $B$ such that $OB=b$\nTake a point $R\\not\\in OA$ such that $OR=r$.\nConstruct the circumscribed circle of $\\triangle ABR$.\nThen the ray $OR$ intersects again the circle at a point $X$, where $OX=x$ is the fourth proportional of $r,a,b$, i.e. $ab=xr$[/hide]\r\n\r\nFrom $(1)$ we get $x=\\frac{ab}{r}$\r\n\r\nBut we want to construct the segment with length $x=ab$ . So it's necessary to choose $r=1$.\r\n\r\n[color=brown]P.S.\nIn fact, independently of the length of [b]r[/b], the multiplication works if we set as unit of measure that length of [b]r[/b].\n\nFor example, suppose that [b]a=20cm[/b] and [b]b=30cm[/b]\n\nIf [b]r=1cm[/b] then we get [b]x=600cm[/b], and $20 \\cdot 30 = 600$, ok!\n\nIf [b]r=10cm[/b], then [b]x=60cm[/b], and $20 \\cdot 30 \\not = 60$\n\nBut if we measure the lengths using as unit the length of [b]r=1dm[/b] then we can transform the lengths:\n[b]a=20cm = 2 dm [/b]\n[b]b=30cm = 3 dm[/b]\n[b]x=60cm = 6 dm[/b]\n\nNotice that $2\\cdot 3 = 6$[/color]", "Solution_2": "we can easily employ power of a point to solve this problem" } { "Tag": [ "search" ], "Problem": "Here is a a problem:\r\n\r\nIf we have an n amount of points in a 2 dimensional plane, where n is bigger than 3, and these n points do not lie in a line, there exists a line which contains exactly two points.\r\n\r\nIs theorm is Called Slyvester's Theorm.\r\n\r\nSlyvesters Theorm: http://mathworld.wolfram.com/SylvestersLineProblem.html\r\n\r\nAnybody know how to go abouts this?", "Solution_1": "[hide=\"A quick and probably incomplete proof\"]\nAssume for the sake of contradiction that the statement is false. Consider the shortest possible distance between a point and a line. Say it is between point P and line m. \n\nLine m must have at least 3 points on it, two of which must be on the same side of the perpendicular from P to m. Call them Q and R, with R farther from the foot of the perpendicular than Q. But clearly the distance from Q to line PR is shorter than the distance from P to line m, a contradiction.\n[/hide]", "Solution_2": "[quote=\"E^(pi*i)=-1\"][hide=\"A quick and probably incomplete proof\"]\nAssume for the sake of contradiction that the statement is false. Consider the shortest possible distance between a point and a line. Say it is between point P and line m. \n\nLine m must have at least 3 points on it, two of which must be on the same side of the perpendicular from P to m. Call them Q and R, with R farther from the foot of the perpendicular than Q. But clearly the distance from Q to line PR is shorter than the distance from P to line m, a contradiction.\n[/hide][/quote]\r\n\r\nI think it can be modified to work, but there's some ambiguity. Maybe you mean the shortest possible distance from any line containing at least 2 of the points and any point not on that line (the point exists because not all the points are on one line).\r\n\r\nThis is basically the same idea. Let $P$ and $m$ the be the point and line, respectively, with the shortest distance as defined above. Let $A$ be the foot of the perpendicular from $P$ to $m$. Then there are at least $3$ points on $m$, two on the same side of$A$, say $Q$ and $R$ with $QA < RA$. Let $B$ be the foot of the perpendicular from $Q$ to $PR$. We have $\\triangle RBQ \\sim \\triangle RAP$ and $RQ < RA \\Rightarrow \\triangle RBQ$ is smaller, so $BQ < AP$, a contradiction.\r\n\r\n[img]http://i71.photobucket.com/albums/i132/paladin8/1-27-07.jpg[/img]", "Solution_3": "for the sake of contradiction, suppose that for every pair for 3 [or more] points, they lie on a line, anyway, take two points, A, B, then an unspecified point, X, then ABX is a line for all X, contradiction, since that makes all the points collinear", "Solution_4": "paladin8, that is indeed what I intended - I didn't word it very well.", "Solution_5": "[quote=\"player\"]Is theorm is Called Slyvester's Theorm.[/quote]\r\nNo, it is called Sylvesters theorem. You even gave a link with the correct spelling!\r\n*posts this for the simple reason that a search for \"sylvester\" does not give this thread*", "Solution_6": "[quote=\"Altheman\"]for the sake of contradiction, suppose that for every pair for 3 [or more] points, they lie on a line, anyway, take two points, A, B, then an unspecified point, X, then ABX is a line for all X, contradiction, since that makes all the points collinear[/quote]\r\n\r\nSay what? If we take the easiest infinite counterexample, the lattice points, you can see that if we take two points, there are plenty of Xs that aren't on the line connecting those two points. I don't think your logic is complete.\r\n\r\nHopefully I didn't confuse anyone, but the infinite counterexample doesn't disprove the theorem because it wants a finite set of points.", "Solution_7": "The infinite counterexample is irrelevant, but Altheman's logic is not sound. The assumption by contradiction is $\\exists X$, not $\\forall X$.", "Solution_8": "[quote=\"t0rajir0u\"]The infinite counterexample is irrelevant, but Altheman's logic is not sound. The assumption by contradiction is $\\exists X$, not $\\forall X$.[/quote]\r\n\r\nThat was the point I was making by the infinite counterexample -_-" } { "Tag": [ "Euler", "modular arithmetic" ], "Problem": "1.) Prove by congruence that the integer $ 53^{103} \\plus{} 103^{53}$ is divisible by $ 39$\r\n\r\n2.) Prove that for any integer $ a$, $ a^3 \\equiv 0, 1$ or $ 6 (mod 7)$.", "Solution_1": "[hide=\"1\"]\nSince $ \\varphi(39) \\equal{} 24$ (for Euler-Fermat),\n\n$ 53^{103} \\plus{} 103^{53} \\equiv 14^{103} \\plus{} 25^{53} \\equiv (14)^7 \\minus{} (14)^{5} \\equiv 14^{5}(196\\minus{}1) \\equiv 0 \\pmod{39}$\n[/hide]\n\n[hide=\"2\"]\nYou could just test every one of 7 possibilities for $ a \\pmod{7}$.\n[/hide]", "Solution_2": "1.\r\nSince $ \\varphi(39) \\equal{} 24$ (for Euler-Fermat),\r\n\r\n$ 53^{103} \\plus{} 103^{53} \\equiv 14^{103} \\plus{} 25^{53} \\equiv (14)^7 \\minus{} (14)^{5} \\equiv 14^{5}(196 \\minus{} 1) \\equiv 0 \\pmod{39}$\r\n\r\n\r\n2.\r\nYou could just test every one of 7 possibilities for $ a \\pmod{7}$.", "Solution_3": "That was a direct quote removing the hidden information. Please edit/delete your post." } { "Tag": [ "conics", "ellipse", "geometry unsolved", "geometry" ], "Problem": ":( :( :(", "Solution_1": "The line V3V4 cut the tangent at P.The bissector of V3V4 is the major axis line support.V3V4 intersection wih bissector is point M.Draw the circle center M and diameter V3V4.Draw the tangent to this circle through P.Through the tangency point T draw a paralel line o the major axis that cut the given tangent at pont J.This is the tangency point with the ellipse.Through J draw a perpendicular to he tangent that will cut the line V3V4 at point K.The circle with diameter PK will cut the majr axis at the points F1 and F2 that are the foci.", "Solution_2": "thanks :lol:" } { "Tag": [ "calculus", "limit", "modular arithmetic", "induction", "number theory proposed", "number theory" ], "Problem": "A sequence $c_n$ is constructed after the following rule: We set $c_1 = 2$ and $c_{n+1}=\\left[ \\frac{3c_{n}}{2}\\right] $ for every $n\\geq 1$. [Hereby, [x] denotes the integer part of the number x.] Prove that\r\n\r\n[b]a)[/b] This sequence $c_n$ contains infinitely many even numbers and infintely many odd numbers.\r\n\r\n[b]b)[/b] The sequence $e_{n}=\\left( -1\\right) ^{c_{n}}$ is non-periodical.\r\n\r\n[b]c)[/b] [Calculus fundamentals required.] There exists a real number G such that $c_{n}=\\left[ \\left( \\frac{3}{2}\\right) ^{n}G\\right] +1$ for every natural n.\r\n\r\n Darij", "Solution_1": "Only a) and c) up until now:\r\n\r\na) Assume first that $c_n$ is even, $\\forall n\\ge n_0$. Then for all $n\\ge n_0$ we have $c_{n+1}=\\frac {3c_n}2$, and we will eventually run out of $2$s in the expansion of $c_n$, which gives a contradiction. Now assume $c_n$ is odd, $\\forall n\\ge n_0$. Then for $n\\ge n_0$ we have $c_{n+1}=\\frac{3c_n-1}2$, which is also odd, so $c_n$ must be $M_4+1$ for all $n\\ge n_0$. We continue like this and we get $2^k|c_{n_0}-1,\\ \\forall k\\ge 1$, which is absurd.\r\n\r\nc) It's equivalent to showing that the intervals $I_n=[(\\frac 23)^n(c_n-1),(\\frac 23)^nc_n)=[a_n,b_n)$ have a non-void intersection. It is, however, pretty easy to show that $a_n$ is increasing and $b_n$ is decreasing (from the definition of $c_n$), and that $b_n-a_n\\to 0$, so we can take $G=\\lim_{n\\to\\infty}a_n$.", "Solution_2": "b) I'll use the following lemma: If $c_{m+1},c_{n+1}$ have the same residue $\\pmod{2^x},\\ x\\ge 1$ and $c_m,c_n$ have the same parity, then $c_m,c_n$ have the same residue $\\pmod{2^{x+1}}$. I won't prove it: just take separate cases when $c_m,c_n$ are odd/even etc.\r\n\r\nAssume there is a $T>0$ s.t. $c_{n},c_{n+T}$ have the same parity $\\forall n\\ge 1$. Assume, furthermore, that we have shown $c_1\\equiv c_{1+T}\\pmod{2^k},\\ c_2\\equiv c_{2+T}\\pmod{2^{k-1}},\\ldots,c_{k-1}\\equiv c_{k-1+T}\\pmod {2^2}$. We know that $c_k\\equiv c_{k+T}\\pmod 2,c_{k+1}\\equiv c_{k+1+T}\\pmod 2$. \r\n\r\nWe apply the lemma for $m=k,n=k+T,x=1$ to find $c_k\\equiv c_{k+T}\\pmod {2^2}$. We apply the lemma again for $m=k-1,n=k-1+T,x=2$ to find $c_{k-1}\\equiv c_{k-1+T}\\pmod{2^3}$, and so on until we find $c_1\\equiv c_{1+T}\\pmod{2^{k+1}}$. We continue in this manner for $k\\to k+1$ indefinitely. We have just obtained $2^m|c_{1+T}-c_1,\\ \\forall m\\ge 1$, which is absurd.\r\n\r\nThis induction seems really nasty, so I wonder if it's correct. What do you people think?" } { "Tag": [], "Problem": "Use the pattern given to express $ 100^2$ in the form $ a^2\\plus{}b^2\\minus{}c^2$. Then find $ a\\plus{}b\\plus{}c$.\r\n\r\n\\[ 12^2 \\equal{} 8^2\\plus{}9^2\\minus{}1^2\\\\\r\n14^2\\equal{}10^2\\plus{}10^2\\minus{}2^2\\\\\r\n16^2\\equal{}12^2\\plus{}11^2\\minus{}3^2\\\\\r\n18^2\\equal{}14^2\\plus{}12^2\\minus{}4^2\r\n\\]", "Solution_1": "[hide]For n^2 where n is even, we can see that n^2 = (n-4)^2 + (n/2 + 3)^2 - (n/2 - 5)^2\nThus, for n=100, we have 96^2 + 53^2 - 45^2. Adding these up we get 194.[/hide]\r\n\r\nEdit: Correct answer now.", "Solution_2": "for c, don't you mean (n/2 - 5)? that would make your answer two less.", "Solution_3": "Here's another way to think of the problem: \r\nThe sum of b and c is always 2 less than the first number, so in that case b+c=98\r\n\r\na is always 4 less than the first number so a=96\r\n\r\n96+98=194 [u]ans.[/u]" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi,\r\n\r\nthere are many nontrivial groups $G$ isomorphic to their direct product $G \\times G$, e.g. the free abelian groups of infinite rank, $(\\mathbb{R},+), (\\mathbb{Q}^*,*)$, ...\r\nThis leads to the following question:\r\n \r\nIs there any finitely-generated nontrivial group $G$ s.t. $G \\cong G \\times G$?\r\n \r\nI already know that such a group must be infinite, non-abelian, perfect, and that every proper normal subgroup must have an infinite index. But this doesn't really help :-)\r\n \r\nAnd what's about the generalization: When $G$ is minimal spanned by $n>1$ elements, then of course $G \\times G$ is spanned by $2n$ elements. But is this also the lower bound?", "Solution_1": "Just for the last one:\r\n$G= S_n$ is generates by $(12), (123...n)$.\r\nI think that $G \\times G$ is generated by $( (12) , (12) ) , ((123...n) , (123...n))$ and $((1n),(1))$ then.", "Solution_2": "Thank you! :-)\r\n \r\nBut I have not made any progress with the first question ...", "Solution_3": "... anyone an idea? :-)", "Solution_4": "Hi\r\n\r\nCan you explain your last assertion: \"every proper normal subgroup must have infinite index\"?\r\n\r\nVipul", "Solution_5": "$G \\cong G \\times G$ implies $G/N \\cong G/N \\times G/N$ for every normal subgroup $N$. When $N$ has finite index, $G/N$ is finite, s.t. $G/N = 1$, i.e. $G = N$.", "Solution_6": "Hi\r\n\r\nI can't understand the statement that $G/N$ must be isomorphic to $G/N \\times G/N$. This will hold only if the isomorphism takes $N$ to $N \\times N$, and that isn't likely to hold.\r\n\r\nVipul", "Solution_7": "Yes, you're right ..." } { "Tag": [ "calculus", "geometry" ], "Problem": "I have been looking over some calculus to try to stay warm (2 years since AP) and came across this problem that I don't quite understand. It is probably pretty basic since its from a calculus review and not a physics one, though.\r\n\r\nA tank full of water has a vertical side which is 5 feet long by 3 feet high. (The tank is rectangular in shape and its pipes are parallel to one another.) Find the fluid force on this side of the tank.", "Solution_1": "The pressure at depth $ x$ is $ \\rho g x$, assuming that the tank is closed from above and so the atmospheric pressure doesn't influence anything. The force on the area $ \\text dS = a \\text dx$ is $ \\text dF = a\\rho g x\\text dx$, and integrating from $ x=0$ to $ x=h$ gives $ F = a h^{2}\\rho g/2$." } { "Tag": [ "geometry", "3D geometry", "pyramid", "perimeter" ], "Problem": "A right square pyramid has a base with edges each measuring 3 cm and a height twice the perimeter of its base. What is the volume of the pyramid?", "Solution_1": "The area of a pyramid is 1/3bh.\r\nThe area of the base is 3^2=9.\r\nThe height equals 2*perimeter of base=2*(3*4)=2*12=24.\r\n1/3*24*9=8*9=72." } { "Tag": [ "ARML", "HCSSiM", "summer program", "MathZoom" ], "Problem": "Paul Dreyer just sent out a message with several pieces of information. One is that the teams won't be staying in Tonapah Hall - instead, we'll be in cluster of dorms out near the Tropicana Ave. side of campus - Dayton, Boyd, Rodman, Williams, with check-in at Dayton.\r\n\r\nAnd he included a program.\r\n\r\n[b]Friday, May 29th, 2009[/b]\r\n3:00-3:30 P.M. New Coaches Meeting - CBC A106.\r\n3:30-4:00 P.M. Coaches Meeting - CBC A106\r\n4:00-4:30 P.M. Proctors Meeting - CBC A106\r\n6:00-7:30 P.M. Team and Power Rounds - Assigned Rooms in CBC\r\n8:00-9:30 P.M. Talk by Cynthia Vinzant, HCSSiM - Student Union Ballroom. (Refreshments provided by MathZoom)\r\n11:00 P.M. Students must be in Residence Halls\r\n12:00 Midnight Curfew - All Participants in Assigned Rooms\r\n\r\n[b]Saturday, May 30th, 2009[/b]\r\n7:00-8:30 A.M. Breakfast - Wilson Dining Commons\r\n9:00-11:00 A.M. Individual and Relay Rounds - Student Union Ballroom\r\n11:30 A.M. Tie Breaker - Student Union Ballroom\r\n12:00-1:30 P.M. Lunch - Wilson Dining Commons\r\n1:30-2:00 P.M. Super Relay and Awards Assembly - Student Union Ballroom\r\n\r\nHey, Paul - are we really going to get the tiebreakers in before lunch? Color me skeptical on that point.", "Solution_1": "[quote=\"Kent Merryfield\"]\n1:30 P.M. Tie Breaker - Student Union Ballroom\n12:00-1:30 P.M. Lunch - Wilson Dining Commons\n\nHey, Paul - are we really going to get the tiebreakers in before lunch? Color me skeptical on that point.[/quote]\r\n\r\nAccording to this, I think the tiebreakers are after lunch.", "Solution_2": "We definitely will if that schedule is right. 1:30am tiebreakers will be a cinch. Just wake up the students who we think will ace the contest on Saturday.\r\n\r\n\r\n\r\nBut seriously, I acknowledge the timing will be tight, but we cannot wait until after lunch for the tiebreakers. Being a little late for lunch is okay. Keeping the other three sites waiting for UNLV tiebreaker results will probably get me killed.", "Solution_3": "All right, all right. So I lost a digit typing it. I've edited.", "Solution_4": "SFBA stayed in Dayton in 2006, and I remember it being pretty nice. It'll be a longer walk to everything, though. We also won't all be in the same building, which is too bad.", "Solution_5": "Yeah, and it's a longer walk to the In-n-Out. Here, broadly, are where the teams are:\r\n\r\nMacau: Rodman, 1st floor\r\nColorado: Rodman, 2nd floor (with a couple of coaches on 3rd floor)\r\nNorthern California: Rodman, 3rd floor\r\nDesert Southwest: Boyd, 1st floor (with one coach on 2nd)\r\nRoseburg HS: Boyd, 2nd floor\r\nOregon: Boyd, 2nd and 3rd floors\r\nSFBA: Boyd, 3rd and 4th floors\r\nSan Diego: Williams, 1st and 2nd floors\r\nWashington: Williams, 2nd and 3rd floors\r\nSouthern California: Williams, 3rd, 4th, and 5th floors\r\nUtah: Dayton, 3rd floor north\r\nPhilippines: Dayton, 3rd floor north", "Solution_6": "Argh I find it unfair that the UNLV people actually get tables; UGA has everyone in an auditorium using those little armrests as tables :mad:", "Solution_7": "At Iowa, we get clip boards! you can snap them! they make me abnoixious", "Solution_8": "What are you talking about? On the individual and relay rounds, we get clipboards too. Always have, even when we were in San Jose.\r\n\r\nWhat's different about the West is that we do team and power on Friday evening - in a classroom building, one team to a room. Isn't that how you do those, albeit on Saturday morning?", "Solution_9": "[quote=\"dnkywin\"]Argh I find it unfair that the UNLV people actually get tables; UGA has everyone in an auditorium using those little armrests as tables :mad:[/quote]\r\n\r\ni think we should protest, given that we are on the same team :P", "Solution_10": "Yes, that's what we do at UGA. Rooms for team and power. Auditorium with the little desks for individual and relay. I assume that it will be mostly the same this year, but I've heard something about not being in the auditorium until awards. I don't know where the individual and relays will be.\r\n\r\nTom", "Solution_11": "At Penn State we get boards... without clips. That's obnoxious.", "Solution_12": "[quote=\"worthawholebean\"]At Penn State we get boards... without clips. That's obnoxious.[/quote]\r\n\r\nWhere can you even buy those things?", "Solution_13": "There is something to be said for how annoying the non-stop click-click-clicking of 500 lapboards gets after a few hours :) \r\n\r\nAt Iowa we had desks, which worked nicely. (But that was well in the past... dunno how it is today.)", "Solution_14": "We definitely had clipboards as early as 2002 in Iowa. I was at one time considering composing some music for several clipboard players, to be performed immediately prior to the individual round. Never did, though.", "Solution_15": "UGA this year I've heard will have lapboards. Just like the good ole' days, when the Southeast went to Penn State. :)", "Solution_16": "Just to be clear: generating and Sly Si may be reminiscing about Iowa in years past, but both of them will be in Las Vegas this year. I see their names listed on the housing reservation list for the site, both listed as coaches (for San Diego and SFBA, respectively).", "Solution_17": "[quote=\"Paul Dreyer\"]Hello, all!\n\nTwo quick things.\n\n1. Due to a scheduling snafu on UNLV's part, we'll have to delay lunch from 12:00 to 12:30. With the added individual round, we might have been running a little long anyways. Depending on how the day goes, we might do the super relay pre-lunch as well. My goal is to have everything finished and folks on the road by the original schedule (or close to it). This schedule change will be reflected in the programs you receive upon check-in.\n\n2. Cynthia Vinzant (from Berkeley and HCSSiM) sent me the title and abstract of her talk on Friday night. It should be a fun one.\n\nHercules, Hydras, and Really Big Numbers\n\nWe'll witness an epic battle between Hercules and the Hydra, whose monstrous heads grow at an alarming rate. The longer Hercules takes, seemingly the worse his plight. A clever Hercules might be able to survive, but can a foolish one? He might just break arithmetic trying.\n\nPaul\n[/quote]\n\n[quote=\"National Weather Service\"]\nFriday: Mostly sunny and hot, with a high near 97. Calm wind becoming northeast between 5 and 8 mph.\n\nFriday Night: Partly cloudy, with a low around 76.\n\nSaturday: Mostly sunny and hot, with a high near 99.\n[/quote]" } { "Tag": [ "function", "geometry", "topology", "algebra proposed", "algebra" ], "Problem": "Well ,I came up with this problem in the last week ,but couldnt find a solution for it (This is probably as usuall ,Well_known )\r\n\r\n[color=blue]Problem[/color]\r\n\r\nProve that there exist a bijective injective function from any closed shape in $R^2$ to any closed shape in $R^2$.\r\n\r\n\r\nWell for defining ,CLOSED:\r\n\r\nI mean a shape closed shape (By its usual meaning :blush: sorry couldnt find a better explanation) with its internal area ,and it should be countinous too.\r\n\r\nthere shouldnt be any empty area inside the shape ,All of the internal area must be including the shape", "Solution_1": "I take it the shape has to be bounded. If so, pick a point X inside the shape and map it to the origin; map the boundary of the shape to the unit circle; and map each line segment with endpoints at X and on the boundary to a radius of the unit circle. Thus we can biject any such shape with the unit circle, and the conclusion follows.", "Solution_2": "That \"shape\" need not be convex, so your procedure of mapping rays to rays can fail. If I am not mistaken, there is a theorem due to Schoenflies stating that any homeomorphism from a closed simple curve in the plane to the unit circle can be extended to a homeomorphism of the entire plane onto itself, mapping the interior of the curve onto the interior of the circle (and the same for exteriors), but it's very difficult to prove, from what I hear. \r\n\r\nIt's in that category of results to which the Jordan Curve Theorem belongs: simple to state but which have only nasty, long known proofs." } { "Tag": [ "induction", "geometry", "geometric transformation", "homothety", "combinatorics unsolved", "combinatorics" ], "Problem": "There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \\ldots , w_n$. It is known that if any coin with weight $ w_k$, where $ 1 \\leq k \\leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.", "Solution_1": "[quote=\"orl\"]There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \\ldots , w_n$. It is known that if any coin with weight $ w_k$, $ 1 \\leq k \\leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.[/quote]\r\nAll n is $ n\\equiv 1 (\\mod 2)$ and $ n > 5$\r\nThe first we will prove that n must be an odd number . \r\nExist a set such that $ \\sum_{i = 1}^n w_i$ which is smallest. \r\nFrom condition we have : \r\n$ S_n\\equiv w_i,\\forall i = 1,..,n$ where $ S_n$ is sum of all $ w_i$\r\nTherefore all element of set if parity . \r\nIf all of them is even then the set $ \\{\\frac {w_i}{2}\\}$ also satisfy condition ,this is contradiction from the chose $ S_n$ is minimum . \r\nIt gives that all elements of set must be odd number . \r\nNow suppose n is even then when we remove from the set a $ w_i$ then exactly one of two set after split has an odd numbers of element ,it gives that sum of all element is an odd number , but the other set has sum of all element is a even . Therefore thay can't equal . \r\nSo n must be an odd number . \r\nThe case $ n = 3,5$ it is easy to prove there is no set satisfy . \r\nWe will prove for all odd number great than 5 by induction . \r\nLet : \r\n$ f = 1*3*....*(2n - 1)$ ( n is odd number)\r\nwhere * can take $ + , -$\r\nThe case n=7 is obvious by some carculation .\r\nSuppose it true for $ 2n + 1$ we will prove for $ 2n + 3$\r\nEasy to check that when we make a transform : chose two odd consecutive and change them sign then we have $ |f' - f| = 2$\r\nIf we move a term from the set ${ \\{1,...,2n - 1}$ then use the result $ |f' - f| = 2$ we can chose sign for $ 2n + 1,2n + 3$ for which $ f = 0$ (not contain the object which had been moved ) . \r\nTake all terms which have sign + in a set ,the others in a set . Two sets have equal sum. \r\nIf we move $ 2n + 3$ or $ 2n + 1$ ,in two case it is quite easy to find the ways by consider two case $ n\\equiv 1 (\\mod 4)$ and $ n\\equiv 3(\\mod 4)$ \r\nSo all all n odd and $ n > 5$ satisfy condition .", "Solution_2": "i think there no N to satisfy;\n(**)it is clear that all Wi shoud have same parity!\nconsider W15$.[/i]\n\n[i]Proof.[/i] For $n = 3$, no such set exists since we would be left with two numbers $a,b$, which can't be equal since they are distinct. \n\nFor $n=5$, we will have $4$ numbers left which must be broken into two groups that are equal. Let the five weights be $w_1 < w_2 < w_3 < w_4 < w_5$. Suppose we remove $w_1$. Then we must have $w_2+w_5 = w_3+w_4$ and/or $w_5 = w_2+w_3+w_4$. The second case is not possible as in this case we would have $w_5 = w_2+w_3+w_4 > w_1+w_3+w_4$, which means the coins cannot be split when we remove $w_2$. Thus we must have $w_2+w_5 = w_3+w_4$. This shows that $w_1+w_5 < w_3+w_4$. But then no splitting is possible when we remove $w_2$ since $w_5 < w_3+w_4$ tells us that the group with $w_5$ must contain exactly $2$ coins. The second coin cannot be $w_1$, and as $w_5 > w_4 > w_1$ the second coin is $> w_1$ we have no other choice. \n\n[b]Lemma 3.[/b] [i]Any odd n > 5 works.[/i]\n\nProof. Let $f = 1 \\ast 3 \\ast \\cdots \\ast (2n-1)$ for some odd number n and the $\\ast$ symbol can take either $+,-$ where the weights of the coins are $1,3,\\ldots,2n-1$. Now suppose that we can satisfy the condition of the problem for some $2n + 1$ and then we need to show that $2n + 3$ is possible. Then define a transformation on $f$ call it $f'$ where $|f'-f| = 2$ which we form by taking two odd consecutive integers and changing their sign to form $f'$. We then move a term from the set $\\{1,\\ldots,2n-1\\}$ then by using the result $|f'-f|$ we change the sign on $2n+1$ or $2n+3$ using $|f'-f| = 2$ to get $f = 0$. Thus two sets have equal sum if we move $2n+3$ or $2n+1$ which holds for $n \\quad 1 \\pmod4$ or $n \\equiv 3 \\pmod{4}$.", "Solution_10": "The answer is all odd $n\\ge 7$. Without loss of generality, assume that $w_1< w_2 <\\dots < w_n$ and that $\\gcd(w_1,w_2,\\dots,w_n)=1.$ Now, note that if $S=w_1+w_2+\\dots+w_n$ then $S-w_1,S-w_2,\\dots,S-w_n$ are all even. Therefore, $w_k$ have a fixed parity for $1\\le k\\le n.$ Since they are not all even, they are all odd. This implies that $S$ is odd. Since $S$ is odd, $n$ must be odd.\n\nThis is not possible for $n=3$ because removing $w_1$ would imply $w_2=w_3$.\n\nFor $n=5$, removing $w_5$, we see that $A_5=(w_2+w_3,w_1+w_4)$ or $B_5=(w_1+w_2+w_3,w_4)$ are the only possible ways to split them into two groups, for size reasons. Similarly, removing $w_4$, we see that $A_4=(w_2+w_3,w_1+w_5)$ or $B_4=(w_1+w_2+w_3,w_5)$ are the only possible splitting ways.\n\nIf we use $A_5$ and $A_4$ or $B_5$ and $B_4$ then we get $w_4=w_5$, absurd. If we use $A_4$ and $B_5$ then $2w_1+w_5=w_4$, absurd. Thus, we must use $A_5$ and $B_4$. Now, if when we remove $w_3$ we use $w_1+w_2+w_4=w_5$ then from $B_4$ we know $w_3=w_4$. Therefore, $w_1+w_5=w_2+w_4$ so $2w_1+w_3=w_4.$ However, $w_4-w_3=w_2-w_1$ so $2w_1=w_2-w_1$ so $3w_1=w_2.$ If when we remove $w_2$, we use $w_1+w_5=w_3+w_4$ then this contradicts $w_1+w_5=w_2+w_4.$ Thus, $w_1+w_3+w_4=w_5.$ This implies that $w_2=w_4$ by $B_4$, so that is also impossible. Therefore, $n=5$ is impossible.\n\nFor $n\\ge 7$ we claim that $1,3,5,\\dots,2n-1$ will work. All we need to prove, in fact, is that there is some way to place $+$'s or $-$'s between the numbers $1,3,5,\\dots,2n-1$ and before $1$ so that the result can be any number $1,3,5$, to $2n-1$.\n\nFor $n=7$, we use \n\\begin{align*}\n1-3-5-7-9+11+13 &= 1\\\\\n1+3-5-7+9-11+13 &= 3\\\\\n1+3+5+7-9+11-13 &= 5\\\\\n1-3-5+7+9+11-13 &= 7\\\\\n1+3+5-7+9+11-13 &= 9\\\\\n1-3+5-7-9+11+13 &= 11\\\\\n1+3+5-7+9-11+13 &= 13\n\\end{align*}\nNow, suppose it is possible with $(1,3,\\dots,2n-1)$ Then we'll prove it's possible with $(1,3,\\dots,2n+3).$ If we are trying to make some number $1\\le k\\le 2n-1$, then take any way to make $k$ with $(1,3,\\dots,2n-1)$. Then, we subtract $2n+1$ and add $2n+3$ to make $k+2.$ Now, add a negative sign before the $1$ to make $k$. If the sign before $1$ is already negative, then we add $2n+1$, subtract $2n+3$, and then make the sign before $1$ positive.\n\nIf we are trying to make $2n+1$, then you take the way to make $2n-1$, add $2n+3$ then subtract $2n-1$ and then leave it alone. Now, to make $2n+3$, if the $1$ is negative, then do what we did for $2n+1$ but change that negative before the $1$ to positive. Otherwise, we can always find two consectuvie numbers $2i-1,2i+1$ such that $2i-1$ is positive and $2i+1$ is negative. Now, if we flip those two signs then we can make $2n+3$ with $(1,3,\\dots,2n-1)$. However, using what we did for $1\\le k\\le 2n-1$ we can also make $2n+3$ using $(1,3,\\dots,2n+3)$ as desired." } { "Tag": [ "ratio", "trigonometry", "geometry", "geometric transformation", "dilation", "inequalities", "analytic geometry" ], "Problem": "Let $ (O)$ is a circle, let $ AB, CD$ are two diameters and $ AB\\perp CD$, let $ E$ is a point on $ AD$ arc. Let $ EC$ meet $ OA$ at $ M$, $ EB$ meet $ OD$ at $ N$. Prove that: $ \\frac {OM}{AM} \\plus{} \\frac {ON}{DN}\\ge \\sqrt{2}$.", "Solution_1": "set a co ordiate system and radius of circle 1\r\nuse small amount ofhomothesy to calculate the ratio\r\nsimpify the trignometric inqualitiy\r\nthis gives us the obvious trignometric inqualitiy", "Solution_2": "[quote=\"thanhnam2902\"] [color=darkred]Let $ C(O,r)$ be a circle with diameters $ AB$ , $ CD$ so that $ AB\\perp CD$ . For a point $ E$ which \n\nbelongs to the arc $ AD$ denote $ M\\in EC\\cap OA$ , $ N\\in EB\\cap OD$ . Prove that $ \\frac {OM}{AM} + \\frac {ON}{DN}\\ge \\sqrt {2}$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof.[/u][/b] Denote $ \\|\\begin{array}{c}\n\\tan\\widehat {ECD}=x\\\\\\\\\n\\tan\\widehat {EBA}=y\\end{array}$ . Observe that $ \\{x,y\\}\\subset [0,1]$ and $ x+y+xy=1$ because $ m(\\widehat {ECD})+m(\\widehat {EBA})=45^{\\circ}$ . Since \n\n$ \\|\\begin{array}{ccc}\nMO=Rx & \\implies & \\frac {MO}{MA}=\\frac {x}{1-x}=\\frac {x}{y(1+x)}\\\\\\\\\nON=Ry & \\implies & \\frac {NO}{ND}=\\frac {y}{1-y}=\\frac {y}{x(1+y)}\\end{array}$ obtain $ \\frac {OM}{AM} + \\frac {ON}{DN}=$ $ \\frac {x}{y(1+x)}+\\frac {y}{x(1+y)}\\ge$ $ \\frac {2}{\\sqrt {(1+x)(1+y)}}=$ $ \\frac {2}{\\sqrt {1+xy+1-xy}}=\\sqrt 2$ .[/color]", "Solution_3": "Thank Virgil Nicula very much. Your solution is very nice. :P :maybe:", "Solution_4": "this solution is really teriffic", "Solution_5": "[quote=\"gauravpatil\"]this solution is really teriffic[/quote] Then post yours.", "Solution_6": "i meant Virgil Nicula's solution was great\r\n[size=200]generalities[/size]\r\nobserve that dilations with $ O$ as the centre keeps the ratios in the inequality for a given point $ E$ constant\r\nso WLOG take the radius of the circle equal to $ 1 unit$\r\ni set up a coordinate system $ O$ as the origin $ AB$ along $ x$ axis and $ CD$ as $ y$ axis\r\nagain Wlog $ arcAD$ lies in the first quadrant\r\nlet $ angleDOE \\equal{} \\theta$\r\n[size=200]solution[/size]\r\nnow we drop a perpendicular from $ E$ to $ AB$ and $ CD$ with feet say $ K$ and $ L$ respectively\r\nobserve that $ EK \\equal{} OL \\equal{} \\cos{\\theta }$ and also $ EL \\equal{} KO \\equal{} \\sin{\\theta }$\r\nso $ \\frac {EK}{ON} \\equal{} \\frac {BK}{BO}$ (kind of homothesy)\r\nbut $ BO \\equal{} 1 \\equal{}$radius of circle and $ BK \\equal{} BO \\plus{} OK \\equal{} 1 \\plus{} \\sin{\\theta}$\r\nand $ EK \\equal{} \\cos{\\theta }$ so $ ON \\equal{} \\frac {\\cos{\\theta}}{1 \\plus{} \\sin{\\theta }}$\r\nsimilarly we get $ OM \\equal{} \\frac {\\sin{\\theta}}{1 \\plus{} \\cos{\\theta }}$\r\nthis implies $ \\frac {ON}{ND} \\equal{} \\frac {\\sin{\\theta}}{1 \\plus{} \\sin{\\theta} \\minus{} \\cos{\\theta}}$\r\nand $ \\frac {OM}{MA} \\equal{} \\frac {\\cos{\\theta}}{1 \\plus{} \\cos{\\theta} \\minus{} \\sin{\\theta}}$\r\nso now we just have to prove that \r\n$ \\frac {\\sin{\\theta}}{1 \\plus{} \\cos{\\theta }} \\plus{} \\frac {\\cos{\\theta}}{1 \\plus{} \\cos{\\theta} \\minus{} \\sin{\\theta}} \\ge \\sqrt {2}$\r\nsimplifying this we get that \r\n$ (\\sin{\\theta} \\minus{} \\frac {1}{\\sqrt {2}})(\\cos{\\theta} \\minus{} \\frac {1}{\\sqrt {2}}) \\le 0$\r\nthis is pretty obvious", "Solution_7": "Let be the point $ K\\equiv AD\\cap BE$ and it is easy to show that $ MK\\parallel CD,$ because of $ \\angle MKA \\equal{} \\angle MEA,$\r\nfrom the cyclic quadrilateral $ MAEK$ $ ($ $ \\angle MAK \\equal{} \\angle MEK \\equal{} 45^{o}$ $ )$ and $ \\angle MEA \\equal{} \\angle CDA \\equal{} 45^{o}.$\r\n\r\nSo, we conclude that $ \\frac {OM}{MA} \\equal{} \\frac {KD}{KA}$ $ ,(1)$\r\n\r\nApplying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \\bigtriangleup OAD$ with transversal $ BNK,$\r\n\r\nwe conclude that $ \\frac {KD}{KA}\\cdot \\frac {BA}{BO}\\cdot \\frac {NO}{ND} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {KD}{KA} \\equal{} \\frac {1}{2}\\cdot \\frac {ND}{NO}$ $ ,(2)$\r\n\r\nHence, from $ (1),$ $ (2),$ it is enough to prove that $ \\frac {OM}{MA} \\plus{} \\frac {ON}{ND} \\equal{} \\frac {ON}{ND} \\plus{} \\frac {1}{2}\\cdot \\frac {ND}{ON} \\geq \\sqrt {2},$ \r\n\r\nwhich is true and the proof is completed.\r\n\r\nKostas Vittas.", "Solution_8": "[quote=\"vittasko\"]Let be the point $ K\\equiv AD\\cap BE$ and it is easy to show that $ MK\\parallel CD,$ because of $ \\angle MKA \\equal{} \\angle MEA,$\nfrom the cyclic quadrilateral $ MAKE$ $ ($ $ \\angle MAK \\equal{} \\angle MEK \\equal{} 45^{o}$ $ )$ and $ \\angle MEA \\equal{} \\angle CDA \\equal{} 45^{o}.$\n\nSo, we conclude that $ \\frac {OM}{MA} \\equal{} \\frac {KD}{KA}$ $ ,(1)$\n\nApplying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \\bigtriangleup OAD$ with transversal $ BNK,$\n\nwe conclude that $ \\frac {KD}{KA}\\cdot \\frac {BA}{BO}\\cdot \\frac {NO}{ND} \\equal{} 1$ $ \\Longrightarrow$ $ \\frac {KD}{KA} \\equal{} \\frac {1}{2}\\cdot \\frac {ND}{NO}$ $ ,(2)$\n\nHence, from $ (1),$ $ (2),$ it is enough to prove that $ \\frac {OM}{MA} \\plus{} \\frac {ON}{ND} \\equal{} \\frac {ON}{ND} \\plus{} \\frac {1}{2}\\cdot \\frac {ND}{ON} \\geq \\sqrt {2},$ \n\nwhich is true and the proof is completed.\n\nKostas Vittas.[/quote]\r\nmagnificent just great!!", "Solution_9": "How often on this forum a simple problem is mishandled\r\n by being solved by some method that blows it out of all \r\n reasonable proportions!\r\n\r\n Fortunately, it did not happen to this lucky problem, yet.\r\n\r\n This one is another candidate for PWW, a nice illustration \r\n of Van-Aubel's theorem in a right triangle.\r\n\r\n OM/AM + ON/DN = OX/XY where X is an intersection of AN and DM\r\n \r\n on OY. Locus of X is clearly a reflection of arc AD in chord AD,\r\n\r\n which is also an arc of circle centered at the 4th vertex W of\r\n \r\n the square AODW radius WA. It makes it obvious when the MIN\r\n\r\n of the given sum of ratios takes place and at what value.\r\n\r\n\r\n Thank you.\r\n M.T.", "Solution_10": "[quote=\"armpist\"] [color=darkred]This is a nice illustration of [b]Van-Aubel's theorem[/b] in a right triangle.[/color] [/quote]\n[color=darkblue][b]Very cool the [u]Armpist[/u]'s proof ! It is the best. Thank you, [u]Armpist[/u] ![/b][/color]\n\n[quote=\"thanhnam2902\"] [color=darkred]Let $ C(O,r)$ be a circle with diameters $ AB$ , $ CD$ so that $ AB\\perp CD$ . For a mobile point $ E$ which \n\nbelongs to the arc $ AD$ denote $ M\\in EC\\cap (OA)$ , $ N\\in EB\\cap (OD)$ . Prove that $ \\frac {OM}{AM} + \\frac {ON}{DN}\\ge \\sqrt {2}$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Armpist's proof.[/u][/b] Denote $ \\|\\begin{array}{c}\nX\\in AN\\cap DM\\\\\\\nY\\in AD\\cap OX\\end{array}$ . Observe that $ m(\\widehat {AXD})=135^{\\circ}$ (costant) . The geometrical locus $ \\Lambda$ of the point $ X$ \n\nis symmetrically with the arc $ AED$ w.r.t. the line $ AD$ and the incenter $ I$ of $ \\triangle AOD$ belongs to $ \\Lambda$ . Apply the [b]Aubel's relation[/b] \n\nin the $ O$ - right triangle $ AOD\\ \\ : \\ \\frac {XO}{XY}=\\frac {MO}{MA}+\\frac {NO}{ND}$ . Therefore, the ratio $ \\frac {XO}{XY}$ is minimum iff $ X: =I$ and in this case\n\n$ \\frac {IO}{IY}=\\frac {OA+OD}{AD}=\\sqrt 2$ . In conclusion, $ \\frac {OM}{AM} + \\frac {ON}{DN}\\ge \\sqrt {2}$ for any point $ E$ which belongs to the arc $ AD$ .[/color]" } { "Tag": [ "\\/closed" ], "Problem": "I cannot find the forum for the class I am in. What should I do?", "Solution_1": "Which class?", "Solution_2": "Never mind. I can see it now." } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Find k such that the equation kx^2 + 2x + 1 = 0 has TWO distinct real solutions.\r\n\r\nWhat exactly is the question asking?", "Solution_1": "[hide]\nThe discriminant must be greater than zero.\n\n$b^{2}-4ac>0\\Rightarrow 4-4k>0\\Rightarrow k<1$.\n\n$\\boxed{k<1,k\\neq 0}$.\n[/hide]", "Solution_2": "[quote=\"Interval\"]Find k such that the equation kx^2 + 2x + 1 = 0 has TWO distinct real solutions.\n\nWhat exactly is the question asking?[/quote]\r\nTo have 2 real solutions, the discriminant, $b^{2}-4ac$, in the quadratic formula, $\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}$, needs to be positive.\r\n[hide]The discriminant is $4-4k$ so $4-4k>0$\n$1-k>0$\n$k<1$ but $k\\ne0$ because then the quadratic formula would be $\\frac{-b\\pm\\sqrt{b^{2}-0}}{0}$ and it would be undefined.\n$\\therefore k<1, k\\ne0$[/hide]", "Solution_3": "So basically, we plug and chug using the discriminant, right?\r\n\r\nThe question is not worded in simple terms. \r\n\r\nHow is a student to know that the use of the discriminant is needed here?\r\n\r\nThanks", "Solution_4": "The discriminant is what determines if the solutions are real, and how many there are.\r\n\r\nLet the solutions be $r=\\frac{a\\pm\\sqrt{b}}{c}$. If $b<0$, then $\\sqrt{b}\\not\\in\\mathbb{R}$. If $b=0$, then the roots are the same, because $\\frac{a-0}{c}=\\frac{a+0}{c}$. If $b>0$, then nothing is strange." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "prove that:\r\nwith a,b,c>1\r\n$ \\sqrt {log_ba} \\plus{} \\sqrt {log_ac} \\plus{} \\sqrt {log_cb} \\ge 3$\r\nthank for Help ^^", "Solution_1": "[quote=\"TUYEN_LOVER\"]prove that:\nwith a,b,c>1\n$ \\sqrt {log_ba} \\plus{} \\sqrt {log_ac} \\plus{} \\sqrt {log_cb} \\ge 3$\nthank for Help ^^[/quote]\r\n\r\n\r\n$ \\sqrt {log_ba} \\plus{} \\sqrt {log_ac} \\plus{} \\sqrt {log_cb} \\geq3\\sqrt[3]{\\sqrt {log_ba}*\\sqrt {log_ac}*\\sqrt {log_cb}}\\equal{}3\\sqrt[6]{log_ba*log_ac*log_cb}\\equal{}3\\sqrt[6]{\\frac{log_na}{log_nb}*\\frac{log_nc}{log_na}*\\frac{log_nb}{log_nc}}\\equal{}3$\r\n\r\n\r\n$ a\\equal{}b\\equal{}c$" } { "Tag": [ "algebra", "polynomial" ], "Problem": "1) If $ \\alpha,\\beta,\\gamma$ are the roots of the equation $ x^{3}\\plus{}qx\\plus{}r\\equal{}0$ then evaluate $ \\sum \\frac{\\alpha}{\\beta\\plus{}\\gamma}$.\r\n\r\n2) If $ x\\equal{}2\\plus{}2^{\\frac{2}{3}}\\plus{}2^{\\frac{1}{3}}$ then evaluate $ x^{3}\\minus{}6x^{2}\\plus{}6x$", "Solution_1": "[hide=\"Solution #1\"]\n\nLet $ S_1 \\equal{} \\alpha \\plus{} \\beta \\plus{} \\gamma$, $ S_2 \\equal{} \\alpha \\gamma \\plus{} \\alpha \\beta \\plus{} \\beta \\gamma$ and $ S_3 \\equal{} \\alpha \\beta \\gamma$.\n\nAnd by Viete's relations, $ S_1\\equal{}0$, $ S_2\\equal{}q$, $ S_3\\equal{}\\minus{}r$.\n\nNote that $ \\sum\\frac {\\alpha}{\\beta \\plus{} \\gamma} \\equal{} \\dfrac{S_1\\cdot(\\alpha^2 \\plus{} \\beta^2 \\plus{} \\gamma^2) \\plus{} 3S_3}{S_1\\cdot S_2 \\minus{} S_3}$\n\nAlso we should see that since $ S_1$ is negative the coefficient of $ x^2$ in the polynomial, then $ S_1 \\equal{} 0\\implies \\sum\\frac {\\alpha}{\\beta \\plus{} \\gamma} \\equal{} \\dfrac{3S_3}{ \\minus{} S_3}$\n\nWhich, if I substituted everything correctly, should make,\n\\[ \\sum\\frac {\\alpha}{\\beta \\plus{} \\gamma} \\equal{} \\boxed{ \\minus{} 3}\\]\n[/hide]", "Solution_2": "[hide=\"Solution to 2\"]$ x-2=2^{\\frac{2}{3}}+2^{\\frac{1}{3}}$ cubing both the sides we get $ x^{3}-8-6x^{2}+12x=6+6(2^{\\frac{2}{3}}+2^{\\frac{1}{3}} )$ now \nreplace $ 2^{\\frac{2}{3}}+2^{\\frac{1}{3}}$ by $ x-2$ to get the answer $ \\boxed2$[/hide]" } { "Tag": [ "conics", "ellipse", "geometry", "geometric transformation", "reflection", "angle bisector" ], "Problem": "Consider an ellipse with major and minor axes of length $2a$ and $2b$ respectively. chose a point $Q$ on the outside of the ellipse, in the direction of the major axes. Draw one of the tangent lines to the ellipse. At the tangent point, draw the perpendicular to the tangent line. The point $P$ is the intersection with the major axis of length 2a.\r\n\r\nWhat is $|OP|\\cdot|OQ|$?\r\n\r\n($O$ is the center of the ellipse)", "Solution_1": "[hide]Let $F_{1},F_{2}$ be the focuses of the ellipse ($F_{2}$ closer to $Q$)\nand $A$ the tangent point\n\nIt is well known that the line $AP$ reflects the line segment $F_{1}A$ into $F_{2}A$. \nSo $AP$ is the internal angle bisector of $\\angle F_{1}AF_{2}$\n\n$AQ\\perp AP \\Rightarrow AQ$ is the external angle bisector of $\\angle F_{1}AF_{2}$\n\nSo the division $(F_{1},P,F_{2},Q)$ is harmonic $\\Rightarrow OP \\cdot OQ ={OF_{1}}^{2}$[/hide]" } { "Tag": [ "inequalities", "geometry", "3D geometry" ], "Problem": "hey guys\r\n\r\ni am trying to learn cauchy and apply it to inequalities, however i need help.\r\n\r\nhow would you apply cauchy to this problem:\r\n\r\nIf $a, b, c, d$ are positive reals, prove that:\r\n\r\n$\\frac{1}{a} + \\frac{1}{b} + \\frac{4}{c} + \\frac{16}{d}\\geq \\frac{64}{a+b+c+d}$", "Solution_1": "Cauchy can be applied in this way: $(a+b+c+d)\\left( \\frac{1}{a}+\\frac{1}{b}+\\frac{4}{c}+\\frac{16}{d} \\right)\\geq (1+1+2+4)^2$ because $\\sqrt{4}=2$ and $\\sqrt{16}=4$.\r\n\r\nDividing by $a+b+c+d$ arrives at the result.\r\n\r\nIn order to use Cauchy, you need a product of two sums, which are usually in terms of variables but can be formed from constants (such as writing $n$ as a sum of $n$ $1^2$'s).", "Solution_2": "Products of sums or sums of products should ring the Cauchy bell. Most of the time, you need to find a tricky substitution or some transformation to the inequality to Cauchy it.", "Solution_3": "how would you prove the trivial, $a^2+b^2+c^2\\geq ab+bc+ac$ using cauchy?", "Solution_4": "[quote=\"sen\"]how would you prove the trivial, $a^2+b^2+c^2\\geq ab+bc+ac$ using cauchy?[/quote]\r\n\r\nYou could be like $(a^2+b^2+c^2)(b^2+c^2+a^2) \\ge (ab+bc+ca)^2 \\Rightarrow a^2+b^2+c^2 \\ge ab+bc+ca$.", "Solution_5": "oh yepp i see. thanks man\r\n\r\nIf a, b, c, d are positive reals such that $(a^2+b^2)^3=c^2+d^2$\r\n\r\nprove that $\\frac{a^3}{c} + \\frac{b^3}{d} \\geq 1$", "Solution_6": "Here's a nice extension of Cauchy (known as Holder):\r\n\r\n$\\left(\\sum a_i^p\\right)^{\\frac{1}{p}}\\left(\\sum b_i^q\\right)^{\\frac{1}{q}} \\ge \\sum a_ib_i$.\r\n\r\nThen we have\r\n\r\n$(c^2+d^2)^{\\frac{1}{3}}\\left(\\frac{a^3}{c}+\\frac{b^3}{d}\\right)^{\\frac{2}{3}} \\ge a^2+b^2$\r\n\r\nCube both sides and rearrange to get\r\n\r\n$\\left(\\frac{a^3}{c}+\\frac{b^3}{d}\\right)^2 \\ge \\frac{(a^2+b^2)^3}{c^2+d^2} = 1$\r\n\r\nso \r\n\r\n$\\frac{a^3}{c}+\\frac{b^3}{d} \\ge 1$ as desired." }